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Diktat Soal Konstruksi Beton Jilid 2
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V.2 Contoh-contoh soalV.3 Soal-soaI latihan
BAB VI. T A N G G A ( "SIAIRS" ).VI.1 Teori dan rumus-rumus........................o........
A. Anak tangga . . . . . . . . .. . . . . . . o . . c. . . . o . . . . . .. . . . . . .B. Tipe-tipe tangga ........................o.....o..C. tvlenentukan bentang ef f ektif tangga dan berat' sendiri
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 13ID. Pembebanan tangga ( Peraturan pembebanan Indonesia utk
gedung 1983)...............................co...... I33E. Cara menentukan jumlah anak tanggarukuran Iubang
pada pelat(untuk tangga)................. i.. o o.... 133'F'. Tangga melayang( t'Free standing stairs" )'. . . . . . . . . . . I35G. Tangga ulir( t'Helical stairst' ). . .,. . . . . . . . . . . . . .,. . . . L47
Vt.2 Contoh-contoh soal.......o.............o.............. 161
VI;3 Soal-soal latihan ........ o............................ z]J O
BAB VII. DINDING PENAHAN TANAH("RETAINING WALL'I)& TANGKI AIR:
( "wertrn TANK" ) .
VII .1 Teori dan rumus-rumus.VII.I.I Umum........................r..............r.. 2L7
VII.l.2 Penamaan bagian-bagian dari suatu dinding penahan tanah .'o............o......... o.......1o o l. 218
VII.1.3 Tipe-tipe dinding penahan tanah .............. 218
VII.1.4 Drainase pada dinding penahan tanah .......... 220VI I . 1 . 5 Sedikit teori tentang TvIEKANIKA TEKNIK yang per
Iu diketahui untuk merencanakan dinding penahani . . . . . . ? . .
.: , . . . . . . . . . . . . o . . . . : . . . . . . . -220
A.I(eadaan seimbang dalam tanah ....... o....... ZZOB'Teori Rankine-......... o.r.............'..... 221
B. 1 Tekanan tanah aktif untuk tanah kohe' sif(c/O)..................... o......... . 2218.2 Tekanan tanah pasif untuk tanah non kohe
)......:.-. ,.:...:..:.. .222sif(c=0)............. o.....,C'Teori Coulomb-. -.............. o L r.....r o... o.... ZZ3
c.1 Tgkanan tanah aktif ....o............... 223. c-2 tekanan tanah pasif ...,....... o........ zz4
D - tekanan tanah untuk tanah yang terendam air( ttsubmerged soiltt )....,....: ..:.. o.......... . 224
E.BebQn tambahan( rrsurehargerr )... o.. r.... . . . o.. ZZ5F. Tekanan tanah aktif untuk tanah kohesif ( c/0 ) 22€
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VrIr. 1 .5 -2 Merencanakan ukuran penampang secara kekuatan batas...o...........o................ 316vrrr. 1 - 5 - 3 B?]?k penampang persegi,turangan rangkapdihitung dengan cara kekuatan batas...... 3lg
VIII . 1 .6 Balok penampang persegi memikul lentur murni dihitungtung dengan cara "working stres s des igr, " ( ca ra elasti" iVIII.1 .G.1 Balok bertulangan tunggal............... . 321
A. Diketahui - M_, ukuran balok,luas tulangan
an-tunggal 1,ditr.,ya tegangan yg terjadipada beton dan baja . 321
" iliilT:illfi;ihfy;ti:;!;t:ililliliii 322VIII. 1 .5 .2 Balok ,"n".n".,n n"I".n, tulangan tunggal
dalam keadaan " balance " , ',over ieinforced,,,
vrrr.1.6.3 ":l*"::ilH]"1.;"";;;;;;";";;;';;;'' 323
tegangan izin beton a baja,ditanya ui<uran balok dan tulangan ....... 323
VIII.I .6.4 Ba.lok persegi tulangan rangkap .. .. 324VIII.1.7 IJendutan balok ("deflection'1 ) ... ......325
VIII. 1 .7. 1 Lendutan jangka pendek ( "short time deflec. tion") .... ...... 325
VIII.1.7.2 Lendutan jangka panjang(',long time deflection " )
VIII.l.8BaIokT ...327VIII.1.8.1 Lebar manfaat balok T (flens) ........... 327VIII.1.8.2 penampang T memikul lentur murni,bertu_
lang tunggal .... ...... 327VIII.1.8.3 penampang T bertulangan tunggal memikul
lentur murni dihitung dengan cara keku_. atan batas ...... 32g
VIII.1.8.4 penampang T bertulangan tunggal memikullentur murni dihitung dengan cara elas_tis('working stress design,,) ..........'.. 330
VIII.I.9 Tulangan untuk balok yang memikul lentur .... 33,|VIII. 1 . l0Penampang memikul momen lentur+ mormal tekan ...... 335
VIII. 1 . 10. 1 Kotom dibebani normaf tekan sentris,dantulangan gesernya berupa spiral .. 335
VIII.1 . 10.2 Kolom dibebani normal sentris (tekan ),dantulangan gesernya 'berupa sengkang ...... 337
VIII.1.10.3 Standard ACI utk"lateral ties",tulanganmemanjang ...... 337
VIII.1.10.4 Diagram interaksi Mr, _ p^ ... .--.. 338
VIII.1.10.5 caya normal tekan maksimum dan eksentri .- €
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Dava dukunq untuk tanah liat oclav".
Tergantung pd konsistbnsinya (kekuatan geser)yg-ditentukan sbb :
1 . Standard penetration test. ( SPT ) t
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rl
dari test ini. akan diperoleh ,hasil: hubungan N ( " standard ps
netration resistance" ) dan tegangan izin tanah.
cl,
fr=q=f-=tr=B=
Keadaan konsistensitanah
Sangat lembek
Iembek.sedang
kerassangat keraskeras sekali
dt"rr"h A
(kg/cma I
0r00-0r300r3o-0r600r60-1r201''r20-2r402r40-4r80
4 r80
2. "Unconfined compression test".
Percobaan tsb diatasrakan kami bahas pd IIEKANIKA TANAHT s€ri Delta
s-ienis Pondasi.V. 1 .3 Jenis-ienis Po'
A. Pondasi danqkal ( ishallgw foundation') .
A.1 Pondasi tapak ("isolated footing").- bujur sangkar.
,;,persegi panjang.
- lingkaran.
02"2 - 4
488 -15
1s -3030
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@ "strap footingo
rl.
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@ Pondasi terapung ('!f roating foundation,,) .
B.3 "steel pile'.t,erbuat dari ba j a .
seperti pipa baja baja profil H
@DaIeIm Kongtruks i beton I ini kita')sl, ini'. ''.''.,.',
,*ru
B. Pondasi dalam ( "deep foundation" ) .Berupa tiang pancang ( "pile" ).
or.;:
B. 1 "Precast conCrqte pile " - beror-gga berongga
tiang dicetak dulu, baru dipancang.
(disebut "precast prestressed concrete pile. ) .8.2 "cast in place pi1er.
tiang pancang di cor ditempat setelah tulangan dipasapg.seperti Franki pi1e, bore pile, strauss . .
tt_Jtkalakan
"sheet pile"
bahas beberapa type ponda-
"t stra
'-;";;Fti
i .'!'. .. . ,,,'L&'
i 'il' !*i;.' , :'ii.l i' -"#:i.i
!.' o \ ",.- .?
'6ue11T S{PA.r
.6ue11?p-req )tr1T1 pd .lesndiral qprEp"e 6ue11 derl-deT+ rstpar
de66ue6uau 6up uElnrlerTp 1dp 6ue1ur1 €ite! uep uaftou uE6un11q-zad'lsepuod 6ue11-6uer1 selpTp r1ede1 lsepuod Da
' ( uossr'pc , uri$oq rntur$s ,
'6uBcued )tp1) Ts'pu 6ue11-6ue!1 epede>1 ne1p. qer{Eg qeu
,-:4 epeda>1 {Tpq ue6dbp uetrnTpsTp sn.req qedel 1="pud 4pd e[:a>1aq 6uert rnluar uauou ,uep 6uelutl r leurou edeg {:
:
z'L. uelprerts-:ad Tlpr{aTau
, etl
-ue?'qsl upqaq' l.pqT)tp r{pl,tpq qfuel rstpal upp se?e ueq !,, l l
,,-q-upqaq depeqral ue>16un11qradgp snreq >1ede1 Tsppuod [. L
'1(Ll qPq rgd :Tsuarlar)
;' r sepuod leTad SeeJueru T eqel=qi
r6ulqcund,
leqTltp, 6ur.lrur
t. rt*
. 'uP{a+ qp;app e{ dnsn.fiuau ?nqasral te?artde1e1 tsepuod 1e1ad uprncupqe{ ue{
qeqa^f,uaur )tepT1, TpB[:a+ 6u1r1u ]tela6' ( r. a-:rnTTpJ uoTsserduoc rear{sr ) . I
: lsepuod eped ue1e6e6a{ ousruplaH
s' [ 'A
.E
.Z
r,*)
il*n
'TSrnl1n.r+s urTert sTuasTuef ,. J.A
IiLi$r :
'l,r
i' \
6
1.4 Luas bidang tumpuan dr pondasi tapak atau jumlah tiang pondasirli
dan penempatannya, harus ditentukan berdasarkan beban-beban
dan momen luar disalurkan oleh pondasi, sedangkan' tekanan tanah,i
.dasat'kan prinsip prinsip Mekanika Tanah.
2. pogctasi telapak derlqa.n bid.anq atas vq miring agaq bertanqqa.2.1 pada pondasi telapak dng bidang atas yg miring atau yang berta
ngga, kemiringan bidang atas atau tinggi anak tangga harus se-at kekuatan yang ditentu-
116ra-trrratr'dalam peratulian ini.2.2 pondasi telapak dengan bidang atas yang miring atau yang ber-
tangga yang dihitung sebagai satu kesatuan monolit hrs di-ikian rupa sehingga benar-benar terjamin satu
kesatuan pondasi Yang monolit.Gambar :
bidang atas- yg miring
1 e n t U r.Momen luar di setiap penampqng harus ditentukan dengan meninjausuatu penampang vertikal yang memotonE seluruh pondasi telapakkemudian momen lentur akibat gaya-ga,ya yg bekerja pada seluruhbidang tumpuan pondasi pd satutikal saja, dihitung besarnya.
pihak dr penampang ver
b-b
potongan a-a
bidang atas berbentuh': . r r-
- Eangga.
3.3.1
Irtgmen
potongan
[+
(g=) >1epuadf,o1 rsrsiE = s Pupurrp
g. =
>tePued qB:e Telo1 ue6ue1n1
ertuql qa T
'Br(Uesrs -rpnT ue16eq eped rpqasrp snreq' Tnl uEp' ( >{apuad-re1 le1ad Tsrs ) g = rpqar 6up uoro{ T{
-e{ nPlP tlloTo{ nquns depeq;e1 sT.rlaurs 6^:rn1e[ nlpns pd plp.raur'rPqesrp sn'rPq qsl ( e-e ueEuolod pd ueuour up{eun66uatu ue6uap uE
6un1tq TT spq TrPp ) qaro:ad1p 6^d '.Tn1 ' >lepuad qprp urTp ue>16uepas
-ad 'rpqor qn-rnras eped =*u'fl; rpqosrp snrpq 6uefued ;iqprp,rrp !
'Tnl'qe're z rueTPp Tn{rueu iypp 6uetuBd r6as-rad >1ede1a1 tsepuod pd ?. E ,{f .1e1ad r{njrnTas eped
p?e-rau Jeqasrp sn.rpq - Tri+. , qe:e z urp Tn{ruau 6^( .re>16ues :n[nq
>1ede1 lsepuod pd upp r{E-re L rurp Tn{rruoru bx 1ede1a1 tsepuod Eppd t . E
:toI
;
,.:
1e1ad ue6uap *o.or"ilJ:;'.Ht;:l-81 rsepuod >Jn?un, p[eq uendunl 1e1adTde+ upp ruoTolt e{nu 6ueprq eJelup r{p-6ua1-qe6ua1rp )telaTral 6uBd c-c uB6u-o1od eped r{BTppe np[uTlTp 6uB,{ uauo6
dI
I,f
;,i*
Iri
'c
Tn{ruau 6ued >1dae1 lsepuod {n1un ,6u
Tpulp Tda+ uep 6urpurp leraq s1re6 p.r
-p1ue qe6ua1-qe6uelTp {elaTra1 6ued q
-q uebuolod eped ne[uT1Tp 6uB.d uaurory . q
, ..1,
**'
'eleq 6urputp
PlPq6utpur
' uolaq 6u1pu1p nelp ruoTolt Tltplt, ruolo>1
Tn{rueu bi, >1ede1 lsepuod {n?un, p-e up6uo1od eped qeTepe ne[uT1Tp 6F uenrour . p
: qqs qeTppe 4ede1a1 Tsepuod ueEuecue-rad ruerep ne[uT1Tp snreq b[, unurxeu rnluar ueruow z. e
eq
-8
Penielasan : misalkan
t@
l+l
pondas i tapak berbentuk persegi panjang tr * B
tul. pelat arah pendek (arah B).Setelah kit,a hitung tul . untuk mo-
men potongan a-armisalkan dipero-Ieh = 6r1cm2 (utk lebar pelat =1m)
dan. kita akan pasang tul . g1O.
maka dalam ialur selebar B :
tuI. dlm ialur B 2= E-T-fdimana s = L/Bjadi tul. daLam jalur
2= --L/B + 1
= 16 cm1g
0 ,1920 btg.
atau gla +zu cm
isa luas tul. = (611) 4 16
= 8r4 "*2dibagikan diluar jalur B tsbrmS
sing-masing sebanyak ry ='4 r2cm(pakai g1O sebanyak 6' btg atau
g1o Pcm)tdlaqan arah paniqnq : (arah L)
jika L/B ) 2,maka tul . arah panjang diambil 20t .tul . arah pen
dek, yang disebar merata sepanjang B.
jiXa r,,/4.(Z,maka perhitungan idem untuk tul' arah pendekrha-
nya disini jumlah tulangan disebar merata.
{. Teqanafl qe8er dan,penv4Luran teganqan.
4.1 Teganan geser beton harus .memenuhi ketentuan pasal 8'. 1l ;11'7
, 1 2.7 PBrt 71 .
4.2 panjang penyaluran dari tuI . harus memenuhi ketentuan-ketentuan y9 tercantum dlm bab 8 PBr. Sebagai penamPang kritis penya
Iuran tegangan adalah potongan b-b' spt tergambar tlibawah ini:
1
Zn
3Untuk menghitun?Wa lintang daiam setiap penampang dari pon-
dasi tapak yg terletak diatas tiang-tiang pondasi (tiang pan-
cang ) , seluruh reaksi tiang dari setiap tiang yang titik beratnya terletak pada jarak 1/2 D dimuka penampang tsbiharus di
(6rlem
B.
2 | (ql
misalkan diperolehpakai g1O sebanyak
4.3
fi Lsto = ue{elaT*at: ereluaruas
fr tEr0 =:, uP{ElaT-radr: dela? upupqequad
: n11eA.LL.ZL uep [.[[ lesed [4rfgd 1e:e.(s rr{nuaruaur
sn.req {eluo{ 6uep1q eped ue>1e1a1*ad ue6ue6e1 ,ue>1e1a1:rad n?EnsuBe.rpluB;rad ue6uelp lsepuod eped ndunueu 6ueI uoTo)t-ruoTo{ pppd Z.S
. ue6ue1n1 qaTo InltTdTp snrpq {Tf,Eaede6 qnrnras =l;:;l;ll'lll;',,:;l:T;X;:-"ffinlo*.0
;::JpTeq-Tp upue{a1 lnier.au ,Tsepuod epedt rrrlr.TesTp ledep snrpq uoTo:t-tesep epeed e[:a>1aq 6ued -rnluaT uaurolu uep 6ue1u11r 1eu:ou pnuas [ . S
's
(
'6uecued 6u
-eT+ 6uT seul-6uT spru r s{pa: = U
U g : p-p ue6uoqod 6ue1ut1' 6ue1ut 1 er{e6 {nlun ue>16un1Tq
-.zadrp )tppT1 6uBdrueuad 6ue>leTaqTpepe.raq 6ue.f, 6uer1-6ue11 ue>16uepas
'JTs-rerp 6A qprapp
-Tp epeJaq 6ue.d 6u a
-eT+ rs{ea.r qelrunt = 6ue1ut1 edeg
@" Lr..
-@
: >1ede1a1 rsepubd qerqpq 6uep-T!l eped ( uPrp{6uTt p,{u6uedtuBuad p>[T [ ) , Tsppuod 6uer1 -ralaueTp = q'Eu.Bdtueuad 6ue>1e1aqTp a z/L uetnpnpa{ eped Tou TpTTu uEp 6uedtueuadP{nurp (l Z/ L ue{npnpa{ eped qnuad TeTTu erplue .xaTurT rselodralur
ue6uep 6un11q1p sn.ret{ nlT 6uedrueuad urpTpp Tp 6ue1ur1 ez(86 up)tlpqJ>1e6uerl de66ue1p sn:pq 6uBfr 6ue11 rs{Bar TrEp uer6eq, nB[uT?Tp 6ue,{
6uedueuad depeq-ra1 6uet1 Trpp erplue up{npnpa{ up{npnpa{ {"i"il rt
'n1T 6uedueuad rueTep "6ue1u1f Btre6 ue>11"eq14e6ueru {e1 de66[etp snrpq
'ne[urlTp bE EsudurEuad 6ue>1e1aqTp qTqoT nple q z/ L >1e.re[ eped tpl-er-ra1 edule-raq {T1T1 6uB^d 6uet1 detlas Tf,pp Ts{pa.r upp,
'lnqes-ra1 6uedrueuad uleTep 6ue1u11 er{e6 up{1eq1>1e6uaru an.rnl de66ue
tq,_o
{q,-0.,n:
1
,, dPcATTd6uet1eled
10
g = koefisien yang
dalam keadaan batasdari PBI tabel 10 .1 .2
pembebanan tetap : Cl-perletakan batas = 0r5. dbkCperletakan batas = o'6'O'bk
5 .3 Apabila tegangan perletakan melampaui batas-batas yagn disyaratkan maka untuk memikul tegangan perletakan tsb harus diadakantulangan, baik. dengan meneruskan tulangan pondasi ke bagian ygmeneruskan tegangan perletakan tsb maupun dengan memasang tula-ngan khusus.
5.4 Panjang penyaluran, untuk menyalurkan tegangan tarik dan tekansesuai dengan ketentuan PBf ' 7 1 bab 8.
5.5 Luas tulangan untuk menyalurkan tegangan perletakan harus diam-bil >O t5 B luas penampang kolom atau kaki kolom yang dittrmpudengan minimum 4 batang tulangan
5.6 Untuk melnindahkan gaya-gaya horizontal (gaya lintang kolom)kepada pondasi, jangkar - jangkar harus dipasang secukupnya.
6 . Pondasi telapak yanq meurikul kolom bulat.Untuk menentukan tegangan tegangan dalam pondasi telapak yangmemikul kolom bulat atau kaki 'kolom bulat r s€bagai bidang mukakolom atau kaki kolom dapat dianggap sisi sisi dari suatu bu-jur sangkdr yang luasnya sama dengan luas kolom atau kaki kolomtersebut.
@ryh%-\
;{*^aama
diambilaa
.g
.g
7 . Tebal tepi mi-nimun
7.1 Pada pondasi tapak dari
7 .2 Pada pondasi tapakht Vl 5cm untukhr p 30cm untuk
beton tak bertufang, diatas tanah, tebala
dari beton bertulangpondasi tapak diatas
tiang pondasi.
:
tanah.
la
pondasi tapak Fdatas
0cm
TT+PT
4-y'(tucg =
TTque ) uoleq... a.. a a a a'a a
=
lnruTfes+{-lqtl qaf oradtp ( Igd
I-;
z' r' oJ reqEl lrrrTrdq 3u6eeq1ur, ) =dqz
.r{'(qz+q+E)z =: sued rasa6 pTq sBnT
(t{Z+Q+e17 =
Z({+q+q+P) = suodt{' ( t{Z+
el z' 6Q2 - ( q+q l ( q+r ) ollsub -d
O - tE{Ttrerr ef.eb!: BTlsrTe usEpraT-mTEF
Blupru Tequl/TDEuTl ) rl W' I qe>16ue1 TrEp TTqupTp T uep g
7, ' g ollauq-?-=u
.olleu D uslnlueuoE .zuep g qaToradTp' r66ugqas
'nTnqEp uElnxuexTp T nElE g u"p'gr{U-.r'saq rnq.la{Tp
tlEuBl12 Eup,rTpqeuEl,
) fr.# =wue1pd tot = rTEltElTp lsrpuod lEragTsepuod lrfed [email protected]
'uEDunlTqred tl
'(ELre{eq {EpTl ueuou,sTrluas urq
J'9'J'A.{roBt TBEpUOd Ur.{ruecueral g- [-A
. TUT urrnltrad ueEuap TEnBeB bI uenluaxat up
Eluaqal Tl{nueueu SntEq Bnraueu tedet lsepuod upeupcuared Z- gqEUEX ?tTue{eu dlsuled eped uE{
-resppTp upp Ts{nrlBuo{ u3p qEupl lprTs-1e31s ue6uep Tpns-as srt{ tlPuel ueue{el uelEequred leua6uaru uede66ue de11as L . g
'snreuau 6ueI 1e1ad nlEnE qeTo Tn{TdTp 6up.t 6u1pu1p nplpuroTot I TrEp qTqeI EuEurTp ( *uoTlEpunoI 1rrr,, ) +Tter lsepuod
nplp (,,uoTlppunor 6u11eoTrr) 6undeeal lsepuod {nspural snraueur >1ede1 lsepuod tnlun n{ETreq lntTreq upnluale{ - upnluaaex
Q=
Q+
.E
rase6 6uep1q EuTITTelt
tl
' ( rnTr[ Tsppuod] snrouau ttclrl Tsrpuod . g
12
4.
-untuk Iseagqan b3ta :
P,r-q;Ltto( a+h') ( u+h ) 'T. = rni"c bpu 2 (a+b+2h) .hi.diperolgh h = ......
Check terhadap geser.Penampang kritis adalah
. 1.
Dt_t = gnetto'(,1'./2',:7=Bt,b 0 1875.b.h
catatan: r:
-gffid1m keadaan ba-tas
,rit.*ilf,....r;, . t,{.
:
,lpotongEn 1-1 :
<"2 . ,disini b adlah L., ri_
bu ( ltre*gdaan Ualps I .
tanqkah perhj!,.qnqqn :1 . tentukan ukuran pelat
. idem spt terdahulu.
2. grr"tto=* = -.--..'
3. l{enentukan h 3
at,au
Kolom
b
*Tuo
i ,,gonq.n, pgla potgngal 2r2 untuk arah B.
a:ft.dpt 'di f,en1q,q5an= . . . (amb1t| S= 0 atau O ,2 )
B u n d a ra
,j"rf ' '!: ;. momen.,,pada potonggn 3'i3 untuk arah t..,, 'r -. ,"
' ': 1,^ , I L-b\zi Mt-: ;:' Qnetto'(ltmeter)( 2 t
tIaaaa.ao...
.6. Check berat sendirr qondasi ( setelah ukuran diperoleh )
Berat sendiri pondasi = L. ht . 2r4 = ..... ton' check apakah (r Ot P ( taxsiran semuila ) .
pon{asi.
\
,!P -gnetto t 7( c + h).h) =
TapSrfin,th=o....'
4. Checl,c oeser- :, penampang kritis: potongan I - 1.
ant
* u,it'.,,t T-.,
' .1 :e6>1ues :n I nq tpB I-tiau ralsue.rlTp {epT? repunq uoT
6t ) 'uolo>1 e{nu eped .rnluaT uaruory
(--
0
@aq _z ) dqz
JLI 'uq 2 -, ( t{+ L tz lll L
(t{'(q + tuz)
. ollaub _d
'lqaTo;ad1p q
=
vt
:q
1L
ue{nluauan 'E
lavl=
o1laub . Z
..... = ZU
d tot = -rTs{elrp M BueruTp
qP"'!> ,t"01 , = qeu=lo
'Tsppuotl 1e1ad uetn{n ue{nlua1 ' I
' uebur1lrqred qelbuel
' ( efra{aq {P? uauou) sTr+uastupqaq Tn{Turau'repunq uoTo{'.repunq {n+uaqraq Tsepuocl leTad Z' 9' J'A
' lsepuod TJTpuas lP.raq {caq3 ' 9
k
I
'nTnp u={nlualTp zj7
- -T' ( .ralaul t ) o1laub Z-ZL
'nrnp u={nlua+lp t-t
l"r'(.ra1au t) o1laub LZ- ,. t= z-zw
@no-
= 'J nPlE
c' zggS'0u= Z" )L i t =e.dusenl 6ue,{ .re{ffi [nq
.re>16ues rn[nq TSTs euBs
TpE[ uaru qeqnTp -repunq uoTox
.a*2 )
= q PuPlUTp
7T'(q; c
L
td' q'5'0C
q'q'gIg'0 ee \ */Z
IsfCI
) o11aub
9-l
= L- Lo
CI
's
t:Sl
14
=
trkmuka
t'I
",, 0r2
'!; ':' t' Rt + RZ
segmen yulq diarsir
G3 *?r,qpe*o
kolom I ' 'i
*,').enerto, {tr (3 ":
(Rz -.,R1 )(3 *r'- 2 Rt R2
dicari, ingat: lebar pondasi
T ksrom =*TLsilahkan lihat contoh soal.
Titik berat segmen yang diarsir terhadap ttk pusat lingkaran.
')
= r /q fL^trro ,a*rl -r /arc 4$,e*rl
t/+7c wl *? l'
o,otn3 * n,n, * n?lR1 *'R2
berat segmen yang diarsir thdkolom :
0,6( "3
+ RrRz + *?l
x
'ut,,
Rt * R2
(3 n? 2- a ,'1,
i.'d
,.lr
,'f:
Rt
f
Gaya keales padaiir t
= +TMomen :p,ada qpka
-1 -
= JL ,_24 r (Kz
i7c=--: 20 :netto
penulangan dpt
5. Untuk jelasnya
.6 . 3 -Pe1at
Rl R2-2 Rt 2 )
2R1R2
2nfpada potongan muka
Rt
? 2 n2)1
)
v.1.'
dasi berbentuk i, kolom berbentuk buiurkaT atau p,grseA.i r menli..Ful., beb+n ekqentris.
Kita perhatsikan dinding penahan tanah ( "retaining wa11," ) se-
perti tergambar 3
gaya.. horizontal akibat. ' tekanantanah (,=PH ) akan menimbulkan mo-
men guling ( "oyerturning moment" )
yang, akln dilawan oleh momen pe-nahan yang ditimbulkan oleh beratdindidg, pelat, dan tanah yang me
nekan pelat pondasi. Untuk itulahpanjang L, dtbtllt lebih besar da-ri pada L2.
r# l#l
8(a
€ = rT
= e/fi PupluTp
E',T' 16 f = u
rt)T g/ l=a
'tuoTo{
-{a eped
d=ax
W
sE depeqral a sPATSTrluasepP-raq ueDte u qPuPl Tsteau
Z/A' E
x- ( l{=) x qPrc uauou uep sTrluas drn{Tuau TUT r{ErrrPqTp TsPpuod
(+Tr)T=b,a'8 rc ;v=b e'dg ' d
@F'lsepuod upeupcua.r
Ter{ ) '{Tf,p1ra1 uEtPr{Puel uel6eqas Etpu T
,z6 p/s tg .r"saqasue{a1.ro1 r{PuB1 E{eur T
ouPtalral qPuPl P{eu T t = o E Tf
'd e[ra>1 s.r6 ue6uap 11dru1;aq qeupl
Ts{Peu'sTirluas{a qpTepE uorot tplaT ,
Euprurp, ruolo>1 s;E eped e[ra>1aq d edeg3 qqs qeusl ueqeuad 6ulpurp {n1un
?a
r?(" ozJ
Tsppuod._ue>treque6 plT{ 6ue;e>1ag
T'8a'd
=Vx=W
?,"I gPuEruTp 9= L*r
xT..--:-I
-xl A' ,il
V =b d
F',
t6
M .xVII:
Y
jarak ttk yangjarak ttk yang
ditinjau thd sbditinjau thd sb
I
vx
x=y=
ft{I
%-
Vt1.7 Pondasi menerus untuk memikul dinding beton ( "wa1l footiLanqkah perhitgngan j'l . tentukan lebar pondas i .
berat pondasi per m, = W
. taksir W = 10 Z p
( P= beban dinding untuk sepanjangpeter ) .
__ p + W ./_.?
0 tanah : B(TI \ 0tanahB diperoleh
2.
3.
..t=PYnetto B(1) o'
Check geser 3
h ditentukan dahuluperlu dicheck ) .potongan 1-1 untuk 1
... t/^2
( geser pons tak
meter :
Dt_1=enetto( I /28-1 /z a-l/Z h) . I
0rg.b. h
pqnjang pelat = 1 meter.
o (1/2s 1/z ul2 . (1)
dicari (ambilf,= 0 atau 0,21= 208 A utama.
7tbu
penula+oan : Untuk
Mz-z = + gnett
tulangan dapattulangan pembagi
Jh
l.
uo a)
,\
I
Io'u: z )
( q+ [ e+q+ |,'t z
z$ + LP)o11"ub -d
q '(t{+[r*q*[.) z
=dqL nEaE
ndo -,= '! : I TUOTO>I
z$+ L=)n, o?1arb_ [.d' I L uPruBTPr{ eped Jpqueg
td .( E- t'o*i:"ut, -
o = td0 = xpyxwe
(e x)la E'z*'
= xeu W lpe[.....-x
of,.f,.eu- g'x' r ' D
1---- xeu w
o?laub g = *t,.I
t
'TsPpuocl leTact qaTo fn{TdTp
'P?P.rau lelTs.raq lsepuod leT6d eped e[rarlaq 6.d qeuel ue6ue6a1 e66urqas
' l sepuod 1e1ad le.req >{T?T1Z a,
I a aluelTnsar e66utqas edn.r upT1Turapes
'O uep I{ tuerbelp ue{nluauanT'g=
olleubzd + [d
of,fau- ' !' D UE{nlUaUaI{
7 t_- t
T'zdue6uap +TdtuTf,aq Ed uep
uP{nlue?Tp q uep P
a,
'€
TrecTp +dp T E{Ptu( ' ' ' =g e6:eq nT
-np TTqUP ) Tnqe?a{Tp I{eTa1 E P{T Iqeuplr)
ffi =_o
'lsepuod Xelad '{n uetn+uauau 'I' urbunlTrlirad qP{buPT
td=( Za+ ta ) t0 t rTs1p4Tp lsepuod lerag
Zd = Zc uroTo{ pd e[:a4aq 6^[
! a =
t J ruoTo{ pd e[ ra>1aq bx ueqag
ruoro{ luoror 1*- zJ uepl3
tpuod loTrq eduel '5uef ued lbasracl {nluaqraq TsEpuocl leTed J'8'1'r.,1,
*d.
,*
i
'uoTvfrsvqNod8' L'A
ol
D,2L
It^I t(,kL_
III
I
I
I
-l
I
::t
ulF,a
nffi+h-3t
-frts-u2
:
'-i xFl-f-t--
-
f:E+h--al./l{-n
Kolom 2 :
f-.F
18
q {rf( r In,,,
D.r'u2 Ynetto, u(u2
TLbp,'- 2 (ar+h+arr-h) .^P '. Qn"ttoiur+r,)2
at.au T--="Dp
h diperoleh =
( Zon
langkah ke 4.:
palapangan.
, B-a1n=*1
2
2 (ar+h+h+a, )
6. Penulanqan pe]-at pondasi. :_
* dalam arah " longitudinal', ( searah L ) .berdasarkan bidang momen padaditulangi untuk momen tumpuan,
,r dalam arah "transversal " ( searah g )
untuk 1 meter lebar pelat :
MZ-Z= 2 gnetto.bi . ( 1 ) dimana kolom 1
Mz-2= * gn.tto.b? .(j) dimana bz = "+....ooio korom z
* jangan lupa check tegangan geser pelat.JL
V.1.8.2 Pb1at pondasi berbentuk trapesium, tanpa balok pondasi.
Pada V.1 .8.1 telah anda ketahui bahwa'iika P. dan p^ tidak sime-, J -1 _?,
tiiS" terhadap tt.k berat pelat (besarnya ) ,maka ttk berat pelat takkasi r*=ultante p" dan p^.
t2berintpit dengan lolDengan bantuan"pelat pondasi berbentuk trapesium, hal ini dpt digtasi (ttk berat pelat berimpit dng ,resultante gaya p t dan p 2l .
q*
+ilz
{oTeq
leTa
E-P '1Od!c
Ppuod{oTe
1PTd D'J
tJ 3 [email protected]' ppP Tsppuocl ,toTpq, bupLued lbasrad
t'8'['A ?dS.g uep V r{aTo.radlp
(unTsade.rl 1p-req {?1 1e6ur) (Z) .
X+e=z- |"d l__g = xEl:J\
x = [ luoTo)t pq1 D'J >1e-ze[
qeuel . T
(l) ""'( Ea+z a*L al z
=B*V
t. (a+v)fI{EuE1J
) --
=A Ea*Za*[aza g [a afuelTnse] Ts
e>toT 6up lldrutraq ( e .31 1e1ad ler-aq {11 ;e6e g uep V ue{nluauau . I
'uebunllqrad qeqbue.l
(za r ta) got = .rrs{e?rp' l sepuod T-rrpuas 1e-raq
Z llroTo{ eped Teurou B.dB6
t uroTo>t eped f Eu.rou edB6
'TJecTp ue{e g
' TnqeaeltTp
rlnluaQ- E-g.[-AqeTPpp e.,{u1n I uelas qe>16ue1 - z(Z) uep (t) upprups.rad Trpp
"" tffilf = (x + e)
1e1ad 6untn pql C.J >[p-rp1,
: rPqrrg({, '
E-=d.
Z- = cl-
= taUPP V
T,q,Bilil
{r
z a*{a=u
I
I
6t
20
tanqkah perhitunqan.'l . L ditentukan dahulu't B akan dicari.
a dan b taksir dulu.1 . 1 Menentukan lqFa,si resultante ,qava:qava vanq bekeria.
Jberjarak x dari as kolom 1.
_ - Mt*M2*M3+p1 (O )+pZ ( 1r)+e, (11+12 )
Pl + Pz *P3
1 .2 Meneirtukan " eksentris itas ( =e )
1.3 check:'apakah e (r,rO L, jika yA ----) oK gaya.
tdk ----) uk. diubah , sehinggae ( l/e L
Ptot.l* berat sendiri 6 Ptot.l'"1.4 g1 <6+
.L
dimana P. = P. +EOEAI I
berat sendiri€ , L diketahui sehingga B
l.lenentukan gnetto 3
a.P total.
L', ,
penulangannya, berdasaruk. balok dan penulang-
Pz + P:
ditaks ir 20 t,dapat dicari.:
2.
*" 3.
Ptbt"I , 6Ptotar'€t-:netto
B'.L L B. L2
Itte4entukan bidanq momen, lint
-
.\
Q2
.
oo
4, Merencanakan ukuran balok persegi, danmomen max, linta.ng max dapat ditentukanan.
JN,
'qs1 6unqnq6ued {ofeq TnTpTEru ueltnq'rsepuod 1e1ad TnreTaru qeuel a>t upltqppuTdTp ruoTo>[ upqag
. 4ede1 TSEpuo d, T ue>16unqnq6ueru,,del1s,, )toTeg
edulsepuod 1e1ed -rpsaq-:adrp ledep {epTa >1ede1 t sepuod p}tT I ue{eun6lp ,,6uT1oog de-r1g,,
"buT1OOdd r r 1 S; 6'1.A
'Teos qoluoc leqTT ueltqeTTs elusela[ {n1un' (g qe-reas ) eueln ue6ue1
'9
n1 Z0Z ue{eun6;edr p
(? e>l'Tnqele{Tp g
qe>16ue1 ) Tnr{e?a{Tp=p( nTnp up{n1ue1 )
1e1ad JT1{oJla Tpqal=q
( r qe-reas ) T6equod ue6ueTnl'ue1nJuelTp ledep 1e1ed
z(&) '['o11eu [b 1 = x,ru w
{nlun : uElP+Pcue6uelnuad e'E
z's
['(Lr-p-"4
t{'['slg'0q-,I xptu
CI
.o?1au t5 =*r*c
'XetU UaUOI^I
q' !.' 6'O _nq ? no-,o -- flXplX - 1' LtE+so
L
'.re1au I3 runrurxeu -rase6tunrurxeu rasab : p-e uebuolod eped tcar{J [ . E
t_Jq x p
J-=
t sepuod xoT eq-, uP-rn{nP-e' aoclf-e'+o
!,l
r-,,,i"
Ii
rl,,rgt,
T
I
1'fft
p
uP{nluaual{
22
]
'Luas pelat pondas i dipil ih sedemiki_an rupa sehingga tegangan yang bekerja pd pelat tsb bersifat merata danresultante gaya_gaya kolom berimpitdng resultante tegangan yg bekerjapd pelat pondasi yg berasal dr tanah.
I . It{enentukan uk. pel_at londasi.uk.. pelat pondasi diatur se_demikian rupa sehingga lokasi re_sultante p
I *p2 berimpit dng resul
tante gaya keatas (pd pelat) yangbe.ra sa I cln r i tnnalr .
berat pondasi ditaksir=l 0t ( pl +p2 )luas pelat pondasi total.
=B ( L - +T.- , -P total+ berat pondaSiE,anah
(*)Pt *PZ ber-lokasi gaya resultante
jarak ; dr as kolom 2
x = Pl 'L+P2(o)
P. + P-r -2berjarak dari as kolom 2 sebesar
titik berat 1uas pelat gabungan
= (8.Ll )(L+1/Z ct 1/Z Lr )
B.L1+ B.L2
2. cari 9netto 3
enetto = n#r= .....l '21
idem spt dijelaskan pada V.1.6Perencanaan balok "strap, ._
Beban terbagi rata yg
9netto.B.l = ...B = lebar pelat dan 1
1 meter.
persamaan, =n' , " = (B 'L1) (L+1/2 c1 1/z tt )pt+f;
harga B(L, + L) ambil dari (*).akan diperoleh harga harga B rL 1 ,Lz
3.
4.
-. t/ndisini adalah panjang pelat diambil
bdlox " strap,'dinding.
ap nPle Tsppuod toTpq eduel up{pupcua-rrp. Tsepuod toTpq ue6u
ledep 1e1ed Tsppuod.
6ua1-qe6ua1 , eped nlTBr{ .r;l;t::":;up6ue6al pupruTp TspltoT eped edu>1rEqas gsl ueTauaqraquad . e.reluauas
ueltT?ueqraq unTaqas u o[ x ru of TPd-pcuaul snrpq uproce6ued p{prtr, ueroc-a6ued r{rTa+as 1e1ed pd {p1o.r_{p1
-or e.fiugpe[I"* ue{repulq6ueru {n1un
-uTT upp ueuou {nrun Ts)ra.ro{,";I;'::::J,*11,..-il;T:::=rT;llsepuod 1e1ad I{aTo Tn{TdTp uoTo}t ueqaq pue-rp{, qe;e 7 urrp :rn1e[ lsepuod le6eqes Euepuedp ldp 1e1ed tsepuod p{pru rpseq pru=s ruoTo{ ueqaq ueqeq uPp rnlere+ uoTolt {plar {p1aT euprurp uppppa{ pd .
r,{Puap senT tog TrPp qTqaTspnT upp , TTca>[ qpup? uTz T
ledecueu up{n1:edrp 6^ue6ue6e1 e{T I up{pun6lp
'ueun6ueqtsepuod 1e1ad1e1ed rsepuod
1e1ed TFepuod
'l.uopepEi6g]JPr =',* fl
I
uoT+PpunoSrlBIilll+PTad tsEuoa0 [' ['A'Teos qolUOCIPI{TT uetr{pfrs eduspTa[ {n1un
'6un1TqTp ledep ue6uelnuad
(zr-znf-1,
I
o
I
I
I
I
[.. [b-
'r'?r-2i.. b
z,zb- (zr+zr).b
tu/l..... =l.fI.o11oub_biil
Z^ut/+ 'r... =
EIc
=IarI
ffi'l
l+ts{
e,
-Z
vl
I
24
Ada 3 cara perhitungan untuk pondasi pelat :
1 . cara konvensional2. cara "finite differencer'.3 . cara t'f inite element,,
Dalam konstruksi Beton f inikan cara ke 2 dan ke 3 akan
hanya cara ke 1 yang kami bahas, sedangdibahas pada IVTEKANIKA TEKNIK V.
terqambar : (tanpa balok pond""i)- Beban-beban kolom = p rs/d
1 . Menentukan lokasi resultan_te gaya-gaya vertikal.
pusat Statis momen thd sb X = 0
Pelat ( P. + . .+P 12) .y=p1 .y l*p-Z.yz
9min D" 12
': + ..t=
Stati s
(e', + '
. 'xzX=.
: : : : .
- P rz'Y 12
momen thd sb y = 0
.. 1Zl .x=Pl.xl'*Pz
+ ... + P 12'x 1z
g maxl_n
9*.*2. Itlene4tukan eksentrisitas : .* dan
"ya x (lokasi resultante
pelat ) .b i (lokasi resultante
sat pelat ) .
1
= 2
1
2
gaya berada
gaya berada
dikiri pusat
dibawah pu-
perianii.jrq, tanda untuk harga z
Lokasi resulranre eaya s [:].\\
"ta=\\ u.,r"r,
3. Ulc_qraO_ _pgfat pondasi.
e* d,an "y :
dari pusat pelatdr pusat pelat
dari pusat pelatdr pusat pelat
e*"*= +tr#+ry++(d."nahdimana M* = na"l"t . "y
', = Ptot.r '€xI-- dan I__ dapat dicarixysehingga B dan 1 dapat ditentukan.
\I
,, aqT T ' TsPpuod ?eTad {nlun ( sP )Ja1ua3, = +-l: ur?Elrc
xo '(---'- rrr + rr + r leTad qpupl rs{ea-r =
TBlold : {caqc
( rrrlprad spnr ) ( r"*lr"uu)= rrr 1e1ad
eped efre4aq 6^ qpupl rs{pa-rZL* --d +
1 rrlelad sBnT ) ( rrleledu ) =rr
1e1ad eped e[.ra>1aq 6f qeuel rs{Ear.[ [a*8a*9d*Zd=rr 1e1ad t1n Te1old
z - rr+e1eduz6 + t6
,uffirb: II +rTad rnTE! rped
1 r1e1ad spnT ) ( r+eTedu, = r
1e1ad eped p[:a4aq 6^ qeuel rs{ea.ro I
a* L a*?d*
[ d= r uauEas >[1nTe1o?d'
' ( Pler 16eq.ra1ueqaq ue>11pe[ elTt urnTsede.:r1 upqaq )
ZI= eerad- xplu- ' s b+DxBlub
se+.8 T6BqTp
r{aToiradlp q
,,,ds+Z)ry. = naq2
d
: ..o.. = qATO.Iadlp q
,r'u: Z> = n6e2
6a*9a*Ed-rn ppdZ
>pn Telo?d
uTub
: IIr lEf€duautm EpPd
uTur
t6t6
Q+
H4
W
(Z/rl+Z/c+TlZ+q+c=suod rasa6 6uep1q 6uTTTTaX
66urd
: rlbbuTcl ruoTox
'r lrTact rnre[ rpPd
TUT TPT{ ruTp )
'JPsaqral q TTque
l
g' 1:n1e[ E
,,.{'rrW.
9Z
'Tstpuod 1e1ad feqag 'I
il
26
6. uenentukan.diaqram, {jqn P utk masinq-nasipq jqlur'belat..,
Dengan cara Cross atau pendekatan yaitu M }ap max = 1/10q. 12 dapat anda tentukan momen dan lint3ngnya.
' .**'i r, ,, rF : I
7 . Penulangan : idem spt pada pondasi lain.Untgk jelasnya silahkan lihat contgh soal
v.1 .1 1 Pondasi tianq ('pile Foundatio+" ) .
--
--
V.1 .1 1 .1 "Precast concrete piIe".A. Penielasan lunum :
Tiang pan"cang yg di cor dahulu sebelum dipancang dinamakan
"precast concrete pile" (tiang pancang prekas).Jika sebagai tu1 . dipakai.baja pratekan(prestressed wire),di-saTplng !'r11. biasa, tiang demikian dinamakan i'precast pr€s-tressed concrete pile " ( tiing pancang prekas pratplan ) .
{ '(ri'-
Bent\lk. pen+$pang tianq -pancanq p.reka5 :.{lr
::;l'.:i::::: w @ , ,{.berongqa :
::il::';.n,*Kami gambarkan tiang
g^"^W",benrukbujur sangkar.
angkur
latbaja
r1 '
I
.1
Pada jarak 2D (D=ukuran penampang) dr ujung tiang bawahrsgngkang .dipasang lebih rdlat, dan pada jarak D dr u jung a-tas , sengkang j uga dipasang Lebih rapat . 'h'''
Tianq pancanq berbentuk seqi delapan.
elat
baj
baja
- se.ra{ qeupl rpdecueu Ted-ues 6uecuedlp 6uetr 6uecuBd 6uet1 qETepp aTTd ,,6uT-reaq 1uTod,,
' r burreaq lurodr, slua[ 5uer1 6un>1hp-Etreq - p
XfNXg; eped TcuT.rad;a1 qTqaT ue>1se1aI Turp{TnltTd elep 6unlrq6ueu >t1n p.rpo ederaqaq EpV
' ( r'svoNoduE{E 1 6uecued 6uer1
t nlrad Tl.reraq .ralaru 0 t =-au 0t ueTepas ( se.ra{ qeuel
uo? 0L gg
uol 09-09
urc 0? x 0?
ruc EE x Et(uo1) d6uet1 '>[n
'6uBdtueuad ueirn{n eped 6un1ue6;a1 6un>1np e^dep e^6u;resag
' r +seca.rdil buPcued bupTl bunlnp er{eg - D
' 1 6ue11 P{nu V
eped)eTpasf,a1 b[' 6u
PqnT urTpa{ ue{{nseu-Tp (e[Bq Trep)fIBq
' uTeT 6d 1 :e>16ups -rn>[1n'Tn1 e[eq
:[>16ue
'Eb.buo:raq , repunq 6ue5uEt-5uere {n?un1e>1ed1p 6u,e?t ue6unques ualsTs
'6unquesTp 6ued ,6ue11
6upcued 6uer1 6ue[ ued ue>16uepes -ra1
ledues 1 buecueuau ue{e elT{ ue}tTpsTp'buecupcl busTl eped uebunques uolsTs .
fl
'6uedueu-ad sBnT ZZ
6updueuadsEnT I Z/L I
6updueuadsEnT Z V/L t
6ue I ueuauue6ue1n1 t
0?ET0t(urc)
unuruTru cl
s[-E'0tg'o l-g' Lg'L1 :e1eu ) T
#*,
u9lsTs
'uB6ueTnl talauerp 6uetl 6uetupd ue6unqnq ue>tTraq Tue{ TuT qpnpqrp
28
Rumus
Qult =
ct l_mana
" statis piIe,'A( 1 r3.c.N"* /.o.*n*o r6 .7,,.o.*tr )
A = luas ujung tiang bawah = b x br__._I +I A !-uL_-I T
c = kcihesi tanahr\T ^!- -ra t a2N.= ctgg (
2 "o"2( + Tc2
ar:aq z cos'(q5o+6/21
N= +qfl ( *of
-1)cos- glapisan tanahkeras
+ g )-1)
= e(3/4rL ' fr/z) tggaK
D = dalamnya tiang pancang../!
T = "unit weight' tanah yang berada diatas lapisan tanahpenyangga ujung bawah tiang.
b.
mq. I = "unitharga
2sanz ; jarak
tanamnah.
p/= faktor yg berhubungan dng tekanan tanah pasif dlm zone ff &III.
keras -
Qult = Bebas batas yang dapat dipikul tiang.b = ukuran penampang pada ujung bawah tiang. '
rr"r n.i l-,, r - QuItP = daya pikul izin tiang=ffits.F = "safety factory" = faktor keamanan
diambil 2,5 s/d 4, biasanya = 3
.
. "Friction pile adalah tiang pancang yg mengandarkan gesekan selimut tiang ( seluruh permukaan badan tiang ) dengan tanah.tiang jenis ini tidak dipancang sampai tanah keras.
Untuk lapisan tanah berupa pasir dankerikil dipakai rumus sbb:buft = 0 ({.2*q).K tgfr(rumus Ireland)dimana 0 = keliling tiang = Q a
weight" tanah (ambilrata rata dari lapi-
tanah sepanjang tiang.t,tk berat tiang yg terterhadap permukaan ta-
c+s_x gnEle
(c + s) u - g
ET,'a = c,.rauueq uea?s $ adtra {n1un
Itz = 3r rauuel{ do;P, ad^d+ {nlun -
D ue{euleu A/a$/a+s)u=s
'EuBcuedTp 6uBA 6ue11 6u
-n[na{.rauueq efruqnle[ 1e:e[ = Lt
( 6{ ) .rauuEl{ :=1": :
}iq'll=g
g,Y//-
' ( raunuEq Tf,ep
,,dpc 6u1a1rp,, 'qeue1'6uet1 epci ) T6f,aua ue6uelTqa{ = T( uc ) T TE)t ede:aqaq >1e^f,ueq
5s -rTq{eiro1, .rauuletl u 1n:1nd1p lees eped 6ue11 ef,u>1nseu = S
( 6)t ) qPuPl ( uPqel cr{ep 1 r acuPlsr sa.r |' = u( ulc 6>t ) r rauuetl,' ueln>1nd 16:aua = g
.T + SU = E
'6uecuedTp IPES eped 6uet1 qaT
-o up{n{eTTp 6^( Eqesn + r6raua ue6uelTqat 6up 6uecueruad IETP,r f,au
-urpq,, upTn{nd r6.raua uetpupfruau uele[ 6up l{aTo.radtp srtueurp snunu
lPESsTueuTp snun.I qaTo uE{n+ualTp DUET+untnp eAeq
'UIqNOS TTseq qaTo.radlP tuep d
I
9e+0'J ' v'd
'6ue11 6uTTTTa{ = 0
(lnc/b4! Ts>tTrJ TPTTu = tqeApq 6ue11 6un[n 6updrueuad senT = v
(rutc/b>11 sTuo{ TeTTU = d
TerTu TreP ue:tn+ualTp 1ep butTlntnp e' aTTd r uoT1cTf,J,, uep ,,6u1.:eaq 1utod, TseuTquox
d
'urqNos
s = TTqueTp er{ueselq 'uEuPtuea{ irol}tEl: = .{'s
rG
'c,{' s
=d lTto
-Je>[aq 6ue^,( dela1 1,]Ts.raq 6ue^f, ,,a6-reqcrns,,':;I :'::r:: =
( ,,uoTlcTr] TPula?uT, ) qeuel :rasa6 lnpns =b
fr
x
:5EE5U
0 'J + q''d = 0
SL'L = Te.raleT I{Pue? ueue{a1 uaTsTJaoX =
30
=Rll='1 tiang S . F
. S.F,1 "safety factopl' ambil = 6
D Penoenqkatan tianq pancanq pre\as , j.'fd. diangkat pada kedua u jung. ' 'r' :
r'i:'
Itl
q
max = l/a q Lz
= berat sendiri tiang
b. diang
c. diangkat pada
d. diangkat
6
tiang
Mmax= l/a q L2
tak 1/l Ljar
tu.
dari ujung
M max = l/18 q L2
dari ujung (pada kedua ujung)
M max = 1/Sz q L2
dari ujung (pada kedua ujung)
l/q L
Mmax1/s L
M max = 0,021 q L2
'J Jarak tianq ke tianq pada kelompok tiang ('bile frouo" ) .
@s diambil 2 ,5D .- 3D
D = ukuran tiang.
I
II
"pile cap"(=Kepara tiang)adarah perat pondasi yg bertugas 'mengikat tiang-tiang menjadi satu kesatuan. Te
ba1 "pile cap " minimum 30 cm ( pgf ps . 17 .g .'21 .
salah satu
pada j arak
diangkat pada j arak
Catatan :
COooo
oo
.6urpuTp rn{rua*ffiu sf,q'e 6up5ued 6uet1
6u1pu1p TntTuaru@
sl8'sL8'
sL
slg
o
I
6uer1 g
6uer1 6
6uer1 L
@
6urpurp rn{rurau "i, -Pq ,[ 6uBcupd 6ue11
oocooc 6ue11 g 1
oo ooo6uet1 t t
6uet1
nele cooo
-oOoo
tt
s[?
6ue11 E6uerq '7
ry+{
I 32
fti meencanakgn "pile cap" (Kepalp tianq).\-/Pile cap berfungksi utk mengikat tiang-tiang menjadi satu kesatu -Errr r dan memindahkan beban kolom kepada tiang pancang.
Dibawah ini tergambar denah "piIe cap".R terletak pd jarak .* dari sum
bu Y(grs berat pile cap) r.y darisumbu X ( grs berat "pile cap." ) .Gaya yg dipilul masing:6.sing ti-ang ditentukan oieh rumus 3
M--.x M--.yp = W + "Y-l= + "x'xY
W = beban total yg dipikul tiang= beban kolom + berat sendiri
pile cap ( taksir ) .
= jumlah tiang= W. eyx= W. exy
=jarak tiang yg ditinjau .thd sbY
= jarak tiang yg ditinjau thdsbX
pi 1ecap
Jn
T pilecap
n
M
M
x
v
"Pile cap" harus direncanakan memikul momen lentur, geser pons,geserakibat lentur.Kita tiniau dghqh "pil-e. cap" dibaEirh ini :
lom Pqrhitpnqan ltomen lentur.potonqan 1 -1 :
ditentukan Mt _, akibat gaya tiang dan berat sendiri pile cap.
Perhitunqan qeser ponF :,
'Prbp = #*ft(zon
D*7 = ++f;fl*<z bP'uLbpu 4
Catatan : h ditentukan du1u.
+
Perhitpnqan qesgr . lentur,.Potonqan 2-2 :
DT6 = d]6r5;5(ri'
atauDruo 5fr;t(e*b,-,
.'{rrI
F"t+
P-E= DiIE-EE-e
=w
dBcATTd
EBcafrra
dBATT
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
,1
I
d
I
I
I
I
I
Ir r
:lIrl ll
I
I
I
r
,+
I
I
I
l?alrl.
I
I
I
''lI
n
I
I
I
I
I
I
I
I
afuetelal:ad
uolaqlET6u1pugf,e[ eq uoTo]t
r TTE
rl
34
5. Jika
{
kritis untuk
Ke1iling kelompok tiang =
Gaya geser pd potonganb-b = 2Pl P4 + pf2p2dimana :
nemoto
I I 7r5cm5cmp, = R
Pt=R( cl)
ct )( *1
6',
l-!l_2:6( *+z-rl-:6 c2)
cz ( ts cm
Jika C 1 ,C2 berada !
i. diluar potonqan b:b( bukan dlm bagian ygdiarsir ) .
r,=n(] + * cr)P2=R(*+.,)(** czl
b. didln potonqaq b-b(da1am bagian yangdiarsir )
Pr=R(*- * cr )
P2=R(+ * cr ) (* *czl
: paya dukung kelomp.ok tiang pancang tyg diikat men jadi satu kesa-tuan oleh"pile cap"lebih kecil drpd daya dukung 1 tiang dikali-kan jumlah tiang seluruhnya.
m t-i ano Jarak tiang ke tiang = ss diambil antara Z rSD 3D
D = ukuran penpnpang tiangm = jumlah tiang dalam 1 ko1om.n = jumlah tiang dalam I baris.
(r +
2 (rr( (m-1
2 (m +
B) .
+Bt)s +
n-
= k = m.tts
2
) + 4 D
(n-t)s)2 + 4
2)s + 4 D
m tiang
Lt= (m 1) . s
f, = Lr + lo + lo = Ll + D
JumlahBl= (n
B = Bt
tiang- 1) .
+D
' .res
-aq eues 6uer{ [eTxe e^f,e6 Tn{Tuaru, >loduibTalt ' , nles eped ( aT Td irallpq
{nseutra1 ) 6uecuBd 6ue.11 pnuas ' Z
: . TP{T?
raa 6ue11 11;adas ( leuuou ) TPTXP
er{86 Tn'ilTulau ledep ,, e.T Td Jralleg r ' I
' ,, oT Td .ra?1pq,, up{-Bun66uaru 6uefr 6ue11 >1odtuo1a>1 up{puecualau {1n upde66up-uede66ug
t.pfps uo? Sr0 tesaqas
Tp.ra?pT ede6 TnltTruau elueq Tp{T1-raA 6upcupdlp 6upd SA=rrra 6uetg,';
'6u1r1ru EuecuEdlp 6{ 6upcued 6uet1 reqed plTt qsl ! p:a1
-pT Br(e6, Tn{Turaur tnlun e.dueselg 'qeupl ueue{a1'edua6 ?eqTtp ,er{u
-TpsTru( lCiuozr.roq ) TeraleT e.f,e6 et-ra>1aq 6uet1 >1oduro1e>1 eped pltT1'
'qBuBl trrsTBpa{ DuTrTu buPcup(ITp
--+
T,tr 'H' srrsq I urPTEp 6ue11 qgTun I' luoTo{ [ ruPTPp 6uet1 qelunf
'6ue11 a{ 6ue11 >1e.re[
'6uer1 6uBdrueuad ue.rn{n
l leterap : e.f,uuenles ) g 6+ cire ,(I
=U=tU
e
=O=0
' qs1 re>[6ues :n I nq
5p 11ue61p (q )L ) rol)tPJ
u'tu' 0 60 -t = u'([-ur) + ur'([-u)
'a.rrEqqET as.reAuocuPptups.r ad
b : g uP{nluauau {n+un uTeT snunu
TsTs = q puerurp O? uB6up{Ptu re>16ues :n I nq {nlun
'rcpunq buPcuPg buPT1 tn+un sE+pTp snun-r : uplelEDu'y = 6uBcued 6uer1 qelulnt = X
6uer 1 eduurelep =JT' ( .rTpuos
Tp l6uu11 lnurTTas uep
ueeqoc:ad TTser{ qalo:adqeupl e.reluP r uoTlcTf J,, = t
'6ue11 >lodtuolaq 6uTTTTat = d : BupruTp
q )l 'Ir lu
o? + s' ( z-u+u) zI-T'J' x'q'z tT'J'(x'q,)l ) . _ ..^J,,rA?^t, :ffi = ffi = 6uet1 lodtuola>1 TsuaTsTJJg
'' gt
t
' ( ,, aT Td .ra11pq r )
36
Metode
1.
2.ltdtode
Eiya6g dipergunakan untuk meeencanakan kelompokMetode grafis ( Culmann ) .Metode Aba1itis.
Culmann : ( cara
,
t iang tt batter tt . G'
grafis ) .Diketahui p bekerja pd pile cap.akan ditentukan P
1 ,p Z dan p3 .Er. grs kerja p1 dan grs kerja p berpo_
tongan di a.b. grs kerja pZ dan grs kerj. pg berpo-
tongan di b.c. hubung a dan b
d. P diurai atas p1 dan pab talu pab
.,,:::ffi;H inlll.l; pa+a masins-masing tiang.
Metode kelompok P bekerja pd jarak eof gravity" ) kelompokP diurai atas pU danakibat PV :
",Q1Va pd masing-masingkan dengan rumus 3
, "t.a J
p=n + s_-_'1' z- tDari P ygio'berarah vertikal akan kitaperoleh komponen horizontal gaya pa-
D'Horizontal masing- masing tiang.5 p..
tl
untuk jerasnya sirahkan rihat contoh soar.v-1 -11 -2 "cast in prace" concrete pire.
( Tiang pancang di cor ditempat ) .
Mula-mula tanah dibor sampai kedalaman yg diinginkan,/direncanakan ,laIu dipasang tulangan dan dicor.
ile.
'ab O\Analitis : dr C.G
tiang.Pv
tiang
( "centre
ditentu-
1 . Daya dukung lebih besar dibanding dng2. dpt dipakai utk pondasi bangunan yang3. Tidak adanya getaran yang keras akibat
tiang pancangdekat denganpemancangan.
prekas .
laut.
H.H. 0,
rfP.
l-. A, p,F'TC KooE
oUI
o
5
-r- ,,aTTd TPlsopod,,
dg,r51tr-l
cd.Ftt
tr ,, aT Td .etPCUn,l,, TTOI{S |' STUe
'(aTTd T{ue.:rd :tloluoc) fesaq
-uau 6ue11 T{E}t e66ulqes ' {nqunuedlpT'P.: 6up )tnqurnlrp nTeT -repuTTTs
a{ ue{tnss,ftp 6uTra{ uolaq ue{npv :' ( ,, eTTd u.ra1sag,, pd : qoluoc )
uolaq Trep' lenq-ra1( dTcuPT ) ?ncnf a{:ed
-nJaq ledep ( ,, Tplsaped,, I 6uet1 T{ex1pr{TT ) -resaquau pdupseTq qpmeq 6uet1 6unCn'TUT stua[ Bped
'rad^+ TElSapOdu E'g'ue>16uns6ue11p uerooa6ued lPPs
pppcJ '4nqecrp edue[pq .rapuTf TS e^dueq [ 'g 6up ptups TuT sruof,' ( uaclAl ssaT TTar{s" ) racl^l paspcufl, , z'E
' I{euE1 ueT epa{ tnseru ue>[e -rapur T
Js e66urqas 6uBcuBdlp qs1 Ta-rpupr{' (,,aoqsaTTd,, ) dtcuel {nluaq-raq
6uBd Euegtr nledas teue6uau tedues(,,TTar{s Taa1s,,) pIBq' -rapuTTTs uIeT
-epat ue{{nspurTp b^ eIeq 6ue1eq qpTEpE ,, Ta-rpuEU,,: uElelPJ( qpuel urpTep 1e66u11ire1 ,, TTet{S,, ) qs1 TTeqs ueTepe{ ( ntnp 6u
Esedlp ' Tn? ) roc Tp uolaq uep ' selea>1 {Tf,p1Tp ,, Talpupu,, nTET
' qsl ,, TToes,, luTpo{ ue{{nseulTp b^ ,, Ta.rpupu,, upnlupq 6up qeuelureTepa{ 6uecuedlp b^ ( ue>leupcuorrp 6ued 6up rpnsas eFu.:a1aru
-ETp) (,,TTar{s Teo1s,,) e[pq repuTTTs ue{eun6lp TUT stuaf pd
' ("$ag6a pa sp.}r ) " TTar{S " sTuaf [ 'g
'o1 >1ede1 )
,, TPlsapod,,
: sTuo
l{ltr I
Lt. -
e TpEIueu ue>{TSe>{TJTSpT>tTp eTTd ,,aceld ur 1s€O,,
38
D. 'Franki pile".Franki pile termasuk "cast inAda 5 jenis franki pile :
1 . Standard franki piIe.2. Franki tension pile.3. Vibrated shaft Franki4 . Composite Franki pile.5. franki Grave1 or sand
Kami jelaskan Standard Frankippda TEKNIK P0NDASI
"
place " pile .
lIeto4e konstruksi Standard Franki pile.
( a ) . silinder ba ja ( "steel driving tube,' ) ditempatkan pd lokasi ygakan dipancang, adukan beton kering (slump=O )dimasukkan kedalamnya, laIu alat penumbuk dimasukkan kedlm tabung tsb.
(b)- Adukan beton tsb ditumbuk sehingga silinder baja akan ikutmasuk kedalam tanah
(c). Pada kedalaman y9 telah direncanakan, silinder ditahanragartidak masuk lagi kedalam tanah. Pada saat itu adukan betonYg ada'diujung silinder dipaksa keluar dari tabung 6ng ja-lan menumbukpya terus menerus , akan diperoleh dasar yang membesar
(d). Turangan dimasukkan ke dalam sirinder baja.Adukan beton kering ( slump =2 ,5cm ) dimasukkan kedalam silindersambil ditumbuk, bersamaan dng itu silinder ba ja dit,arik keluar dari dalam tanah.
( e ) Franki pile s iap .
pile. ,
coloumn.
piler 1rang 1ainnya akan dijelaskan
\
Di tabelkan pada halaman 39
dapat dipikul.ulangran-beban nominal
' rpseq .ralauerp f,oq 6uet1 {nf un r s>[n-r?suo{ apole14
ue6upduel
T uTrx
'luc E
'aTTd ,.ace1d uT lsecu rPqueo
L'0 = TTqueTp 1---- qeuel ;ase6uelen{e{ 6up ( uoT?cTr} ) uple{aTad ele6 ue6utpueq:ad = 3
' ,, PouTe.rpun,, qeuel .:asa6 uplen{a{ =nJS
'6ue11 lnuTTes senT = -Vi-qv
=o 'nc 'c + Qv ' nJ '6 = O
: alrd T{uerd bun>1nP PAPP
9LL0E t0tI
927,>t9ZZ>T5ZZ>I
ZZbsZZfr LZZbg
0E9 009009 - 0080E9 - 00s
0090ss00s
I
060LOE
u<
9ZZ>I9ZZ>I9ZZ>T
\
9 1frgetf,s91il?
OOE OE?0E7 00?9Lt OEt
0E700t0EtL lj'-
(uo1)TnltTdTp ldp bi
TpuTruou uPqaquolaqnlnu6up f ueueuue6ueln'1
( ruru ) Tpe [ 6u-8.1+ .ra?auerp
(urx)eIPq.rapuTTTs-relauE Tp
$l
eT n1uola
aTTd,i
69-
,00
(a) Sillnder baja dipancang sampai lredalanan yatrg diingi$tran.( b ) Bor beker ja nrengeluarkan tanah dari dalamnya.( c ) tulangan dimasukkan lalu dieor
e.7 eilinder baJa dikeluarkanatau
e.2 elllds baJa tetap dldatr,an tanah :
Hetode konEtruksl umtuk bored pile.d,i,Fneter hecil (30 cm)rdapat d!gunalrao bor tanglan ( 'haad auger' ) , kedalanan, narinun 6 meter.Pon{tesl demikian kita naralsap SERIU'SS.r' t,
?utrp,Egan qptl**buFr{:0.i.1#...-:-.
dipasaoE or5 B lnas pgoailpang bored plle pada hedalamaa 2 / 3
'.j
Pada tabel beritqrt kant Butip&an ealcir satu lrasil teat trnah'-:uotukmenertukao daya duhmg" izin bored pi.le.
80100
tr0115
1e0
125
f50235295
3t0340
370
ur1 tge6'L = ,(J- \ 9trg. + = vv " z.fro z t __ _ L
Z \.o1?eub Z_= Z-Zw L'rtoffi:g )'- t.1e1ad -rpqeT .ra1au L >tnlun
'axo + ( 9'6=,to* )rwc7s4 zg, v =
E L'0?z '010 =
nQ ?=J-
uol zlt.g' L L = B'z' (glolo s L'o- L )9t's =
T '(t{ Z/L L Z/L - s Z/L) offaub = t-tqL- L ue6uolod qpTepe sT+Tr)t 6ueptg
ffi.P1e1ed 6un[n eped u]cE I
Tpp[uaru 11ca6ueru nTeT uoTo>l uee{nu.zed eped rucg Z = 1q TTqurp
ruc V'tV4 rl qaTo.radlpq' ( qZ+8' 0) Z
( ru+u B'o+9 L 'o ) ?o'g- ( or ) 9' i .) 0E t
. Zw/q ?0, B =
ol1aub E, [ = o1laub
Z*/1 9e 'g = ,wc/bl 98E,0
lxB,Z x wZ : ue.rn{n TTque
z*/4 s ># 'Jalau Z = B TTqUP
qeu,*g) f,ff=f uol E = dg0[ = rsepuod ?p.req - M
ueqnluauaw
"dqr = naql --J/ ( e+E' o ) ( r{+t, 8 ; o't"b-td: leTad Tpqel up{nluauel^l 'c
gtI
,9
'v
re
uo? 0E=d
" 6ue I ued t6tsepuod rnr{E1o>tTo: tES
082'002000,:0e
T'Bd,
of"f.eu-D 'q
_._7F' : qeare1. 'tJL'1c1e1e1 lefrsroq) uo? 0e = d
e sElEq uPlPn{a{ eJecue6uap' lsepuod ue6uelnuad J upedue:Ited
vz n e[eqSZZX uoleq n?nw
-vtc/64 gt0 = I{eue1,
(.
-as;ed {nlueqJaq r sepuod ?pTad , >1ede1
'TEOS r{oluoJ t{OlUoC Z' L
12
M3_3 = Z gnetto.lTl j
= + (5,36)( ry ) = 3,5443 tm. I i .:'
Penul.Fngan- pe.laF : ht = 20 crl --r-2 h = ' 15 cm
MZ_Z = 119363 tm'
fcu = = +ti;{t ( 1,:,:,,1t(225)I = 0 ,2 ( secara teoritis kita tak perlu menulangi pe1at,
dengan tul . tekan -----.f f - 0 , tetapi paqa prgteknya tul . tekan &lanrhil Eebesar 0 ,2 ttl,fllaitk I
dari tabel diperoleh g, - 06 1 e5
ld^5-
{.
Drf=
3
U !!r
= 0,05125 (100)(15) ,
A12(9r94) = OrZcn2 lfi63r5443 tm.
' ,!.1
Deravaratan PBI s ps .17 .4
dalan jalur'B! ,ttir - Er' /...
:; .., .z
2cm
:
10 - 11r3tcmZ')i :"
;
!
( pakai-2$=1
15 = 3109
tr= 0,2dari tabel diperolehi' q = 0, 1 1563
[ - 0, 1 1563( 100'] ('ts1 +tV'i*']*fr"/! 2(0 5l(225)ba
(Pakai g12i'- 7 -lBr76 "*2 15'r53em2sisanYa =
tulangan tekan
Chec ali berat
' ,i ji'ri';
t 6 ;t,tern2 )
= df r 3em2
,(pakaL gl2 - 30 = :f'l7on2)= ai2 (18r75sm2) = 3i152em2
'tbakai ilI,,i 12,5 q fiiroZ"r2)f.
h-=20cm,B-2meterlL--E
, berat gendirt , :r ' '*., j') 2 i,8 . ,i: ._$UL_;l s*,^ ,
"':t';".Catatan g Jika berat sendiLri "qFbih besar dari ,iaksiirrie
2r8?.2
3,5443 .1000
18r76 em. .i .l'r': I t.i.: r
j. j 'i ,
ditrlang lagi. | ',i
. .n.t t.,1.la makr rperhttungan
a: ooo LW=n92
.UO1 LBL,ZLEf'0Z/ L g
Z, L) BOZ,9 =
z/L) oflaub = [-[o
@.ru3 EL TpEtuerl lrca6uau nTpr , tucoz = ruoro>{ uEp>{nu;ad pd 1e1ed Tpqe?
z*/a 0Et> = n,dq -) *" *b nd -L
2>;nc/64ys'
= (V'Z) (Ef 0'0(B)(q z/L e
nq+
Et ,\iz'6'a
'p
rJ
z(v'z)zv/l 802'E = Z'- "--
= ol1aub .qc - 0t
ur7rz x wV,Z .{n pqoc
_gtu Vt'Z(a (-- g \ Z- -n ) t + oE -!
@.p: qEtrpf,
e :e46ues ;n[nq {nf uaqf,eq rsepuod 1e1ad E}tT [ , L . ou Tpos pped
,*
@l.
N)
(Jl
-E-o\I
N)O
--T
N)I
Ot\)I\t
r-{ul
HO
: uebuelnuad esla>15
'd.
il.
Ii.
44
Pequlangan :
1Mz-z = Z gnetto
= + (s,2oB
ry)z(ttzt4 - at3\2 = z,g7og tmTI
(
)(
15 = 3 r43
2(o,s)(1)(22516- = or2diperoleh dari tabel 3 q = 010925
A=(0,0g25)(100)915) W= 15r01 c*2 (pakai g 12 7,5 = 15r08.*2)
Ar= Or2A= 3om2 (pakai gg 15 = 3135 "*2)
sketsa penulanqan :
Soal 3 3
F{l.R{
rn
Plada soal 1 , pelat berbentuk bu jur sangkar, kolom berbentukbundar diameter 45 cm.
Pertanyaan i penulangan pelat pondasi secara kekuatan batas?Jawab :
Gambar pada halaman 45.a.ukuran pelat = 2r4m_x 2rhm (idem soal no. 2)b. gnetto = 5,20g t/*2c. ht = 2A cm pada permukaan kolom lalu mengecil menjadi 1 5
cm pada ujung pelatd. check geser OK.
91 2-7 ,5
'tu g'LKa<---- s gt/rlsz'os stzt) a
p{Eur ( up{e1) sTuaIesq ue6ue6a1 1 ue
-ut>1a1 ,'';*::,;':T;ri":":r;u=':un urlo
: qPAPe3 rPqueg
ueupqequed 'ue6uelnued upp rsepuod leTad . {n J upe.f,upl.rad
'ZEO e[eq , EZZ,N uolaq nln6 . ,wc/b>{g, g=qpup1 {1n uTzT 6un>1
e^fiep'ur1 0t = uauou u=p uo1"o, - d Tn{ruau >1ede1 rsepuod nD3 I f,eos
( ruc ,ZEt't = gt ;g fr te>1ed1
(rtuc1 l',Vl"oBoz
ffi
: uebuelnuad es4ar15
Z*a LtZ = V Z'O =rV=. g Ztfr 1e>1ed1
(Et)(00t)tIt80'0 = v Z
ElEgo'0 = b
ruc 6?,t[ =
qaTo-ractlp- z'o =
:,1 g'e =
L' z(Z'oruc$? =
strul l0g, z =
z'Ll(8oz'nl f ' z-zb{
(---- rucgg'Gt = q
Z(gnl al n/L = Zq
'etuPs snleq er{usen1
, ' re46uesqeqnTp -repunq 6uBdulpuad rnfnq TpBl
L,.' rc,
' de1a1 lpJT s.raq
\aN)r@
Lsee)(t)1s',0)z
I
I
I
I
46
CobaB=2 5 meter.P+W,. +6e.fq=-(1IE:t
'-
44 ,., 6(0,25).=
-
\r =---,( 2 ,5r- - ' r
-,= rr04 (1 + 0r5)91 = ),816 t/n2 ( %an"n ( 8= V^lL ----2 oie2 = 11 '164 t/n > r;;; (=8 t/nz ) ----> uk. diperbesar.
Coba B = 3 meter.44 ,q=-(13'
9l= 2r44
42= 7 '33
Jadi ukuran
! 0,5)
tt^z^ ( Ft.n"ht/mz ( 4,"nun
3m x 3m dapat
3.
dipakai.
Itlenentukan tebal oelat :
q, ( 0 r45+1,275 )+q^ ( 1,275 )
-
{,518 t/n2qIl ( 1 ,275)+gZ ll 1275+0 ,43l
' 'u
- gnetto,u (or3 + h)2
a;
gnerro =- *(4,518 + 5r 2521
o-=.J
=
Q3=,.=
2
"bp,t -:
di,mana
I
tl
I
I
I
1
3m
= {,BB5 t/n2(ambil harga rata - rata).
r. --= 1,s(*91-l,Il14 .Ig,'sbpu- 2(0 t6+ZfOrffi' 46 15 t/nZ '
= 4r6s kg/cm2(t*onro (=15)
oie'rr----) OK
jadi h diambil = lScm, OK
, M2_Z akan ditentukan.,
ct='44
,2= 5 ,1295 t/m- .
qt(1 q275 + 01075) + g2(3-1 ,275 - $r0?51
- ;1.> .,.
"'$i'.' '
-..-s ?
,{,)
-e (S'o+ tlfu =xEtub
: b pr{1 T6pT {ceqc
(uer.E',m)uq?= ro+Etro (v'z) (€6(t) = rr-r,1r1 Tf,Tpuas lEraq
'urcgI Tpefuaur ltca6uatu ,tucge = 1,{ ,t = g! Tsepuod TJTpuas ffi
, .1e1ed 6un[n eped urcE L Tpe[ueu11ce6uau nTeT urcgz = ruoTo{ uep{nur;ad eped 1e1ad Teqel -z
resaqredTp 1e1ad reqol e{pur tQ* 2 ( tQ2 e{T[ . Ig,6 = de1al'qaq {lnnQ+))rwc/b>t g9,g =
'g -reqaTas {ntrun uo1 ggg, gZ ; (f ) (gt, t , ffi
: uElelPJnQe
J-
Z-2. =u
3 Z-Z ue6uolod ep@(atucLB'Z=l g'LL I il le>1ed) ,urc gL'Z = VZ,O =rV
(atucgL'?1. = g ZLb) Z*" 6'tl" =
ffi (sz)(00t)staeo'o =
gfg90r0 = b qaTotedlp
(szz)(L)(9'0)z bz'o =9
SZ=nC
(Et)(00t)gztLZtA - v
I
tw196
' tuc 0 t = f q (---- luc gZ = r{ ueqrpe t
sn-rEq ?pTedr -:psaq nTET-re? ) aruc gt gZ =
s6't =
'tr6q TPqaUedTp08 LZ
(szz) (s0)zgZLLZ'O = b r{eTo;edrp
.LE, Z +
. z'0 =q(szz) ( L6(s'o)z | -'
I
=nJ EL
rul\u
[ 10'g -Iz'g76
(st'[)tirse+
7Xv=l^l
L
L'(E6et
z( 9e' L
S-Et' L)
L?
)(9621'9)
: 1e1ad reqaT roloru L >tn1uc
TU
Sketsa penulPngan :
SoaI 5 :
8-17 ,5
uT'
Pada soal nomor 4, tetapi denah peJ.at pondasi spt tergambar( tegangan tanah harus merata ) .Pertanyaan; penulangan pelat dan ukuran pelat, secara kekua
Jawab 3
P dipindahkan ke as pelat pon-
yang saling meniadakan dengan M.jadi pada pelat pondasi hanya. bekerja tegangan sejenis merata.
f=- (d."nahBz Eanan
BZambil uk. pelat pondasi
menentukan
gnetto = no , = 6t94
(2 ,4)tFenentukan tebal pglat :
Lupu='.,iull*Hr;li' 1.5<., so
diperoleh h )0,1g4 m
2 ,35m
= 2r4 m x2,4 m
t/n2
o|,earI J-e
tan batas ?
M=l 0 tm
(rr I t?r :tru,
ue.rr sltPl >Z
rI@
8-V Lfr
?=)(
9Z0 + st
( -uc ZA' V=) Z Lc
_ruo LZ'61 z'
g, ZL-
: uebuelnuad psla{S
uol g?g L'z =
) (v, z) (v, z) = T-rTpues 1P.raqtu V'Z = g
rxcgl ( uc gZ = 1q
(---- -tuc S't = UZ'O =rV (
re (02)(00t)2t80
b qaTo-redlp Taqel
,d
xo
gfr 1e>1ed
B V LfrO8 LZ
t LL =
['0 = v
T-TPP
(szz\ (sZLSO l,, O
0
0z't =OZ=nC
ergg'E = (L)z( e'Ll'v6'g-z- zt'l
[ {n1un : 1e1ad uelruelnuad
oz' (ozv\6'0 nq7=J-
L
= (?'Z) (Z' Ll V6' g =
.Zt
6un[n eped urcE t
6t-
1e1ed -rpqaT.ro1au
* 2) ,wcysa ?6's =
uo? zLg6'61
E'(r{g p1- a+B+ )o11oub=L-to tttt - [ I{ETepe sTlTir{ 6ueP1g
'1e1ad1
11ca6uau nTeT uc gZ = -I{ (---- tuc 0Z = q, BqoO
50
boal 5 : Diketahui pondasi tapak memikul Momen
P.M*=12tm , MyIl5tm;P=30ton. Iriutu beton RZZl ,
arah x dan yt dan ga
baya normal
=3 0t 1a _U24 tegangan izin tanah = 0 r6kg/
"*2 pelat pondasi berbentuk bujur
sangkar. Pertanyaan ukuran pelatpondasi serta penulangannya, jikadiminta bahwa teqanqan vanq beker-ia pd pelat pondasi bersifat mera-
tp (Fpkan) ?
Selesaikan secara kekuatan batas.Jawab :
M.,= 1 5t4l
12=30
15=30
I
I
ex
Ir{
=-J =P
!t= 'f-=
P
0r4m
0 ,5m
t;I
I
I
I
E. Irtenentukan ukurar:r pglet l)gFdasi.
.
'il
$
iir
fl
h
O= ry(4un"nBz
- Eanafi
33)iB- ( 6 --*-, B> 2,35 m
ambil B = 2r4 rn.
b. Menentukan gnetto
gnetto = 'o u = 5121 t/nz(2 ,4lt
c. trlenentukan tebal pelat pondasi.
Txn = Pr, - gn.raorrr(or4 + h)2
--Yu 2(0r4+2h + 014).h
2(0,9 + 2h) . h
h > 0,133 m -----) ambit h = 25cm ---,,+ = 30cm lalu mengecil men jadt 20cm pe
da ujung pelat.
I
II
@I
-nc}-(rl
H,o
'ue.bupf nuailEslats
=o: Tnqurl 6upr{ ue6ue6a1 TTequa{ {caqc
(uejrTs)tE?)+e(uo199?,t=(?,Zl(#-)(?'Z)(v,Z)=TfTpuos1P.raq
luroz (---- urcoe = *q
,uc?,g,V = g'ZL gb te>1ed (---- -woEgrt = UZ,g=,V c A^A' Zg-VLfi te>1ed)_ucgZ,gL ,Z = ffi (92)(00t)s19,0=V
glg,O = b Taqel Trpp r{aToradlp JUJ L'
Z'o = ? .t)l
' o'v = = 'rfrul tt9g,g = (L)-(8,Irolfauu + = z-zw
CL: 1e1ad uebuelnuerd
( rtuc4Z'6L =
NO:
''l ,'
uo?
+ L)rwc / b>l
T.61, LL =
gz' 0?z'6'o gL'v
(v'z\
flQ ,t.*
9, Z L_B
l9 a
: 1e1ad :esa5 :1caq3
52
SoaI 7: Dtketahui Pondasi taPak
=2 Otm
Jawab 3 tida erata.Lanqkal]__perhitunqan :
€r. eksentrisitas :
e-_ = t1__/p = 20/30xx'
memikul P rMx rMy
P = 30 ton, M* = 20tm r ry = 20 tmukuran pondasi spt tergambar.Pertanyaan il.Tegangan tanah yg terjadi pada
titik 1 ,2 ,3 ,4 .pondasi tsb
P rMX rMy
h ukuran pelatkuat rnemikul
TI
Il1
It
2 . Apaka
cukup
dkm
lJa
Catatan 3 + berarti tegangan tekan.c. Akibat adanya momenrmaka jika momen M* dan r, dihilangkanrP ber-
pindah, tempat ke titik A. I
i
c.l menentukan lokaEiqaris netral aki,ba! P bgk,er,ia di A.?r GRS NETRAL +----) Z, Z
xr z (2-xt)=7t5:2215xr = 0r5 m
I
Fr=o ,667m
a1T-egangan tanah
yxp M -v It'Ib.q=f;+++-,Y--f
x Y, 3 1 rr\( 2)3 = rr333 mdimana I* = r, = fi,.b'h- = E (21
A = (21(21 = 4 ^2
riril P /A(t/nz I x(m) y(m) M*'Y (T*(t/nz I
I,l .x/ IYcY(E/m'Iq
1
2
3
4
7157 ,5-l ,57 ,5
+1
+1
-1
-1
+1
-1
-1+1
+15r0+15r0
-15r0=15r0
+15r0
-15r0-15r0+15r0
+37 r 5
+7 ,5-22 15
+7 ,5
gangan tarik.
OrUfl\ -I
,5m
.:*::
ru ?98'0 = (L0L'0lz/[ =
]a Z/L =
?, qs at ?, )t11 lere[ = Zc eueuTp
" c'd, ld uetn+uaual{
z'/1 6s?'8 =V829, L
ffi==r= =?6 =Zb 7Z"c' w
vgzg'L
t6 PUPIUTD = '5 tr z EA'c' w
. -w/a gzt Zl = c?g?,gt L
ffi=z
' r- = [r"tz'c r w
C---- z-2, us pql uep
'p
lrt
/i>
z'/1 Et,' 9z =m==Ec ru 0g0r[ = ggL'L _
-U Z
ru1 [6'BE = (L6Z,t)0t =za'd
=
9gL'L - L0l'0 + [90'[ = ]l =
z Z qs a{ t tq? lere[ =
lil
Lceupurp
ru L6Z'L = LLV'0 (J?(u '(199'o-t ) )
X qs .rp Lgg, g=tr" upp
+ ot) =2
Jl - a
I qs rp ur Lgg, g=*a terp[raq d TsetoT: t'€I tTlTl bulsErrr-bu1seu epuP{n+uaua}J g. c
TP1o1 '-rT,ru tgeg'L - EtSz'o + LEE'I
?ur Et6Z,0 = e(Lot,0)(gtV,Z) rxstl'z =
zrtz + zzl'z ,
, = | == - EIBI PlPr rpqoT - rEeaf uEp nL + Pc --L --r
rul 0 L' a= Jf = lBi6ula 6up 6up[ucd r6as:ad de66ue (---- pzqp?c
,((t9o'[) vLu [EE'[ =
i + LaL'0)((re0'r)( ezl'zlll + E(re0,r) (zzl,z,) T =.
z( (gL) f + Ja){v e6t1t6as senl)+ E a[ pc T = '-'T
ru LALTO = Ugr| = ep ='ia
ru [90,[ = {= aL:,L-
tu g'[ = ptlu S'[ = c[
@
r !.I
z-z
\ j..
.., l
itffi
I
?a
5{ i\rlokasi p
.1 ,769m
tekan.a r7 07m
7'
9r\ tarik ( abaikan )
.8 ,459 = 14 ,802 ,ton
dx .o'x
,768 x).M .xzI ctx
z
luas B . e2(2,475)(0,707)(luas A). ql2(1,768 x) .
1,769
t 2(10,707 J
I)='2=
Pt =
-
ri1 ,768/ 2 M
-z= 28.531 ton
Ptot.I = pt + p, = 43133 ton ) n yang bekerJa (=30 ton).jadi pelat pondasl kuat memikul p yang bekerja.
_catalan : yang tidatc kuat memikul gaya yang bekerja adalhh ta_. nah, karena tegangan iti ttk tl = 42 ,25 t/m2=4.r225kg./cn2
tanah tidak kuat ) .soar 8 : oi.lt"t"r,ri-at;;;;; i..".'..0.r-20cmr memikul beban 60 t,/m.ke. kuatan tanah izin = 1,2 kg/cm2. Mot, beton K225, baJa u32.
:
si dan ukurannya ?
Selesaikan dengan cara kekuatah ba
,758-x )
@lh.t
tas ?
(rwc7?'0t= S'L
(z(18'og = g'L
'1e1ad 6unInT6BTTeqatredTp
|li 1e>1ed) aurc ZO'OI = V ? AZ : t6equrad ue6ue1n1 {n]un
,Icr.?'01 = s,L - otfi 1e>1ed) ,uc ro?:,":Til::t Psle{s
zzfr) ,urc80'0s= ffi( 0E ) ( 00 r ) srE zL, o=uELqZL,0 = b qeTo.redlp Teqp? TrEp
66'Z =
pd urcgz rpe[ueur 11ca6uaur nrrpr rucgg=1q +- ucog=e eqoc
rI Tpp[ ,:eseq nTeTf,e? qrseu (---- Zr"
g L, gg =08 LZ
(szzl (s9zg0z' 0
0)z-b
(0r)(00rlszeoz'oqeTo;adtp Taqpl
z'a
=VTf,EP
=3.1
= -rI(szz) (r)(s'0)zt,
6e 'Z =0v'urc 0? = q rppf uau 1e1ad 1eqe1:ad p1T>{ edulsele6u
5ru {n1un e66u1qas rre{as 1ede.r {prp[:eq up{p up{n1:adtp b^ue6uelnl E>teu'qsl .rpseq b^ z-zw Ter Tu ueltrleq-red ElTlt e{T I
ru1 so'zv = ,((z'o,tz/L _ (grz/L).01. z/L =
( L, z(ez/ t, - az/ L I
' (ruc Z' S( ue4lesB eIes ede-raqtllcQ[ = q TTque (--;- ruc
o1?aub g = Z-ZW
t
: uebuelnuad ffi
(q s'o 6t?,) ott
q TTqurP qeToq PpuP )
Z' S (q qeTo-radrpq' 00['6'0
000 t' (qE'0 6'z) .oJ- s-T= (q z/L (Z'O)Z/L (9lZ/r)0r
' (rtr z/L P z/L s z/L)'o11eub
z*/1 oi = oE =
nQ-tJ-
-.= [-[collaub
'c
'q
qeuPlg>#* {'rolau t - T {nlun
'rsepuod 1e1ad . n>[n uE1n1u6ufr - p
iii
ri
.lr
J. SE
Ielou 9 = g TTquIp tu
: qPlrPf
-S*|?
rt:
56
Soal-9 :
{,,r. \:
Diketahui pondasi jalura
t,A,. I
{:"
tergambar.
A dibebani 60 tonB r' 75 ton
6t.nar, = or8 yg/cm2
jarali kolom A ke B = 3r5 m
Mutu beton R225 rbaja UZA
Pertanyaan i penulangan pelat pogdasi secara kekuatan batas (pem-bebanan bersifat tetap ) .Lanqkah perhitunqan :
Pondasi ini tanpa balok pondasi.Er. Menentukan qk. pelat pondasi
Diusahakan agar lokasi resul,tante Pa dan prberimpit a.rrgan tt,k berat pelat.
PB. 1. %=
1 ,.94 m
1/Z L - a
(1/2 L
'r. 11
sepertiKolom
75r
x.= 75 ( 3 ,5 )50+75
x=ct
0-7 ,5 |
1,94')
!:r - |
i r , lll
:ffir r I ll
@ t'vv
'p
lul Lt' L L
(9'E)
,(et'[)(Lg'L)rrul
Z, 6ZT L , OD
Lg' = g.re^aTTlue{w
9L'VZ+ = xeu
= xP 1-- x'ru xWP
g'Bv + xog + z*gLgz'zL- =
([8'0-x)0g+ z*(s'E)( Lo'L)r -=*w
ru1 S0'B = (E't)
,( te'o) ( Lo'Ll+=v -ro^aT T 1ue{w
W
W
I
8'lE r
ttl' t iL o, L=o11.lub
9L'VZ I /Eof d
urs' € i tu
'(T rlPrpas) "rPuTpnlTr,qe.rs ueTPp o'w
z*/1LO, L9L + 09
TSE
v6'
: ralau Z/ L €
uetnluaualdol1AUL
=IJ
'c
ol4aub ue{n+uauan 'q1e1ed up-rn>[n Tpe I8'o-E'g - q
9'9lZ/ L P
r{aTo:redtp T
puod
E'[-lr-(t'g<
renorTlue{ ( *vg' t =s't 9z' L v'9 = Q
\ wgz'L = v6't (v'glz/L = e
tu?'9 = T TTquP 'wz'g ( q"1o:red1P T
rug'E x ulg'f =
xo (wet't =l utB'o =
llrE'S = T TTquIp
6uelued nTeTra+
' 6ue I uednTPTf a1 -re^aT rluP{
TpEfuau TbPT qPqnTp g
@ur TEeT geqnTP A
g't- g0' z-g=Q
v6'L (8)Z/L=eru
ru g'z
(----llI8=TTTquEgZV'L < qa.To.radtP T
= nTnp ue{n1ua1Tp g
T'go
lC r
f *vv' z
I ug o'z
Gt+09)aot+(et+09)
58
Kolom A :
Tuo =' ,lf
Kolom B 3
-Lbpru
dari kedua
1,5(60)-1,5(7,01)10,4 + h)(0,6 + h) (tso2(014 + h + 0,6 * h)diperoleh h> 0 ,202m
1,5(751-1 ,5lZfll)(!/!_r9_(0,25 + ilz\ rsO
diperoleh n) O, 3 7 3m
hasil tsb diperoleh h> 0,373m ----) ambil h=40 cm.
€ . Penulanqan p-e]:Ft fr.ah logitudi,r]p_1 ( Fqqra.h L ) .
Catatan:h=40cm40Cu=
lt1,l 2(0,5)93,5) (225')
untuk K225 ->L*b,, = 12 kg/cn?A min = 0 t258.b.ht = Q ,252. 350. 4535912 untuk lebar 3r5 meter atau 12
= fitzAr = 012 A = 7rg1 a*2 .o... pakai
atauil12 50
350
= 39 r 375350
3s
10
\7 912
= fitz
Iulomen harga llomen
batas ( tm )
Cu5
q A
(c*2)I intangbatas ( t )
Zuu .r(kg/em"
)
KantileVer A
12 ,07 5 10 ,2 0,2 Amin
=39 t375 50, 1 95
50,1 950r9.3r5.0r4
=47 r7t/m2=4,77kg/cm2
(r*.--->oK ]
3s912
=39 r 55Lapang-an(max)
37 ,14 5 ,82 0,2 A min=39 t375 0
35912
Kantilever B
26,055 6 ,95 0 12 A min=39 ,375 68 ,7
68,7 -019.3r5.0r4
=5 4 ,5t/n2=5r45k$/em,2
(e*uo -) oK
35912
Lg'o r v/L T v/L Ig'o
( sB?r'1nuad
( r.{Prrreq 1 1e1ad'1nuad qeuap
'uebuPlnuad Psla{s
'repunq uoTo{ >[n1un urepT
:5uefued lbasrad uoTo{ tnlun Z'l'09 - ZLfr ueteun6.redlp e^fiuesTs qe-roep {n1un
. EI xaAaITlEe{ upp I luoTo{ sP Tf,Pp T V / L -reqeTas' -reqasTp er{ueq TUT ue6ue1n1
(-tucg['9[ = L; ZLg -re>1ed)-ulc 69'91 - -z non,
'(0802
ffi (0?)(00t)Ezgt0'0-v
)1e 1adqeuep
a.t,lql lmirl
gzgtg'0 = b r{oTo;edtp
z, a = g(ezz) (L)(e'0)z 1,
ILvt s ==n3 0v
rul6g ['g :
(L)z(t.w-\g' t'
L) z(9828'[ )( Lo' ')r
=
ol1aub g_ =
Z-ZV1L
(
Z
0
t' ?v
leled.reqaT.ra?au [ {n?un
= ,(OS\ U +/l,
c-___ qeqnTp
'rrepunq uoTo>t ne[uT1 E1T]t L'.J
cit-ztfr
:+,i I | ' lL, irlll
,0
LE@-
' (g qe:e ) Tes-roAsuerl qp-te lpTad uelrueTnuad
FI
:- 50
Soal 10 : Diketahui pondasi jalur (tanpa balok pondasi) dimana pe-
lat pondasi berbentuk traPesiumoJarak kolom 1 ke 2 = 3r5m
n_L = 60 toni P2 = 3O
tTtanah = 0r9kg/cmz
Mutu beton K225 | baja U24.
Pertanyaan; tulangan Pelatpondasi secara kekuatan ba-tas ?
a. Dtenentukan A dan B :
P1 + P2 + 108(P1 + Pz)ll=
T orr
2(60 + 80 + 0,1(140))(e + B) 4,5
A + B = 7r6m
L,2p.+A\a+x=T( A+B')
< f tanah
( g t/^2
...... (1)
dimana ; =P z'LP1*P2
,80 ( 3 ,5 )
60+80
( 0 ,5+2 ) =
28+Dari (1) dan
,2H^+A\\T,12 167 . . . . .diperoleh A
B
4 153
A'=(2')
(2)
= 2153
= 5 r07
m
m
b.
d x:rl
menentukan gnetto 3
60 + 80gnetto =
= 8,22 t/^2llenentukan tu1 ,dan D :
pada ujung Pelat A:q = (8,221(2,53)
= 20,8 l/npada ujung Pelat B !
q = (8,221'(5,07 I
= 41 ,68 E/n
C.
(8,22)(5,07)
=41 ,68t/m10r98
,697
43,
11 9,68I
I
I
32
@
1
@
)dz,sr), d t-/m
20
t 2 ,296m r
-
I
9 ['E8=UIIO L' V=( g'0
+ \ez'z)vgs'0
+ tg'Z
tulgs'E?=Ue
6uede'1
1EE, I,L=(20'6vls' L
Z L, ET=
N Lfrgz
ru[8'Z=(E'0)?9S'0
+ tg'Z
L6g' Z-- L
. ruoTo{-raA
-eTTluex.rg'lt=uTlx v
1 1 6ue1eq6ue1u t 1
Q= ) 1e1ad-reqaT
rrc E, =1q 1----lllc0 ?=ttr
: TeuTpnltbol qprp urpTpp uebuelnuad 'a
TTque (---- rxl t'0 4, ,a
q ' (qZ + e '0 + t'0)zz(q + €
: g tuoTox
'1e1ad Teqalue{nluauaur {n1un : suod resab {caqc ',p
O\ (ZZg)g'I-(08)Et = ,raa2
gt ' 68
g'vT
It V x,.
-=
x8' 0z
uI1[['E (- Erg =x {Jnx g ,x, (*b-. Z
8e'L r+ +r*'*b i =*w
ueleTnquad uf,{
89' 61=edusn:et1as1
lgz'oz(-
lxbrao't ?
E' 0=x
)"f =9'0 ) *> O
9t , 6g=b4-n=x
E,? XL(FT-f8:r,Z=
L)
E,Ifr-Jrcffi
+
ru1 BE,tr- =Xpl.u W
tu 96C,'Z*x t dTp
o=xp,/xwp (-- xeru r{L69t Z +-- 9'0=x
0E + xog-,xl'01 + rxelL'0= c ( E'o-x) 09- ,
-*b , | * z*( B'o af =*w (
zt.' og(--- ?=xz0' 6?- +-g ' 0=x
09x-7x b+g'oz)xi = Qv )") E'o
(x+'T)g'02
-w/1 z L' 87, =^b +- Sr0 =x
TEs5-r7
x-=D
T
0 1---- 0 -x*ti ' x
B, o z-xb\t* rx ( B l rrrf=**
]6'01 =*O(- S' 0=x
g=xq +- 0=x7X x' (xbElz) = og'0 > x>
x-bW
xo
xx
a QQs ue{TaqelTp q'n buepTg
I,lomen(tm)
Lebarre lat(=b )
Cu q A
(cm2)
l intancrbatas -(=t)
TbuQ<g /cn2 )
Ket.
Kanti -leverkolom2=5 ,1 1
tm
2,53+0,564(41=4 r7 85m
15 l0 ,2 A min=31 ,64
I r , s ( 60,132) ='90 r 48t
go,4g.ior0;eTa%;6)
( 40 )
5 ,25( z *h,,,
OK
24914=36 196
I
62
CAtatpn : Ar utk lapangan = A ,2( 83 r 3 )= .16 ,7.*2 (1ggt A="15 ,4 atau 914-40)
1
z=
2,53{rff},0,*rr.. -
x 40
= 2153 + 0 1564 x
A min= 0r25B.b.ht= Or25?.b.45tu1 ' utk bagian kantilever kolom 'l = zlg14 utk selebar 2g1 ,2 cm
arau frt q 28L428
= fitq _. 10tu,. lapangan : 54fr1q utk selebar 410,7 cm
atau g14 - 410,754
= fr1q 7,5_ . JJ
tul. kantirever kolom 2 = 24g1 4 utk selebar 47g ,6 cm
atau g14 ry= g14 _ 20f' r :
1Mu_" = itt,22)( 1 ) ( 2,393_0,1s)2= 20,67g tm
#t;rl = 3 14
diperoleh g = 0,09375
= 40 t56( pakai 2A
frta 5)Ar=0r2A =
A=0,09375(100) ( 40lW2
cm
fi I O = Q,0 ,2 .*2 atau
8 , 1 1 2"m2 (pakai frl0- 9
= 8r73cm2 1.
2(0,s) (b-)ImAr rrtk kantilever kolom 1=0,2(31164)
= 6,33 (5fr14 = 7,7 atau 914 -'6OiA' ntk kantilever kolom 2=A,2(31 ,6i)- 6,33 (5fr1L _ 7,7 atau fr"tA iOOt
(281 ,140 rrow-e
'qerlpq sTdeT uebuPrnuad qPuap
1116 ' 0=
OV-?1f,
0L:VLf,
z-v:fl
9- 91fr
SZ
r0E
Z9'Z = oE
(et0tb
9|fr
I
glg r
I
I
I
I
Eil
( ruc .Z
: 1e1ad uebuelnuad EEla{Si .6
) atuc ?Z'Z = UZ'O =rV90'ZL - 91fi 9 re>1ed;
= S7'001 'ZSZ, O = unuruTllr V
nEf.e uc ,Z
z*' 9z' L L
z'o -- 0(szz)(r)(s'0)z )ffi g0'g .==nJ 0,
ru1 g?' g =
)(r) (zz'8\ f = Q-QYI
txv/ L
64
Soal 11 : Diketahui pondasi jalur seperti tergambar.
10-9
14-100
mutu beton K225 , ba j a tJz 4 .Pembebanan bersifat tetap.tentukan ukuran balok pon-dasi drr. penulangannya ?
selesaikan secara kekuatanbatas ?
Jawab :El.
Coba dulu a = 0r5 m
b= 1m( tersera h anda, asalkan (t,s m).
jarak ttk berat pelatke as kolom 1=,1 / 2 L_a
1
= :;(13r5)-0r5 = 6r25mZ
jarak lokasi resultantegaya gaya thd as kolomi - E.
910-30
I
II
A----rv4-- -JI
'ha--- F A-ru
IILrl
I
I
I
lo.*I
L4m 4m b\_----..:)ar.
- f=6 ,3
1+,
,
I
If -frt- -1' n-*_4-uGFI'-
'dE- -_
A ('-
l, 4* .l 4m- 4m mI L= 1'3 , 5;
I
o.''mLn9o'u*
g 6' ?,V
E?,8 [
I
| .o'on
gi' 9z'sv
@EE,?E
61:
vz' vrul L'g
69' zg
gLo'
€'B
?65'
I
I
,w/t69?'
tu10 Z=
?09=
6sc'g= 6Zt.'A + e6'g =
ru g'E I
z*/' I'z'L ' u"ur*g,)
=/ a .T8101 d g
'uol 0? = (OE+0?+09+08)t0e'T8101 d z0z
, (uplta?) sruaIas
ri lg
urlor_rlpur+o'..hTz tu1::pI
LI,' L E
gz9'gggL'9c
.J
= 6zt'O t6'g = z6
ZgL, I I
z*/^ [ 09' 9
z(s'tL)9'z(9'Z).9't.L) _o11aub921',0'002'g002
' ollaub u=tn+uaual^J 'qru g'Z : Tsepuod 1e1ed ue.rn{n lpe[ x
t
g
(9ZL '0 ) ( 002 )9
ru g, Z = EI TTqueru gzt' Z 4* qeTo-radlp
-(g'e [) gc, (9'et)(s) r o? + ooz
(r) (B)lsepuod 1e_raq + Tplol d
xPtu-=b
= -rTS{e,1 lsepuod r-rTpuas lpjrague6ue6e1 TnquTl (---- T g/ t) a
tu gZ'Z = (grgt) + = T -9.LL
ru gZLrO = gZZ'g gLt'g = a
ru glt'g =.0 s+0 7+0 g+0I
I
l+i1 roe,E
- CO -
(zL\0e+(e)
;66
Bidanq U dan D ditabelkan sbb :
Denah x o'x Mx Dx
1 '_1 0 (*( o,s g*=5r601(13r5-x) + 5,259x
13r5
= 51601+010487x
x = 0r5 -r--)g=5 ,625t/*2
Mr-^-.,-= l- (5r601)Kant 2
(2,5)(6,5'2 +
(5,6 25-5,601 )(2, s ) ( o ,25) (1/3',)(0,5)= 1 t7528 tm
i
catatan 3
iEar pelat=2 r 5m
D.Kant'
= *rr,s) (s,Go1
+5,6251(0r5)
= 7 ,015 t
1-2 0,5 (*(4r5
Q*= 51601 +
0 ,a487xM*=(5,6011Q,il+
x2 +(5,501+0,0487x
-5,601)(2,5\ . 1/6
*2-50(*-0,5) + 25
=7100 125x2 + 010203
*3-50*+50x=0r5 t 26r753tm
x=4,5 --) 31137 tmM max ----)
1 4r002x+0r0609x2
-50=0x=3152mM max = -38137 tm
D = 14,0025xx"+0 r 060 gx2-50
x=0r5 -)-42r98x=4r5 -) 14124
2-3 4,5(x(8r5 qx = 51601
0r0497M.-=7100 125x2+o,ozo3x
*3-50 ( x-0,5 )-60
( x-4 ,5 ) +25+4 0)
=010203x'+7r00125
*2 l1ox + 360
x=4r5---? M =' 81621
x=8r5 ---) M :56169
M max-l
0,0609x' + 14,0025x-110 = 0x=7161 m
M max = 42169
x = 0r0609x2
+ 14 tO025x110
x=4 r 5-) -45 ,76x=8 ,5-) 13 ,42
&
'ruc 00ti1
PqdO *q
00 L 'if;tuc(---- aOL/Ovg'L6 = ruc
uErn{n eqoc
L?Z'0 =
?n
q
., L?Z' O =
6ZS'0 =
,r('i!,l1
g:*
tftr -'j = V * 9L 99rq/
_-U-J
LZ
,utc/6>1 00?t
,rnc / 6{9 t
o4 =urruTp
o)o=e
tul 69, Z9 = xeru W
=U
= nQ* vz n
*tn
.T: '4"
'g
';
-., ,,
.t. l."t
0
tt'0r./r,/,+-g't[=x
9?'El 6--9'?,1=x002-xs zoo' I L+
zx6o90'o=xq
0 z6'0 c-- grt[=x? 69' 8 (-- 9' Z L=x
SLI,L+x00z-zxgzl00,[+
ttE ozo' o =(g'z[-x)0E-oz+
OtL + xgg[-
Z*9Z[ 0 0' t+rxt O e0' 0=xWxL8r0'01t0g,E=xb)x
g'EI
)E'zr
I
r l-?
sE'?E (-s'ZL=x89'92- # S'g=x
':
0st-xiz00,r t+)t
r,2x6090'0 = *o
,Lk
xCI
| *A l0r0g- = xeu h
I tu gz'ol - x
I o=
I on t-xszoo 'v L+
z* 60 9 0 , o
I < xeur !{'90v '[ [-=w (.- g'ZL=x
69' gZ-=iN (-- g'8=xOtL + x09[ -
Z*92[00'L+ t*t020'0=( s'g-x ) 0?-0t+
,, 09e +. x0[[-Zx
SZL00'L+rxE0Z0'g=fwxg?o'0+ [ 0g, E=xb3'll
)*)s,B7-t
D{xx-b
xqef
:aeg
,,w:
qIUtttc(dt{0)+JoM
!"r '{ItnocOtd
O{C\E(d
UtcL.(d
A4ofr0)a
lEn()cO.td'
O.rSTEdq I
rboqO{da\tt E
dUtCcdl,l4 Itnlclolal
I
I
oooo
Ulcd
J4tncoa
tntr(d
.YUtq0)a
tnU)G-
.F{
+JtJ4dOt{O ts.
rO
FI
ob.<s
ooc\.F{
+, IJ1do!r-Q. ts.
l/JLn+)l,14dot{FA{tS.
LN
r\Io
ts.<il
COl\
.F{
+JtJ4(dtSl{FATD
otn+JJ4r(dol{O{ ts.
ro
f-I
ot!.on
oulN
.r-{
+JtJ1rdo11F-O{ ts.
tr)\r\
IotS.oo
rn
I
ots.N
5ECI(Ni
.!
v5c\ .o;fu
^.flio\ C,)
\ ti-HoroE
*fi
I
VQ\o or
tg5.afN
^gir-(llt-
---tN=
^c*N
I
ovflN
e*fN
l.rA 8lc1 or\o+N5F- .orl
I
qo(dEt{
^3$\05c_. ,1Q*N
C.O.tt
V6Hc\l o+J
rFlf\ ,A
tuIT
Ul51Cvdo+r(dC+,'r{ dF1-Ol
sfc\roo
(>r\sI\tl\0
U
oo\0' \on:c\*
Osr{\o@r\o
I
Ocn
oN
or\@(,\r,n
I
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Ln
rnF
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N<E
or,o
@
LONts\N
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on
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on
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\olo lrn\lNot lts..- lrn
o\oC'r
tnc\tnN
tn
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@
Lnc\tr.N
oonn
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LNC\ttnC\
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c\r\oC\
tr)Ntr.LN
Nr\NN
LONtS.r-o
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ooon
rnNts.r\
ril.Or
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rnNts.\o
LNNts.rO
VnEc\
tr)tlr)r.o lcrl.lts.o\ lr.$l-
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rOC\ts.o\
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r\r\\o(\
ro(\ts.rn
tl
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tr
or)@Ln",oC)
@LN@O
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cn@tcoo
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ta <r
osf()
rlf
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osf
osr{
osf
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ocs
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aU
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r.n
sfOr
na
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r\
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f-Nti{ no
tn
ro
trn\T\
E(dUh,xqvgod(do+Jo6
.Q .A
N
orN\oc!+
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osI+
rotntnf-LN
I
rOrOOr\sfl
I
LN
f-oaOrN
+
tnoOosr.Or
I
LNooolr}@
I
LNooOosf
I
rnotnf-
I
o\or\
I
o\@c!
+
troEoE
jcdJ(
E
'.{ O{HE.4AJ4 +J
E
O{dr-l
E
qdO{CEd5,Y +J
E
'F{ O.tr{E'-l 5,14 +,
i
E
ad,
F{
E
codEC5d+J.Y
t
Q',.tE r.tA .r1+J A1
E
O{d
F{
E
trO{ .dEtr5d+J J4
E
+Jc.d
'l1t
lEntrgod-o +J
IN
Ion
IN
tfI
rn
sfI
sf
5g
.Flad
rl5
oO{
J1o
r-{(s.o
cdttttr(d
F{
acoA
a
o
-w/a6gz'g= c olloubt
TppI,s
E
4ere I
I
01fr V 6ue>16uas rerled
07 '6'61080e ' gEt
q' nr2=SP ot*J 'sv
,rrq.qt''Q*2>
ueqaq nq+z (
uPqaq tQ*Z(rwc/b>t6,6t =
nQz
! Z-L 6ue1uq'q'6'0ngr
J
NP ,.J+
)tg ,O , q. o{Z
-urc LLroZ = -==,999?, z (05) (}v)zl9g'Z =
-U1c/6>1 0g0z = n**J 6---- cgzz = {q,J (----
q' q'b=V
= uTtll Vun.
fr
06
{oTPq uern{n
= *'?vzngZZ>I
=nJ
IuT
ruc Zrg =
.'uc gse t 3, = G
uu oti
'Aedel nTpTraa
z(Lt aL+'E = "v'6uedueued E
ge>1ed1p 6ue>16[as
'de1a1
( [ t=] PJe?uauras
(Sr6=1 de1a1
s'! orfrv 1er1ed
0? '6'6 t=
c
st0)z
69 -
001/0n =3 Uele?e3
- 70
.
h = tebal manfaat, dicoba = 15 cm.
D*.* = g1 netto. 1/2 (2r5 0r4 h) . 1
= 61259.1/2(215 014 0r15) = 5r103 ton
T = 115(6.103).103 = A, ?a ba{,bu .- i00. 019.15' = 6,78 Xg/cm2 (Z*bo(=915)
M max = 1/2 g1 netto . (#i2(t) = 3,45 tm
15rcu =I / 1,s(3,4s).1l/t f, = 0,2
= 3113
q = 0111252(a *2)*22s = 1g r25 .r22080
(pakaigl4;8=79,42(pakai fiA 12,5 = 4,02
"r2 )
A = 3 ;55 .*2 ( pakai 0A 12 rs
OK
2,em,
= 4,02.*2)
dari tabel diperoleh
A = 0,11(100)(15) .
r E 3165 a*2tulangan pembagi = 20t
Sketsa penulanqan :
or
{&',tl,',
tonqan b-
6-30
l-q,
1 2g2s
il8-1 2 t
14-8 61il,u
'ru z TTque uur e'f
-t-ru ZtL'g = +-!. = t8'zL I
' T ) Trep -rP saq qT qa T
TpltedTp +dp {81 Tdp+ ) rug g LZ' g nelp
(tt z/L L'v)[T'g =
I.8, Z L
7 g'z/tg'zl, = "i.(---- ru B['E = [T Tpp1,
t6'L = Z't= g TTqurE
{!
}i
|ji:
#
mili!il
,:ii (zt +
BUArB{
z rxoTo{
_tu eB'ZL = _ruZZ
179[ = tL+08+09 =TE?o1 duol I L= ( OA+09 ) t0 [=-rTs{P1Tp gsepuod Tirrpuos leroq
TsPpuTAltTad uern{n uE:lnluauaru - p
: qpi[m( de1a1 IEf Ts-raq upueqaquled 1
e selPquelen{a{ erec 6up ue{rpsaT5g'?Z,n B[eq, SZZy uo+eq n1nl,\i
. eduue6uelnuad upp
lsepuod ue-rn{n I rjEEffi}-tstc/b4 z, L =QEue?3 c
uI g, V =urof o{ a>{ uroTo}t {e-rEC
'0V/0V = Z-L ruoTolt upirn{n'rpque6-ra1 r1:adas '
,,6ur1oo; de:1s,,
SZfrV {---- 1 ue6uedeTuotuolu {1n ) ?- t Eueluaq ueEuap up{prupsTp
Tde+ gzile re4edlp r V {ln edusn-rpqas z Z-SZfrg f gzils 3 t-Z uendunl uaulotu +
TTqurpTp 1 gzfrg ; L-z upndurn? uetuoru +
lu8['t = [t qaToradlp'.ra1au g, Z = eqoc g
( EB' Z I ) gZ6' L r{aTo.radlp
= (zl + tt)g Euprurp
'10g=ed '10g =Ldtsepuod rnr{elatTo : Zt Tffi
1 Eueluaq ue6uedel uaruop{
7r("T + 'T)g= 8Z6t
('T z/L L'v)'T'g(zt + tt)s
to 7.', rt, (2,0ffi =
z Z ruoTolt sp depeq-:a1 Z + [ Tsepuod 1e1ad 1e.raq {11 >1e:eE
tu gZ6tL = =
depeq;a1 Za + ta aluplTnsa-r rsp{oT >1e:e[ = x
* = senT:redrp 6d Te?o1 lsepuod 1e1ad senT VgL L rL -
I10 g'
{
Z luoTo{ ue6uolod EpPdi
fi
e',owE
?iJc lF{adl I
ue6uelnuad {n1un--: requeb uebuB:alax
- 72
b. ltenentukan q ng!to_:
co
., = 60 + -80=netto 2,5 ( 3,4+.2 )
si. :
c. 'l
'l . 0rg. h
8,0775(2,1 - h)o;90.h-
h >0, 1813
M max = \ enerto (
1O,77 t/^2
( g,s
cm.
h ambil = 2 Ocm.
Tinjau pondasi Z (pondasi 1 juga boleh)
2 ,5m
untuk 1 met.e.r 1eb4r pelqt :
D = o I () q n Amax 'netto '1\lt5 0'4 h)'(i)1A,77"1/2(2,1 h)5,383 (2 ,1 h ) ton
Lbu1000 ./
Cav
lebar strap beam diambil = 4Ocm.
\
-&hf
l
atau h. =t 25
akc.2 Penula an arah t lurus balok "strapr.
+ (z,s 0,4))
,1025 = 5,937 ton m
/c,r = ---?: -.--- = :-_?-I /"*+l!erlJ!-qqI ' 2(0,s)(1)(225)t [ = 0,,dari tabel diperoleh q = 0,0675
. A=0 ,067s ( 1oo ) ( 20 , aLL*#4 = 14 ,6cm2 ll12-7 ,5=15 ,a'l
c.3 penulangan searah denqan balokstran.. ambil 20* = 2,92cn? (pakai S, _iu = 3,35cm2)
al, Penulangan balok rstrap,, -
Lebar balok diambil = 40cm.Diagram M dan D tergambar pada halaman 73.
= + .1a,77 .1
+vs, tz =( g'0 ) Ez6' gz
' LZ =uwc/64 00r[ = *{ (---- VZ o
,rtrc/b4 gL = ee ,- SZZ x: sTlsPTe PJPJ
?EZ,6v- =( ?'o l6Er -926' sz
lgL'gz+=(?r0)0s t- (z'.t.1926' gz
gt Lv
I
I
I
I
i
I
i
I
I
I
t
I
W
,1r'I
w/9Z
1L
0'0)(E?i.9n
w/lOZ=? , O/ Og=Zb
tu
'gz-ogL
w/t gz;'92=s'zxLI'o[=b
w/lgg[=V'O/09=[b
,,'.(',t l' ,o t,T!
CL
0?
'z=g
74
0a = 0 ,529
= 0 ,247do=7s. o ,52g. ( 1- L#,
h = ,Ll+ dimana M ada ; ah rvr max = 54,85 rm
54 rg5.1 0540
95 ----) ht =
T"= . oa-t-
nFn
l
I
= 0 1247
ambit h =
Penulansan :
M lap max
D max
Check geser
91 ,47
105 crrlr pakai uk. 40/105.
5,4, B5 tm49 ,23 ton pd pertemuan korom dan balok "strap', .
1,5(49,23) 103
tinggi balok kita perbesar.ambil ht = 120 cm.
T - 1,5!42;?j).103 = 1g,65Lbu m
-4bu- t- \ -J Juq
019.40 . 95 = 21 ,59
A.^-*SUAUs T .basu
u= 110
l*g/"^2) 4.*b,, sementara (=11)
< L*bm, u beban tetaP (=24)
xg /cm2
pakai sengkang 4 penampagn g1O
A" = 4, 1/4 7C (112 = :,ial cm2
= 3r141 . 2080 o ?tr18,65--10-.- = 8'75 cm
tuk lapangarl .
I t 2( 0,5 ) ( o, 4) (225)(| = 0 r4
A = 0,08167 (40) (110)
= 3164 )I
)rq = 0,08167
2. O ,5.2252080 = 38,8 7cm2 (Ilzs = 39 ,z7cm2 |
€.A'= 15r55.*2 (4925 = 19,64 cm2)
aman 75.
'T.rTpuas up>ln1uotr e
'tll 9 =tuoTo{ a{ uoT o>[ {E-ref
,
? 0L = 6a i-: 9a
looL=8d=tu-su = 7a
4 o? : ola = [d
: luoTo{ er{e6-edp6
0v / 0v = ruoTo)t uern{n
i=P
: E I T.PoS
]:="
J'
t-rl@ulg
L
f.==' ( T
"*luod {o'i. eq eduel ) +eTed t sepuod rnqe?a}tTq
g' L-
E I-BAZ
'I ruoTo]t tpdues ue{sn_ralTp SZfrA ue6uedel uu6ue1n1 +
(_ruc Zg,V=\ Ztfr ? 4---- cruc Lggrt = VggL sEnT
ZE's 6utdulBs 'Tn? Euesedtp (S. t'6'sd rgd ) T66ur1 )toTeq >{nlun +
luEffi
z'z'+odL-L'1od
szfr
ztfr<-
szfraf,
tfr
NLfrV-izga
s +or i lor-T ?o
ol, +ool 10
or tl v
EL-ovba-otfiv
76
55r 00r 00r
potongan a-bm
Jawab z 9 ,4*1 ".le-
P total= 830 to
P total = 830 tonKarena beban-beban kolom tidak sama
beban-beban kolom tidak samajadi ada eksentrisitas.
El.
.[trrr.r, = 0r5 kg/cn2
mutu beton KZ2S
bala U 32Pertanyaan penulangan pglat pondasi seeara kekrratan batas (pembebanan bersifat tetap ) .I
I
I
v
A
!_30 + 108.g30
-b-
I
I
I
-ltrr
aTr"l,_
I
Anggap reaksi tanah pada pelat pondasi terbagi rata du_Iu.
r=Ptotul + b.s pon,Casi
A
ambil. a - 50cm ; check (B)(L) t(1s)(13) ,
untuk as ArDuntuk as 1 ,3
4.r,.r,
192 ,6192 ,6
Statis momen terhadap as 3 = 0.
= 613 m
0Et'e0vv' z
( S= ) qeuE? g> <-- Z-?1'.?
6gt't
rus'g-rug'g-lxg' gtug' g
rug' 6+urg' 6-rug' 6+UIE, 6-
pcqP
( zw/t)u[,x}IT?T.I
'^ gglLo'o + xeog?o,o + 9t,t =
^ gr:gLvt +
6VZ
?ur gS,0ttL = E(6t)(Et)
?tu 8s'8 Lvt
ru1 6' ggt.
rul 6vz
x %ffi#+er,r= b
+ = CrB'+ =
+ = es.r.\ =
= tr". Tp1old =
= d". Te1ola =(6t)(tL) T'I
0 E g Tp101d
uof gtg = Te1o1d pueurTp
x[,
= t(tt)(6t)=(EV'O) 0es
= (e'0) OeB
z*/a 9t't =
xr
^Wx
W
r+-xA' W
r (t)(s) E-
;4r@ld='r
' 1 1e1ed lesnd' 1 1e1ad lesnd
'qpuel
selE r{eToqas )
b qe>1ede >1aca6uau uep
ru t'0 =
: b uptnluauall .pg t.'g =
d"
uPuP{ qeTaqos) ru e?rO = 6 tV'6 = *"
: a uEp --a.selTsTrluasta uetnluauan .c'.ro1au eV.'6 = x
088
-w? - +- 6r J
ll
lrA".t- I rrr
,l "l
-h-f
(er09+(8t)SS+'0 = V sP pql uauou ST?P?S
X ,'=8 t ) 0v+(zt ) 06+( fll fll 66 s106+(z)(e)00r
$,I:
78
€. tlenentukan tebal pelat pondasi.Check kolom pinggir no. 4 ----)
,4+h
Pa = 100 tonKeLiling bidang g€ser= 0 r 4+h+2
(0,7 + h/z) - 1,g+zhLuas bidang geser = (lrg+2h).h
h > 0,388 m
ambil h=4Ocm ---_) ht = 4pcm,
0 ,5 +a ,2+,
- T-_l
f . Membacri ndasi lat adi 3 alur dalam arah memania
= 4r031 tm2
= 3,601 t/n2
t/n2
harga rata rata.
282
qt t=
g'' ''_
Untuk
Jalur
Jalur
Jalur
Check
4,292 (3,5',) +3 135 (g r5)13
il = = 3,119 t/*23,369(3,51+2,44(9,5)
-
= 2,690
tiap jalar , harga q diambil
r ' Qt = = 3r7oo3 tym2
rr: grr= 3,119+4t031!21690+3,6a1 = 3r36o2s t/*2
rrrrgrrr= 2t69+3t5a1+1 ,35+2r44 = 3ra2o2s E:/1112
i ql(Iuas jalur I) + grr(luas jalur II) + grrr(luas
h lurttotar
?'a3,7003 ( 19 ) (3,5)+3,3602s(19 ) (6 )+(3 ,O2O25l (1 9) (3,5) =" 830 = 830 -:--) OK.
,
J Et-
III ) .
4rzg2(9,5)+3,35(3,5
830
t@,rt' u
tr,.o'.Lg
.1@l, 6el
I
II
'zg sfI
r, r r rt 18?, !
9E'Z
w/lvz' I L
9L
nt
,E8'ZL
La' z!t"
ul
I
I
I
?'!'vi'gt
lu9
+8?'[91LL, LLLAL
: qqs requebral q urp rt uerbelq
'uol g'g7e = dI
'Ts{aro{ uE{nIPTTpuol 0[E
19 9=
= (sE)#tf=t'u
(oor)#+3 =8d
(oor)#+f =ed
(ss)#+f =zd
6t- h
-
s'g?t
uol gv' I 9
,roi LL' L L
uo1 LL' L L
uol gl' lg
[=
[=
w/l tz' g L
uol E'9?E = 515,ft51: = P1e-r-e1e-r d
Eteu Tp?o1 d # Tp1o1 Ts{pa.r eua-re{
= SE + 00[ + 00t + E? = TP1o1 d.., uo1 tgt = ( 6[ ]E[9 L' LyZ = Te1o1 rs{pa6
'oz = (9)(gzogt'E) = b
: II rnle! nELuT,f J'J
w/t Et9[
100 !=
, 80 -
Karena pelat pondasi ditulangi drm arah memanjang dan melintangrdanpembasian jalur adalah dlm arah memanjans dan-r"ii;;;"g;-;;, harsamomen dapat direduksi sebesar 2OZo**M lap rnEu( = 80t x 72rgl tm = 5gr30 tm
h = 40cm, b = 600cm (Iebar jalur ii)40
[=dari
= 4 t97
0 r2 *!t t.,|
tabel dipijroleh q = 0r04ZS ..."
A=0,042s(500) (40)e$#*#3-:.I - BZ,ss cm2
(gakai s0 gtz = 90,4cm2 atau rl2 _ gg8 = g12 _ 7.51Ar= O,2A = 16,51cm2(15g12 =. 16,95cn2 atau trfZ _ lOt**M tumpuan = 16,44 tm x 0,g = l3r-152 tm
pakai A minimum = O t25* .600 .45 = 67,5 cm2(pakai 50 112 = 67 t}cm2 atau g1Z _ 10)
A,= 0r2A = 13,5(pakai 13 g12 = 14,69cm2 atau !i2 _ r!5)iheck qeseir : D max = 0,g x 57,0g t = 43 1664 aon
--- "'-,
Zo" = *f|ftf,}|f;ft@ = 3,17 ks/cn2 ( Ers,r(=s,s)--) ox
f.2
=4 0t
(3r5)t/n
P. total = Zg0 tReaksi total = 12-rg5r05(r9) = 246007 tKarena p total / reaksi totar maka perlu dikoreksi.
p rata-rata - '280+246 ' q7
=263, 0 3ton
Pto = Pt = W (40) = 37,s8 ron, 7
Pl = Pq = 253'03 (1oo) = 93,g4 ton.280
)b' s= (3 ,1003 )
=12 r 95 1 05
=100t =100t
i s= 3st#= 1 3 ,84t/m
'(S - etfr nplp
Zlfr OL 1e1ed ) ,uc g?' gL =c
gLgo'0
Zr" 16, L = ZLi L 1e>1ed1
(0[ ZLfr =
tFE - zLf, nele z"ss'68 =_ruc gLEr6t = Z
gt'6sg 'tgz'o = 1.{'q'
w/l?g ' e I09 u,1, Or
08 LZffi (0?)(ssc)
= b'r{aToradlp Taqpl T
z'0
lt GL =L90' 0=v
ep
I -'r
"rl ;
96'E =0?ulc0? =t{ 'ulcoEE = q .rpqeT
( 09 zLfr nelP
Zr"gLB'L = UZtO = rV
ZLfr SE reqedl
7EZ,I =,urur V te>1ed
q .irEqsT
gttzg = xpru q : .rasa6 {cqac
lE,tS =68. 99x9, O=Tprt ueur TstnparTp (---- ut16gr99 = uendrunlulll, g?le , Ztrtg, O=Tpts[ueur rg:inpe.rTp (--:- lrrl €, Ze = ue6uedel : xEnr W
. splETp rpque6:a1 q upp W rue:6e1q
'zs9g'0t
gg'P?'II
I Z,E
68'
@0t.
@EL, L
lgs'IE=o tuT
btl
vt
.utn\
lgs'rE=ta I
gzz'(s't)(E'olr.,000['([s'tE)E'T
6'g 1I
I
I
}V6 't6=
I
1?6'E5=?d
Ar = 0r2A = l5r3 "*2 (pakai rg g1z = 2013 "r2 atau 012 zor
f .3'J41ur fff : tulangan disamalran dengan jalur I.
g. Penu1a"perat dibagi atas 4: Jalur spt tergambar dibawah inl 3
82
;.'-:*l:--
:
l
Statis
x=
g=
g=
9r=
momen thd as tcolon l=
5133 m :
6 5133 = 0162 m
*-t 6' ' FA B.L2
#8+ffi
0
gi=A ,5'=27 t
6t/n
gx
It{ dan D
F{l
=
ditabelkan.
3, 46 I 5+1, 020=4, S3 I St/n292= 3r4615-t r070=2r391 St/nZ
I
i
l
= 27 rlgg - 0r9g77x
, (szel (91 (9, OIZ0oorffi
:)
z'o =
=?0relJ x B0g Tpefueu
gs,'e ='w1 t.vtl.L"l .,Ts{npaJTp (- r.-- ru1
0v
V1reVL =xpur deT h@
6Z9 ' g.g
IJiE98'Eg
Ltl' vv
v0''t t L
LLV, EL
glE't
L6O, Z9I
, Et'5tt
@g'z LI?8'0
: o urp14,rU@
t.I
t
1 L6g' Zg=Q f -- g' z.L=x
1 LeL' ??-=o' (-- E'9=xo0z-x6g L' LZ+ rxg8E6?, 0-=
. ru1 g?' 6L L- = xpru W
ur9, Lr g = x eped Xpu 14
L 1.9' Z = !{ (-- g, Z L=x'00L +
x00 z-,xg?6s'E [+-x zgng L ,0-=cu
(s,g-x)oot-09rx00 L-,x sv6s, E [+^x zgvg [,0-= c - t.--'
9'ZL)* ) 9'g
:l 6Z9' 98-=O (--, S , 0=x
00 1 -x5g I' LZ+ rxgE6?, 0=
ur+ ?E'oL-ru+ ?0 , t,V L- = xpru W
llrg6rt = x eped xeu W
'0s+x00 t- rxSn69, E L+ -xZ9V9 [,0-= Ct
(s'0-x ) 00 t-,xgi69 'E [+ -xzgvg [ 0-=
-.-:']1r
g'g)x )s'o
I
x6g L'LZ + z*ggE5rrO-!
Z*iV69' E i+ t*Zg7g[' 0-=
(XLL86, O+68 T' LZ-681' LZI
ztii Zx' (xLLB6, o-68 r I LZlf =E )") o
ox
Wx
TeAra4ur
rQ
Et[ :
Dari tabel diperoleh q = 0,0g625.A=o ,oBG2s ( sooo ) ( 4o,W = 167 ,s4 "r2
,tEl-
diperoleh g = 0r04062
912 16,95cm2 atau gt2 - 40).----) diredulssi men jadi 0 r gx
rE 69 , ZZ3 ton .
= 916 - 7l
0rB x 7O,g4=56rG7tm
86 r529
o,04062lroo)(40) w78''s
"*?" t:'r: ,::^gr
6 atau st6 - t {
A,= 0 ,2A . I 5, lgcm? (pakai I 5Check geser_: D max r g6 rilZg.t
Zuu = { tglkg/r*2(.2*ou,i=9e5) :_____) oK
@, =40t =55t
.l
-,=50t P total = l{5 tonJ r-
: r =pl( pi.i;q ( 1? )
- 6rt|13 m
e = 01413 m
{
9.' =2t=9 x3r5I t/n
,58x3 r 5,03 t/n
otx
M
'i tt .
= 9103 + 0r326946x
halaman 85.
rt:
hi
lll
I
(In-3 t7o=13 t
47
r)(92
= #hm- ? 6il4s)(0,,!r3), ) 3,5( r3lzl3r197 - 01607
92= 3 , 187+0 r 507=3 ,7g4 t/n2
= g
!!'l3 ( 13-*J+i,3,r27gx'
dan D ditabelkan padaBidang
9 Lt. L0, 0 = b r{aTo-radf p Teqpl Tf,pe
;'l 0t'u1 Lg.rLE- _ 6otzL
xB, 0 TpeIuaru Ts{npojrTp
6S' gt' 6Z
9S9'
,1st
It'er/g' gz,' 'lirl1rft,
I
I
60'ZL
gz'vt g'I
: -rpque6:a1 Auep W ue;6e1q
o
gt" r
1 l?te?+=O(- g,ZL=xV' 6Z0 =G (-- S r
9=x96-
xtO'U*rxf ? €,gL' g=xq
ru1 60tZL_ = xpru W
ur?0r6 = x eped xpu W
(9'g-x) gs-+xor--xg L g, ( ?+tXZ tVgO,0=xW
9, ZL
,.\,> s's
?-xt0'6+ oxLltg1,0=x c
rul ge 'Vg--;-xEffrvJZ L , ?=x eped xpu W
gz'vt-=!{ (-- Er9=x0Z + xgr-
Z*9Lg'V+t*LV?90r0=(s,0-x)0?_
Z*9LS'l+t*l VVgg r g=xry
g'g)")s,o
'1 9gg'?=q (-- E r 0=
xe 0' U*rxLlt.gL,0=x
u1 9f [ ' [ =!,I +-- g'0=x
,*LV?S0'0 + Z*SLg,V=*'f '"',tulu
f .,*' Ib z/ L=xws'o)x )o
? qaupp W 6ueptg
tA - 0,07375(3g0)(40) W = Bt,5 7 cm2
-, VV
(pakai fi Tte = :r,u3cm2
atau g16 _ fi$ = g16 8)AJ
l';,i:'rl-=_ru:rr,^:*l (pakai 18 912 = ro,ri.,n2 atau g1zI tump max = _3a ?q *o, ___.,, direduhsi menJadi o,ll, I,,lr=
..0 , li
rCu =
t rt(10
co
rI
,Ca!}*- Pemilihan jarakdaerah lapangan
f, * ,-,)Dari t,abel diperoleh g E 0 ,0 3437 ( ambil utk Cu
f,,= 0,03437(350) (40) W = 38r9.*2(pakai
= 5 r 53
12-5
rg12-
f V=f
tulang&n r harusdan tumpuan.
20 )
-27r424tm
= 5r5l )
21gl 6=4 Z,Zl em2
atau fr12 _.90).menjadi 0rg x 43141 =
34,729tks/cn'<f *b,., ---) oK
atau 01a- #= flte t6).Al= O,2A = 7r7g cm2 (pakai fltZ = 7,9lcm2
,I
.,. ..-.r ...a.n
.
"'1' Lbu="?i61n6l3l'atq"9g= 4,13
gl2:50
012- I 0F1z-4s
S gtz-40fl 1Z-7 ,5
disesuaikan untuk tulangan
.ffi='*,.,..
ryv-zltu
3-C
'10d
I
ff'
os-z t
Lf,
fr9
F&
i
88--
Soal 14,:'Diketahui
60r
0r5m
pondasi pelat untuk denah kolom berikut :
.I
4,; +| 50tI
I
I
I*bIeo t
7 20 + 102 (7 201
(10)(18,5)
bI sot
( 4"n"t,5 t/*2
6m 5m
1l'llkuran semua kolom = 40 / 4A .
Otanah = 0 r 50 kg/cnz ; mutu' beton Kzzs , baja vz4 .-.__- J ysJs v_z
Rencanakan pelat pondasi berikut balok pondasinya, dengan earakekuatan batas ?
Jawab : denah pelat sbb :rtr{h*tL r
pnbll= 1m
ambil= 1 m
ambil = lm
\ttk berat pelat
Lanqkah perhitunqan :
1 . Menentukan ukuran pelat pondasi.Anggap tegangan tanah merata :
Ptot"l + berat pondasi
pondasi diren
, /-ffi$ rstq
$* luas pelat
4,28 ( s ____) oKjadi dapat dipakai uk. 10m x 1gr5m.
'v
nW
Zt f;
{U
fi
I
III
lI
'rUc $Lt9 = LZ.OOI .tg7r0 = c
zrh 6Et =+ gtg'tl
?,{'q:tg'o= UIIUTUTIU .4.V(s27,) ( r )(9'0)z
Z,Z
vtc LZ
epedqP{TaqPlTp
=nJ'(ulcg =TTqutE uoloq lnurf as I wcr-,Z = y, (____
. 0 6 ueuPTer.{
a'p'cq'tTT E TPP(---- 1ETuPburTnuad
Blel- e1e.rTTqureTp b
vlgz t 5
(tt'v
: ppup? ueTIueI-rod 6up Tensas qpTepp
x- [,
;fr uen;fo eped
uep + ppupl reFiiA5r ?t6o'o + xLlazo,0-z6g,E =
%+!{%Sf B€i =b
?rEe'g Lzs =E( s, g[ ) ( 0t )gf =fr,
?rtg't?st =E(ot)(s,g t)+ =*rur1 g0t =(gtrO) OZL=xe.d=frW.
'u? tv L = (2, Ol OZL=rta. d=xw
.,u g8[= V L'uo? |Z,L = TplOl d eueuTp*rTfr,ffi;;4; p,crqre 1e1adt ; =b
lnpns{T1T? pd ol?aub Terru
tu Z'A = n Ztlru EIr0 = [rg EZtg
z.'t
u ['g =
+( L r )09+08+0t ) +09+08+01 )
TTBP
Trep
uPtnluauaw't i =ex=A
=^
9[0?+09+09luoTo{ .sEruoTo{ se
s){x
tvtere[:aq Te{T1raA edp6 aluelTnseu
'z
vLg' tg€.2' e
OLL,Igvg'v
s-E-
s
s
gz'6-gz'6+
9Z'6+sz'5-
p
3
q
e
( rw/t1b(ur)
^(ur) x)t1,1
i4
9C
Penulangan pelat
Tipepela denah
Ik=f
xMomer besar
nomen(tm)}Iusl ,5M(tml
Cu f q a(em 2) tulangan
o
l*= 4m
1=5mv
1,25 M._ux
0,0,18ot2'4x=jr348
51022 4166 0 0,0475 1 113 912-7,5=!4r 1,6
M.ry 0r03gq
12x=01650
31975 5,23 0 0r0375 8:r92 fil2-12,5
= 9106
Mtx
-3r349 -5,022 4166 0 0r0475 1 113 p12-7 o5
M.tv-2165 -3r975 5r23 0 0r0375 8r92 pl2-12,5
o
1-x]=v
4m
5r5m
1r40Mt
x0,053q
fx=31696
5r545 4r43 0 0r054 12085 fi12-7,5
=14116
Mtv
0r03gq
fx=21650
3r975 5,23 0 0,0375 8r92 pl2-12,5
M.t,x -31695 -5,545 4,43 0 0r054 12r95 912-7,5
M.E,v
-2165 -3t975 5rZ3 0 0r0375 8r92 fil2-12,5
o 115Mt
x0,0569
fx=3-906
5r858 4131 0 0r057 13156 fr12-7,5
l=5mvL=4mx
Mtv
0,037q
L2x=21581
3,871 5r30 0 0r037 8r81 fr12-1215
Mtx
-3r906 -5rg5g 4r31 0 0r057 13156 fr12-7,5
Mtv
-21591 -3$71 5r30 0 0,037 8r81 912-12,5
o
krrb= nl1*= lm
4Mt
x0r054q
12,,
$.e:s
0r353 17 16 0 q mtn 6r75 912-12,5
=9'r06
Mtv
0,01_gq
L2x=0,083
01124 29 ,6 0
A=20t x 6175 1135 "*2tu)o
!{.t,x
-01235 {r353 17 16 q mrn 6r75 912-12,5
M.t,v
-or0g3 4,124 29rG A=203 'x 6175' = 1135 cmZ( pakai ff-l20 )
Untuk pelat tipe e disamakan dengan tipe d.
j ':eqtue6-re1 1ds qoTeT sr-re6 pTod
'ma=dltoTEE
-:- -1(2j-@
i
92
5.1 Balok as 2 :
9z=
Berat sendriri bal0k tak diperhitungrca"nngi ni1.ai momen yang diperoleh.
Krc=+=0,88r*"I = + = o,657Er
KHr = f5 = 0,7z7ilr
.,:'.llomqn_primer : 'l
M g r71g. 1'-FE T- !
1
,= q 1* = (4,359)(4)=17t436 t/m.8,718 t/n
karena akan mengura-
ql-90
= '2( * s r*)1
Z 91 =
I y't", =
f&"r=) 4r,"=
ilr, =
0,9O-Jf0-;87 = 0,s4s
o ,667AIe=m;667 = o,4ss
0 ,657W0,4780 ,522
'I"=+'t#->l
(1-
*-r-rt
3(2T'
l) = 2,906 tm
* /'tdimana/= i: ? = 0,4
'I. -'tt'a19's2 (1-2(0,4r2*(o,a)3)= -21 ,0259 tm
ErMir, = +27 , OZS} tm
W (1-2 ,1,, + tf,t3t'[" = -,[c =
Mflr= -*l* =
= -42 1621 tm
i7,4l8ls,*12 lt-z ( 2 )2 + r 2.?12 5,5 Y .E-)') = -34r443tm
6/ov = {n
-T'.ffi6 frrz,o I'ruco? = eqocTp q ! -I-/or= v, J r-
(
"L r-€ ) # =xEuI r^l
VZ
u
"o + beO
rffir*
hrt: qqs snunr ue6uapueqeq leqT{p xpru deT !{ Turffi
'6ue1uaq qe6ue1 eped Tpp[ra+ dB66ue
[0['8?+ SLO'ZF 88LA'Z?+ 8906'Z-. g0 6'Z+
L?z'o = '{,.-ffi -L)ezE;o ''st
t "P
6Z9'0 =
LZ =u
,wc/6>1 00? t = 'Q <---- VZn p[eqL-
,vtc/b>t SZZ = ! e---- EZZ>1 uolag nln14'STISPTA eTECAS
ru? lgtrg? = xeu drunl W.rul LgEr 9Z = xpru dBf W Tpp1'
'ur? Lge,gz =
trrfft
ue{nlualTp uenTsaderl
906'Z- g06'z+ zol'}tr
=y
v- uffi + (eo6'z+zot'8r) f =rH dPTw
utl .sgvL,LZ+ =
trtll t t) ffi + ( tor'B?+B Lo'?,?l i =He dPTw
. . q \ vz 'uI1 zL Lt'' oz+ =
( r(Z) ?- E )
ffi + ( 88l o' z?+Bs oo'zl7 =3d
dETw
It.L'z+ 6lt'g+?t.L'z- 691'gL-
ztg'g+g7g'E+
g69'?-gE6'l-
L0s'g-90,ZL
gz'v-9Z'v+
906'z-Ltg'I t-t?v'vt
69z'v-elv'vt-
0 [ 6'E-LZg,ZV
960'L+LZg'Zt
s'gSS1O' LZ
OZL,VZ+8SZ0t LZ-
.iI
E5: saorS
94
unt,uk balok GH dan Hf .
48 r '101 .o : *ru+2) (17,436)=6 9,744t
0
D
: Ukuran 4 0/g5.M lap max = 26 1361 tm , M tump
D max =.Jgr73 ton.max = 48r101 tm
90= 4 r29
Dari tabel diperoleh q = 0,05g33.
. so,os833 (40 ) (9 O)? (o 's l22sAo,o = 22 t72 .*2 (sgzs
AI = 0r4A = 9,0gcm2(ZgZ5*= g,g2.rn2)M tump.
24 ,55 cm2 )
Cu= 90= 3r18
t = 0 14
' A=0r
Ar =
geser :
diperoleh g = 0,10667
10677 ( 4o ) ( e0 ) 'z,0r;*6rrr, = 41 ,s4emz (sf,zs = 44 ,r ecm2 )O
' 4A = 16 ,61'6cm2 ( AfrZS = 19 ,64 cm2 ) .za' #d 1336 is3'
1000 = 17 ,s3ks/cmz2z*b,,
Q*b"', u beban tetaP (=24jadi seluruh Tru dipikul ke tu1 , geser.
turangan geser berupa sengkang. Y=oer'
,
sement"- I
ra (=11) I
I
beb.
7 ,436t/nD = *rrs,744) + 42,078;48,10J
6
= 33 ,868 ton.
7 ,436t/*
*rr,s+r,s)
*rr1 ,026 ) +
38 r 73 ton.
( 17 ,436 )=6 I ,026t
48 , !9?-2 ,eo6
1
8r102
1_L5 . 26 ,361 .10002 (o, s )l 07] ( 225)
ii
UO1 LZg,LI = iZEI8,E gEV,LI = E
# #f=s 'ur1 gLLg'tL =
6u (sss'oz+eo6'rtf -{r)(eE:u'LLtft = EEE
906'9e' 0z
@TT
Eluaq qe6ua11p Tpp[ra+ de56ue xeu deT W
W rl{
906'z- 9o''z+sEt'02-
906' z
sgE'oz+
: uauour
906'z-
TTSEH
906'z
00
00
0zz8'9-
0zzg'g
lvg'Lt-906'z- tg'? !+
0Eg', [-
E,O
0tg'tr t+
tl9' L L+
Es'?[- 906'zt\Isorc
E,O
TW- - !{1- _
.{^ = dtt!- =
'nI1 906'l-, = ( [); ' r:#rm =,ww^ = ,xxw : :autrd uauoru
= wrx
= xrx
,,b
\rlI
'uI1 tSr?[
'uB6uEdeT {1n E I
Z*" g Lt L' e=Z(
Z,(v)(eEt'Lt) + = zT;b
E,o = ^=)y' = ,rT/
9;s
xT---hl ,tt
(nf rs?1vLffi
==I^-
: BSOTJ
w/t gett LL = ?' 6gt'
! rsnqTrxsTp uoTsrJ:aotP.recas 'etru 1,!{ ue6unlTr,lrad
v = ("T b f I z = ,bw/4 Bt['B = rb z/L = ,,b
3 g sP {oTPg .
g|fr uep uendtunl {ln g'L.- 1lfif'luo['6=ffi
S IIS 'l= V:010 v 6ue>16uas X '
=r nPrr S*t,'v
Z'S TPOS
1e>1ed
=
sED
+'v laL
il;'
'i I
i.'1t
I{j
i
i96
M lap max = 11 ,6275 tmM tump max = 20 r 355 tm
D = 11 ,621 tonIulenentukan ukuran balok: secara elastis .
M = 20 1355 tm
cobab=30cm.
h =o<os@ = 0,247 //W5 = 64,33 cm.
ambilfu=70cm-;*-Lpengl,anqaq
, iM tumP = 201355 tm,
ht = 75cm --) ukuran balok = 30/75
E 3 r29
M1
diperoleh q = 0,
A=0,100(30)(70)
,4 fi= 9 tOg
n * 1115175
7A
100
2(0,51(_22512080
"*2 ( 3gz2 =
tm.
':; cr. :'iI
I
I
,l= 0
anga
'Cu =
A
aP
(
l
= 12 ,72 "*2(pakai 7g2Z = 26 t6l cm2 ) .11,4 .*2)
4 ,36
diperoleh dari tabel 3 q = 0r0i667
A=0,0s667(30)(70) a$ ffiEI = 12,s7cm2(pakai Aflzz
=. li r21 "ro2
) .At = 0r4A = 5r148e*2(pakai ZfiZZ = ,?,6.*2).
sengJcang Zbu = = g ,2z *g/cnz ( d',rbeb. te-tap(=9r5).
. pakai sengkang prakt,is 0g- I 0 untuk tumpuan
il}-t S untuk lapangarl .sketsa penurangan barok, pada halaman 97 .
q-q
97,fi7,
g'L-0
szb
E[-0Lb9ZfiZ
0
I
I
EW,
OLNLfi
szfrzt+-flt
o
ISI-OLf,
I
o o
ZZfr
0r-BZZfrZ
9ZfiZ
€=Z=[SVXOTV8
O =J =$ -V SV XOTV8
98
Soal 1 5:
,6m
,5m
,5m
315x3 5
b - Apabira akibat gempa, trmbur momen pd dasar koromtm periksa apakah tiang pancang cukup kuat serta
Diketahui pondasi tiang seperti
,5m
tergambar dibawah ini.kolom ( beban mati+beban hi-
dup) = 160 ton.ukuran tiang pancang 35 cm x35cm.Daya dukung izin tiang 3
-akibat pembebanan tetap =50t-akibat pembebanan sementa_ra = 65 ton.
Mutu beton ( tiang, kepala tia_ng kolom) adalah R225.Er. Rencanakan tebal kepala ti
ang ( "pile cap " ) serta penu_langannya akibat p.( tebal selimut beton=7 ,5cm )
sebesar 50penulangan
pile cap nya ?
e. Rencanakan penulangan tiang pancang (panjang 1 tiang = 12 m).Jawab. j€r. Tebal pile cap dan penulanqannva.
Taksir berat sindiri pile cap = 5E p kolom .160 = gt.
Beban pertiang ( ada 4 tiang ) = lStE 42ton daya dr:kung jzin.geser pons 3
Iuas bidang geser= 4 (GO+h ) . h
Lbpu= aTuh.I.T (r*'n,,
ffi(lsh> 40 cm.
ambil h=42r5cm --) ht=4 2 ,5+-l ,5= 50cm.
tabel pile cap = 50 cm.check seser :potongan 2-2 r ht = 50cm .D Z_Z=2 Pti.r,o berat sendiri
---- J kepala tiang .
=2(42)-(0,5) (0,5+0,3- * h)(0,5+1 ,2+0r5 )(Z,A')
= 82 ,449 ton
'?{
IiI
i!i
i7*, I
I
i
e-p pe6uolod
6-? r
I027,
s' LZ-Z L
: uebueln@
(g'tz ztfr =
Zffr nele Z*rS0r6 = Zffigl Z*" gL, L = V Z, A = rV
(6 vtfr =L- vtf, nplpurc ozz OZZ I
Z*"96''9t = Vtf,?Z)rwciL'gt, = E9 .OZZ- tgZ,O = uTur V
'unuruTtu b
{.,, =: LY
6uB C uedas ti
TPqAIe?T{
(r
, rrri rr' rr r r, o'r'ro =!
ooo['(9tIot'vZis\ /' I
I
Itj
iIt
i
.
. XO PrPluauas ng 2+
. uo1 t lv?,, zg ='tuc$g = lq
( [.[ = ) e.rP?uauos *Q*2
rut gLLol,lz =
z(8'o)(z€?,e )z/t-(€,a)r8 = t-lww/+ ze?, e =
w/l vrz. zrz- Egro = bb = dBc aTTd T.rTpuas lprag
=nC
uo? ? 8=d,
.rB6e ue{Er{psn6uau sn:pr{t6ue1n1Tp snrpq edullre
)rr"1s4 ?g'ot = = rez(v' z,l (z,z) (9 L82,0-Br0) (Eg'0 l-(zntz = z-za,(---- ruc g' LS = { TTqure ,T6BT Tpqa+radTp q
,wc/b>1 ggr[[ =g'zg'ozz .6, o nga='J
'uo1 ?,L62' zg = (t', zl (z,z) (gzgz,0-g,o) (gr0 l-(zvlz -z-Za'rrrcgg = rq +--- ruc g, Zs = g pqoc
'Teqa?:adlp dec aTTdqPTnlT {nlun .6ue>16uos
1e>1edtp tppTltsepuod 1e1ad eped 1de1a1 ,6ue16uos
66
[=) Pr'luauas tQ*2 ( r*r7ur 69, vl - =
.e_z
100
@ P kolom = 160 t I beban sementara..Iv[ kolom (akibat- +1mna) = 50 tm )
lP=16Ot- M.x=g+ v
mF
Pz = P4 = 52'88 Pt tiang untukPenulanqan pile cap.
r \
+|l21 ,17I
, .ll
I
ler i2,
0 rSm 1 2m 0,
tebal pile cap jadikan 62r5cm =. 7 _ 1r05_(1_24r313).1000_
Lbu o ,9 .220 .62 ,5= tor55 t<g/cn2 (r*b,,
mentara ( = 1
n = jumlah tiang.x= jarak tiang yg
hadap sumbu Y.
akibat P :
,ilitin j au ter-
tiang 1 ,2 ,3 ,4 tertekan kebawah. 158 tseDesarT=42t,
akibat U :vtiang 113 --) tertarik keatastiang 2 ,4 -) tertekan kebawah
D =D = 'l 68 50.x-1'3 4 *2**2**2**2
=42- r#h = 21 ,17 t50Pz=P4=42+ aT6;6-I = 62,8 t
check q,eser : potongan 2.2Dz-z=2,(62,8) - ( o r 5 ) ( o r 5#0 r 3 -1 /z]nl'
(0r5+1 t2+0r5) (2,40)
= 124 ,247 ton.1 05 . 1 24 .247 .1000
zu,, ffi = 11'45kg/2cm.
)E *U,, beban sementara ( = 1 1 ). -
agar tidak usah ditulangi sengka
Dg, pelat harus dipertebal .
h ----, ht = 70 cm'
D=2(62,8)-(0,5)(0,8 -( 0,6 251 ',t 12 ,21 (2 ,41124t3 13 ton.
s0ZT=42-
1+P3Pz*Pa
1$ 'l[e sJ' -!ah=57 ,5t
1
z
beban
1)--)sgOK
i
I
Jsoyl
( zz,ge TPltPd )z*'6?'o I =
EgE?,|'o = b
.uA Z6Z, S !{ Z(Zl ) ( ? G?,, O) f =
ZT b B/L = xelu w
ur6)t gSOr6gg =ut1 gEg5ggr0 =
z(zl)(vGZ'ol tzota = xeu w
wh6z.O a lrZ.S€r0. EErg =b eueurp
zl b lz0'0 = xetu !t
'ruc ?,6Ltl = (g?r0[) VrO clgLZ ,^^t,^^
6 (ot)(EE)t.EtzL
v'o =(9ZZ)(88,0)(E,0)z A.
66'Z = =
=lV
'o=v
3l
"ri'ur1 Z6Zrg = resaqrol W TTque
'uele{6ue6uad up{resepJaq r6ue1nlTp 6uecued 6uet&. bupcupd 6ue11 ur5ueTrNre4
q g0'[ = oq , hI E6r[ = tW ! g,O =
or{'c
. url E L6V' ge =?,(g'o) (969'E) Z/ L-(tro Ig,sz',t =
w/l 969, t, = (v' ?,1 (2, z,t L' o
dBc af Td TrTpuas 1p.raq
' 6edure61 prpluauas ueqaq 1nlun(9, LZ ?,1fr nBlP 7,1fi9) z*" L'L = V
=oD
t-tlt
=
=b
: uelElrcZto = rv
(6 ntfi = # -rli nele
ucg7g 6ue[uedes f Lfi gZl ,wcg,
gf . = OL. O?,2. ZgZ, O =urur V
unuTuTru b qaTo.radlp
=^3
8L, L =Z9
?9,EZL=
3. 7fo L* 'rlg.
sz7,) (z'z'0
(g' olz000 L' (eL6tg€ ) E0' L
'dtc aTTd ue6uelnuad
"102
Karena pengangkatan tiang tsb.berfungsi sebagai bagian
-
tarik,ng kita pasang 3frZZ pada setiap
dilakukan dengan salahmaka untuk penulangansisi.
satu sisinyatiang panca-
l0-1s
D=7 0cm.r? -a
fr 1o-L,2D=7 0cm
910-7,5plat baja
angkur baja
pancahg utk pilar jembatan.Dari diagram sondir diketahui bahwa kedalaman tiang pancang harus_lah 20 meter.P=2800tonrM=250tmMutu beton utk pondasi K22S
ba ja u32Pertanyaan ; -jumlah tiang pancang
-pile cap berikut tu_Iangannya.
Catatan i tiang paneang berfungsisebagai "point bearing,pile.
sepatubaja
angkur baja
SoaI 16 : Diketahui pondasi tiang
mI'
d, rer- si ter-tarik.
F 8gzL
P=800t
=250tm
f ^oo rla-a '
'uol E'ts = +h = TnltTurau 6uet1 L €-- uolo0g=d lpqTltp
xo (____'uo? 0?g = (09) (
oE?rg[ =qfi 6+ cr= = 3u* c.rP= o
u u 06-i_r-.@ 6-t=U": dnor6 aTTd lsuag,sTJa
uc0?t=0E+09+ (OZtlZ =gtrrc0SE = 00t + (OZl)n =
0E + 0S + s n = T
TsTraq EuTsBul-6uTseu sTrpq g 16ue11
rloduolerl TsuaTsTJJa uep e[ra>1eq 6^6ue11 1
^, 108 d v] T6dE = TEFild
tir% =
' utco, x
I L ) L, 0 e^fiueq 6ue11 >1odulo1e>1 6un>1np e.f,eq
0A'0 = t1'gt -[ ="6g
'ralou Z. L = Oe = sp alt 6uer1 se >1e;re['16ue11 S upt
urc0
'Y"oE
T T qurE
S [ = 6uer1 qelunt TTqup. 1 6ue11
ueuou eIuepe puarp{)
Trpp >1e^dueq qTqeT snrpq >t
'({ =)
6utreeq lurod.
'rTpuos TTspq rp ue{n?uelTp 1dp 6ue11 tg re151_uol 08 = 6>1 0000g = (0g)(0? 0?) = 'rwc/bt1 0E = ErL/SL = "Q,€ Tpe[
'9, I upupuea{ -ro1{eJ TTqurp pf T}t
,wc/b>1 sL = ( rea ) =q, {, <---- gzzx >[1n pueurp'sq-O. 6 -
.--- - lEuedupued senT ) = 6uet1 [a0, Eqoc 1 re>16ues rnfnq Ouedueuad )
-iEpapf-,.a.a
. a:
-EOl-
- 104
akibat U = 250 tm B
tiang deretan
akibatP+M:masing masing
1 ----) Pt =
2 ----) PZ =3. ---) P3 =4. ---) P4 =5. ___? p5 =
tertekan kebawah tr
rl tl tt
Ivl = P1(2,41+yr(112)+pa (1,2)+p5'( Z, :a1PZ t Pl = 112 .. 2r4
D=
'D=
p_ = p.-5 -1
M = ?rrf2,4)+0,5P1( 1,21(21 a 6 pt
250p-= -; = 41 ,G7 ton-1 5
dipikul oleh 3 tiang masing-masing= 1/3(41 ,67 I =.1 3,89 tonkeatas dng gaya tarik = 13189 ton
rr u rr = 0.5(41 ,67 1
6,9456 t9451 3 , gg
1
2
akan tertarikllIt
ttt
It
t,
4,5,
tekan ='it . -
.t ----) oK
tiang pada deretan ke I ' ,. , .,', ,l
5313 13189 = 39r41 t5313 - 51945 = 461355 t53r35313 + 51945 = 601245 t53 ,3 + 1 3 ,89 = 67 t19 t, < F, ai"ng ( =80t )
i
(2(2+h)+2(1+h)) . h
P pons = 800 ton. -
150 t/m2
h>0,8s 1 m
n 2-2: beratq= 0rlD 2-.2=
7"bu
sendiri pile ceip = q
1 x 314 x 214 = 018976 t/m201 ,57-0,8976(1,35) = 200r36
ton1 r 51200 i 36 ),' 1000 r g Zkg/" 2= 9,
< f *O,rbeb.semer:tara (=11 ).
B-2) -1r35
I( 67 , '.191
20'l ,5 7 t
t 6ue11
rell?9"
I
,,/r''ao1- .,/ ,/
/,/,,,
ar/,/
t 8t,'62 = ztbnplPi 0t) Zurcg?t.
9Zl ewc6z = UZ' O
= 0O[' 08g' tSZ'O'Zt O =
rft ue6uo1o6
t I s,,e =
-1-23) ( 8's ) ( E ' ot z
/ ' -t
00 [
=V
3
n3
'(lu 8'S reqotas)uq 6W,Sg[ =
( ?,JEE , [ ) 7,/ L- ( L' Ori::'r;" =w/l Zleg't = lrz x grg x tlrg
'u1 Eg' ggz =6J ' L9 +E? ?,r og+EE, Es+Est r 9 ?+ Lv, 6e
€-Elnl
=b
=d
'( g qerruTTAur qErP ulelepATTuPburTnuad. (8, LL 91fi =
!+= gli nple _urc6 t, g0?E'c
(6
nElE ru I t e 6ue[ uudas ?,wc
L
ouc lS,0B[ = #Z c Wzggg0r0
Z,O=( Ir Lt.L -
,'€ reqaTos t?n )u1 gZL' g Le =
z(6't)(9L68,01 ?,/L -(2,'a) ( Et L' Og t )+ (t' Ll Ls, LOZ = [- [n
w/l 9L6g,o = b
' ( T qprp 1 6uef ueularu qprp ueTpp[- t ue6uo1q1
t = 91fr61) Z*,szfr = ffi =ez
9r [8[ = szfr Le
(00[) (0?E)
= b r{aTo.radlp
v
fr
Ll',9t = rV
r e>1ed 1
29590'0 = v
Toqe+ Tf,PO
(ur
o,,:f,
E0 t *
3 ctec eltd@
1Ix.rt
105
Penulangan pile cap.
120
Soal 17 : Diketahuivertikal
. 2,5m ,, f
16-17 r5fr 1 Z-22
40x20m
120 120 12A
120
120
tiang pancang spt tergambar terdiri atasdan tiang miring.
tiang
=500t p=500t bekeila pd sudut 7So deng_an grs horizontal.Pertanyaan, gaya yang dipikul mAsing-masing tiang dng cara grafis( cara Culmann ) IaIu check dengancara statika.Dan jika tiang pancang akan mengalarni penurunan axial sebesar I cmjika dibebani 50 ton maka tentu-kan besarnya pengeseran horizontal( " lateral displaeement f' ) .
t 6upTl lodruolo)t {1n
rell?e,, rlodtuoleq >{?n
'oSL uTs d = (Zt tlz + (6soc [n t)t'0 = A3
'g upp
6ue11 6uT seu-6uT spru TpltTl.raA B^dpE = Z t
' t'Z' L ,,aTTd6ue11 6uTsptu-6uTseu Te{T1-ran e.de6 = [n
p{TlP1s PrEoas {carlcTq
6ue11 1 ! uo1 S8 = # = S=l
6ue11 1 t uol E'EL[ Tt,. = e=Z=1 dno:6 : ,
: dno:6 6ut seu 6uTseu .Tppf
uE:tTesaTasTG : qPliEf'
'uox €'gz
'tlol 8 ' [€
t1S8
t
0?rOLL
1 001
oZZ=0= UIC?'E =
Vd
= uclrl = 8a
qeTo.radlp
= tucl : BTBTS =[
9d' ?d 6uTspu 6uTspru eped up{Te.rnrp 8a
t' 4'
Z 4'
I q 6uT splu - 6uT spru eped upltTp jrnTp Va nT eT
'€d uep Va sPlE Te.rnrp + 00S = ct 'Z}tT?T1
eped nlTe/t TEltTl.raA 6ue11 >1oduro1a>1 leraq {11 , V }tT1T1 eped n11er{
':t- 't.' Z, L 6u1:r1u 6ue11 >1oduro1a>1 1p-raq {11 rsp{oT up{n1ua1 . I
: qe:1bueT qe{5uET
TP
' (uuHrrnc Ercc) sTlexb E.recos
'6ue11 g[ = 6ue11 -TeAol TpE[Euutre g 6unpue6uaru dnor6 d=Tr . Z
'uPtpuEcuardereq ,, atTd .ra11eq,. ue6urrTuax ' I
. : urlPlPC'dec aTTd {n1un
u01
uol
- l.0l
l08
9 vt eosg 6 v,, - p sin 75o:H=0
I
3(3 vr9vt
Momen
(3)v1
sin0=pcos75o
fi)
( 2'lthd kelompok tiang No.. 5 = 0.;...
"":ni;'"i"'irxl ,],=o)(3r
+ (c vr
coso + vz = 2ps[n.75o o...... (3)Vt cos0 = p sin 7So 6 V^vr cosg = + p sin 7so ?'u,
cos0 ) . 2 + vzfi )+v2(0t27
(1) :
vt9
disubstitusikan ke persamaan(3 ) zzt( + p sin 7so 3 ir, .'ur = p
diperoleh Vz = -F sin 75o =' -17
(3) 3
sin 75o
I.iarSa VZ disbstitusikan ke persamaan ( 1 ) .(1) : 9 Vt eosg = p sin 75o-_ G VZ
= 500 cin 7so 6(2g,4) =
100.sin-17 = 28 r4 ton
31 2 ,56 ton.
750
V, cos 0(2):9Vtsin0
Vt sin0
= 34 1729
= P cos 75o500.cos 750
cosO 34,729sing W
34,729--= 37r59cos 22 .4gl
= 14 ,379
2,415 ---) 0 = 22,Aga
ton
9
Perbedaan stdtika dan afis :Har cara culmann . cara statika
37 r58ton28 ,4 ton22 ,4go
Diketahqi penutunan vertikal tiang untukt : i;*{ i:iii{: *u"ins,ians
P = 50 ton adalah I cm.vertikal.
vtvze
vt
cara
37 ,9ton28 ,3ton220
GF:
r'i
Ze+ Zl+ ZZ+ ZL I E= ZrZ
6ue11 qeTrun[ = u 6ue11
= utz wg'Z = a >lodruo
d selTsTfluas{a 'z\JJZ =
lu[ + wz + tut + tu?=X0=
g rlodurolerl pql uoruolu sTlels. (g.C) 6ue11
doduloTa{ lEreq {11 ue{nluauoru ' I
: qeAPI,
'6ue1q t sPlE TrTp.ral dno:6 drT,r
e sTlTTeuP E]Ec
-as 6ue11 6uTseru-6uTseru pd eIra>1
-aq 6ueI e^f,86-er(e6 : ueBr{ue1.rad
16ue11 >1oduro1trrg
aq lproq {?? eped e[:a>1aq {epTl) -
' dec aT Td 6un t rt
Trep ralou Z,4e:e[ pd e[;a>1aq ds TUT t{pr{EqTp reqtue6f,al ads ,te{T1.raA 6ue11 upp ( 6uTrTur 6u
-ETl) u aTTd .ro1?pqr splp r-rrp.ra1 6uprl >1odtuo1a>1 TnI{e1e{TC(
'ruc [9'0 - dec eTTd
TpluozTJortr upJasa6rad
rxoggg,
oZZ=A
-ruc ssL, o = |fo =
6ursptu 6utsptu upun.rnuad =
' .'08 = (
9l = ExS =UIE, O
o = 0'J Pq1
: gI rros
,3
1009=
- 601 -
' ,, oTTd ralleqrl
- I,l0
PV = Psin 75o = 500 sin 75o =c.yu vertikal n";;"*u"ing_ma", n 82,96 tornssroupjp+ry
Gaya vertikal pada group I (masing _ .mas___ . . ,2
= ssfoT + t '=-o;,;'::,tians) '
Gaya vertikal pada masing_masi
= 3! ,'; ;::;:"::nn tians sroup 2
Gaya vertikal pada masing-rnasing tiang group 3 3=% o= 32,rgton J
Gaya vertikal pada masing-pa"i= 4g2 t96 ( 4sE- t':'rffi:";
= 1'rif a1;r,r.lJr, "'l',;;r.:,Catatan ,: Akibat M = pv. e
3.
maka
ke54
3
2
1
kebawah
kebawah
kelornpok tiangakan terangkat keatas'ttrtr
tetapakan tertekanakan tertekan
32r19 tso ) =pcos 750
Gava horizontal cos 75o di ikul oleh "batter lle' .3 ( 48,3 tgOE+ 40 12 tgo +
0 = 19 ,660Gaya
= "a**::": v9 bekerja pd "batter pile,, (masing2 tiang).E6;E = #kfuE. = 5r,3 ton
Gava :,.,.|)r""::::"" pd rians vertikal (rnasins-masins tiang).
soar !9_-i Diketahui denah bangunan 3 rantai seperti(gambar pada nafaman lll).Diketahui : mutu beton K225.baja U 24
bangunan diperuntukkan kantor.tegangan izin tanah= 0 r43 kg/en.pd kedalam.n .lr'mukuran batok as t =as ;=;"'-;' _]:baLok as a=r=a=, ----.
*-----) 25/50----*__i ZS/50
I
tergambar.
I
I
I
UIE ,
e lpdwalas Tsepuod nple.rnTp[-pdTp qe>1ede r-rrpuas ue{nluaf de:req , t sepuod
4\Y//
lov / ov
t/ 8ulll
j
Ee/stI
Tn',,t--l -c
:
,P,
--rO
rsepuod TE{ue{PuPcua-r
E TeluETI rPlueT 3
3 rEqurE9
t ueeduelrad:.
ruoTo{ ub:n>161
lu9 .t' tuE' S ," tu
/0s/ s7,
-: 112
Jawab :Langkah perhitungan :
E.
tebal pelat atap = 1 Ocm , pelat lantai ,l =2=1 2 cmpelat atap :
berat sendiri = 0rl0 x 2400 = 240 kg/m2beratplafond+penggantung = 7kg/n2air hujan (taber 5cm) - 0,05.1000 = 50 un't;,berat lapisan kedap air
;Abeb. mati = 31g Kg/n2
= 1Og kq /n2q total = 418 kg/n2
berat sendiri = O ,"lZ x 24OO = 2gg kg/^2beratplafond+penggantung = TUnt^,ubin teraso (tebal 3 cm) = 72 kg/*2partisi (pembatas ruangan ) Uiauk
beb.'mati =Wbeban hidup ( utk perkantoran ,
9totrI = 717 kg/^2b. Menentukan beban
oad).
beban hidup
pelat lantai l=2 :
as pelat
Beban yang dipikul kolom adalah semuabbban yg bekerja pada bagian yang diarsir pada gambar sebelah ini.(termqsuk berat balokranak balokrkolom,pelat ) .
fpehtkolom ditabelkan :
I
1;
Lantai
3 ataatap
Nama kolo beban pd treIat (ks)berat kolom
as 1-A= 3-A= 1-D= 3r-D
as 2-4= 2-D
(0,25)(0,5)(2400)(1,5+2,75+2+2,75t,=2700
( 1,5+2175)( 2+Z ,7 5 ) (418
= 8438r3
(0,35)(0,35(3,s)(24d6)= 1029
12167,3
(0,25)(0,5)24001(2+2,75+5r5) = 3075
(5r5)(2+2,7s(418)
= 10920r3
0 r 35 ( 0 r 35 )
(3,5) (2400)
= 102915024 ,3
as 1-B= l-C= 3-B= 3-C
(o ,25 ) ( 0,5 )(2400't (2,75 +3+4,25 )=3000
(2 ,7 5+3 )(4,25)(418)=10215
(0,35) (0,35(3,s) (2400)= f,029 14244
bw' sLLs 6w' gLLs 6>t E
szQts 6tE v' t tlvz t b>tsv'6>{9'E
tltleoior 6tr ' sjorc 6xr ' slarc 6xE bzov
: TUT r.{eAeqTp 1e1ed qeuap epedue{sTTnlTp luoTo{ uPqaq_ueqog
g?, L IvZ L3-Z=8-Z s
T
V
,I
o
,l
v'gL6LSJ-t=8-E =3-[=8-[ s
S,EZI,L9a-z =u-z se
ew6vQ-E=V-tQ-[=V-[se
L, V6CLZtvt,LL' 9 L,ZZ =
( LL L)(E,E) ( T+E L' Z
EtTEJ-Za-zsp
L' 998 LZvvt. L
LI LZ?LL =(LLLI(SZ'V
( E+91, Z)
000€
J-E8-EJ-[s-[se
L=Z
Te+ret
9'0gLt,zvvt. L
9'LeLgL ='( LLL)
aL,Z+Z)(E,E}ETOEA-Z =
u-z se
8 t E8 rtlel =( 00?z ) ( 9, r )(t'0)(?,0)
VLVVL =ULD$L'Z+ZI( gL' z+g' L I
OA LZ
q-to-[v-tv-Lsp
97,, EZ9 L L
6Z0l =
(00?z)(s't)(gt'o)(9t,0)
9Z,6LZt,L=
(8rr)(E's)(E+sL!zl
SttE=( 9'E+t+s['zl (00vzl
(E'0) (Ez'0))-zs-zse
deleIe1P t
Te10?uolo{ . qeq
(6)t)ruoTo{ 1P-raq
( 6)t ) 1pTed pd upqaq
(6)t){oTeq 1P.regtuoTo{ eueNTe'$qI
tt t .-
1t4
Pada perhitungan lvlekanika Teknik untuk portal as A, B, C i D , 1 ,2 ,3 ki_ta akan peroleh momen-momen jepit pada pertemuan kolom dan pondasi.Jika momen tsb kecil harganya maka untuk perhitungan porra"=i dapatkita abaikan.Daram soal ini, kami mengabaikan momen jepit tsb.Pondasi kit,a coa 2 tipe 3
pondasi setempat.pondasi jalur .
Coba pondasi setempat :pondasi tapak bujur sangkar.
as2EB:p=7?r4l1tpn.berat sendiri pondasi ditaksir = 1 0tp=7 ,241 1 ton.O=W<4,5 t/m2B'
B >n- r, * ____, ambil 4 ,5 m
as 3 B : P = 571975 ton
0r= 57 t975+2,797 t(n,, -___,B-
as 3 A: P 49,023 ton
G ae"..o,?3.i{q,?gel ( a,s -___,as 2 A : p 61,3r3'aon
B>3,76m -) coba 4m
B> 3,46m-) coba 3,5m
3r5
Or= 6.1,,3,2P!6 ,1325 ( a, s _--_) B )3,87m _2 coba 4mB
kita gambarkan denah pelat pondasi sbb !')
3r5
'tlv'rL
BV' t,v
w/+o
1SL!!i
' f P?ol. d.90 [ = f,TstelTp
gsepuod T.rrpuas leraqI
'L.g L6' La-: "
'. - ''i ri'C
w/l szo' v =
: g,n bepTq"
(g't) (e[) 'ii
0eqnluaual{
= o1?aub
( 9 L6' Ls+ L Lv ' z L+g L6' Lil-
g'l )(s) (Er)
IPTAd eped e[;a>1aq 6upd ueqoq
TIe{Tlxa^ e.de6-e Aeb aluelTnseJ
LvtzL+gL6' LE) t'l(-- rut E eqo3
(r) (B)=n tsepuod TrTpuas ?Praq +TE1o?d
'P1P-r r6eq:a1 IPJTSJaq lsepuodpteu l t sepuod 1e1ad D' D 6up ltdurr:aq
T se{oT ) s T:aauT s upqaq {e?oT Eua.rB{
(916'f9+ [
utL = T
E,?}
Tr
lEtq'C'g'V sE 'qe:e [ : rnteg Tsepuod
TTca{ 'urrpd
(r{TS,ar r=l:ilt*:;::"I l:]:: :l:1":-;::J'::#t:;Ia?T)t p{eu ' V L L uEureTeLITp ; sepuod 1e1ad qeuap ue{r?eq:rad plTt p)tTf
soo'8Y I 9L'Zgvz'
(9'E)gz0'?=b.tJ + U + .! +.I .|.1, .t, O U
. ue{TeqeTp +Tde[ uauFr]I
lgt 6t Lg
?t t .
116
tlkuran,balo-k pondasi : coba b = 40Cara elastis z R2Z5 ----; FU =
u 24 ----rd =
,n=Ir{ max = 591005 tm
h =Nofrdimana[ =
cm.
75 kg/cn21400 kg/cm221
7b= 0 1529
geser dipikulkan ke
coba sengkang 3g1O
3. + caDn
No= = 0 1247
= 94 r Sem --} ambi I h^ !r l00em, selimut beton=l,enukuran balok pondasi = 40/ I00.
' - v-'r" I
cara kekuatan batas : M tumpuan = 12 176 tmIr{ Iapangan 5g ,00S tmD max = 431483 ton
Check geser :
Zb.r= = 20 ,13kg1/cnz ) E*uu beb. sementara(=11 )
(f u*, beb . tetap (=24
sengkang.seluruh tegangan
^ A"' 0""u
-s T+u. b
= 2 ,356 .2090 ?Zo;mff = 5,08 cm
coba 4 penampang fr10.A
a=
pakai 4g1O-7,5 untuk tumpuan dan FlO_15 untuk
A"=3. * 7C(lf = 2,356 cm2
( terlalu rapat )
lapangan
g :E 0, 13167
5 1, 3"m2 (1 1 gz5=5 S r77cn2 |nr * )0,5 2cm2 (Sg2 ZS ,34 cm2 ) .
= a,z4lsgr!o?Jp:
'lf,ll, .-' I
...,
gf, 1e>1ed) Z*,VL't =VEOZ - t6equrad ue6ue1n1
(atuczA'V - ll-gb) Z*oVL't = UZ'A = ,V
6 VLb 1e>1ed)
z*rzL, et = ffi (szl(oor )rrBEot = v
= ,{ r'} \tzrv = T=,rcJ'rxogt = 1q nele q = ucgz Tpe[uarl Teqal
Teqal e{Eru ':esaq nf pTra1 'Tn1 spnT 6unqnqraq
z*rL'gz = ffi (sr)(oor) tBgLL',0 = v
, .ji,l
Itii,; liP
Tadlp 1e1ed
(9'6=)dela?' qaqnq
(LggLL'o =u.i
L ss'z =
z'0IZ
3euw
I
x
ET
ru1 ZSL,S = (t)z(9'[)(gZ0'?)f -- [-t up6uolodeped
xo +---
*z )zrc/b>1,28'g =
Et '6'0 '001000['(gEl'g)g't
vthgZ1't=01laub
cz), I
n92
uol 8[['9 =7
( t ) (EtT -e' Ll (szO'? ) =
( t ) (qg -9'L)o11aub =x€ur o L
ra?au [ = ?eTed fPqar {1nz Z-Z uebuolod eped xeu O
a
(-ulct['0[ = SZflZI owcg'8 = V?'0 = rVcne
(,tuc vt., sz=9lfrs) rurc9 L, oz = .=fff;f,,.f = fu = uruv
c ( v,o= S)unururrub
l ( / t
\ t-l, a
.i
ffi
(9ZZ)(r)(s'0
LLL .i
W uendurnl n
+ 118
Jadiiukuran pelat pondasi jalur B = C :
tinggi total = 30 cm
lebar 3 ,6 m
Panjang = 13 m
JalurasA=DKarena bebannya tidak begitu besar bedanya dengan as B=C, makapenulangannya di-samakan . tie beam
13
Antara pondasi jalur yangngikat ( "tie beam" ) yang
$grencanakan "tie beam'.
fl r1 L] I rlI
I
tl
tll- ,
lctitE
C
CI
CC l
oo
.osf
nlsatu ke lainnya kita pasang balok pg
menghubungkan kolom ke kolom.
Tie beam direncanakan memikul beratnya sendiri + dindingyg berada diatasnya ( j ika diatas tie, beam tsb ada dindingbata
bataTie
).beam tidakAnggap ada
erat
ukuran tieberat
menumpu pada pondasi dibawahnya .
dinding bata diatas tie beam setinggi z m
dinding = 250 kg/^2.q = 3 x 250 = 750 kg/m.beam diambil = 20/30.sendiri tie beam = 0,2 x 0,3 x 24oo = 144 kglm.
4EIK="BC 6 = 0 1478
Kea
*1.&'r
-Msa
4EI= 5r5 tr"o = o ,522
,[, = b (8e4)(6)2 = 26s2 ksm.
h (894) (s,5\2 = 2zs4 ksm.
'ueaq aTl uep rsppuod 1.pTad
tu6>{ 0L0E = uendtun?. ,{'at./02
urnurxpru uebuedel W
es
rt
0L-Bg 1e>1ed 6uer16u
6'J frZ leqed r Vnpna*ffi=uTtug
tu6>t g96 = ue6uede1,
Lf,zl z*, 996' t. = , V
(gZ) ( 0z ) tetS [,0=v
.vto = 0 .,
(z'o)(s'o)z I ) I
wl r
-
= tLL, ?
9Z
uendtunl {r.l1un
= 0g0zW=a
(;tucL'9'= 6 c
= - 0g0zGc,7XmTZ
( SZZ)
LV'Z =
-
ug'{
'up6updeT {nlun SL-gfr uep
,' (|Lfrzl,uc6'z L
( aulc V' L L = 61fr? )
aurc Z6' 6
ettgl'g = b I L'1
l
tI
l.
h
H: H
#.
I
IF
It
I
I
i
*it
f'
Ii'
iIr
I
i
i
I
i!,
h
:
*
h,
flii
I
il
fisii
L-OL
'tu6t tS6 =or0E z(9) (r68)8/L =
5[-0[L-0tfivsr-01frg'L-oLgn,.
0L0
T T sEr{
067+5e9+Z L.L+t t-
0r0E
v az-T,ZZ_T9ZZ-
gLV'0vq,77-dzs'o
@l
L
f-1
- 120
B-12 rs u sgzs
polonqan a-a
feamI
(20/30)I
tie
l
';::
lpITsraq eIeaqaq.
'selEq uPlpn{at precaseIuup6ueTnuad +ntTraq
'BIPIUAluaS6ueI upuEqaqruad
3 qElrref
P-P*T=uI
I
II
,fsr-71 [--',2i07
'u
3 qrAPC
'V?,n BFeq , EZZ>I uolaq nln6e splpq
9t' /
. uelenta{ Erpcas , Br{uue6uelnuad
ln{Traq 1 rerl6ues rnfnq ,ledulalas ]
lsepuod lpf ad uern{n ! ueertuelrad
,mc/bt1 S'0 = t{Puelo
uol 9?, =foSUo{ duol 0g =ruoTo>t d Tnqpla{TCI
LT
luoTo{'( [ pTTTI rXOTVg qpq T
olag Ts{n.rl :
s qoluoo pd
4u
EO
-
: I rtos
I
I
lsepuod upteuecuor I ueedue?:ed lffil'etn B[eq ,SLIX uolaq n?nu
-wc/ bltg ' 0' '-qPuel Z)tTTPq >teToq BsTq H
uo1a8 Ts{n.rlsuox lPr{TI ) g I
utg't
Eluox leqTT ) ' Tosuo{ 6ue1ua1 ZZ "'ON
1S=
: qqs rnqela{Tp ( t pTTTI'bp WoTox qeq Tpos qoluoc pd -ffE6S-''suod pq1 Iceqcrp 1e1ad Tpqa1 . a
- uPuP{ r{PTAq
-as >1ede1 6up uerl6urpueq
Tp 6uefued q.TqaT TrT{ qeT
-aqas l sepuod 1e1ad r1ede1 . p
' (o = Jll g?[r0 - ruoTot sB
alt lBTad sp te:tefi 11e1edse a{ ue)tqepuTdTp Telo1d I c
, ru g?[r0 = TPloldi/w=a
'lu1gZ L' g=W TnqruTl<- uoTo{ sp
alt upltr4epuTdTp Tosuo{ d . p
@;
,{W
urr{TlPT TEos - TPOS E.A
I'g
lFrtl
il,4 "ilril'l
l
122
Untuk memikulpondasi dalamkan adalah 25kitar 30 ton.
momen yang besar (=g0tm)rpondasi yang eocok adalah( tiang ) , dimana ukuran tiang pancang yang -urn"J;;;u
x 25cm. Daya dukung per tiang (mutu Kl75) adalah se
0tz-10
[,rTi*l, l:lml
I,J
I
I
I
II
ill0-
\ ls, zrafi1e-,10 |
3o /es
fi,,
H=5 tM=6 0tm
l.l fi,ats4n44
l"lI
I
I
I
I
I
I
I
/5m
lrlrlr
emr I
Pefungtrr
I P=20t IA1fH=st|
,=u0tm
I
ILH KeF.rri
)?l,a,2 P:ai eqnausi '
1 P=2Ob+berat sendjri
ff,t"lom=2 3,572 t
I rltl=G0-5 (6 )=30tm?r?)--r- €- H=5 t
H berarah kekiri
le=z0t+b . s kolom= 23 ,67 2/t\I t ,Iu=60+5(6)=90rmI
t _tH=5rH berarah. kekanan.
fr12-10 --j*F+ -
P=20t
aF
t eAuup6upT
-nuad ln{Traq lsepuod uetruecuar
, Zu/6>t 0S I = 1e1ad dnpTq ueqaq
etn PtBq
szzrl qolaq nlnu Tnr{Ela{TO
r r yl t !!-Tslnrlsuoy lpgTl ) 1e1ad 6up1ua1 €
. oN Tpos qoluoc. TuT qBq s [
. oN Tpos qo1uo3 T TEqua)tZ,-2, uep t-t ue6uolod {1n
.luc 00t = urcgz xar 6urT1 leref Tp ( rpsraAsuprl r{ErB ) snrnr >1e6a1 qprp
lu g=e TTqueu 9'L <q 1-
'selpa{ 1er16uera1 tE? t uBlarap rp6e@
'qPrqpqqt ue{a1ra1 z uP
-larep upp selpa{ 1er16ueral up{pt lerep 6ue11 , tu10 6=H lpqT{V . q
lnre[ raq 6up^f, 6ue11 larap Z pqoc
Tt FcTat ) ur g, ZZ TrEp rBsaq r{Tqalu g. zz
EpPd : E TEoS
1Er{T T uP{r{ET TS
6un1rq1p uauohl
n=O?=6ue11uf p uerl6uepas
6ue11 [a
uEtPr,IESnTp
rus'UI
@ur UIE' ' ( uEuEurPa{ rol{PJ1e6u1 I ,wcy6rtOg-=TTgor i -urE PrT)t sirl {rn"Q,g :urlPlPC
= TTqUE '108lgz' lt - 61 gEZr lE =
0Sr SZ x SZ = 6un1np e{ep. EZ X 57, UErn{n Pqoc
'6uecued 5ue11 uetpun6:ad B1T{ EtpD{. ( uT>16unu
Enreq lsepuod 1e1ad rpqaT( a (---F s ? > B,E
E ?> .. (rTuetae) lrepuod xeladl eprd .6a1 1pe[ra1 re6e 1e.rer(s
ut g,t - W = d/w = a = Eerrsrrruas:(a
ledualae lsrpuod ue:1eun6:adurau rlTrl er(Tt ,11106 = y4 lscle6uaut :{nlu1
.gzi-
'([ pTTTI r uo1
ffi=-lJ :--
124
Y-:::-' c:n:oh soal
sebagai berikut
.IawPb ,: pondasi yang dipakaibentuk bujur sangkarGambar :
4m
ok
No. 3 ( tentang pelat ) diketahui data_dataaa
Nkolorn Ab(ujung atas)= g53 7 rS kgNkolo,o ge(ujung atas)= g53 7 rS kgM Pd ujung bawah = 4sg,2kgm(fcecit
::r$r|,r .!ore.h diablik"i t .'..,".':' " "-".' :':r;1;t'' - .'"-.'.' ", :, 1:..--II::'
adalah setempat I150 x lS0 sm.
dng pelat pondasi ber_
hP'r*,ug 1b -12 ,5t
150
I
7
)
\\:l_, --
\
U1 ,5m 3m
. ( rp{6ups rn[nq {n4ueqraq 1e1ad) Bdu
-uESuelnuad upp lsepuod ! upe^fiue1red
Z€, O E[Pq
9ZZrl uolaq nlnuEr'0 = qpue?, TnqrlatTp
EEX9T
uolon=J,wc/b>1
'rpquE6;1 ads leduralas lsepuod Tnr{e1a{To3 g Tros.'T g/t> a
efru1epTl-tepTlas nElE Xelad leraq {T?T1 6up 1Tdfi1.rao Edtza'ld p{p6-erB6 aruplTnsar rB6e upteqpsnTq
' I I uPp 6 . oN Tpos r{o1uoc 1e.qTT
t eAuuern{n - uErnrn upp lsepuod 1e1ad ue6uelnuad- ,wc/b>1 9, O = r{pue
Zl , VZn B[pq ' gZZ>l
ffit uBBFueAred
uo?aq n+nw
urcol=firuco t=fl
-+0 s=.Zd109=[' ( TEBp
E6d)toTBqedue1)rPgue6ra111redasrn1e[1sepuodTnqP1a{Tqm6E.l [ -oN Tpos rtroluoc leqTT rc
e lsepuod {oTEq uPp '-tsepuod lpTad ue6uelnuad : uBB^fiue1rad
,wc/6>q gL, O = r{EUEr a . ZeA B[eq , ;ZZX uolaq n?nu
tlll Ot = Erl
uI? s€ = zrfi
utl gz = tn
109=Ed1 oL =
Zd
1 \V ='[d
let=tJ
t
sz t
I
@I -@II I
I
I
-@I
I
-f-If-l-
'rPqusbralrnle[ lsepuod Tnqela{Tc. I Tros
i
-! 126
Sqal 7 : RencanakanFlanah =
Soal I !
-i.-=
pondasi yang memikulor8 kg/cn2. Mutu beton
Normal tekan=s0t rmomen=l Otm.K225, baja U32.
=80t
5o /so ftanat = t
Itlutu betonba ja
Pertanyaan i
kg /"m2k225u32merencanakan ,r"!
rap footing " tersebut.
Lihat contoh soal No. l,Z .
YDiketahuidenahko1omspttergambardibawahini:I
:lT:1.:"il:"'semua sama vaitu 35,/35. Drminta merencanakan ponda-
tu beton *rru, olrl';;:.n""ut"tnva)' Jika Etanah : ;;;"r,;;'-;;petuniuk : Lihat contoh soal 14.soat l0 : renrukan a"y. a,rrrnl-;r";; pancang dinamis, dimana dara_da
.. _berat "steam hanmer i = 3 tonjarak jatuhnya hammer terhamasuknya .r.," r.i,
-:J1.:::rlTi"ffii ;:il:,="il::nyak l0 kali adalah 2 cm.
-Faktor keamanan = 6.ilawEb : 55,5G ton.Petuniuk : Lihat teori :
,|
R=
=
F=
s+c
ffiR6
"Htrap b6
E
l,l:H
ffE]
$fl
II
lI
II
I
I
i
- _,- .Z€O efeq ,gZeX uoteq nlnu E6nt Tnr{E1e{Tq
e Tlrf BnaE{ tn?un {ococ 6uEI TEppuod ue{pupcuareu EluTuTO_'uol9g67-uElpqua[ uEqaq-urqeq rrEp Teseraq 6uEI
"rr.* ;;;; . (r"qEE6 lBqTf) dec a1ld TrEp u g/7 1 4ezel eped eg:arlaqi tuol OOt = rlEuEf urur{a1 Trep TBsEraq 6uEtr arruorrroU ,i.J. ( gz t uErrETeq eped rEqurE6)
'rEqute6ral 1ds uelrquag (raTd) ,luau?nqp,. TnqElatTo ITI-E
,t ZtR eFuq , sz,zx
-Ecupd 6ue11
iuPp duc eTTd
0?f07
I
. uElpt6ue6ued lpps pd TnqurTl6upr{ usutou uptresppreq erluue6urT-BUed uetpupcuarTp 6uBcupd 6ue11 . q
uol gt = qeToredfp
,wc/6tt OE = TTque igg x E'o x E,o = -.6ue11 t 6untnp ertep . e
: uEtTlcqradlp nTrad bi Tpr{ edeeaqagr lthFEnTq[
-L-0s/ 0s
*;108 t
uoleg nlnu.6u
uE6upTmrad
'EAUUETn{n
ue6upTnued t ueerluplradoeEt
t.
l*I
'zenp[pq( gZ?X=6upcued 6ueg1uEp dec eTTd {1n) uotaq
nlnu, (rug 1=6ue[ued) 6u-Ecurd 6ue11 ue6uelnuad
e p^t{uuern{n
0Ex0ttI,4,
I
I
I
I
uep dec ..ef Td ue6uelnued ! upBdup?rad
rpqure6xal 1ds 6ue11
LZI :.
-l-+Wi-l--*
+I
I
I
+
lsepuodTN
,n$Iilh
flTiit :
r. l2g
=1 00tm
1 00x200
H=?,03t
0 ,5m
,2m
,2m
,2m
r 5IIt
960
Petuniuk :Lihat contoh soal No. I 6Beberapa ha1 yang perlu diper_hatikan
3
Pyg
1. daya dukung I tiang( 50x50 ) = l2S ton.
2. jurnlah tiang = 3 x l03. effisiensi kelompok
tiang hrs diperhitungkan.
bekerja pd pondas !=2000t+berat sendiri kolorQ.
M Yg bekerja = 400 ( I Z/t) tmDiketahui: pondasi tiang panea. ng yg memikul, p, H n;.Mutu beton k2ZS , baJa u32.
Gaya H tsb akibat gaya gempa.Pertanvaan: rencanakan pona"=i
tersebut ?Petun{trk:Lihat contoh soal I g.resultante p dan H adalah R
f, = g24 16l- bersudut 73 1960grs horizontal.Untuk rnemikul gaya horizontaltsb harus kita pasang tiangPancang miring ,'batter pi[e,,.dicoba tiang 40,x 40.
E-'7 tiang = 80 ton."batter pil6" untuk kelompdk I,tiang vertikal utk kelompok 2 13Cara perhitungan ,,batter pile;silahkan lihat eontoh =o"f I g.0 diPeroleh = 4l 160.D' 1 tiang "batter,, = TS rZ ton.
Pt tiang ( =8ot ) '
AJ
P=800t
+-*-+i-d- il,-wt;rtq-d -d
*orrrr__;_ SgIeSai bab V --*______\-
6ZL
-@
6ueIupuaur uorot
Eap.roq{OTPlrlI
' ( ,, trT Teluoz Troq 6utuuBds srrpls r )
Teluoz T jror{ qe.rp ueEueluaq 6up e66tJefr, . I'8
0E? p/s 00? = aparlue ue{TTp}tTp apa:1dolucog = (aparlur) [ + (apajrado] z
r! ltI
I
I
I
I
tuc Bl p/s lucg[ = apa;1dourcg E p/ sno O?, = apaxlup
! q5l p66uB1 ueeun6e>1 pd 6ufr1ue6:a1 aperldo uep aparlup up.:rn{n
' (aparldo=asTr=) ue{e[uB1
s selp 16eqra1 e66ue1 >tEuV
ffi.v3 Snunr-snunr urp Troetl l.&t
v99NV,IrA 'gvg
q-q
=-ilii Fir
*'p, I
oled rebEQaB
e66ue1\
aparlu
eparld
2.2
J_
130
2.3
2.4
-) bordesI
vbordes
2.5
J Tl)taJ J"T
i_. t
'{oqual-au {pPTl
-Tp ez(ueq nP1e
rEt
lEraqurprD'DuPlITlrtaJJa TTTPU
ueOuap':{oTeq
nledueduel
'qe6ua1
ue+uaq ue1@
ndulnle66uBt
{oqua1
-
3 S reos
re 6udues-rypq ' qe6tre4 ry1eqnduln4p qoto ndtunggp
ff)we I
-r
Ll
eoffi reqq
{oTPq r{aTo
lTda I ra+
-
3 g TEoS( srTPls raAaTTluPJ,, ) tleraqas
W.
I
i
132
t*trdaaatiltfed€rrri.re o.optreCelebar tangga r! bberat 1 anak tangga :
r *gln
berat. sendiri pelat tangga per n,
'i rl
x v lerieltrr
)1* )1 m i c+lm+lm
)lm' (r*
(t* )t" G+x+Im( t'ni c+xty
terjepit pada. dindino
:
qPnq S *ttlatar, r - urc04[ = lLt)0t
-rP.rag uPtEtuel 0t tTqurBTp
IIErqEq 1r ffiI t + ) saProq=- ==;=';;rc?
ul x tul = repiroq uerntfl! EDDupl qpuap uBtrpqutEE EATX l.a
.1n{Traq qe16uc1 up:{nrlET "?T{ TuT snserl qea,refiuau {nlun
2 6ueqn1 uErnr(n ,E66uel ue6ulrlura:1 up{nluaueur {nlun EluTutTq. ( r{Tsraq rPqaT )
rglaur I =e56ue1 reqel ,urci1 = apa:1do a ucSZ = aperlup'lpdruaraades :toTeqreq etruadTl 6,{ e66ue1 lEnqTp uetp
o0', + = splp TpluBT ,oo.o + = qpAEq TEIuET 11ad 1nqe1a11q=-€.-ffi:Im
:x'zu/6>t o0g = dnPTq uPqeq (-:-- uE{{ntunlrad
. 6ucnr ,pTFSaur ,dotsoTq ,Etere6 ,esupp 6uBn: ,e6e: qelo 6uenr-',, zrl6r ooc = dnptq ueqaq (---- lT:fes qeunr, EuEraB ,Ta1oq , ue,rofsar .,pqraso1 ,oto1 ,qETfnrl 6uenr ,qeTotas-
: ucun6urq 1n4un ueleun6radlp 6ueiI e66ueg,
"on, E2Qp-
(umr 0g.l -T) =dnpTq ueqaq {ln ueueqaqruad reqaT
( dnplq ueqaq ) w/ b>t $rutu a L L<
6ueqn1 relepuau {pre[[-0t = uEtP[ur r{PTurnF
+ 0 = saproq TTad T1
1epe.r{do ) }tTPu LtP{6ueT
,-lJ.t,^_llaf",
L_r
ll,l,;il,
--lrL
Tanaoa atds :
Pgtopg,Fn b,-b*
:
+4.00
lrrl
-t a
l2x25cm - 300cm.
ri
langga sb
10x17
=17 Oem
I
T +0.00
,Jarak vertikal - 400 _ l7O = 230crn,rumlah tan jakan (optrede ) r # :--r 3,5 buah
jarak horizontal = (13_t) x Zi"" = 300 em
.'i.
ukuran Lubang utk
I 00cm
'.:
l 170
T,,+Itmg
)
9x2 Scm
1 2xl 7em
=204em
tr
$
H
I$tt!gI
E
5
tr
I$
tI
rti
t
rll
iiIt
I
IH
t
I
3
.
:i
i
i,
I
I
,,
i
iI
i
Iti1
lltl
II
o
lr1lI
I'l*
r
' xnga'sral Erpc qPTPpE er{unxesqeiee ,EueIpTaur e66ue1 ueTps-a1ar{uad epo+au pdp;aqaq epV
t r96 t rJv
feurno1, eped rUgZ&nVS ZNVUq
TIEP Troal uB{rTqurP TurP}t TuTEue.telaul e66ue1 {nlun snsnqy
' ( ,6uTpupT,, )saproq
6u1r1u 6upfi
6uo>1o^duad {oTpq
I{
rEoS
{n+uo
uo>1o^f,uedOTVE
' ( ,,1Q6TT} , )e66up? ueg6eq
'(.srTPlSputls aardr )UEAETaI{ PPill' Tpos r,{o1uoc-qoluoc leqTT ue{r{pTTs ez(uspTa[qlqat
ffi,d
- 136
Karena sampai saat ini tidak ada diktat,/buku '( berbahasa Indonesia )yang membahas tentang
, tangga tipe ini, maka kami akan jelaskan mu-
rai dari penurunan rumus sampai dengan contoh soar.
qr = beban(mati+hidup) pada bagian' 'tf light" 'g" = beban(mati+hidup) pada bagian
bordes (." landing " )
Tipe tangga ini termasuk konstruk-si statis tak tentu derajat 6rtetapi karena keadaan struktur tsb si_metris ( lihat gambar ) maka hanya a .
da 2 harga b.ilangan anu yaitu 3,
xl= momen yg bekerja pd A dan D
xZ= momen yg.bekerja pd potong_an BC.
fita harus menentukan besarnya *
1clan *2, dengan mengingat bahwa legdutan pada potongan :
AA'dan DD' = q = 0
BC = tZ = 0
rumuE 3
xl . f,r, *: x2. tr, + Ero = o
iz' t2r + *z: tri + tri = o
dtO = lendutan pd potongan A a-
f,,,= H::.f';:.:T:.o,,nun A
frz =i}:::.:"I'n]i"",an BC e+kibat beban *2.
akan diperoteh xl dan *2.
Memen,
rumus
Mx
lintang, normaL pada Tasing-masing potongan ditentukan olehrumus sbb r
'*o' * **'
'ro + "r,
BrAr cr
@D
D
Mv
Mz = *'o + *r,
*t + {*r.
' xl f *'r'
' xr + n'r'
x2
x2
*z
x2
Qx=I
.i:'!
potongan
o*o * o*, xr + u*r.
eTpuP6un 1T q jred
'ue{TeqeTp ledepureTep e66uTqos TTca{ u e6:e,ry
ffifi----f+l
' ue{T pqpTp
TUT >1o[od ueT6pq
(D
ttlmr,Xur.=WIx
ttrr.mXulr=Wo
f eqT>{p ue6uolod n1ens eped X qpJe ueuou = *W
z @Lx ' zN+ [x ' [N+oN= N
zx-'^o+lx. too*no=f,0
uE{T pqPTp
-t/-
@.:1'K-( qe TUT )t\ ,trlr A,nn /
6zx
Ix
,b
/;, )/i-;/
zx
tct
,, b upp
ru
138
7' e= lintang i M=penprunan Rumus :
atauY
I
Norma I
bat xr = l satuan :n^rp;Q_ -* I
-, >.- I :I:";-'": ;u' beban vertikar
*1=' d' r;r =^;
= u
I H^ . h -1 = 0
l41<_-n-_1 ^1
'ol = 1/n (-: )
l
l
I
i
N
I
I
)
coF0h
s in0h
r+h
ti{tiau bagian bordes BB'C'C :
sinfl
/o soc
)o UTB
,Y soc
)o uTs
0=
0=
0=zx=
7,x_ =
0=
zN
,^O
t*o
7,2D{
Z6W
zxn
zz
L=z x
6uTs =
0
0
0
duTs Zx
0
= 'zN
= '^o
= '*ozz
=i W
Z6=W
zx=n ljlroc rEfi soc Zx
ffi'utnlta I= zx rtqr{E Trurrouef,e6uPp uauou-uauolt ,6ue1u11
,)grSOC'L8
fb soc * -tr'
/g soc * =r'
(^ uT" tt,
1 ;psoc Lu*,
)
;obt */p ul
" '"rr ,=,hsoc
t *u
0, rr,_
lo'soc -
-t
xr__r{ _
. bsoc
lzW
t6W
IxW
tN
,^O
tto
8n
.l3u1s '
tI
i<;u,lg,/"
Jw
IT
=)O uTs'
; 6[[ *
0=
3 grgrV\l "1rI6TTd"
uE-rDeq nE[uTL
l{0
I{omen-momen dan normal akibat,fliqht: ) .
<T;
h*[, f ,h -;
x
{
vi^.= gt .L-
--rJzMg=o-'i^'n + (g'.r.)(L)-1/2 q,.L2 = ooHi=# (ts- )
'io= # (-t )
Nto= H'B sin/ = sin/. ";.*'oio=-Hf'cos/= ft$ eos'0;=o
o
'io= r sing' H'B^..ro
= ql'L'n5- r sing
*i^= 0 -'; M,, = 0o
N; = ,io cos + g,.x. sino(
= +# cosa( + 9, . x. s Ln{o'*^ -Hi^ Ji"o7* g,.x.eoq\zo -o.r
=- .{ 14
# sinff+ g,.x.eos0/Qt" = o
'o
'io = tio lo"o1 ' r
= q'.L2 -,
-- r cosA/
* 9' .x2
' = +# x teo/- *e,.xzr sinO(
txlHI
\,Y,\\
-'4
H; eos(o
srn6{. fu )
. x. tgo( - * g' ,*2sinV.r =-o'iill
Mr
Mrzo
= (Hi
=Hi,o
=Hio
dh.. (q"T\ trrrfll'i"Urrl I
Vi = .g". b
Lllr= 6
g".b.L Hi .h + l/z qr/bZ = oo
Hio
Hfio
= q'rrb(3.-E + bI
irI+f,f,-f = r,
I I 'g Z
TSJOI UAUOU =ft 'grttrn .tr{ .
=*, . {*w . tr{ = *rj : Tpp[
q
&r
EqxI>1^d
sp
(q Z/L)o soc--r
1 + = ^'{ru ue>l6utpueqT: d,
puare{ ue}tTeqpTp +
!{'T'{ +Fm1+ +{*w..,^/^ =n3- .ur{TeqrTp Teurou uep 6ue1u11 lpqT{E
uel6uepas 'E[ps ueuour ?EqT{p r{EIrpE ueq6unltqradlp 6I uElnpueT: ue+npuaT ur{nluoua}I
&r' esp
(qz/ [+x19',,b-&6+ x
,r\"uT s
)o socqz
Iq+rzt7[q.=F'=+ x)q',,b-
o')ouTs $n
r4z
mo+ Tbsoc $r,
q,z
mEZ1q+tzJT;E
oot'fuuTs Hu = |,vt
o o-L')osoc $n = *rn
.I
.I
/outs'q'ub+Fsoc
x= ,r[r[
o= ,,N
o-=fo
o=
*ttO
oz= ,rW
=o^
=
ttw
o
)osoc [u t K
""$rr4tr ,trrr r"?,*A'
^soc,.fi: l;
,L1t{6
N iuIS' q' ,, b
7rQ soc'cl',,b +^urs
)t.JSAC'Q',,b
0
ouT s .r
6ur s
r4,z
mo+ XuTs fln-
ox= ,,W : 0
r4z
mo,
/uTs r' flnr4,z
mo,
ours' flnq,z
m
lla
q
J" uelbeq nEfuTt T T
= ,,N
/soc o$r-
=od,,o
o=oIo
o
tDr
fisoc
saproq ueTbeq nPLuTIl
ruI
- 142
E r* ,. f,r* Mri 'Myk 'Gr-[r* = J"rri '*n ds *"/iri' Myk' 2
Il/Mri 'M*n ' ds +
/"'-, ' ds
EI x
1
ds
1) ds
1) dsE1
EI
f,r= + j"r,,,
bagian "flighg",,
ds + [t2tl"t1 '
1) ds1
z (t'+v
+ u2*1"I;(t'+
vbagian bordaE
dimana untuk bagian "flight" -___) dx
harga-harga M
bordes ____)
*1, *r, disubstitusikan
cos O(
rdgpersamaan diatas.
+ +r) &,+o+o.... !.... (a)
1) ds
ds=
ds=ke
diperoleh -
Er irrl ,7,*L= rEeFq-
qo()z.&-,2 sind2h
Jt'+ t)
+
Ifr "i"(t? lr
1I.*(Tr+
+ ,Xtrv
t tz= ,[**, - **r. ds *.,fir, . *r,
ar-Erlrr= r '!r*r, iL*+,)#
untuk bagian "flighg,,
untuk bagian "f1ighm
untuk bagian bordes '''
't.
'str'
Jrr u .t,. .. i.l
o "?**rf '' rdil +
o
-?*rr"'i ( ;i . r ).r ds
ft',-^?*IP," h4^ . -t-..,t-k &Lt*S ij iii'4:{ri rs .'t
r( f +
v
= o + oTrcos
q'tz. +, { + r) Ffu- ry-v #fl &g+ .......... ;ffi ffi
Ijif
,,1r{0TT},,ubT6Pq {n1un0+0+
0
I
tr *! l9'sp ,oiro,,"nrf* "r',oT*,,'*^rf - oz"J ra"r I
, T-,(], ",t.t)(q+rz)2u1"--r-{15.ffi-=
. r o+o+
+-,, - + l f .rrour" . ra;#fu-r )()rurs I tfro-
4=:ot 19+x)q',,b-,1061'x'(q+rzl S+6 lf ,voe* ,f,t =xp 'q
sp (t + f, ,f ',"i*, i'^,f t sp'to::*""n'f = 0r"J Ta
s saproq ,rrro=q eppd "; +eqT{B ot.s uep o} j usp{n+uauan
,i: v
w=0+0+^ (a) ................... o........ " ( t + * )
r
o' ,g rs
,bsoc tt +xp \9 '
zb soc-- xp
TJ, ueT6eq ,,1t'l6T{nlun
+spt Tr,ul t sp (t ++ t f toT^tl'^tft
"4I{6TTI" ueg6eq eped r b +eqT{P oz:o:3 u={ntuaua1;
r3rtII
uep
+ (g TIII\\D ' *, ' ;psoc T
I,ZITU
' x- ' a'21 I (c)r
dfrp r'(r ++ )
Ifip ,' z(
gsoc ,f,* +
rbI
f0i 'r(frur=y'zh +
; 144
EI f,''
Dar: i
BI
EI
EI
BT
EI
EI
20
(a)
q".b;r 9O-EOf4
(g) kami
L
(2L+b) ( 3 + 1) ...... o... (g)v
tuliskan sekal i Iagi :2/r+
=="ra ( t' + 1)
s/d
f,, ,,:
t,,r=
f,rr=
f,"0
6' z,o
(,,d ro
Crozo
3 cos
-Er D^
+( ;rr + 3)Y
-qi-r,1-{-sos-{
. L coso(+T +( It'+
vIt'+
v
1)
q 'J,3 r q:.L.-r224 coso( + * 4- cos 4 ( 1)
Ka rena I etrar1i maka I-_/I
x
I( t' + 1)v)s.:=!'-L-- + -+# sin (tzr,+b)( 3 + r)12 coso,
r -y9l-.1.J*se*(1aLrq 1 :l +1)
4r -v
tangqa djhandingkan dengan tebalnya adalah besar seka-. mendekati 0.
v1*v 12
I xI\v
r b3
*,'-, o
seh.i ngqa rllmus-rumus diatas men jadi I ebih sederhana sbb. <' -- L- * r1=U{_r:i Dtt 3;;=o( " ?-h--*
(CEI btz = Er 6r, = ryd-(' 3 TE r L coso(EI 022: =1;=- +
i--YY-D_--(
= 9.1-L3-- + q'.L.r2 coFl(Er oto = 24 cost 4 l:lr-
nrf;zo = { '';=rcoso()
rrf;,ro =-bt*;*, + s:'# sin( (2r+b)
nrf,':ro=W (2r+b)
I
it
i
I
[.
IH
I$
i'
( alE- psoc v- (,fl+ t Izro"z! aL 6) ffi
=
("5)(ozZ
wI
"3 ) -) ,,b =
,"j)(ot('y )(oz
,r9 )
"3) Prsoc va' v !pzsoc. 7g
)o soc t rQ 2l E
z!Prsocl ')osac 70 Ue
,osoc 8?)(t3
)(0[,,9't (02 ,3
)(0t8
)'( ou,9
-rt" 9)("j
,r"3)(0t,9
) (0',g
( Fzsoc gla e +
qn,"-;smi-ffi+
(,bsoclzLl
,bzsoc gj' Va
(
g.9 ' Pzsoc tT'Rztg +x.soc I )
I.I 9tNuT"rrt2Ttt
,"5 ) (0r,3 )-('3
tr )'b =
"'9 ZL r [ = (rts,
t t'
= ("3)o soc 8
Za' .<' X1
)o soc Z6 L
--;518I
-T oc#+ -
ztv
,ZZ=t
)
,3
so3 -rZ
trrS
= ,"3(
.8t I t'?
?"
z!) ,b = (z'3 )(otS )
lcbt = +T '/= .r +---- l/a =
0 r'zr tu
q Z/ t =-r : 1e6u16uatu: selerp ueeuesrad
,ot,S ,zzS rrrS ,LL! e6.zeq-e6:eq
z1 ttr_ zljZ3 .\Lj ,r! .oz3
Prsoc ?T'.\ soc v9 +W
=Zx
,rJ .r tj ,L3=
= ory ; ,rS .zx + ,rS. rx
= ot3 + ary.zx + ,rJ. Lx
: upeues jtad
Lx
0
0
/ofoc Za' Z& PZ
( fsoc {2t8 L+ )c,soc t+ &ate f EPf,$ = ,'y ) ( o3 )
8
Prsoc'79' ?T'rb8
ZI,)osoc' el' la 1L E
ue{TsnlTlsqnsTp oz ,9 , o [,3 ,oz ,9
aa r
W
146
( t"2il(( f,"ro)(
1)= ,,,T2rr'a + t-'n ""or2,,( + ry *. Y1 .r4 "og'x )
I I ,L4 "o"2r( V 5 .r,Acos2or
(F,
t,
[,
f,:
)
( -g' .L2 )
.LZ
xi =
( -e,, .L2 )
setelah xi ; xi i xi , *) diperoleh maka momen,pada masing-masing potonf,an dapat ditentukan.M*='*o+'*r'xl+**r.xz
M=lt4.t.*Y=*ro+'rr'xl+*rr.x2
ro)( [ 2)= W (12 cosa(+ (8eos q).tr )
( =q' .L2 )
( -g" .L2 |
I intang r normal o<I
(f,"2il (
diperoleh1)
xi
.t
ua#
M*=M, +M-o '1
Ql = o*o + o*1
QY=oro+orr'xl+rrr.x2N =No +Nt '*1+Nz .x2
penulangan pelat dapat dicari.untuk jelasnya silahkan lihat contoh soal ttg kasus ini.
. xl + *r, . x2
.xl+r*r.x2
3 rcT +qcosa/ +tgTL T3 ."or*Scos/+ G 7L/ +s 7c V3
"or24. A.L2
3 cos q,Scos ol +A Tc T +g 73 "oJ
-B.Qt
= -c.g" L2
-4 cos o/ -g TC t8 coso(+O lCV +g tC Vl
scos a(+etctr *iffit= -qt'.D.L
'(r[-[ up6un{-6ua1 TUT Ter{ ueTpp ) uegaq
sE 1 s >1a.f,o:d 6ue I ued upnl-es:ad eler 16eq:a1 ueqaq = b
ueueqaqurad sp Tjre[-1reC= [u
.T o z('u + u) : =
.renr qpraqas Treg-1rlg=tuPTEp qpTaqas TrE[-f.lBf=
zuo
ut'u
")
V Ij*Jr
-JS
! rseloNtnqPla{Tpqe6ua1-qB
{tn TdB;TPCTTaH,T
suassnc u\r qaTo
EET@ Trpp,.slroddns paxT3 rllTn srlets TecTTaq ro; lreqcuE{TTqurE Turp}t 'TUT adT+ e66uB1 seqpqulau tnlun
. 'TPTpET qE-rP rruTp TpluozTJor{ -rasa6 ede6=- Hs
' 6ue1uT T aul ue
-6uo1od qprp pd ,"=.i e.fie6-'u,I
. ( ,, l sn-rq1 ,, ) T pu.rOu Br{e6= "
'UT sJOl UoUrOllr= Je
1--TP.IaleT qpre Uorlrotlr="'W
TP>tTllaA qp.rp Uouo,r="^' ( >{ec
und) 3 Tp ue6un16ua1 '6un66u
Js 6ueplq urTp rnluaT ueuou =ow. TeTpPr
I{PJE ueTep TEluozTJoq er{Bb = H
. ow uep H n11ed efiu:esaqunreq 6ue^[ ro1{p} z ppE edueq e{eu e66ue1 166u11 Trpp5ue1 eped ue6uolotued up{n{eTau 6up uep , s1:1auTs upqaq'9 1e[Erap n+uo1 {e1 sTle1s b^ e66ue1 tnspurol rs-rrp1s
'uopuoT 'auo-rg , plT .reuuefr, rg raTlng qaTo ue{1Tqra1Tp
'gl
G = modulus elastise = modulus elastisG,/E = 3 /7rt1z
g=
0=
1-,\--I
beton tekan'.beton tarik
sudut yg dibentuk oleh Ers. sing_gung pd lengkungan "helical " terhadap bidang horizontal.sudut antara grs vertikal dengangaris yang melalui suatu potongan (pd denah tergambar).momen inersia polar penampang tangga (b.h) = kt .brh3
( r- }4Pt4 )\ !
12
= 1/12 .b.h3= 1/12 .h.b3
dibawab ini adalah persamaan telah diturunkan oleh( "comparison of Analysis of herical stairs " , concrete and .orrstruotion Engineering, Mareh .l960).
*r, = Mo cosO + H.Rz .e Egfr .sin/ q nf ( l-cosg)*',=Mosin0.sin0-H.R2.gtgil.sin0.H.R2.sin0.eos0
+ (q R? sing - ,i Rr RZ g) sinfl. .
Tr = (Mo sing -H.R2 0 cos e wg + q R? sinO-q RrR2o )cos/ +H.R2. sin0 sin/._ __--F.
trr, ,= H sinO cosg - q Rt O sin/tn, = gRt e cosfr H sin6 sin/tn, = H to:Dan menurut Dlurqan &, Scordelis("Internal forees in uniformlloaded,helicoidal girdersi AQr Journal october 1960") :*Mo(ffi(m+T*)+s
cos,fi(t- tr )) + q.R?
m) + H.R2(- tr k qg + sk tgfl+m sin g.
( tr (m + ry sins)+ ms + n.s. p,=oRl.
tr)m)+H.R2(tr ry ,03 o2si,13- + 2k)+Mo( trry(q;S| cosfi )"12
m(s-ftr.i(s-
k + s k +(s-
^2? sin20 + 2
) +qnf ,GJ'Er
k) + 2k tgfl (s_ fifl** "or2g (tgg
(n-k) + sk + i t'r' sino + 2 n)
tr))=0.
il
F.
i;
'Teos qoluoc
AEdPp t "( Tsua.rala.r
/rt t t za/)a
(q) "
|u'
(6soc-[) lrU- 6uTs fr6+ g. la.n.Z, **3 qqs 6un11q1p *W e6req
' 09 L p/s OS I uerueTeq ule:6oruou eped
'r;i
lpr{TT ue{qpT r s TUT adTa e66uB1 e^f,use1a [ {n?ug' [, I p/s I ExEauE (-- Za/Lx
9 [ p/s g, 0 e-relue e---- q/q
o0I p/ s o0z E.rEluE f ---- fr
o09 e P/s oot e-rPlue (---- / +3 e6req {n1un €p 'Ze '[" e6req-e6req ecpqTp
n{nq Trep uptTTqure Tllret ) 1n{Traq uB:6otuou pppd
! q,/q t fr 3 ralarue.red eped 6un1ue6:e1 EE
Zr'r' tE *=' '
+ osoc ' lu'b' [B =
'qaloradTp H uep on
PcEqTp ledep Ze !
la'b'ze = H
Zr'r'[t = ow
uep (P) ueeues:ad TrEp
= zu'H' ,{ + ow'g
+ zu'H'.{ + ow 'g
I'x,t{
qEIalasI P e6reg
Tla
: qeToradlp (q)
.......... lA.b.d, la'b'N = x
z.--d
'b' c 'rr =
lu
'ru'
n 'T: qqs {nlueq
(E) .......... lA
fu qqs {n1uaq uETep
o+ l{ 'gadp (++) ueeuesrad
o+ !il'v
=X+uPrEp
'b'o E
zu'H ',{ST TNlTP
zu' H' E
|u fr:b.r.fi iE
'b 'J = zu'H 'E[. + ow
sTTnlTp ledep sP?eTp
q )ztt c--fr ruls +* + fi rsoc
ouTs osoc 0
oz urs t o,z soc o +
'v(+ ) ueeuesrad
=Q
=S
=U
=ll
={ EueuTP
6VL
1s0 Harga a Utk Rt /RZ = 1 .00 dan g = ZOo0.8
0.6
0.4
0.2
0.0
4.2-0.4{rO'.6
-0.g
-1.0
-1.2
-1l.4
=1.5
-1.9
-2.0-2.2
-2.4
-2.6
-2.824A 3m 360
- ) * f(derajar)
ra. RI/RZ = 1.00 i fl = 2oo 2a. R'/RZ = 1.00 i fl = 30o
1801n 180 240 300 ,3€ff+*f
I
I
i
I
I
1
0.8
0.6
0.4
0.2
0.0
,\ 4.2
- I -0.4(lI -0.5I ,.,
_1.0
-1.2_1.4
.-1.6
-1.g-2.0
_2.2
4,4
-2.6Qrg
120 180180 240 3004lf 120 240 300
_+ *p3a. RtlRz = 1.00 i fl = 40o
./ b n
,/
\I 4- \ \
\ v \ \N *\
)N \ \_l \ J
N \\lN \t\
N\\
-) \ I-/\
\\'\
\
Int' iitiz = 1.oo , g = r;l
di"1g=fi ! g0' [ =ZA/ La ' E
%'z-a?
c'(:'[-'[-'t-.L
'[-'tr.F'0-
"0_lIlE
F0
0
?'0'0
'0
ot-y' i go' [ -za/La 'e9ooVfr i Eo'[ =za/La'PE
4*i*ffi.-
oog i o0? I ooe !
noz=fr uep Eo' [ -za/La {1n I e e6reg
Z-
Z_
Z-
?TL-
t-t-t-t-0.
r0-
r0
0
0
0
0
F
48fr.- mr
0.6
0.4
0.2
0.0
,.tl:i
lh:;-1.0 i
;.zl,.,
I
-r.a1_r.t
I-z.o L
I,.rl.-2.4 L
I
''u l-,-r l-
-:.0 L0
0.6
0.4
0.20.0
_4.2"^0.4$lr-u
-3 [+.a_1.0
-1 .2
*r.n ]
-r.a I
-r., I
-2.0 I
,4:z:q I,.rl,., I
{"00
/G wz
1g=fi i 00'[ =-U,/'U'q,
1e=fi i 00' l =Za/La 'qZ
1l=fi t 00'[ ="u,/'u'qE ="u,/'u
ooz=fi t oo''[ =28/La'qI
lr,
IHItq
t;$
*,dF6=S+;"- wz
9'Z*
r'0-.FZ'
0'
Z'
r'0
'0
?'0 y
'[-'[-
0'
u'
9'
0'
9'g'
0'
t'o'
'0 8'
0 lB' Fr
''L lE e'-a
p? 'I.L
'It
o'z
Z'?v'z
d+*
\\N(I
I7/zuN
!
I\/,,r1
z,,/
.\\'\
/1./7
,zI
r:n'\\/,/)7gL
l
q\f*
\\4,K
154
!illr;
lLP
, 120 190 240 300 , ffi- +*,Flgt^, P /p : I nR ,1116-l '
2.2
2.0
1.9
1.6
1.4
,s 1.2
f'f 1.0
gl o.B
HI0.6.q' l
0.4 l
0., I
o.o I
-0., I
-o-n I{.6 I
+.a I
-,.0 L
I
-t., I'-1.4 I0$;pw
loos=fr ! ot'l ="u/tu .qz.!IV
d4 ?- 0r " mt wz Bt 07,1 09 0
oot.=fr l 0 [' L =zu /la 'q0 [
[rgr=, , ot =zu,ztu'qrrl
1Z=fi i 0t.[,=ZU/La.q6
V'L-
Z'L.
0'[-8'0-
9'0-
7'0-
z'b0'0
z'0
?'0
9'0
8'0
0'IZ'L
v'L
9'L
8'L
0'z
Z'Z
IN
tl\
*Kl.?l.^2\$K
(m4,
7\\ \\
a2:ZIr{lE.q
Eo-t\
\-t27JF
Na7
Z'L-0'[-B'0-
9'0-
7'Fz'b0'0
z'0
vt0
9'0
8'0
0'IZ'L
v'L
9'L
8'L
a'zZ'Z
v'z
d?*8'(9'(
vrc
z'c
0'c
z'c
7'C
9.'0
8'0
0'Iz'lv'Lg't
8'I0'z
Z'Z
v'z9'Z
8'Z
I r=p o [ ' L =Za/La >t1n Ze eo:eg I
rjll# j'
i['ii
155
0
30
60
90
120
150
180
210
240
. 270
J,,oNl-$ rro
360
0
30
60
90
120
150
-Laor
C^ C, ,h C, d, "taaaaaa@O\Af\rOoac) a'r C, ,!, il, rl,
aaaaN)OOOI
u,a|S
$,t'+,SJ\)O@O\
iilrLtsAtobb,LirbbS[[il, rl, ,1, ,L ,1,aotaa@o\AN)O
C, C" C, C"aaaaO\olNO
c,a"t*t
d, r r .[-L,riF bbb, tiubbS[[
&ry-a}+
.A C, C,.. C^ l, ,!,a, ,a, a a
180
210
240
270
J 3ooxrl-.b 330
360
Harga a, untuk R1,/R2= 1.00 i fi=20o;30o;40o;50o
trura ae" '-)
(.!aqqqqa9qgo!a..o.qqC) ????TTTTTCT'TTTTf
09tN
O EE\sS',
00t
0rJ
ovz
OLZ
08[
0E r
OZL
06
09
Qf::::'r
0 aT Y
o!aqqqfTTTY
oc!sr aaa str \g @ aal
1T S'l f ,l,l s
'e eb.req\0 @ o c! t(,l \oo o c! sr ..\o o c) (\ <r \o @ () c\ <I \o @ o a a t a a t a a a a a a a a a a a. o' t a a a a
(Y)
0eqsa_
OE€
t 00e II
0Lz
0vz
OLZ
08 r
0s r
6zt
'06
09
OE
0
o0g 'oov=fr uep 00'[ =za/[u rt+. Ee e6'reg
I C,
r58
60
90
120
150
&c)at\)
**t***tt'
**ttg*br"**mharga a -.
-
.t &,taaaFN,O
t****t8*
*s*t*g**t* a**
AAA,.U&&c,b,Li-rbbb,I.h ,l^ C" C^ .tb,LtrbL
&, ,,1aao@
0
9
120
. 150
180
210
0erq
0EF^ ool
OAE T i
I
i
,? ()
STa
=t?
N
O
str \0 @ o c\l <rr
???TTt'9qqqaqqTTTTTTTScYagqqqq YYYTTT?,I
qqolaqT??TT
$rooo .aob,ITfY
\p@o(\t aaaa
TT?f
rooo .aa
?"lY
0vz
OLZ
08 [
0s I
OZL
06
09
0t
o
OLZ
8t
EI
ZL
6'
L p e6-:eq
++??T;Tfril+fifi?qaqqqc,t T??T?TTge\
\0tEA
,ot I
o0S uep s}i=g l g0.[ =ZA/La >t1n 8p e6.reg
Harga R 1/R2 il=2oo
150
60
x)
1n
150
180
210
240
?frI\b
aN)lE
&&!aoo\
Iaoa
N),L.t t ].aaaao@c a
,1, fuaa
'sN)ao\
A .t "t ..t ,,L ..t C" C, C" ..t ,1,[email protected]\ A N) C) @ O\ A N) O @
6, C, C, C, C, ,!'oaaaaao@o\FN)o
,-:,
&,CrC^ C, &..h.t t I f tt" ,t C",b fut (,,!, fu (, t ]. I ]. J. S&6too. a a_ a a a a . . . . . a_a- ! a_t a a a a a a a a a a I iO-o @o\ F N o@ o\ rr\) o@ cn F N) o@ o\ F N) () ac arr;F iu b boin;r i.,
\
r)
.t/
/ / I /
f// / , il
.t, a
4 v/ , a -
/ 74 7
, - / /, 11
/ ( 1
\ \ \ \ \\ \
\
\ l' b \ ,L .F
t) I
0
30
60
t,;t,,l,*L,;
l,;l*
\r:oN
360
l,
-90
Bfr L+Z I frZ
( atucV L' L=92-9b
( aulcgl'Z =
tQ*ZS rwc/6>1 98(,[ = ffi = nr7
'6>1 Ez'g€,t. = 00[+(8,[)(06)+(EEr[) (gL.) = g
:.rasa6 {caqD1e>1ed ) auc
V LV, O =VBSZ = TTque l6equed . Tn1
ztucoE'0 + zr.gz'z = gfrL + zlfr e !e>1ed,i
z*, Lt.'Z = ffi (ZL)(sZ)?Ool 0= v
Cl'-: uebuelnuad es+afS
9t9't)(gt't)Et = +lda[ I^r
= dnpTq ueueqequad rpqaT
( #;r1?50,0 =bJ (st't)(sz'Q)
0 =3
| ,orLF#4' ,
\ Zi't = = r,,J)
: e56ue1 f@'tu6{ 09t = (Er.E'[) 00t + ur6>t g'zoz = Tplo1 ?Tda[ nr
' (6urrrer = e66ue1 s=lequrad -re6ed Tf,ep Tese,eq ) b>1 00 trpsaqas lesnd:a1 ueqaq Epe de66uetp e66ue1 {pup 6un[n pd
'ul6>{ g'ZAZ =(EL'o -. sLg', L)(E, t)(06) +(9t, tlL
'ruc sgt = uc gt - rucorr'l t
'tuc At LgL =ruc g, L + tucgg [ =JT1>{eJJa -rpqaT
w/bX S9t = gL + 06 = Te?o1b
vt/b>t gL = 008 x gZ'O = TTb
ru /b>t 00t = 1e66ue1 {n1 Zun ) f e66ut1 qeun-r >t1n dnpTq upqag'ut/6>l 06 = OOVZ x gZrO x E[rg =
e66ue1 {eue r-rrpues 1p.rag uls'
rucg[ = 0EI x 0f/L =.;rT sltelrp e66ue1 {eup Teqa?
rucg I
r_L) P : qtltEft
e seleq uElpn{a{ e.rpc ue6uap qs1 e66ue1 ue{puecuau'VZn e[eq , gLLN nlnw ,ucgZ-apa.r1ue ,wcL[ =apo.r1do
1e66ut1 qpunr {1n up{eun6tp qsl e66uea J urg, [ =eTs.raq ;pqar-PuPrurp' ( -zalarrlue>[ ) qETeqas lTda[:a+ e66ue1 adT+ rnqEle]trq: I TPoS
: Teos qoluoc .? tloluo] Z'TA
(srr)(sz'0)(9'0)z l(09tls'Ll
L9L
162
soal 2 : Diketahui denah tangga seperti tergambar
dengan dinding.
1.
Pelat tangga dicoba 20 cm.
dibawah ini, dimatangga menyatu
antrede = 25cm, optrede = 17cm. Tangga digunakan untuk ge_dung bioskop. Mutu beton K2z5rbaja u24. Rencanakan tangga
:": ui, cara elastis? dikerahui: regangan izin tanah= i',n,
:ff],,
tangga
1atngga
PoIa pembebanan spf- tergambar.Beban pada bordes A adalahkarena bordes tsb kepunyaanngga atas dan bawah.Sedangkan pada bordes B, bebanadalah e, karena bordes tsb a_dalah kepunyaa.n tangga atas .
1
7qta
uf-4"
Jawab:
I2
. Tpuas de66uelp
I uep V 6un[n euputTp
up{p1a1rad g sg?e {oTEq le6eqas ue{eupqrap-asTp ledep edu1lu{a1p)tTup{aW ueEunlTqrad
vt/b>tt"o8=( e BLJ)| =b f =V sap.roq eped ueqag'rc LZz = l[xt[ = e6 -:l
*6ue1 Te{Tlra^ 166u11tlgE t =
tr+ Zt =apa;r1do r{eTurnc
QQZL =c7b J1'
-rt31 rtrar tttr'?rrrn^66E = apaluE r{PTunft
ut/b\ggtt = w/6>t90p + w/6>ttgtt =
TTb+TOb_Tp1o1bw/61 S0? = 00t x Se ,[ =
TTn
'ur.gt,[ =. ( TuT adT+ e66ue1
{1n ) dnplq ueqaq >[nlun JT1{aJJa -rpqeT
' ( ueuuqeqrued up.r
-nqerad 1Br{T T ) Zw/ b>t
zw/ 6>l
lrlg
@o
= ruc I TPqal uE{npp
= ,ur/b>4 tZ S' Z = rucg, Z ( Tp6a+ ) uTqn
vt/6X ttg =
/=EI
irrli;irirli
ZxZtL | 'rt11l
. , ,' I
I
I
I,.:l
I
?r
:
uEqaq
1E.roq
'Zr/brt 00t = 1do>1sotq 6unpa6 {?n) dnpTqrtt/6>tt8t[ = ut/6>lvoz + w/b>fim + w/bx9of = Tp1ol rrrpuas
7
lP.rAq.l
*=r"" l
6x voz = [s x Sfo ,ru red,, ,
6>1 [E = (tg)(Srll(L1,0+SZ'Ol = e66ue1 p{Eup [,t1n Tpe[
Zw/6rl tg = Tplo1
JZ
09
00? Z. S, 1.-EL = 00, Z:g, L. fiffi =
ut/6tgog = ?xg'gL = e66ue1 {eueb Tpp[
'B66uB1 {eup t = fifi ppp ra+au I ueTeq
,6>r,s.sL = oovz x s,[ * (,Zi"Oi?eZrOi =
e66ue1 {eue I TrTpuas lerag
E9 [
i'i'i1 lrliltlr
17 86ks /n
- 164 -
Penulangan petr-at tancqa :
M max =, 4169kgm, tebal pelat tangga=
i
l'r-^- 17
I
1 1,575. 1400
ls=;''.diperoleh 100n = 16,01
100 (21 = 010076
82 \ v'
ambil lm x lm
sebagai tulangan dipergunakan9A 15 (tuI. tarlk) dan g6-20'( tu1 . tekan ) .
Ra _(893 ) ( 1,I) (4,5-0,8 )4 r6
,(1796)(3)(1,5)
I = 2s28 kg ( ? )
Ru=38seksi t f
8s9
20em -+ h= 1 Tcrr ( selimut beton3 cm).
mutu bet,on "::t. -:t
{a= 75 kglen .
ba ja u24 ---) 4"= I 400kg /"*z .
n = 21
A = 0,0076 (157,5)(17) = 2O,3 cm2
pakai. l5l14 utk selebhi 15z,5cm atau 014 - 1..flJ+
atau /14_10. tul . pembagi = 20tA - 4,06 "r2
(trg-tO = S,O3cm2)untuk anak tangga dipakai gB - 20
. .. / , . ." _ i".:.,..
fu = 13t'f-i%.;6;17 = .1,647ks/c.2
( 7or=ers/cn2t---, or, .
Penulanqan pondasi : t,'lKarena ujung B dianggap sebagai sendi maha pada pondasi tidak a-'
da momen, harlya P = 3859 kg
^J,3.8 5 9 kg
az-vtfr
oz- \fru 07,- v L
'{
e66ue1 leTqd0 L-gfr0t
AZ
e66ue1 IEUE
tI
I
I
lJ
n
t
I
I
i
I
I
0
{I
I
,il
il
c
dtl'8i
l\518
I
I
I
t,,
4
J o?
lJ \.lrIJ
T
s9 r
165
qT17I6k2=89 3kg /n
jauh dng ta-dengan tang-
k berbedadisamakan
d
1
ta
I
rjaanny
'914
yg bekea tulangah.
I
I
-:roi1f-?0l1lr
-J .A tsI l'^ I-l ar-IMr \l--| fis-1 tr
il14-1t
anakaw
I
ila
eba -bebbawah rm
I gab
{lr
ffilt
Krn bngga10 I
;to\
*Fl$)914-fia -z
;Jtl.iL
:h1
4
7FrJI
g
1
i
detail B.
WMlffr"w1l!
914-10
e66ue1
e66ue1 1e1ad
B66uB1 {eueffi
'q-q ue6uo1o6
LIEZy.
I
Itl
-L
PUixI
l-
I
I
I
I
I
I
I
I
I
I
I
I
I
I
e6Que1
I
I
I
I
I
I
I
lrotI
I
00'0
UE?
t
,, ,tit=9[*r
ltl _t_ I I
u ZL. L+
P.rec ue6uap B66ue1
_ut/b>l 009 = dnpTqc
sap.roq 1eT a
: requrEg
'1a6a1 TiraqTp {ppT1 e66ue1 {eup p/vtr{pq Tnr{platTqe sTlspI
0
@-
b:{t
-a
e66ue1 {eup Tn{rurau' TuT r{P/qeqTp
I qt
up{puecua6 'VZn e[eq , SZZN uolaq nln1l1
upqag .log! = apa:r1do ,tucgz - aparlup
b^ e66ue1 {oTEq 6up ue{pupcua.rTp e66ue,tr
:eque6ra1 ez(uqeuap bfr, p66ue1 rnr{pla{T6l: g Teos
{i
E 168
il.awab : menaksir ukuran balok daq pelat :tebar perat dicoba = gcm, barok tangga = 30 x 30.
;1 meter'4 anak tang{a
adaGambar ! r
I
rI
I
I
I
I
:r-16{,
Il.IeTencanahan anak_tanqqa :Untuk 1 meter: lebar
1 anak tangga = o'25 I o'16Z .2400 = 4g kg/m
Karena balok tanggadiuraikan dlm arah
bersudut 0
tegak lurusg
dengan grs horizontalbalok dan searah balok
maka harga
25
q
qr n,Pada ujung anakdr railing atau
tangga kita dapatyg lain sebesar
q1 = 105 cosO = 105
= 88,4 kg/m
"'l-etakkan beban terpusat ( berasal100 kq : mj-salnya org berdiri).
P tsb diurai dal-am arah begaklurusbalok=pcos0
= 100 254' ,2kg
Beban hidup :
untuk 1 m lebar= 0,25 x 500 =
diuraikan tegak
tangga125 kg /n
Iurus balok = 12
-.L-,l
I
I
=TTft.li,
tangga :
( arahnya vertikal kebawah ) .
= ( o, oB ) f,urt2ffiuz .1 .z4oo= 57 kg/m( arahnya vertikal kebawah )
Qtotrl = 48+57 =105 kg/m
5 = 105,3 kg/m
J
;,+I
I
I
0oz=g t
frsfe
rpsaq etru6ueluaq Euaret J adT+ e66ue1 qpTppp 6un1tq plT{ 6ue13 pbburl {oTpq ueburTnuad ueleuE-uarag
8T
II
I
I
I
I
llTPJl1t;Ir
I
I
I
I
I
lI
I
I
I
I
I
rlI
I
I
I
tillttlltl
I
I
I
I
I
SZ
z*" ?g'o =
LZ = u
wc/b>1gOf t=e{C- VZ n
,wc/6>tsr=t*- SZZ x3 dPlel uruPqaquadurcg 9, ge = ouTs ( E-g t )
' ( TUT qeTaqas .requle6
'(tuc[ = Uolaqtuc
gfi 1e>1edrp 6uer16ues re6eqas(rutc6L'O = 01fr1 1e>1ed1
(s6'8t)(89'EZ) gOL00'0 = v
eor.ooo'o = **+4r-t =m
Z8?'[ = f?uOOt qaToradlp
=0)
0
gl' g
E+ osoc(E-sz) =
Ile
=EJ
leqTf ) e66ue1 teup rpqaTlnurTTas 1e6u1) utcg6rgL =
8 + osoc'(utct rucgl) =' 1 ue16un11qrad Tp
e6n[ e66ue1 1e1ad ) )toteq teue JT1{aJ}e Teqa?
ur6{g8E'88 = ( ?'O lZ, t8*z( 9,0 I ( L' t5 r )f = ]rutt^ vt/bl L,t6l = ErSg[ + f rei = rElolb
z'osoc ( E-sz
(00?t)Es9z'0( 988'981 LZ
69 r
: {oTEq rtEue uEutTntrad
170
Beban vanq bekeria pada balok vanq memikul anak tangqa:Berat sendiri balok = ?rS x 0r3 x 24AO = 360 kg/nberat sendiri anak tangga + pelat per m'
dalam 1 meter ada 4 anak tanggaq = 4 (105)(1,5) = G30 kg/n( ingat lebar
, tangga = 1 ,5m )
beban hidup per m' = 500 x 1 15 = 750 kg/mq total = 360 + 630 + 750 = 1740 kg/m.
Bebqn yanq bekerja pada balok yancl memikul bondes :tebal bordes = 8cm.
berat sendiri pelat bordes = 0 ,08 xq = 360 + 288 + 750 = 1398 kg/n
Pqrhitune?n ilekanika tekniknva :
1 ,5 x 2400 = 288 kg/m
anggap D dan G perletakan sen-di.perhitunqan disederhanakan :
q diambil = 1740 kg/n sepanjang bentang.
M max diambil = +T q
= ]: (1740)(s,10
= 4795,g kgm
Menurut PBI ps. 1 0 .8manfaat balok T utkan kekuatan :
b*(bo + ro/5
(:o + s25 /5 --) (, 7r cm
bm(b --) S 150 cln
b* (,bo + 1o * bk/Z
(:o +525/10+15 o/2(ts7,5cm.
ambil b = 135 cmm
AIan ditgrltukan ,momen max va-ng dpt diptkut oleh tut.. tu-nqqal :
I
I
I
adaan
L2
25 )2
ttg lebar:perhitung-
' Penulanqan ba.lqlc. t_ :
3o=8
ba lancel:
dalam ke
9, 1 400m
91= 1 3 9 8t g/t e2= nalkg/%q
0 r 75m
Balok tanqqa Sbg balok T
1+; 0 1529
6us:t&xas*r'Iffilrppnd pt+I ueurp >topq fp4 (
,vtc/b4 LV,g =9,2* q2
7(0t) '88 ., -,= 00 ,. ?Ti6rE L6O'v ==,'t'
o
. fr +9?,0 ,c -
-
+t =feuewTp
sEra
,wc/ 649=y
Q-ttl
,utc / 64 6g' Z
3f +s?,0T g'z
oe)
160' V
3.
zq''q f.'W
ozr4
+ (8)Z/L - Eg =
+ o,{ z/ L r{ = z eupurTp
+ =[
-Es)( #ff;ff +r) Er'B'srL' f+ o.{
f -q)(# +tlqe,.o,{.*q fo
='W
;nc/b>1 BS,Z =% =q2
rur ZZ,6V = (8-t0'Self lgv9 o( q-xz) 9
'1e66un1 16ue1n1Tpqs1 .f, {oTPq e{Pru
8- (Lt0',8z)zl9n9
, (oq xz) gE
Zq
ru6:t V l, 6gt = rs-ro16>tt , EggE=( 9 C,'g) ( E9 t.L,)Z/ 1-6ue1uTT
gg x 0t up-rn{n t6as-:ad >{oTpq 6qs de66uerp,l {oTeg
u6{ v L' 6gt-edu Z/ L -rpsoqas g upp o 6untn-6unCn e{ ue)tT6eqTpru6>t gz,gtL = ,(st,0)(gz,s)(00g)) z/L = rs-ro1 w
lxgZ r g=T {nlun
' ( Ts-ro1 ) .rtlund Tuere6uau up{p qs1 {oTeq p{pru e66ue1{oTeq ueuP{ nele TrT}t qeTaqos eped eetueq pperaq dnpTtl upqaq e}tTf /
\
\(z*"9sf8 = 61frt. te>1ed)
,urc96'g = ffi -v
w/ 6>tO0 S=
(-- U^ )ru6>tg '6Lv = rnqplatTp br uauoru puarpx
ru6>t 9, Igt?t =utc6>1 9G,09[gt?t =
+ (B)f
: 1e66un? . Tnl r{of o Tn{TdTp ldp bA uaurou
( lllcs = uolaq AnurTTas : 1e6ur Irurnu r. 1-- (rrcg-) oq (*rt€g,g1=(EE) 6z9.0 = r,{ .qd = x
LLL
ffi,IYiIi"r'ti
i
172 -
Untu* - memikul torsi , pada PBI disyaratkan pemqst^angan tul . meman jangM. . U.
selrias :'!:-:!2 .fr.atr(36.9, 14.100 ) (20+20+48+48 )= --'T
cm
d+pasang pada sisi vertikalSketsa penulanqan tanqcra tipe C :
2= 1,87"*2 (pakai 4g1a)
masing-masing 2910.
r%_B
1g 10 g8 -25 1g 10I -258-10 P
sengkangfia-ts
6 -25 w-1.53fr1e
potongan a-a
'\/
/c,
F--in-d3g 1e
0 t-gfr6uB6ua
6tfit97,- gsL-fi
3 TTETEP
sL-}fi6uas
Z O T-
0 [-8
q-q ue6uolod
g TTplap
sL-gfre16ugs
fi u.,
e
7rt
:*o L-Bflo0.[ - g0
6tfitsL-gf,
6ue>16uas
174
Beberapa catatan untuk tenaqa tipe C :'l - tulangan pelat tenaga searah balok tangga dipasang 06-ZS ( tul .
pembagi ) .2 - tul - pelat tangga, tegak lurus balok tangga,untuk tangga tipe A dan B; tuI. disamakan dengan
sgrl 4 : Diketahui pd denah perat disediakan lubang utk tangga rdima-bersih ) .na ukuran lubang tsb adalah 3m x 3m ( ukuran
Diminta merencanakan tangga yg disokong oleh balok tengah ( tanpapelat tangga ) atau balok samping ( jadi anak tangga terjepit,pd balok tsb ) .Diketahui tegangan izin tanah=O,45kg/cm2; beban hidup tangga= 400')kg/n'. mutu beton K2z5rbaja U24. Ketinggian lantai | = + 4.00 rr.
aloktengah2m
Lebar tangga kita ambil = lm( anda boleh mengambil 1 ,5 m ) .antrede dicoba = 25 cm.
2 0 +,a = 60 -___) 0. - 17 cmjadi antrede = 25 cm.
dipasang g8 -'l O .
tipe C'.
tipe A
optrede = 17 cm.
{
jI
( Z*16t' t
'gz
-ZLfrt te>1ed;,
t6equled
0802
ue6ue1n1
(e)(se)
rB6e
?t ['00=
{edTp gfr TP
urc?0, t
f\( LJ1
qas
=[
1,1
(szz ) ( er 0Tz
v6'z =
+ z(L)(9,Etz)lTdeI ueuo6ru6>t gLL, LLZ =
w/bl s'stz =
=[*B = epe:1dOc7JT,
= 002 = apa-r1ueurc LL = ope;1doruc sz = apa-r?ue
(r) 00r
T Bfolb
(uelinpe ) Tseds= uroE,Z 1e6e1gt'0 x B0r0=
r-rTpues 1e-roq
= d leesnd-re1>[eup 6un[n pd
: e66ue1 {eue
De t'0
Z
L
w/bu, o?t =
Sf,0xLZx[-uroI Teqalru/6XLZ=Sf ,0x iZ x S,Z
w/b>t Z, Lg = 00iZ x
e66up1 1pue'64 oo t
eleb TjreqTp e66ue1
tfr:'-?,
eped e [ -re>1eq
+
7,ft
,{
0V/ 0Z BqocTp {oTeq uprn{nucg = eqocTp e66uu? {pue TPqol
ucEe = fTqup ,e66ue1 {pup -reqeT
'ruc gzt = szL -i L ) =rB1Bp qe>16ue
'lxc9[ = ApaJl-do [ + urc/lp epe-r1
-do tt 6upsed ElT{
. E,VL =,LL -m=e pe -r.1do qe 1un t
rxcl ?Z=>1e6e1 qe>16ue1
LV' Zft;ilz
'r *{
ulcIzz=L
T*rt [ =0
EE
qq6L{q g
'zlfL=o uc
SZ=E
(9-ZZ:) (Se_q ) rE'o ) z
6xoo[=d
I
S[=LLx6
176
BiasanYa balok tsbbanan tembok diatasnyabawahnYd r sehingga utk
12
q
PenulangdD belof tan a (utk tan ati A).
g6 -2spotonqan a-a
didlm tembok, sehingga mengalami pembeo
juga ditahan oleh tembok yg berada di-ini kita pasang 3g16 dan sengkang g8-1*
beradatetapibalok391 6
sengkang g8-153g 16
Tetapi jika diatas balok dan dibawahnya tak ada tembok
perhitungkan terhadap, berat sendiri balok ; beba-beban
dari anak tangga berikut torsi '
Sketsa Penulan
ang g8-15
0
teAuntuk pondasi, balok ab diikatkan pada sloof pondasi '
Perencanaan Uafof tanqqa tipe B :
ts4= H* = 0,765
qt= 37 ,4ocos(= A,794
maka harus diyg berasal
Balok def dicoba = 20/40
pelat bordes ambil 8cm
Beban vo bekeria Pd balok : efberat sendiri balok
0,2 x 0r4 x 24OO x -fu,= 241 tB kg/n
berat anak tangqa :
dalam 1 m ada 4 anak tangga
berat 1 anak tangga.
= 0 r 08x0 r 35x 2400x1+0 r 35x1x (2 rs
x 24+1x21 | + 0r35x1x400
= 236 ,55 kg.
ut6>{ L, Lgt = xpru W <j--- ru6 [, l. =x (_-__ 0=xPxI^[p
,x(18 t t ) c
L' Lgv x[tg , 60l L
-(xggL'O) g,g?62xSgLtO = hbl x =7ui L,.Lgv r"au
+ -xZ6S- = c+ xl't?g- =
*W
^ pueurrp
+xL't?B- = *I,{
ledep {oTpq . Tn1 ue6unltq:ed {n1un
u8'gv6z + =
'au
,(z) ( ?8 r r ) 3-ctJ {
C,u+(z)L'tt8-
T0 ='w3
: Ja ue16eg
'>[T:p1 Tpu.rou
u
{
.a
a
It
lt
L
PI
tS
p{TPqp PlT}taq )toTPq pd
,81 L'6gt{r I
rr-5Pt-z.'( -> ) 6)t
0=L'tgv (€grL)
rg ggg, t
b>tt' L LZ
u'Jut0L, LSV+L! Lgv-
%
: Apoqaard
0
99-8'gv-
L9',v6e+Lg'v6t-
T,, L6L+Lg' zt -6' LL L-9',E6j
0e E'gg+ L9'v6E+
,*\
z(z)(?8rr)= Z( L, (?81)
3 rsnqTrlsTp uaTsTJaolt
Zy;
,t/b>t z6e = (oo?+o avzxrr:l;' ,':' , ='::rl;'lr;":t-:::: w/br z6L = oovz x vro x 'z,o = {oTeq T-rrpuas 1p.raqvt/6X VgL L = Z, ZV6 + g! LiZ = Zb Tpp[
w/b>t z'zv6 = w/(6>tg,Etzlv = e66ue1 >[pue ueqaq
1
or--wd
,JIW
ZLt-r
L
fa=" -Wd
DA=' -wd
sLL,o = -gg.g,.#+ =
vgz'o = l"y
Jo
ulcc I
: ssorJ
: pa {oTeg
v6
te 'gg
z(t9'[ ) +r(z)UJZ I u[ov/o\t0v/02
: Ja {oTPg
i
i
178
Penulanqan bal,ok -: uk . 20 / 40AA-
M lapangan maximum = 387 ,7 kgm
35
['
1,5 ( 387 tL_
M tumpuan = 451 t7 kgm
3s
.l0 = or4
. 12 b hml-n = .pvau='l 2(20)(35)
2080
= 4 r04 .*2(pakai 3g16=6r03"*2)
pakai 2g1 6
2cm
5 r 03 "*2 )
Beban 1 anak tangga = 56 kgL=56kg Dlm 1 meter ada 4 anak tangga
jadi q = 4(56) 224 kg/m.Itl torsi = 224 (2)(0r4)
= 179 ,2 kgm
( inqat : panjang balokjarak P thd as
0,4 m).
=3+0,45 + ht/b 0,45 + 40/20
Mt' ut
u=. =,rrj A
= 9r0lI
f a min = 4,04
) (Pakai 3916 =A ' pakai 291 6
Ar
ana'k tanggaq
ef= 2 m
balok=
40
*Er2 O- au'^ t1,5(89,6 ) . 100 ( 1 3+1 4t30+30 )
2(20S0)(14)(30)0, 5 7 cm2(pakai Zfl10= 1 157.*2 )
,l0 r4
g
Jika beban hidup hanya bekerja pada setengah anak tangga (dikiriatau kanan balok tangga ) maka balok mengalami torsi.
l-**tI
M torsi dipikulkan pada perletakan d dan fmasing-masing sebesar 1/2 (179 ,21 = 89 r5 kgm
4r0b
= 4,06 ' '? 1?'':::-1 oo
lzno"* = 3,40s vs/cn2 ( t*b,, ( = 12')(40cm)(20cm)' ___, OK
YE
balok ditulangi extra pada arah memanjangnya 3
tah+
-f
f'b, = 6;} dimana Y= 3+
k-2!-rlmeman j ang
WH'
0z-sfr
t,{/
^L8T
e-e ue6uo1o6
/9 1frt,ooL/z
J9 1f,|.
91frz.EI-gfi
6ue>16ues
6LL
'g adT+ e66ue1 ue6uelnuad
180
SoaI 5 : Diketahui tangga
denahnYa sePerti( '! f ree standingdibawah ini :
antrede = 25
optrede = 1 I
stairs" ) yangmelayang
tergambarcm
cm
l_d.TI,r
lebar tangga = b = 1r5m
jarak railing tangga bawah
thd railing tangga atas= m EB l0 Cm.
Beban hidup = 500 Yg/cn2
Tegangan izin tanah=0 r 6kg/
"^2. Mutu beton k225.baJa U 24
Pertanyaan i rencanakan tangga secara kekuatan batas 7
J.+pab,.: h = 180cm = 1r8m
arc tg h/L =
r = 1/2 b = 0,75karena m = 10 cm
=. r/L = 0,75/2Langkah perhitunqan :
b=1 5m
t0xtg=ft80
; L=225cm = 2 r25 iarc tg 1,8/2,25 = 381660
m
kecil selrali rnaka diabaikan -
,25 = 1/3.
l. Menentukan ta a untuk
.
Ii
ba<tian ml"ri 'fliqht "pelat tangga ambil = 18 cmTebal
berat I anak tangga(0,18)(1,5)
dalam I meter ada 4
jadi q = 4(81) = 324
= * (o,zs)(2400)= 81kg
anak tangga.kg /m.
96?,6 [ '0 =E?0gt'e L
ffientrg?09 l't L
'e L
=g
ogg ' gE
)orsocsoc
t9(t + (t./L\ZlZ(t/L) I6 + -9a9 + )osoc Iiq2soc (E *.RZIZlg
szs0,0- = s{€?ff+
E?091,I,L - (( E/Llzt €699'ggsocn- (e / t+ t ) o99' strsoc z(t/ Ll lt 6l z(E/ Llz
\orsoc e&A6 + 21 9 + soc 8
( .J21Lt -7p soc n ( g+ L I lsor6) /(zE69o'o =
hrsoc grQ a't
2?ils?09 r
o99'Be soc (t / Ll6 + .Q.U-g +]losoc
tI
099'BE zsoc t(t./L\
)e socl! €.
zt.vL'o = Wla 6 + (t/L\ z-L g + o99'8e soc I
v= t
?t- o99'8t Zsoct(E/'Ll 21 8L
)orsoc
+ .,99'88 soc v + (t./L) f,r E
tgeLG + .(,zg +!soc I v=V )o? soc.Q, 1L 8[ +! soc y + -a,I E t
1EIITT } Gg'v ro1{Prol{eJ uE{nluauol^l ' ( Troa+'w/6>{ 868 [
'w/'6>{, B'eo6[
w/6x 8'tEL I
tn/ 6>t
= TBlo?b :
= Te1olb 3
= 8'6zg8'6zg =
a?
saproq uPT6Bq {1n-6ut;tru 6^ B66ue1 ueT6pq {An-
i rPlol uPgagl Tpeltw/6>t ogt=S' I x 00E =TTb : sap]oq uep 6uTrTru b^ e66ue1 ueT6eq pd-
Zw/bX 009 = dnpTttr ueqeg 't
urcS L - TTqupTp sap.roq Teqal: saproq eped ueqaq ue{n+uaual{ 'Z
+ VZt. =. 6urrTrrr 6ue.f, ue16eq >{1n TGb
(00?z)(E'[) oY 8['0 = ,tu radrucg[ = TPqal : p66uB1 1e1ad
w/b>{ B?9 =sap.roq {n1un Tob TpE[
611 8V9 =64 oovz' E'[' Bt'o =
sapJoq T jrTpuas 1p.req
c8L
r 5. Irtenentukan harga Xt i XZ akibat beban hidup 1 X/m
Aklbat q'= lt/m yg beker ja pada bagian "f lightr' .
v' : -A. 
'lu g, L = .reqaT {n?un 6>1 Z, 6G E g
E'90? 0LEt + 6ZLZ + 6'SLgZ +g'0602
(zo,Lt6-t fu
+
( E' L ) 86t t + o99'8t soc +
urs (sz'zl(g'toGt) + o99'gt soc ffi
o99'8E UTS
o99'Bt
0*[x.#+(Frrs.q.,.b+#)+1}u1s..x.,b+,$soc*lzx 'zN + [x'lN +
oN
= zx .'^, + Lx '^o + o^o
E' I reqeT {nlun 6>1 g, t S t L+
+ t.'t.glz '- 6'v?Et + g'zLgL-
N
f,o0=0+0+0
uI
z'szt + g'Lcat
o99'gE soc (s't)g6EL + o99'gE
o99'gE soc (gz' z) g, Eo6 t
zx
ru g'Il'zLgL
(8'9t6t-)()99'8t soc+ (z'Lt6-) (os9',8t uTs
sz,z = T = x euerurp o + ([x E+ ) +
1r!r soc'Q',b +zr.uTs # I -Otx,oc'x',b +fuuTs ?-" -) =
urs - , (gtLlz s i-,
+ oee,Bc uTs ffi
=
-'*o+[x. t*o*o*o=*oreqaT {nlun ru6>t S' gE I t =
6'Evz g' Legt + e'vgzl =
6fl;l+oee' Bt urs ( st' o )
8, J(zo'tt6- )
o99',8t uTs
s rcr,nl (q'l)? = ( + o99'gt uTs (gl'o)z(92'Z
) g' to6 [zrrfosoc + Lx')ouTs
+ +/ou1s r' # +-purs r i#;" =Z'
zx '"^ + [x' "^
+ o'*
= 'wurr.ri$' 1 reqaT >t1n ur6>{ Lr 6[g?+ = 6,60zl + 6, ?08 gr g voz + 6, tgg1 =
(8'986t-) (oS g'iltuTs-)+ (zo',Lt6-) (os9',8t soc 6ffi1 * o99'8t soc'
(g'L)z E['o
- E8 r
( S' L+( EZ'Z ) e ) E' ['868 [+ o99'8t sqc'9L'o' (I'Llz ^
zffi= '{
184
Jadi pada polqqgqqj-A ' untuk selebar 1 ,5 m :M-- = -'l'a75 kgm
xM__ = +4519,7 kgm
vM_ = +1135,5 kgm
z
Qy=oQr= o
n = 8399,2 kg (normal tekan).Gambar :
11. l-len€4tukan monen, lintang, normal pada potonqan B-Br
* = 9, perbatasan bordes dan 'if light".M =M +M X.+M .X^--x *o *1 , *2 z
= (Mr +M" )+(M .X.)+(M X^)'xo xd' *1 , *22
+ t frts(t xr + (o)-xz
=0-0+0-q".b.b/2+o+O= -t3e8 (,s) ( ff I =-1572,8 ksm
M =M +M .x.+M .x^-'Y '-Yo :-Yl --Y2 --2
- 1M',Yo + M"yo) * rr, . *, + MYz
= ,o'i*' .r coso(- nHSu .r cosd() + ( * coso/).xr+ (-sino( ).xZ
(0,75)cos 38,660+ 1€bi#fl (0,7s)
cos 38,66o * t# t"" 38'55o(*937'02)+(-sin38,56o){ -1936,8)
= 1567.g + 2046,8 - 304,9 + 1210
= 451918 kgm untuk selebar 1,5m
cr
r' I
li
i
'tu s' I
5V'90?
o + ?,0'LE6 - J8r+re +
.reqaT
0rEt
o99'BE
{nlun 6>1 9g'EZLS =
+ 6zLZ + s9'o6az =
uTs (E't)g6Et +
(8'Ll-Z (o99'88 soc)(8'tlzffi+(+099'88socry =
itII
t*lr
I
Ii
I
I
tII
i
f; .
II
l'
x.(o)+[x. #l*( louls'q' ,,b +)6 soc
#fo;F + rb uls'x' , b + losoc Ig " ) =
(,,1eorrJ,, ue16eq :188il1 | zx(zr.r) + rx (tn) ; ,?*']'i-, =
zx ' zN + [x ' [N + oN = N
o = 7'x.o + [x (o) + (o+o) =
(zx.'oo, * ([x.t^o) + t"fo + ofo,
=
zx .'^o + [x . too +
ooo =
oo
ru S, [ .reqoT {nlun 6>1 60, t6g t- =0 z'gzt + gv' ttgt + zt, t.gLZ- 0 + gv, zLgL- =
0 + (zo'LE6-) p3+ uTs nee'88 soc (E'r) 85Er +
LulDzW:"zx
'(o) + (tx#F-) +
-)+1fisoc.x.,b+)ouTs f.=
-)) ? ( ( ,{Csoc'g' ,,b + buTs - tl Z[q.+rz-]ffi5
(zx 'o*0, * (tx ' t*0, * tofo +
o*,0) -
zx.'*o + [x . '*o * o*o
= *o
ru S, t reqeT {an u6>t ?E, gE I t + =SE,ZL|L 6,EVZ 6V'LEIL + VE'VilZL =
(8'986t-)(o99'8t sgcl+(zo'LE6-) o99'gE uTs Hfo +
o99'eeuTs(s L'l,ffi +o99'BE urs(Et'0) --!8'Llz ,, (gJT cYY uL u lD Z r " nTi;ffiTfr =
(((x1 1,bsoc)+(tx)( urs + )+(hurs .r 1o;#F +)ours .r #l =
7r(zx ' "'w(+(tx '"'w)
? zz"x'-w+
o('
oI,,W + ,W)
ro'z zW+ W
+
Ixa
- 186
Jadi pada onqanMxM
vM
zQx
Qv
N
12.
= _15;72 r75
= +4519rB
= +1135r54
= -1893 r 09
=0
B-B ' untuk selebar 1 ,s meter :kgm
kgm
kgm
k9
= +5723,06 kg. (normal tekan).
MeneBtukan lngmen, lintangr normaltengah "fliqht" (x= 1 r 1 25 m) .
.
an dite
M=x iis02*(?{?s)2 ("t ,125r.:s 38,66o' 2(1,9)
+i3egjl.f:*16). (1,125t rs 38,6Go
+ W t9 38,660) ( -s37,oz) + o
cos 38,660 ( # cos 3g,6o0) (-937,021+(-sin3g,6eo) (-tgg
= 1567 ,91 + 2046 r8 304,87 + 1Z1A
= + 4519,84 kgm utk selebar l 15 m
Mi= W (0,75)sin 38,6Go +
sin 38,660 t (t-3J sin 38,660) (-9 37,ozl+(cc 38,660) _ 1930,8)
= 1254,34 + 1637 ,51 - Z43,gO 1512,4= 1 135 ,54 kgm utk selebar I ,5 m
o*=-% sin 38,66., + fi903,g) (r,125) cos 39,660
% sin 38,560 + (139g)(1,5) cos 3g1660
- sin 38 r 6601rg
= -1672,45 +
= -220 164 kg+
m
6r8)
l3e8[l!sl(6) (0,7s)2(1rg)
(-937,02)'+ 0
1672,45 2193,32untu'|< selebar 1 ,5
- +oegst,s)(r,t2i + ry ))
(2409,51 - 1204,751 + (3145,52 - 3931,gg) - 46g,51-50r 10 kgm untuk selebar 1 15 m
1637 t4g + 325,2
@60'E6Br90'€zLs
E?T66e8
ts'ge[[
.tl
6q I L' LgoL+ = N
o =do6>1 v g' ozz- =
*o
tu6>t Vg'gtIl+ ='wu6{ v8'61.9?+ =
o^
ur6>{ g0 t'08- = *W
tu S, [ ;pqef os {n1un 3
"1IIbTTJ" rFbual-qebual Tp ueSuolod pd Tppf'ru g, [ .reqaTas {nlun
s'90? ot€t + L'6zLZ + 96,Lt.tL
uTs (s' L ) 86t t
(szL'L18't06[
64 LL'L9aL+ EE '0602 +
( z0' Lt.6
(8'L"lZffi
(8, LIZ
g'Iffi
+ o9e'8t+ o99'8e soc
zl8't05 [=N
=oo
vg'azz
L, ZLgI
L8L
+ o99'ge uTs+ og9'8t socz( sz
0
lm
:
Potonqan B - C (baqian bordes) : fi = 0o
Mx
Mv
= *Z = -193618 kgm
=M+M.X.+M.X^Yo Y1 t Y2 z
o'.L2 .n g + q"..brj2l+b) r rj-T r sr- ZI'-- r sln
0 karena g i 0o
M- +M- .Xr.+M- .X2'o '1 | zz
0 + 0 + O.Xl + sin g. XZ = 0
o*o + o*, ' xl + o*r'xz = o
oro + or, ' x' * or, ' x2
, -o; 'H' cos g - q" ' Q!3Ltb ) cos g,
= 1903,8(2,25f I _ 13gg(1'5)(6)-Eo'!E
-2677 ,22 3450 + 520 ,6 = -55 06 ,7No + Nl .Xl * N2.X2
, q' .L2
s + (fr sin9).Xl + 0 (x2 )
=
Qx=
Qr=
N=
+ ff-91 'rr + (o)x2
.t *, (-e37,ozl + o
kg utk selebar 1 ,5 m
gl + (.- ) xt + Q= 0
karena g = 0
jadi pada potongan B
M* = -193618
'y=oM, = o
Qx=Q
Qy = -5506r7
N=0
C bagian bordes bekerja 3
kgm
k9
-' vv 6' 01Tr) ' b -L v6L'
vt.'v =sI
zg'Lto? = (zE'0) g'B6sz[ = t"'tN
tuZt'O =€0'0 60'0+gZ'O =p 1qZ/L +t.-*t"u gZ'O = LZ1'O + E[0'0 + 09['0 = Za + [a + o" = t"
- ru LZO'O = (gl'0)gl'0 = *r{ Sl'0 = Za
ru rro,0 = (8r,0) .,laO*4q).1. r = 1q .,+ )zctc = ra
(, dct
llrggt,g = zo,o + 0?1,0 ='o, + to"
= o.
uIZO'O = TTqure'u90,0-(B[,0) Ot./L = ?qOt/t='o,nl
tu 0l['0 = =I ="o" nW
u6)t 8'86921 = Z'66889'[ = *N
ru6>{ g'Z,1LL = (gtt[)g'L = *W
r VV ue6uolod {n1un .' ( "lQbTrJ" rIErEasl 5ue.[upuau uEbuETnuad
[=
Ee-rJ
-
ZE
-L It
=_3
vv6,o = "5(ggE0'0 = b
I
lrr.Bg'Z
6too'E6gt- = *o
6>1 go'tzLg = N
ur6{ ,S'EE[[ ='Wru6{ gr6LEv =
f,^
ru6>{ L, Z Ls L- - *w
sgTp64 g, Lg?t =
*O =
6ue1u116>1 Z'G6eg = N = (ueltal) Tpu-roN
ur6>{ SrSt[[ = 'W = TS-ro1 uauo6ur6>t ti arEv = ^* = ( 1.{
n11ed (-6uedueuad lapuadral TsTs uenp
-a{ eped ue6uelnuad {1n ) -rnluaT uauo6ru6>{ ;LLL =
*W = (gv
qp.reas ue6uelnued {1n ) ) rnluaT uauowru g, L -rpqeTas {nlun : V Tp
ffi',,1e6TT],,-ue16eg @
saz(:0gt)(E'0'tz
6Rr
- 190 -
=qbh
1=-1 .794
Penulano+n di, B :
= 1r5 (1572r81
No = 1r5('5723r0S)
2 ko' O'ur*
Cl au
(o,os6s)(1so)(1s) W = 7,67
(pakai 7g12 = -l ,9t cm2 )
a
ia
A 2m
235.9 ,2 kgm
s58 4,59 kg
M 2359 .2u z53Y,l,==#=Ar275====o" N-- 858 4,59lu*or= 0 ,02 m
"o = to, * to, = o'zgi m
ht 0r1g
1.
"1= cfz t fftl2 ht = or 032 m
"Z= OrO27 m
",, = 01354 m -----, €",, e 01354 + 0109 0103 = 01414 m
1015
{'" ;Lt=
!11 =-= 115261- 0,951 #
= 4 1621
I q = o,o4e)q, = o,esl
0,414
A =;-:EE (rso)ilsl2(0'2!(.?251 (0,049) = 7,Br cm21 ,5261 2(0,51(2
2080
(pakai 7iltZ = 7 ,91 cm2).
+ti! .
'ZLf,E EuTseu-6uTspru TsTs pnpa{ eped6uesedlp 6uefi ouc gL' g = Ztfig 1er1ed c
( 0rr)(sr)(0802)zl0'g =z(ovt + gt )oot' (E'aelns'T
anPrl- ',{. {z
I n1- = 6ue[ueuaru V
3 TS.rO1,n
-au {n1un 6uBfueulatu E.r-lxe uebuelnl eduepe up{?pJredslp , rgd'Eue1ueq qnrnras ynaun Sl-gft 6uBx6uas tBxed
ruc 6,SZ =. (t9'VL) B!
ffim=
Tn{Trueped
Z*e [0r[ = Z(09,01 U' 1 6uedueuad Zl gfi
.s n/L 'Z = V
6uer16uas 1e>1ed
ns-J'q s_nP=T = E
f,v
,rnc/6>1 lg, V
(8t)?(09[) 1Q,Q
;nc/6>t Ee' o = ffi L 62, t = +,t=
"q,2
s [o?!)(gr) 8! *
! 6ue>16uas
'xo (----
'g__'q nj 2==n2.--
rr.W
ednraq :11und ueEuelnuad
3 TEtTlranEuBdueuad qe6ue1 qe6ua1 pped
n(rwc16t0t= ) *9*2) wc/6>1 EE, [ [
1 + s?,0+ t = *r5T - + t =rA
6Zte =
nq2
8t0Er+ E?'0
9'ZTpluozTf,or{ Euedueuad- qe6ua1-qe6ua1 EpEd
6Zt e =
l5t
3''TS.rOl UatUOW : TSrOl UebUETnUAd
ffii'
192
Jadi :
Penulangan searah "flight" 3
untuk potongan A-A' : jumlahuntuk potongan B-B' : jumlah
Untuk memikul torsi i sepanjangtas dan 3912 pada sisi bawah.
Penulangan pelat untuk memikul M__:_vM = 451
v
tulangan tarik = TgtZtulangan tarik = 7912pelat dipasang 3fltZ pada sisi a_
Sepan jang bagian : f 1ight,, .
anggap pelat berupa. balok denganukuran 18 x 150 em.
cu= 150 - 10 = lo,g
kgm
T,,
I
Aminimum=12.bh=ry=,l415.*2(pakai14g12.=15,8zcm2|o*au 2og
srsi bawah= 7g12Resume :
Penurangan pada potongan AA ' bagian ',f right,, :
tulangan tarik = (7+3+7 ) = lA/12 dipasang selebar 1 40 cm
atau fitz- # = fi12 - 7,5at
tulangan tekan 91 2 1 5
Penulangan pada potongan BB' , bagian ,,f 1ight,,.tulangan tarik = g1Z 7,5tulangan tekan = g12 15
Penulanqan baqian bordes :
Untuk potongan BB , : tulangan tarik fr 1Z -l ,5tulangan tekan fitZ 1 5
tulangan pembagi dipasang sebanyak3g12 tulangan atas; pada pertemuan bordes dan
',flight,'
9 ,B
ffi
sr-e66uB?{EuB
e66ue1
SL-Z L
sL-efr
L_ZLfr
9, L-ZL
s L-8fr
r,/t
9 l.;ZLfr
I0' 0+
o
o
d
(z*
Et\86---fr-fil\lr
8ff"-*_
8'[+6--89t-ZLfrg'L-ZL1'I ZLbT.
S, L-ZLfr
SZ_ BfrBf,L
@/ 9L-ZLfigl.-gfr 6ue>16uas
g' L-
ulcg S='li / L rxogg="1?/L
194
Soal 5 : Diketahui tangga ulir ( "helical stairs', ) yang denahnya se-
F = 24ao
b = 150 cm
tebal pelat tangga = 20 cmjari-jari lingkaran dalam (Rr)= 2 m
mutu beton K225, baja U3Zrencanakan tangga tsb ( tangga digu-nakan pada bangunan utk perkantbran )
diketahui : tinggi lantai= +4 . 0 0 rn .t'''Jawab :
-
Lanqkah perhitunqan :
Er, .MenentukaSr harga R r/R, :
R1= jari2 as pembebanan; Ro= jari-jari lingkaran luar= 2m + 1r5m = 3r5m
^R3R?R1=i (ff)o1
3 r53 23mJrf, Z
R2= jari-jari as pelat tangga
= jtno*n, l = *rr+3r5) = 2,7s m
R1/R2 = # = 1,A25
untuk penentuan harga u 1 ." 2ra3 nilai R1/R2 kita ambil = 1 ,05
b'Menentukan besarnYa g (=sudut yang dibentuk oreh grs singgung pa-da lengkungan "he1ical " thd grs ( bidang ) horizontal ,
sur = 24Ooplnjang bur3500
24oo .2 E .2r75 = 11r52m
g = arc sin # = )o r 3o:- 2oo
i
= 2r818m
fr=2oo i b - 150 =7rSh20
h = tebal total pelat
= ry = lzoo
=,1,
.2 7C Rz
3600
c'ry@Lihat nomogram halaman I51 utk Rr/Rr=l r05;
di.peroleh .1 = -0r05
L2cara memperolehnva :
cari angrka 120 pd absisrlalu tprik grs vertikal keatas,memotong lengkungan (ambil kira-kira lengkungan T rSrla_tarik grs horizontal kekanan, akan diperoleh harga ;, )
perli ter'gambar :
xtzF$i! t ! \'rl{'t' -- is- t ir
urcg [-0ruc g'L7 = (E€+09) i = E
tuc 9t = 0 !
' ( >lapuad nTeT.ral >tepTl )
urc €t E
<A, : rcquprgt=TE tuc09=o=Tpe[(---- tuc er = lli# = o
L + ruc 9E P qq ez Pqoc
luc
xo
tuc
6,7 Tulc 6z = ruc;iE = 'P
ruc 6'vt. = # = Te
09 P qq.tz Pqoc
= e66ue1 {eue qeluln[ruc 09 = o, EqoD Z'p
t'€!.=ffi=apa-r1do
ruc L8 e qqt
sv' vz =
+ IXC
09LgV L
resaq.radTp n1:ad apa:1do qeTnlT {n1un'uetuedu {epT1 Eserel qs1 e66ue1 T>tTeuau p1T{ p{T I e6
-6u1qas'4epuad nTpT-ra1 (,,asTJ,,) apa:1do 'ue{T1eq:ad e1T}t p{TI
: ru€TPp r{BTaqas
ogz'B = 6Z
ojvz
apar+ue
=r'\71nPns
ruclgE qq t + urcoE e qq 8z eqoc
' qq ?t'62 = w =
e66ue1 {eue qelunt Tpe[
ruc tllc 09 = o"
0 E =JPR{ qPTaqas apa-I1ue Eqoc ' [ ' p! Eqoc
-eqocTp apa.rldo uep apa-r1ue .:reqaT
o09E luc L9tL = M (Ogtl .rt Z =renT uere>16uTT 6UTTTTaX
o0gE tuo 8t8 =(oOfZ (0021 2l: Z =ureTep uere>16uTT 6UTTTTaX
,.
reoE
aper1d?r 'pE?,0- = te qaTo.rodlp
'8E L uPtuPTPg ue.rbouou leqTT6' L'= Ze qa1o.:ddTp
'lgL uerueTer.{ uer6ouou 1eI{TT
g' L=rl/el o0 ,t---d ! olz=flr s0, r=zu/La {1n
g, L=r4/e, o ozL=tr ! ooz=fi! g0, L--za/La >{1n
96r
[#fl
is:
196
ubin teraso tebal 2 15 cm = 2 15 x 24 = 60adukan tebal 1 cm = 0r01x2100= 21
Mt-r
€. Beban vang bekeria pada.pelat :tebar pelat (termasuk anak tangga)= 20 + + ( 16 cos g) = 27 tsz cm.
berat sendiri pelat = ry x z40a = 702,9 kg/*Zcos 20' :
kg /^2kq /^2
- beban hidup(utk perkantoran, tangganya ) =3 00 ko /_2- ri;L;";Hi:ffii i;:H
rerbuar i.-"'"'= 'oijii,
lebar tangga = lr5m ; jadi'beban total= 1,5( lOB3,9)+35G1975,9kglm.f. llenentukan M^ dan H so
re,05 (1st5,sr(2,7sr2 = -747,14 ksmH = u2.g.RZ = 1,9 11975)(2,75) =10324, 1. ksg. It{enentukan harga_harga }t r, ; M, ; T, ; Sh, ; ,r, , nr,
Dari point f diperoleh , Mo = -7 47 ,1 4 kgm
H = 1A324,1 kgg = 0o s/d, 12Oo
R2= 2r75m ; R1 = 2rgl} rn
g = 2Oo =q.7L = 0,349 radial190-
q - 1975,g kg/n
= -7 47 ,14 cos 0 + 10324 ,1 (2 ,75) . tg 2Oo. sin 0(2,Bta)2(t cos o).= -,"n1 '::=';;
.n + 283s1 '?75 fr'"' 7r ts 2ao sin -1s 6so
'87
gatatan : nilai 0 yang disubstitusikan ke' dalam dera j at .
", =
To, =tn g ' sin fi-u.R2 . €'ts/ c4io. siaf {.R2 . sin g. cos g+ (q.R? sin 0 -g.R1R2.0) sin g= -747,14 sin g.sin )06 10324,1.(2,7, l. 4.ra tg 20o.
sin 2ao 1a324,1 . (2,751 sin 0. cos 2oo- 180o -r
+(1975,g .2,8182 sin g 1975tg. z-75.2rg1g. #ou.tE).siazo.= -255'54 sin g 61 ,69.0.cos 0 26679,a7 sin 0 +5366,59
sin g g1r4} g o.
1975,9
hapersamaan ini adalah
l
.rur wrTETTu pr{1
'-'s u=p -Qs TprTu.Jf
]ir-W ue{eun6.rad e1T}t 1e1ad
'o0z UTS 0 uTs ' I'VZ|,L - oAZ soc''il uTs
'ooZ uTs ' 1t' "o:t
3 qqs ue{TaqplTp Tpu.rou ' t s-:o1 , 6ue1ur 1 , uauo14 e6-req-e6.reg
e6n[ ]tcaqcTp {oTpq 6uBfueuaur' ue6ueTn;up{eun6.:ed plT}t {oTEq rasa6 up{aca6ua4
TeTTU ue{eun6.:ad e1T{ Ts-ro1 up{aca6ued'Ju
d T Plu-rou +
bue I uptueul ue6ue1n1 ue{pupcua.rau {n?un
0 soo L'VZ;1L0socH
0 uTs E0'IEsE 0 zt'16^08 r
2t' " 6 '818'z '6'9L61'0 uTs'H fi soo 0'lu'b'0 vz'tt. 0 urs gv'106-
=r-IJ='S=
I-II=S
=Ju=d
' 818' z' 6' gL6L ooz soc ouTs' [ ' nzt0 t -'fi uTs 0 [u'b f, soc 0 uTS H-
oE L' LSZ 0 gV' 69 L 0 uTs [ 0' t 66t.Z
0 uTs 0E'0E56 +
0 tLt lsz
o0z soc ( 2L.
+ o0z 6+
L6L
0 uTs 69, ??LtL + 0 soc 0 gV,69L O UTS gO, ZAL- =
o,gt 'o0Z UTS '0 uTs '818'Z 'L'lZt0t +
T 'a|gt z '?l'.2 ' 6' gL6L ours'- ( g t g'z\' 6'gL6L
'0 soc' 2l' oO: ,gL.Z .LrVeEg L - O uTS.VL,LVL-) =. 0
Jt
'fi uTs'o uTs Zu'H +
+ f,61 o soc o Zu'H o uTs ol,t) = Je
E'0682 t -v' 006Ll'zgLg-gL' I Lgv'0s6gz-z' 0?,v v -o07 L
L,I,69ZL'g'Lgg?0I t' [ 6e Iz0'v6L6z-E9, O?E+oo6
L'96t0t-Z' L Z,IZL:Z9LSgt' gzg?'01092-92, ZE I [+oo9
6'L?Bg-t'vL66'0r695t'6g'9zLE t-L6'tv-ooE
00L , VZI,O L00VL. LNL-o0
( 6>I ) o'u
( 6)t ) 'u,J,.
(6al qs( tu6t ) J;*6rt ) Ju
W
J-( ru6>{ ) *W0
? lgg
H. Penulangan nenanjang (,U_ ) Itf
Tebal pelat tangga = 20 cmlebar tangga =1 50 em
M _ _4420 ,2 kgmtfP- = 12390,5 kg(tekan).tfsecara kekuatan batas :
Mu
Nu
eole
o2
= 1 ,5 ( 4420 ,2) - G630,3 kgm= 1 ,5 ( 12390 r 5 )= 1g5g S ,75 kg
6630,319585,15 = 01357 m
= 1/30 h. =t'
=e+c=ol -o,,)
= 0 13770 ,20
= Cf,z
= 1(7)( r )2(o,z)0,227 m
0,15 h.c
"o + "1
", + 1/Z
0,006m ambil =
4,377 m
0,02 m
eo
eoq"1
= 1,885 ----) CZ = 7
,#t,'.n. dimana
= 0,15(0,2y.=
+ "2 = 0,377 +
rt = o17
L ambil =
1L = A,7(0,03 m
0,227 + 0,03
L
240 1_)360' '"11 ,52 m
11,52) =
= 0 1634
0 ,025 =
(2,751
8,06 m.
m
0,709m
"2e
u
6
au
Cu=
ht a = 0,634 + 1/2(0,2)
= 2 r8o
?D=
I
0 12
1
-<-_-.
1-c h,ue
au1 - 0,ggg =040 r709
]I q = 0,14375) 9" = o,BB8
= 1r334
A = ffi l0 ,14375) (1s0 ) (,r7,5) 2(0,rl=),!zzs)' 27dI-= 22,81 "*2 (pakai 15g14 = 23,1 "*ri"Ar= f,.ta = 0,2(1,334)(22,g1) = 6,0g"^,)(pakai 4flta=6,1Gr
' unuTuTru v epedTrep TTcolt qTqaI qaI6:ad1p 6ued V puarp{ runuTuTrx V leledrp e6n[ oOt = O {nlun
.rw StL = V Zr0 = rV v,l
'(sPlE 'Tn1 ) ,urc SrL = OZ.OE[.tg7-,rO ='rtr.q.tgztO =UTur V c'! o0 = 0 {n1un @5'
(SE1e -tn1)
z*' 9E' I
' ( qe'u-eq 'Tn1 )
-tuc Lgt L c
900' ?,
t,E 6' O
gtt[0'0
zg' t.
v L'zg0t
.Z LE, A
L6Z'O
t0,0
6LZ'O
t
v L'gvz' 0
8?0'07,0'0
gza'0
sg'6t06t-tg'0 ?g+
($e?p lrq)ulcrq, I
Z ,VJ '
r{PmPq' Tnl )ruc ?[r8 z .r v
V6LI L
gE 6'0
Etgg0'0
I e6'r
L 's6ss
I r zv'oI
[ 8?E'0
I r0'0
VZZ, A
t
. 68'9Ll, 0
?60'0z0' 0
vL0'0z'v6Est-izi zE [ [+
(qeAeqTnl)
z*' vgi o
se?E' Tn1 )
_lllc VZ t tc
tgz'zBE5, O
,LEV€0'0 =b TTqup
,g's
6'IZ0tgtnE'0
s6gz'aE0'0
ZLZ, A
t
vgrgge t'0
9LZO, O
z0t ogt00 r 0
E8, I LL9-gg6 ' gg-
0L L' OZL L-
V Z'O =rV08 LZ
Ws'Lr)(0sL)b+
nEta n(To I -tT
Ib
(2,0 =Je6:eq1
( u6>r ) t'". oN
"-1t{? 1to=te"Tron "e+'A+ A = A
1q s['o -2,1q'
f
, iqgl tzclr= L e
v'T
tcZJ
. trr/o " zo Lo o A+ a=_a
ru za, o =1o"(u) ";;"; =to.
N E, [ = tN
w s'[ = *w
(SZZI(9'[)(E'0lz
o06 = 0o0 t=0ue6uera?ex
661
oOt =Oo09 :=f, UPP06 =€eTnfi t qqs ue:f{aQeqp
200
i. Penulangan untuk memikul Ir{ :,f
0= 0o
g = 30o
0 = 60()
g = 90o
g =120o
ang9ap
i. Penqecekan torsi 3 T' max =
ftril.*itabe lkan
0
;15126,4 kgm
-26010,4 kgm
-2g794,02kgm
-25950 ,4 kgm
5 Ocm
0 r 45
2 = 3 r 33
IEbu
-l-l
Pd tepit-
Zbu -
+
6
0,.4s + 58
b6-;
ll
b. F.1:
1,5(1391,31).1002A(17)(140)
A*s . O-au-6'r -su
4,38 Xg/cn2
frA 2 penampang --)a AQ= 1
s
,IY= 3 +
Y= 3 +
Penulangan puntir berupa sengkang':M.t.u
----) ,r,____) *r,
----, ,r,
----, *r,
--*-| M' nfsbg balok.
1391 ,31 kgm pada 0 = 90o.
Pd tepi penampang horizontalM.Y+
b.h.'t.2 22 115(1391r31).100j l JJ
( 150 ',t (20f1,58 xg/cmz < z*b*ro (=30)
penampang vertikal 3
2 22 1r5(1391r31).. 1005r55
-
20 ( 15 of
,0 1 "*2
Keterangan ,'9=600 = 90( g - 300 0 = 1200
M,, = 1r5 M
[= q,(20)(145)2(0,5) (22s,',)-
27 80
2(o,5xo,2lQ25l
( harsa f,= 0 )
390 1 5,5
4 '92
0r0425)0r98 cm-
pakaiAmin ,= 12 ,51 cm
22689 ,6
6 ,45
in imum
min inum
12.20. 145
-?7-m-_12,51 cm2
44691,03
4 ,60
0r05)11t74 cm-
pakai A min 1r
= "l 2 ,51 cm2
l( lqo .- ,
paka i
0t-tLfr= 'vLf,vtleqed
az-v | fr
+0[ -vLi
0L-vLfr 0
OZ-V L fr
i;'hl;I
rereo
[or- v tfr
I
oz-v tfr
oz-vtf,
oz-v Lfi=vtfit
1e>1edoz-v tf,=
h-v16
r{P/vlPq . Tn?
sE+P 'Tn1: uebuPsPuad
vtfidtvtfrtz
vtflttvtfrt
vtf,ttvtflt
vtbtV 1fr9
V LfrI.
VLfrL
( sTaT.roal )q/{q 'Tn?
sElP 'Tn1
V tfr te>1ed1q
zg' v L
tB' L ttg'gl
6Z, OL
LB' 91
9e 'o I
ILt' 6
L6', L L
gL6't
gL6'6
( -uc ) r{/{q ' Tn1- L
.( -uc ) splp 'Tnl- c : Tv,ro,r6uB I upurau ue6ue1ng,
gLv'zgLI, ZgLv'zgLv, ZgLv'z'q/rlq -SE1P ' TR1 '=
r s.ro1 TnltTuau >{1n 6up ! ueruaul
Prlxa ue6ue1.4gggz'gggz'g992'g992',90q/{q -sE?p 'Tn1-
JuI/t Tn{Tru
-au {n+un ue6ue1ng,
60'g
L' t.z
18, L
9s' L
vL'g
t9'L8r9'0
rz'tg'L
g'L
( -ruc ) qltlq ' Tn1- C
( -uc ) selp 'Tn1-c
JaW TN{
-Tuau >{1n uebuelng,
o0Z Io06oo9o0eo0uB6ue:a?aX0 eped uebuelnl l edu:esaq 1e6.req
xo
Tn>[Tureu {n1un
'1e1ad 6ue[ued
L 47.
(-;-- ,utc/64
o0 - 0 ePed
z" 96'v =
:h uEbueTn+ tnlun eunsou 'T
g9,9 =ffi =.ez6>1 t'VZ7OL = unurxeu :asa6 @ '{
(0?r)(rr)(0812)z=
1,{' *'*4 z
: qqse.r1xa 6ueIuBurau 'Tnl 6ue1ua1
qn-rnTas >t+n g L4 re>1ed (--
16 ' n1W= 6ue I ueueul V
6un11q elTlt e{eu :11und
I€td ue?Pre^fis-rad Tqnuauap
Be'v '02 ruc zt =ABLZ 'LO
ffiffii
202
Gambar penulangan tanqcra :
W'TM r alntLe+eI sebelah
g1l4-20
,6ogmJ5cm
,
I
I
I
I
sebe lah tda I ar1= aIPxR
1
1g
i
i
24 anak tanggh
;;1"
00'0+ A
8fr
0eT9V
-Lo
sL-9fr6ue>16ue s
,,
rrrcg [ : opajrldo
9Z- 8fr"
rucsf= ruTplxc09=-renT
0z-?r6r
tksl 0r-?r/--\----\
oz-vL#A\-/o r-r 16T-1
e66u>[P
I 0E-n L
sZ'Tfr * 8fr J
tod-Jr
UP?eue
or
/ , .'.'l - ---Tgfr1I ot-vL#t, -
OZ-V Lfl
t.0z
SZ8fr
: c-c ! .q-q" ! B-e uE6-uo+oa
Soal7:Diketahuidenah
i
I
I
I
II
I
I
t
tangga sbb 3
+3 .5{._
Antrede = 25 cm ; optrede = 17 .* u,
-tTangga direncanakan untuk gedung pbrtokoan, tegangan izin tanah
= 0r5 yg/cm?. Mutu beton K225t baja u24
Rencanakan penulangan tangga dengan cara kekuatan batas.Jawab :Tanqqa bawah : antrde = ry = 8 bh
LJ
jumlahoPtrede=$+1-jarak vertikal- - g x 17
9x1
0 .00'
Taggga-alag :
= 153 cm.
j arakj umlahpaka ij umlahj arak
oPtrede =
optrede =antrede =
hori zonta I
350197
=1710 bha
11 1
= 10 x
153 = 197
1 1 ,5817 em + 1
= 10 bh
25 = 250
cm
bh a 27 cm.
cm
Gambar 3
uE{rler rad4oTrq - {oTpq srlpTp lrTed Te6pqos uE:{
-Euecue.rTp ue{E EIurluEu sapfoq lpradTpues : g upp V up{EleTrad de66ug @.2
'!ut rad 6u11:rer 1e:aq ue:16un11qradfp sn:eq E{pu eleq 6urpuTp Trep lEnqral p{Tg 1de1ag, .ue4
Trq"Tp ledep eluleroq 6urI , r[Eq Tp{pdrp 6ut1te: le6eqag : [email protected]/b,t gg,OgLZ = 009 + 9g,0ES, =
TTb + lTb = Tplo1b'u/btt 009 = Z x 696 =
TTb
Zt/Brt OOt = (ueo{o1rad 6unpa6 >J1n e66ue1; dnprq ueqaqw/b>i gg,OSSt = TL'ZL?, + L'OLB + gO' = TEtol T.rTpuas lEraq
u/bti gl,ZLZ =ro,g9 x p = ,ur :ad6r( ,o'89 =
( tg) (zl (L1,0+EZ, 01=e66ue1 {eue t
zvt16>{ [g = Tp1o1TPqal ue{npP
Z*/ 6tt
=?Z x S'Z = (urcgrZ Tpqal) uTqn 1e.rag
w/61 L'oLg = oo? z'z. ff?; =
I ru -rad e66ue1 1e1ad leragut/6>l g0? = ZOI x V =e66ue1 {eue V ppp I lu .rad
6>1 ZOL =00?Z x Z xoLL'o)(92'olf = e66ue1
z/ L [ =061
: upbunllqred qeq.6@t
{1n69zgtO = gsoc !
gg'0oz' v t. -o
Eue [ 1P-roq
'tucQ SZ=92x0 [ 902
9'
urc0l It tx0[
3206-
M
{;
a -2M max = 1/8 q L'1 ^.2= -+ (2150r86)(21-8
l-'rvf vvrr-f
= 10ZS ,43 kgm1Ra = Rs = * 12150,861(21 = 2150,95 kg
Penulanqan :
tebal;pe1at tangga ht = 15 cm
h : 12 cm (selimut beton = 3 cm)
max = 1075,43 kgm
12
{W= Or2
A minimum= trrr)
A minimum
Ar= 012
Tulangan
0 ,25*0 ,25
roo
A=1 r5'cm2pembagi
a3
,:'b 'ht
.200.15 = 7r5cm2 (pakai 1Og1A utk selebar 2 tn.
atau g1o ?+3 = g1o 2ol(pakai fl8 25 atau SgBl
= 20t tulangan utama = 1 r5cm2 (pakai gG - 17 15,
,82sF:E'\ .:\+ B> 72,5 gn
Coba B *, :80cm --.) uk. pelat 80x80.untuk tulangan pelat pakai g8-15( tut. tarik ) dan f,6-20 ( tuI. tekan )
Pondasi ta
1
Qambar tanqqF ba*ah :
+1.56-17,5t,,
IT
=34 r29
urE
tu
w/ b>te9 9=b
aproq )toTPg
OE
(_ruc ErE = c. _ruc t .Z
zL'00['tsz'o
8b 1e4ed , V
I L.-9fi re>1ed 1
= UIIIIIITUTIII V
s0'tt =
ruc 6 = rI (---- ruc zL
ru6>{LeZ'gg = Z(S'Lur6>{ Str66 = Z(SrL
1e1ac[-@
Iru 'vva'o = o'*
l='u'990'o = *t^
cx
f.= 'q
) ( 6ee )
) ( 6ee )
3 Saproq
vvot A =gg0'0 =
_vt/bX 699 = Te1o1 b Z'Zw/6>t 00t = dnpTq ueqoq
w/b4 69t = Te1o1 Tleru ueqeq z t . vrv L--T-'7 i---
ru/b4 LZ = luc I Tpqal up{npe- tlIZ=z
t L ,v 9 L-a-T
w/64 09 = ruc g, Z Tpqa? uTqn- Z t . vr J r, .L--l-.?
Z*/ 6XggZ =00 lZxZ L, A -r-rrpues le.raq-
EE,
: saproq uebungqra4 .,
.T!=r.T
( 89',8892)9', L
'6q gs'gBgz - (s'z)(ggroE[,l f = G( ruc .ZgZ,Z = SrZl-gfr) ,rucgALt1-, = ?Er0L x tg7 : rEeqruad
qqs
'lucr{ETPpe sap.roq uEqag
zL = saproq 1e1ed Tpqal*Q*
2>;nc/64 98'[ =zL'002'6'onq -r=J
raseb {coq3'Tn+ leEeqag Z
(EZ 8f, = dfiZ sfr nele ,r!? re>1ed1 ,
nrol: VL nrd hE-'=, ud*-ug'Z'(vL |tfr = ffi oLfr ne?E
ru Z .rpqoTas >{?n 0 L f,V 1 re>1ed 1
oSoz zrllc ?E'o[ =
ffi (zl)(00zlzeor0'0 =r
zso?o,o-b { ,'t#:s}lL0'9 = = .:O,
ur6)t9t'089 L.=, (s, Z) ( 98,0s [ ,lt =xeur
LAZ
lrrS'[=
Ww/6>{9g'og LZ=ba
ffi'E
208
Balok BBt : dicoba 15/25
P = 1165,43(21 r"BBt 2 I
lrtrp = 1419,02(tl
W
Q*=501 ,75k9/n
99Okg/m
Br=1 165,43k9/n
ry (sot,7st, E 1479 ,02 kg
trr165,43)( 1 f( t 3r0,75) l - 501 '7,2(o '2512
= 786,547 kgm
Penulanqan balok' BB t .IL=CC I
)
fCu = 'O =.-= 3138
It;]ffi'--[= o,o94171 t5) (2ol 2(0,51(2251
A,= 1,224.*2 (2g1zl 2o8o
sebagai sengkang diambil flA 1 5
q = 0109417
A
3 r o 6cmz (3912= 3 r 39cm2 )
zfitz untuk tul.
]
I
;,f
t'
ll
tulangan memanjang
tarik = tul. tekan.1Rsc = i(90)(1,5) +
255,66 kg
5.
Nu = 1r5(1734r68) =to,
'"r, = o'02
3=g4a=0,1ht o '2Ct = 1, CZ= 6132
"1 =c.czr11$12
2602 ,02 kgm
.ht
1.6,32 (#ffihf (o,zl
0r0062 rn.
'9*=501,75k9/n
1075,43+90
C
Penul.anoan kolom :
tinggi " kolom diukursampai pondasi.
Reaks i di B pada tangga AB = 2150 , g G
dibagikan merata sepanjang balok-
=ry=1075,43kg/nBeban dari pelat 3
1' .rqx = Z.Q.I*= + .G6g. r ,5 = 501 , 75 kg/n
Berat sendiri balok= 0115 x 0125 x- 2400 = 90 kg/n
k9
;1
fli
t,itijtl:ii
ij
ll.
li
II
[!,,
:;,;I
jl:l
t,'{r:l!t1,
iI
Ir#
tu
l"II
l$l;
H
fi
F
I[1 :lij
i
i!il
$
il
il
Ii
I
1
I
i
l
I
I
I
i
I
sz- 8fr
Z=92x0 [
.(up>ta1 .Tn1 I 0z-9fr
.({TrP1 -Tn1) Et-gf,: ue6ueTnl le6eqas
' AL x 0f = 1e1ad . {n C-- ruc O L =g TTqup
s L-ef,
0z-89e46uas
t9'[+
' tt-gfiL6 l=tx0[
: selE PbbUPl UbUelnuad requreg
voi o '", ' >42=no_o
I
0z-0tfi
s't,L-9fiI
tu
LZ
ruc t, ss (aZg
T6>1 Bg' vr. ,1 g?'0>(89',Ve.LLl t0[ + ggttt,LL
ruoTo{ lsepuod uebuelnuad'g L/gL TppIuaru TTca]t.radTp ruoTo{ ue.rn{n
UPO 'TsTs leduaerl eped reqosTp 6^ ZLf,V 6uesedtp unurrurru V re6eqag'b TETTU l{oTorradtp {p4 l ue.rrdureT Ter.lTT ) ue;Eotuou Tf,pp
so'o 1---- gzo'o =
90S'9 =
LgZ'0 =
zgg0'0 = zo +
ru e0'0 =
Lqzl l0z) no,^w-r
z'q = z9g0', 0
to'a + a =1q g['o =
f.'r{nd
nr,
?,a
- gfi-t
LiI, ZLfiI,
(9ZZ ) ( 9'0
- 602
t'r.'rIIiij
2ta
Vi.3 SoaI - soal latihan :
ptsdd., soal latihan dibawah ini , kami tdk memberikan jawabannya,
tetapi kami berikan petunjuk cara penyelesaiannya, sehingga pe[baca menjadi terbiasa memecahkan persoalan.
S'oaL l : Diketahui denah bangunan spt _tergambar.
Sebagai lubang tangga disediakan ukuran 4m x 4m.
Lebar tangga direncanakan 190 cm. Tinggi lantai = + 3.60
Mutq beton K225, baja U 32 -
Potongan untuk tangga bawah dan ataS spt tergambar.
Tangga digunakan untuk gedung pertemuan.
Rencanakan tangga tsb , dengan cara kekuatan batas atau ela'stis ?
Petuniuk :
Langkah-langkah yang harus ditempuh 3
1. Tentukan dulu antrede dan optrede, mipal' antrede = 25 clllr
optrede = 18 cm.
Z. Tentukan jarak horizontal, jumlah optrede, jumlah antrededan lebar bordes. Akan diperoleh jumlah antrede = 9 bhroptrg
de = 10 bh, Jdrak vertikal = 180 clllr ketinggian bordes = 1.8
lebar bordes = 175 cm.
3. Demikian juga untuk tangga atas.4. Tentukan beban yang bekerja pd pelat tangga, dan pd bordes.
5. perhitungan Mekanika Teknik utk tangga bawah, dapat dilaku-kan dengan menganggap perletakan berupa sendi.
tangga
sendi
. lnqasf,e? sapJoq leTed eppd| ( rga roqp?'uaTsTJao)t ;e4ed 1 I uep x qe.rp uo,roru nrnp 6u-nltq6ueur ueEuap u={n{prrp +dp , sep-roq 1e1ad ue6uelnuod . g
nelp TirT{ qpTeqes e66ue1 {pup eped et:e>1equeqeq pltT I e66ue1 {oT pq pd r s-ro1 >laca6uaul
. ueuP{e^dueq dnpTq
ednl ue6uep - V
rpuasue{p1a1-zad re6eqas de66ue1p }toTpq 6un I n 6un tn puBrurp
' L 'oN Teos 1ds uapr e66ue1 {or=q. {nlun [ - ril ue6un4Tr,rf,ad . E
' ( -reAaTT+up{ +Tda[, Tpe[ (____ {oTeq eped lTdaF:a+e66ue1 {pup 1e6ur ) e66ue1 {pup ,r"sr=1rru.u up{Euecuau . z
'e66ue1 {orpq eped et.:re>1aq 6upd uega-q up{n1uaf, - [3 qndruellp ueJle 6ueI qe>16ue,1
e l sepuod uep 1e1ed , e66ue1 >1bue
3 {nLunladue6uelnuad ue{pupcua5
sap.roq
" - L LZv
1ds
'uPo{o1-rad
splp uep qe/qpq e66ue1 ue6uo1o4ru S. E+ = [ rplueT 166utg,
'Zt. n p[pq , SZZ ]t uolaq nln14
{n1un up{puecua.rrp qs1 e66ueg,
,''r ..ll*,;::il
::I'],:il:l H:ll11 e56ue1 TnqEla{Tq ffi' 'qei"rEq e66up1 {fn uapT sele e66ue1 {n1un ,I.W ue6unlrqre6 .9
?
s plPe66ue1
e66ue1>[oT pq
r{P/{pqe66ue1
'JPque6:a1
SoaI 3 : Denah tangga spt tergambar dibawah ini.Lebar tanggra = 3m
Tangga digunakan untuk pusat belanja (pertokoan)
- 1-1Diminta rencanakan tangga tsb lengkap balok bordes, kolom
penyangga bordes. Diketahui juga mutu beton K225 | ba ja 1J32.
tegangan izin tanah = 016 Xg/cnz. '
Ketinggian lantai = + 4.00m, ketinggian bordes = + 2.5m.
Petunjuk :
dinding 1 /2 batatinggi 75cm.
,eLanqkah vanq ditemPuh :
1 . Tentukan beban-beban yg bekerja pd pelat tangga- ( ingatbeban dinding vs berfunqsi sebagai railinq harus dima-sukkan ) . Antrede dan optrede tentukan dahulu, sehfffihorizontal, vertikal, jumlah antrede, jumlah optrede da-pat diketahui.
2 . Untuk tangga potongan 1 - 1 perhitungan M. T nya dapat di-anggap pelat diatas 2 perletakan ( anggap sendi ) .
3.
4.
Pelat bordes direncanakan sbg pelat lantai yg
pada balok perletakan ( lihat contoh soal No.
Tangga atas ( potongan 2-2 ) direncanakan idem
ngga bawah (potongan 1-1 ).
menumpu
71.dengan ta-
Z qs1 e66ue1 ue{Egeoua'rau Plururo
, ';nc/6>1 99'0 = I{PUPI uIzT ue6ue6a1
. trrog L =opalJdo 'uIcEZ =opoJluP 'lllc0L e66u1qas (e?Eq Z/ Ll
Pleq 6uTpuTp rp lenq'ra1 6uT T TeU ' Zt n e I eq ' iZZN uo1aq nln'{
'(vLZuBuIeTEqlEqTT)'rP*quebjro?ldsebbuelqeuapTnqPlaltTg:sTeos. ( Z = q,To{as '-6urrpa6 e66ue1 {n?un dnpTq ueqaq lP6uT )
'uroTo{ eped lTda[:aa sap'roq {oTeB 't. L
. oN TEos qoluou pd e.re3 ?n-Inuau ue{EuECUATTp sapJroq lETad 'z
' ( Tpuas-rpuas ) ue{elaTad Z {oTeq de66ue6uau ue6u
Spue{n{BTTpldpZ-Zuept-[ue6uo1ode66ue?{1n.l,.Wue6un1TqJad.I: qndualTp u"#l
: {nfunlad
rug' I
'qrf"Xr= 6unpa6 eped Epelaq qsl e66uBtr'
.g'L + = z Te?ueT ';L't + = [ TelueT uET66u-rlAX
'ruc.B[ = epe:r1do 'ucAZ = apa;lue 'Zt,;l e[pq 'SZZN uo?aq nlnl/{
' ( e?eq 6u1Pu1P ue{nq ) B[eq rs
-aq re>1edtp 6uTTTef le6eqag 'tuoTo{ pd lTda[.ra+ sap]oq {oTeg
. TuT qefi\pqrp :eqrue 6f a1 ?ds er{uqeuap b^ e66ue1 Tnqela{To3?T*
sepiroq{oT eq
214
Balok penumpu bordes , menghubungkan kolom-kolom, as A dan B.Balok tsb harus diperhitungkan thd torsi yg diakibatkan oleh pelatbordes, sedangkan perhitungan tangga bawah dan atas dpt dilakukandengan menganggap ujung-ujung pelat tangga tsb adalah sendi.PeLat bordes dihitung penulangannya dng menghitung terlebihmomen lapangdrr tumpuan arah x dan y (pakai koefisien, tabel
tangga ataspot. b-b
. s Joruou TPos qoluoc lPqT T
'urcgl 166ut1 Eleq Z/ L 6utpurp
sPlP e66uea
00'0
rucg L 166ur1 ElEq
0B r
z/t b
'lucs L 166u1 as pleq qebualas 6uTpuTp ednraq e^t{u6urTTer - ucg [ =aparldo'ucsz =opoJlup '
,ruc/6{89, 0 =I{EUEI uTzT ue6ue6al . ztnB I eq ' szzN uolaq nlnu e{T [ ' qs1 e66ue1 ue{Bupcua.rau p?ururq
.reque6ral ?ds eluqeuep 6ued 6uer{e1eru e66uB1 Tnr{E1a{T€
'qef{eqe66ue1
TPUTP
iLZ -
Il a216-
SoaI 7 z Diketahui tangga spira:l- berputar ?Oo menyambung tangga Iu-rus spt tergambar denahnY?.
L,ebar tangga Z meter. Tangga diperuntukkan rumah tinggal .
Mutu beton K225, baia V32'o '
antrede = 2 5cm; optrede = 1 8cm ( untuk tayrgga lurus ) .
Ketinggian lantai = + 4.0 III'Rencanakan tangga tersebut lengkap pondasi ?
Gambar s
t6----6 00
lKetinggian tangga spiraL = + 1.8 menuju + 4.0
( jadi tingginYa = 2,2 m ).
T450
I( 1'Tl
' Te?u9z T-roq
-uas ueqaq
LLZ
e.,{p6 + ( TTp , uep.lepua{ upqaq ,1edse
lEqT{p ) Ip}tTltaA ueqeq 1n>lrured te6eqas'>{oTBq ,leTad TrTpls6unlraq Auaulnqe
deIB sdeu{es
sef e >1edure1
:e1td( I rre/'l6urtt,, l dedes
( ., +uau1oQP,,1:a1d
ledse uesTdeT + uep-repua{ TelueT 1e1ed
: qqs ( Tde p1
5:a>1 uele[ ne+e1 er{e-r uele[ ueleqruaf ue6uolod nlens ?eqTT plTx' -reseq 6uer( ,, {upJ .role/yr., r{pT epp 6ueue.r uoToX
'6ueua; uoTo{ 1ez{unduau ( up-ro1ue4:ad , TaloH )
qe16uT1raq 6unpa6 undnele 1e66ut1 qeunl 6uepe>1-6uepey. '.ros6u
-oT )tppT1 -rp6e qpuel ueqeuad 6ulpurp e6n I ue{euEcua.rausn rpq elTlt'lTltnq 6ue;e1 nlensTp /,toTe6ung ueun6ueq ue{pu-pcuaJeru up{e. elT>1 e.dulesru e/{qeq Ture{ ledepued uebuap
n[n1es edunlual ppuv . rnqela{ elT{ {n]un qs1 r{Bf eseuBdu6utlued le6ur6uau ,, {uEl Jalelr,, upp ,, TTp& 6uturp? a.rrl
6ue1ua1 6un66ur s rrup{ TuT f uoleg T s}tn-r1suo{ uef ep TO
'ulnun ['['rIA: snunr-snunl uPp Troafi J'ITA
,( "{ue,f, -ra1e6.r B "TTeM
6u1uge+ago )
urv rx9Nv,f, 3 HvNV,tr NVHVNSd 9NIONTO
rlrl J- t--
--E-
lrll'l
)toreq
rrA 'gvg
'{
218
akibat gempa (bisa searah searah balok jembatan ataupun tegak
Iurusnya ) , gaya horizontal akibat tekanan tanah dibelakang din
ding tsb. Sayap ("wing wall") juga berfungsi sbg dinding pena-
han tanah.
VII .1 .2 Penamaan baqian-baqiq suatu dindr nahan tanah.
I'surcharge " negatiP
tapak ('toe" ) .
badan ( "vertical stem" )
( t'hee1" )
,tapak( tttoe t'
)
'dasar ( "base t'
vrr. 1 .3 Ti -ti dindin nahan tanah.
vtrYV,
1 . Tipe qava berat ( "qravitv wall" ) -
a"tUrr* a= beton cor padat ( "concre-temass' ) . Karena berat sendirinya be-sar sekali maka stabilitas dinding initerhadap geser dan momen guling ada-Iah baik. Dinding direncanakan agartak terjadi tegangan tarik.
2. "Semi gravity walI" .
Pada tipe ini, tapak dibuat lebih be-, s€rr agar tak ter jadi tarikan pada da-
sar. BarJan <linding dibuat besar sehi-ngga stabil thd guling dan geser.
I
tumit\)
bBban t&ftibahan( " surchargg,l
;,.,1
,tl
I
_l
.i
l
l
I
i
I
II
s s a.r11nq rl
m TOOQll -)
tlEu
TpE[Uout ( lTurnl) ,, TTaQ,, Ue.rn{n-BqTp EpEraq 6uo>loduad rdB1a1
'TTca{ qTqaTuep , 6urputp Tpp ( ,,oo1,, ) uedap upT6
'6uo>1or{uad ue>[Bun66uaul p6n[. luT, adT,[
's
'TTca{ qTqof TpEFuau qeTorodlp 6uEI uauou e66u1qas raAaTTlue)t re6eq*oS uetnq , sn.reuau 6upI ue{E1a1rad
. seleTp 1e1ad te6eqas upteupcua:Tp ( ,, TTar{, ) lTun1 uep illualsr ueppg
.Ed(u?roJIa?unoc[)6uor1o,(uad6uesed=**;"l:il,.m,::IT:-::(Tetlttu qTqat r1soc,, e66u1qas ,-resaq qrqar rpp[ueu 6utpurp rpqal T1*.rP ulp1Ep ) sTuouo{a {Bpr? qs1 6urputp , 6uo>10^f,uad upnlupq eduel prtu-pspTq 'ra1au 9, Trrp 166u11 qrqar 6ue.( qpup1 ueqeued 6urpurp {n1un
"TTPA 1rO3r@
'6urputpresPp pd (,,6uTpTTs,,)
up-resa6 qeEacueu{n1un,,do{,, up>{T.raqTp
(,,laTTTIr ) e?noA ue{T.raqTp. ue>l6ueTTqTp >1ede1BSETq 6ue^[
: qqs ueceu ede.raqaq ppga
(,qPuPl ..
( rua?sr )u9peq
a,
iI
aaD
flrlffi$.
220
%V6 uTs+ [fi urs- [
'(v xl+l+) q ueupTEpa{ Eped
' ( -relepuau {Tf elTp I{Eu
Ba ) f T1>tE lPf Ts-raq qs1 qpuPl uPuB{
-a1 e{Pru '6utPutP t66u1+ ZZ e:T>[
-ef T{ .resaqas -renT qe'rEa{ T sElu'ro]
-ep-raq qeuel ueqeuad 6uTpuTp e{Tf'(0 = c = Tsal{o}t) JTSaqo{
I
7/S+oe?---> bnl a>1 >1e.:e6lag
qeuel {n1un ' L'g
uelpnlta)[ ue6uaP Pr{u-rPsaq
6ueI uedas .rase6 ue6uB6a1
qeuel upsTdet eltT I tde1e;,
= "rt eueurp
/itiJ
'qEuPx uep
6ue{ETaq ue16eq 6uTPuTp E.rEluE ( ruoTlcTf,Jx 1 E{5s5E-@E-Sffi e6n[
uep'qpuE1 luPTPp sTlseld ueepea>1 qPTTpe!'ra1 e66u1qas uET{Turapas (ue
-unrnuad TureTp6ueu) qertreqa{ sFTquE qsl qeuel ueqeuad burputp .EAqPq
updp66ue epE ,euT{ueu Pfecas t{EuP1 ueuP{4l uE6unlTqf,ad ueTep TG ,
'au1:1ueB Troa& 'g
ltsedd
qeuPl ueue{a1 uaT sT Jaox - )t
LI '.Q'u, = vO
: e[]a>{aq V >{T?T1 PPPd
'Teluozr:roq s1re6 ue6ueP
Z / fi o'9 ? lnpnsraq rasa6 6ueptg c/ p- q
' qeupl :asa6
eures qsl;asa6 ue6ue6a1 uep 6uep1q ede:aqeq
Tpe [:a+ ue{P ' .relepuau e:ecas 'uelalTp_qs1' J T 1>[e qEuEl ubuelal uaT s r Jao{ ==x
q '9' 'rt = b et'ra>1aq v
' ( ITlte UEPPEe q' {=VuleTep Eperaq
uB{e1e{TP qeuel 'TUT ueepea{ EPP6
- ( ,, TTos uT uoTlcTJJ TEU
-ra1uT Jo a16ue,, 1 qeuel lnpns = fi'TeluozT.roq sr6 ue6uaP
Z / fr + oSD lnPns-raq 'resa6 6uePtg
' qeuea :ese6 'ueYdn{a{ ue6uap er(u'res
-aq eups fnqasral rasa6 ue6ue6al uep 6uep1q ede:eqaq Eueluedes rasa5
ue6ue6e1 tpe[:a1 ue{e 'relepuaul ExPces {Tf,E?TP qsl qeuel uesldel n1e'1
-tzz-
T >t
-Ep
-Pp
Eq-Tpr,rT'
IPI
-T:-Pd-T1
222
untuk 1 rneter Pan jang dindingIrlEus bidang gambar ) . '
P: total = +. oi. h
bekerja Pd jarak 1 /3 h dari
( tegak
dasar.
D-inding' tsb tidak memikul
'heban tambahan ( "surcharge" ).'
Dindinq berikut memikul beban tambahan ( rsurcharqer ) 'Beban tambahan berupa tanah yg ber-lereng ( sudut dengan grs horizontal )
OA = k; -[. h
;::"." ,'4.o,* -'-a '"
sif untuk tanah non kohesive (c = 0)'8.2 Tekanan tanah
Jika dinding penahan tanah berdefor-masi horizontal dan menekhn: tanah
maka timbul tekanan tanah Pasif '
q= t. h. +*#
a^,
tanpabebantambahan(''surchargg''}.-Dindinc berikut memikul beban tambahan ( rsuriharqe" ) '
Bana mentpu{rvai lereng ( hr+,Fan vertikal ) 'untuk dinding belakang yang miring, maka tanah
. t ! r --- -I--- t^^-a4-qrra,tasnyaharusdiperhitungkanberatnya'
P
yang berada dia-
i,1
i;
,]
'6ue4ef aq uPT6eq qeuel ueqeu ,..
'-ad 6ulputp ue6u1-rTula{ lnpns = 0
' ( ueqeq ue+ ueqaq ) qeuEl sPl
-e uesTdeT ue6ut-rTua{ lnpns = h'( uTTos
uT uoTlcTrJ TEu.ra?uT,, ) tlBu
e+ Tf,Ep ruPTPp ue4asa6 lnpns = fr'6ue1eTaq 6u
-TpuTp uep t{PuP1 .rasa6 lnpns = 3 .(0 c e6-req) Tsaqo{ uou qeuel {nlun 'IT1{B LIEUE1 ueue{a,tr ['3
'0 # 6ue1e1aq Eulputp uep qeuel ue>1ese6 lnpns6un>16ua1 ue{nq rplep 6uep1q edn:aq de66ue1p .zasa6 6uep1q
'sT?seTd upPpea{ ueTep Pperaq qpuel uP{?eq
-14e6ueru e66u1qas edn.r ueT{Tllrapas sBTqule qeuel ueqeuad 6utputp: qqs uBde66ue ppe quoTno3 r.roa1 pped
'quoTnoc Troafi 'J
t. */= t']"*"'M = (CgV
e6t1t6as 1 qeuel leraq 6un11q nTeT. TE{T?
--raA 6ue>1e1eq 6ulpuTp Prtqeq uedB6
-6ue lPqT{E I a ue{n1ua1 eTnu-pTnw
I
M + _ta /= Telo?d
'qsl P6T1T6as 1Praq {11 eped e[-re>1aq 6ue^,t M = (CgV
e6t1t6as 1 qpuel lp.req 6un11q nTeT
6ue>1e1eq
zr|t 'Px f = Ia: TPtTlraA qsl
6u1pu1p de66uB eTnu-eTnp
soc-'q'4
7ooa(D
Te101
t,ZZ
'le6reqcrns" Tn{Tuau {P? buTPuTP
224
6t =.tr. h cosdimana k
g. ka
a .o=2s.c,ostf,-* g\(1 +
p rotar = *.f .n, cos o. k"coslF+e).cos(o -d)
c-2 Tekanan t.anah pasif* untuk tanah non kohesif lc=0)
-tr TTn,Z':---
.kph. eos0t.
D.
p totat = + .7.h2 cos 0.k pTdkanan tanah akibat tanah terendam air.
sin ( p(l 12
cos(0 ).cos(0q)
Jika pori-pori 1apisan tanah yang"permissible " ( dapat tembus air ) berisi air maka pdda :dinding penahantanah akan bekerja tekanan hidros_tatis yang berasal dari air tanah.TeJcanan air tsb mengakibatkan tim-bulnya efek mengapung ( ,,bouyancy
effect" ) pada tanah sehingga be-rat tanah efektif yang mengakibat_kAn tekanan tanah akan berkurangseberat air yang dipindahkan olehtanah tersebut.Berat jenis tanah yang terendam a_ir =Tr.
0
qdimana k =p
tekanan air+.f.h2
tekanan tanahx- {' h
koefisien tekananberat jehis tanah
cos 0
= [.h. cos o. k
.o"2g.cos(g - 5-l fi
Dw
ft=
f'= #,effendam air. ( disebut
e
4= berat jenis airp=
trrekrir ) '
c
' b '{ LIeTppP Zq
166u11as 6u1pu1p pd etruqnrB6uad
t 'Bpe >[ppT1 t q g66u11as 6utputp eped e.duqn-rp6uad TpE[ ' :asa6 6urPTg irenTTp e[:a4aq ,, a6:euc.rns,r , TuT qe/{pqTp :eque6 pd
' ppe {ppTl Tpluoz Troq r{pup1 ueup{a1 depeqjraleduqn.re6uad e{eu :esa6 6uep1q .renT Tp e[.ra>1aq ,,e6.reqc.rns,, E>{T1'
'qpup+ upue{a? pppd .,abreqcrnsr qnrebua6( +uaurlnqe ) ue+tqure I T1p{apuau6ue^f, uee.repue{ Tf,ep TESp.raq
e[-re>1eq 6^ ueqequel upqoq
eueuTp ueleqrua[ luarulnqe pd
: qoluoc
,w/1 " ..' =b
selp qeuel upsTdeT eped ueqpqupl ueqaq-ueqeq' ( Teluoz
Tror{ s;6 ug6uap lnpnsraq ) 6uu.ra1raq 6ue.,{ sele qpuel uesrdel,i, : edn:aq ledep ueqpqrupl upqeg
zrt
'(fTe uepua-r
:.a', :
.."4X
' ( rab.:rPqcrnsr ) upr{Equrp+ uEqag'u- ^.!, \-z\- ,,!.
'A ,,-1 --
-e1 {P1 6upF )qeue1 slua[ 1e.raq A
,.z\' ,/-
9ZZ
: qeuPl E{nu qEnrpqTp -rTE e{nH
il"Iii
;
226
Pada
tas
F,.
gambar dibawah ini, "surcharge"tanah.
2c.-f
+fr/21
bekerja pada seluruh laPisan'a
Pengaruh "surcharge" Pada te-kanan tanah adalah Pada selu-ruh ketinggian dinding.
-D
- -T-.o f-I-rrTekanan tanah= aktif untuk tanah kohesif
. - 7+rzc crg ( a so+ g/21tl---zl
( 4s
Ka= ctg2(aso + g/21
1 - sin 01 + sin g
'. h . ctg' ( nso +fl / 21
pada suatu kedalaman z :
A" =tr.z.ctg?(qSo + fr/21 - 2.c. ctg(Aso +'fr/21 .
p- = n f (t.z.ctg2(qro + g/21 - 2-c.ctg(aso + g/21a "J
= + .t.h2 ctg?(aso + fr/zl -2.c.ctgt (aso + U/zl +
I .dz
2"2T
pada kedalaman z :
C" = tr.r. ctg2 ( 45o +
dimana k" = ct9
ctg(Aso + fi/21
1-sinE=f+-ffi'fr/21 - 2 c
2(qso * g/21
/ .n. ctg 2 (aso+g/21
o. k'a
VXTNVXUN eceq epue
6u1pu1p upeuecua:adua+ seluldes P.recas
' elTao T-ras I svoNod xINXgtr' uEp HVNV,I
ue{qeTTs eduselaI qTqaT {nlun 'qeuel ueqeuad
{n?un eun6.:aq 1e6ues 6ue^[ qeuel ueue{a1 6ue1
ue4sela[ Tue{ TuT I NO,lUg ISXnU&SNOX eped.H
(z/fl+osvl zh'ra$H
/ urs - tf,uTs+[
(Z/fr + oe?lrb,
U=
d=X
(z/ fi
(z/ fr+ oE? ) 6+ "r
8lp1 ueuP{a& '9
0uTs+[ '= (Z/fr + o9?lrb1" = px 3 uBlPlPc6urs_t (rq z) ^,! +
+ o9?) 6+c'c'z-(z/fr+o9?lrbl,c, Itu;')"!* L;'.Q) -'!
('U(21 z z ueuETPPa{ PPed
='t' Z'l' ,9"'' t-u
'4
,&rd
-rTe
JTl{effe=Y7,4
(z/ fr+ oEr ) 6+c' c' z -.5f'l
zp ( e>['h
'qeuel -tTE
+ tz/f,+gs7 ) 61c'c'z- (z/fl+oi?) z61c'
u epe'
nOlz
)/*
e{n
z'!etTf
=ed
t 77.-
229
1 ..:l
VII. t ,ffiukuran untut t*pe dlndipq penahan .tanah.
wa11'! .20cm (b ]-asanya
30cm).
=1 ,2Q
dpt dibmiring. hfr
s/d
s/d
h6
hTG
@ "counterfort retaining waII". I
1nf.
\ornin
B=0 ,4h h ---
s/d o ,iR l?s
{
han tanah.
l . faktor keamanan unluk stabilitas -
Keterangan faktor keamanan utk tanahkohesif non kohesif
Gul ing("ctertrarringn)
geser( "sliding" )
2
2
115
1r5
r(
\ \cm s/d 30
,3h s/d 0 ,5h
if
Jika tekanan tanah pasif diabaikan didalam perhitunganfaktor keamanan, maka fak-tor keamanan dapat diambil= 115 tetapi jika dimasukkan
dalam perhitungan, maka FrK
- 4o
B=0 r,4h s/d 0 ,7h
i, '.
I
['
^d'"ut"u fuqT{p rtrprpp' (rluauroru 6uT+srseJ,.) ueqeuad ueuol^l'{'
q4=*u rpqT{E rIETppE (v {T1T1 )
';";;, ' nu* 6uT Tn6 uauow . .rpspp 1e1ed ,. s_e TJEp a >1e-re I eped e t -ra>1aq U
' ( U= ) qpupl rs{pa-r = Tplol dM + =M =
nd = ( Tp>tr1-ron qprp ) re+o+ d a
M = Te1o1 ueqeued 6ulpurp 1P-reqSM = pcqp qpupl fp_raq AU--d uep -d
?r+r rp.rnrp td
. ( " luTo[ uoTsuedxa,, ;
-un ue6unqups Epp snteq E{pru.rulg, Tr{TqaJ.9_q 6uTpuTpr PtlIPS,
rsueds>1a {n16up t upd pltT1,-v
'gs1au {+n uen[n1raq 6d-Bd up{n{pTTp snrpq
PIIIPS
ue'ufrqups eped TpeF:e+ -6ued
_zase6 edp6 Tn{Trx , uepeq/Eurpurp upp -resep 1e1ad uenual;ed p.p(,, uoTl{nrlsuoc luTof,, ; 6unque^duad. Ts{nu4,suox .
E' I{puel uT z T 6un>1np edep epedr:rep T Tca{ qTq
-oT snrer.{ 6ugpulp
6ZZ
ue6ue1n1
.respp qenpqTp qeuel eped et_ra>1aq 6upd upue{ofr,
2?O
Stabilitas terhadap guling ( "overturning" ).
gaya penahan thd geser
Untuk jelasnya kami persilahkan anda
VII. I ; 8 Merencanakan "cantilever walI".
s.F ("safety factor't) !& ffi 2rruAkibat, gaya Ph maka dlinding akan bergeser ('sliding"), taPi ditahanoldh gesekan antara tanah dan pelat dasar.
Fr = R tgr' + cr.B + Pp
tglt diambil = tSl s/d O,67 xg0
R = (Pu.+ Ws + Wc)
c' = 0,5 c s/d O ,t5 .ikohesi tanah.c
B = lebar pelat dasar
. Pp = tekanan tanah Pasif
s.E seser= ft )r,ujika S;f geser kurang dari 1'5 maka perlu dipasang"key"
. pelat dasar diperbesar ukurannya.
lihat contoh soal.
l'key "
' ( urc 0?, urnuT uTur )91= (
"r= ) s'lp uolaqJ Jtl- ( 91=
) qeateq T eqa?e6:eq qaToradlp uete
ralau[=qpuprurp\
JJ"1'q' 6'o nq,)ffi='2
: easa6 ue6ue6a1. ]Ja-+=
uepeq e.-p lod eped JT.?taJa Tpqol( OZ' t=0 ) 8f / t=TTqruE upppq ue6ulrTurax
fr rsoc - )vrsoc r
'lllcE, L =ez(ueseTquolaq lnuTTas
z'u
tld upp nd splp rp.rnrp d+ Tosoc
,rsoc = lf pueurp
(8r/Ll rr
lnurT Tes +
'JJtf-@
t9* 2
ftl1
t/q'Ha = (respp 1e1ad upp uep-Pq uPselPqrad ) P-P uB6uolod eped Tpe [:aa 6uBr{ .unurxplu uoruow
yuTsd=^dl;asocd=I{d
: ( 6tuols,, ) .uppeq Tpqal ue{eupcua.raff .V
:
uTtu
)osoc. 14. t! =!:
:
iralau [ = 6u1pu1p 6ue[ued TTqE\t
*r:c
- )o soc
LT,Z -
a
uPt
"tri
- 232
B. iheck dtaFiLitas dindinq'
li-gl
PnIa tT'
'/3
.kta
P total
t = tebal pelat dasar.p
diurai atas Pr, = P cosa( dan Pu = p sin a(
Momen pendhan ( "resisting moment" ) thd ttk A.
= w1 .*1 + w2.*2 + w3.*3 + wa .1/2 B + Pu.B
dimana wt = berat tanah yang menekan pelat dasar.
wz s/d Wa = berat sendiri dinding '*
1 ,*z . .: . . x4 adalah jarak lokasi ker ja gaya W th''l tt'l6 A'
gaYa vertikal = Wt + WZ + W: + Wa * Pu
Momen gul,ing ( "overturning moment" ) = Ph ' 3:Faktor keamlnan untuk qulinq =
Mpenahan n.rJ" ), ,u
.Ih't =h
JikaperluGaya
+ t + b sirtr(dimanap
= +. /.n* . h'2
F.K untuk guling kurang dari 1 ,5 maka ukuran pelat dasar
diperbesar.penahan thd geseran pelat dasar = Ft
F, = R.tgfit + ct.B + rpdimana R, = gaya vertikal
tgfi' biasanya diambil sebesar 0,67 Lgfi s/d Egl
g = sudut gesekan dalam tanah (diketahui dq
tanya dari test di lab. MEK TAN ) '
= 0167 c
= kohesi (diketahui nilainYa).= rekanan tahah pasif = +. tr. h3 . *n
ctc
'TPos qoluoc leqTT
'uepeq eped uebuolod6uB1 'sPlPTp esla{s
@
ue{qeTTs e[use1a[ {n1un 'g'Etups
{epT1 eduleqa+
1e6u11p nlradde11es
eped +ds
eped Prurt{Pq qPTeT
o uep ,{ rue:6e1o
xPul
: uEpPq urburTnuad uPtPuPcuaral{ 'o
uTru
'resep 1e1eduP{ruEcuaraH 'c
' ( Tsepuod ue
Suap efrue.rBc ulapT ue6uelnuad
(# t)#=urtub(g+rlTE=xeub *a9trrd
' (:asa6 6uep1q sn.r
-nT 1e6a1 1 1e1ad 6uB[ued = T
(#rt)T=b
-Eq resep 1e1ad ueJn{n P{PttI- r{_
5't(snreq uPP # =.{
tez
' ( ue{repuTt{ TP
snreq ) .resep 1eT
5d pd {Tf,e1 ue6ue6
-a1 TnquTl e.(u11;e xVGf ,l E{T I'srua[as' uE
- e6a1 TpP',
;):;:Tl;lul^,lll:x g Z/J =a
TP>tT1.raq eute6 tI-tL
',,fio{,, 6uesedgp nplp resaqradTp snr
E, L TrEp 6uernrl upuptueo{ .rolteJ EtTf'
( oEuTpTTs,, ) rasa6 {ln upueueat rol{Pg
uebuelnuacl
TPlo1a
-234-.
lftffirl Perkiraan ukuran pelat dasar dindlaq penahan tanah aqar pada
pelat dasar tsb bekeria teqanqan se'lenia (tekani.
tapadiabai
y h.k,rumus perkiraan ini -ada dLrggapan sbbDalam
BeratP
-. berat jenis tanah dan beton (dinding )
dan dianggap sama =tbagtan tapak ( "toe" ) diabaikantotal = w =Tt1-k).8.h
= + .t.n, .ka
a
pada bagian ba-
berjarak x dari ujung
1 - sin0'1+sin0
resultante P dan hl kita namakan R.
R memotong pelat dasar di tit,ik A yang
=+{nz
t"pelat.
2'o = o
P.h/3 = (ks + + B- r) .h( 1-k).B
=7f-T (g + kB 2 x).h
*.t.n' +*i*fr . + = {*tsitr + k) 2 x).h( 1 k ).8
l?l!.?-i*E)= .,= = B n + k) z x3(1+sLnr)(1-k)Bnamakan ; =d.B
= (l-klg2(1 + k - 2d, ) .......... (*).jika disyaratkan bahwa tegangan tarik tak boleh ada maka
x (3jika i = B/3, bentuk diagram tegangan :
; (gZS, bentuk diagram tegangan 3
t
WfiP
I
}}"
t
tt
z1
'ZI b ol/ I = 1.r166u1d 6ueluaq Inlun) uB6uPdBf n'uTEt-6upA aI nles lrolralunoc lere[ = T
( unuTxrlu 6u
-Br[ TEITU rTqurE ) TeluozTroq qEuPl r{BuP{el - b EUeutTp
za"ti s TTq,::;ti';:: l;:1,'* 6ueluaq {nlun) ur'uedul n. ( .xtrorrelunoc" r{pTpT ertuue:1e1ar1red} uElElalrad ed
-iraqaq sglpTp snrauau 1e1ad lu6eqes ue{putcuarfp u"pEq tetad' ( "QPrB TEcTlIeAr )
tu, - ug'znElt
r{g'0 P/8 qE ' 0
'EuTPrrr
qt'0/$*q?'o
urco z
' sTtuoqg{e qTqeT , f IEA Xrorralunocr urtups
Efq ,ug"TqTqeTeur e.{u16$rrT"1 6uer{ qsuEl ueqeuad 6u1pu1p In?uO
e / J .r ITqUPTP tt
q L, A g/E r{ ?'O = f TqutETp g etrueerlq
r{rg
-r
p u3pEq uiltuEcEersta
U: ,!.L
le /z-)t+ [ ) eg ( {- I }
. ( r ) srad aI uelTsnlTtqnsTp , E/ I
,. f E$+3*t
- rC-- e /g
-EztlI
-EtZ-
'&
t- 236
I.tErenqaqgEan taeah ( 19oe" I dan. *umit ( "hedlf ) .
ilika pada tapak (pblat da"bar)' tidak ada "coUriterfor't'r maka, ''Si-
rencanakan sepertiSelat p.ondasi.
sbg pelat
' Direncanakan seb'agai balokJarak counterfort Yang satu
Tetapinakah
karE-na
kantilever.ke yang Ia j-n - lrr
:
3lIr
"r .i
'1 'n
, i '! I ". .: "..., q+l::
Ivromen max = ( *,d.n'k.) (1/3, h).tulangan dapat dicar j- . "
gleset' , D* = ,, /2-t.h2.]..).L '
, r ,. kapena "counterf ort " beruPaDb"ton=D* Stqg
4bu r..o. (Z*b,,
.t.. ".
,
aaaa.
Ilubunqan "counterfort" pada badan :
,:
balok voute maka
potongan cL -ct
J ika',:
batas )
'f Bos qoluoc lar{T1aur ppup uptqells.rad TUE{ etrusele$ qTqeT {nlun'uetnlualTp ledep ue6ue1n1 6qs
.."' = t-t ueEuolod eped xpu n
'lrorralunoc 6uesedlp e6n[ ,,eolo ue16eq eped 6ueperl-Euepey' (.eol. ) up(rap UBTDBq ppBd l.IolralunoJ
il, =6uer16uas
np.Jntle ,.-;, E =v
T'Itl =,1TnltTdTp bA e,{pg
f = UTET 6uertM- a{ n?rs lrolralunoc {ErPcM =E[ra:1aq 6ued r1euel lerag
'i
relau [ = restp ittqaT TTquV
{
eo- =r-
'( re83{r I rBsrp lBled EpEo rlrorralunocr uBDunqnE
- 238
rr.1.11. U,mutrr.
Ada 3 tipe bak penyirnpanan air yaitu 3
bak yang terletak diatas tanah dan menumpu
bak yang terletak diatas tanah dan'dilumpg,pada tanaholeh .koIom2
'-bak,!an$terta'nar.'[da1amtanah.','Kami akan j elaskan secara sepintas cara mer.encgnakan bak a-.-ir mengingat dElam merencanakan gedung , rumaf;,' ting,gal , dI l ,
pasti ada bak :penampung air ( spt pada bangurtS4 yang mempu-
nyai "basement'r harus ada bak penampungan alr'Vang tertanam,'dalam tanah ) . atau adanya kolom renang, yang' harus anda regcanakan.
vrl .1 .11.2 PeTsvaratan vano trrs kita penuni utl< per dk-t
'l . Beton harus kedap air ( air tak boleh meretnbes keluar ) .jumlah semen yang dipakai i Untuk 1 . *? adukan ,,,bgLon,, ',
tl4nus
lebih dari 300 kg.Pada perencdnEElrr- bak air, tidakdan tegangan tarik yang terjadida yang,. diizinkan.(keretakan dapat terjadi akibat
susut ("shrinkage").expansi pada beton ( pemuaian ) .
kontraksi akibat temPeratur.Diameter tulangan berukutran kecil lebih baik dipakai da-ri pd diameter yg besar" karena akan mengurangi keretakan.Resiko retak dapat diatasi dengan jalah'mengurahgi-perkuatan pada struktur d,gpqan memasang la.pisan.geser ( "s.Ii-
,' -.\,.ding'')berupa''rubberpadl';*i:]....'r-..;"g,,i.,;.,,.''.'-..,r,li.Jika pan jang bak cukup besar', maka harus diblgi atas 'be-
berapa segmen ( bagian ) dimana ma.s ing-mas ing bagiah,'dii-Sam
bung dengan "movement joint" ( sambungan yg dpt bergerak ) .Jika bak digunakan. untuk menyimpan cairan yang panas TITtEt;
ka perbedaan temperatur bagian dalam dan luar harus di-perhitungkEn.Koefisien pemuaiano^
Koef isien :1il":'il-l
"::::l-:::;1",ta,, permuraan
= 200 . 1 0-5 untuk penyusutan pada saat beton
rt
3.
'.. ,{ ": '" . rl
boleh terjadi keretakan,harus lebih kecil daripa'':
,
:
akibat perubahan te*p"t.tur:i i . io-6
4.
5.
5.
mengeras .
r acrJ.rns 6u1p11s,,le6uqas ,, ppd .raqqnr,,
,, punoduoc 6uT TPas AuTo" TsTT
n{Ts B[eq ue6unques>1era6red e^fiuepe
6uep1q eped uE
lseleouaur tnlun @
" f,PQ .f A1Prulr
,,f,oTTTJ ?uTo[,
'uolaq UETPp
-{ex-.}uo{ uz(uepel sueds>1a n -pJp T s
Tsele6uau tnlun @z'e
,,punodwoc 6uTTpasr
' sague.rau rTe qeEacuau
/ l6ueluTrau {n1un eun6:aqbfr, , >111se1d ,1a;er1 , e6eqtual
'rTsTueAfE6 pIpq TrEp lenq?}F
Eaq 5^d 6unques rnle[ r{eTppe @' 1 er{u1se1e6uau1 . Ts{prl
E'o>t eduepr' ue>1de1s:adulau tn?un @ ue6ueln'(.luTo! uoTlcJluocr ) Tsterluot UEbunqurs .V
'(.luTol luauaAou., {rrobraq ledep 6Uef ue5unqrtrpsi .r
aa€'Il'['rr
ire6uapl T=TTpe[Bq 1e1ad
ac) -
240
b. Sanbungan kons!gg!e-1 ( "constquction ioi.n" ) .-= ,,,pacta samh,ungan ini r' trk:nitefr ter'jadi pergerakan antara bagian-bagian yang disambung.
c. Sambunqan terbuka ( "iemp-orarv open ioint:. ) .Kadang-kadang antara dua bagian dinding, ada ruang kosong ( "gap" )
vs akan dii=-s:lgil iSEIil ;iTSl ju3*ii' arau beron -
" sealing adukan semen"+'pasar
bajacompoun
vrr.1.11.'4 Beberapa hal yanq ,perlu .diperhatikan dlm perencanaan bak.
a. l,antii bak (untuk bak yang menumpu diatas tanah)
- jika luas lantai bak.besar sekali, maka lantai dicor bertahapdengan ukuran minimal 4,5m x 4,5m (berupa rranel-panel ). Dan an
tara panel yang satu dengan yang lain dipakai sambungan kon-troksi atau ekspansi.
- sebagai lantai kerja dicor beton tebal minimum 7r5 cm, Ialu c[lapisi aspql . Beton untuk lantai kerja dapat dipakai mutu
181. ( flbk = 100 kglcm').
.
br Lantai Uah Lurrtuk bak vq ditunianb.1 Lantai bak yang ditumpu pada keempat .sisinya oleh balok- balok
penumpu irrti," iit"n""n.xun ""s.g.-i pelat lantai spt lantai ge-
dung, tetapi bgban yang dipikul oleh lantai tsb adalah beratseniliri + berat air.
. b,2 Jika lantai bak menyatu dengan. dindingya (hubungannya kaku="r1gid") maka momen pada sambungan tsb harus diperhiiungkan.
tulangan menerus
-ad EpPd
e[re>1aq 6ue^[ upqag 'rpues ednraq qpr{pq upT6eq 'seqaq nele Tpuasednreq sBlE. ueT6eq 'uelldeE r{aTo ndunlrp TpluozTf,oq l{pre rueTpp
eupurp 'qele Z 1e1ad le6eqas up{pupcua.rTp {pq 6utpulp-6uTpuTq
' TPIUOZ TIOI{ t{E.rE
Tn{Turaur {n1un 6uesedlp
TPluoz Tf,or{ ue6uelnuad
'0 [ Toqel rgd )
ueTPp ) tnqurT+
,.-]]]*,'rl*ffi uebunqwes 7
uauou
snrPq
'6u1pu1p a{ 6ulputp n>{p{ ue6unques eped
(z' v'uauou apqT{e uTzT {Tf,e1 ue6ue6a1
' ( uolaq6uer{' uauou lpqT{e {TrP1 ue6ue6a1
' ( uolaq luPT
Ep ) TnquTl 6uB,{ }tTE+ er{e6 lpqT{E )tTrp1 ue6ue6e1
' (z'?' o L Taqpl rgd )
TETXE {TrP?lBqT{P uT zT >{Tf,P1 uB6ue6a1
o=b
=b
="p
_s9 n.*,e.o sAo
! r\.- t
: 1e:eIs lqnu-aueur snrpq ueuout + Tp?uozTrol{, r1tre1 e,{e6 ?PqT{e r11re1 ue6ue6a,tr
' Teluoz Tr6q qere uatuour eped ut:{r{pqulp1r. p sn:eq 6ulpuTp .resPp Eped 'rTP uP
ue1al lEqT{e up{Tre1 qalo uetqtqaslp 6ueI leluozTroq r11rea er{eg -'TeluozTrol{ uEp
Tr{T1ra^ qpre uauour ue{TnquTueu, 1eq 6u1pu1p'eped r1e uEue{a;L -.a
' uer16un11qrad1P sn:eq qeu
-e1 lEqr{E TeluozTroq upup{al p{€u qeuEl ,=1"p ,p"i-q {Eq P{T1' -ttq EuTpuTp PpPal t[ra{aq but^ uEut:(a[ 'p
' qs1 .6u1pEfp pd ue{plera>l 1pe[:a1 up:{p te1 'Tsupds{a nElP Ts{er1uo)t TtuPT
-p6uaru qs1 6u1pu1p e11[ 're6e uS' L utnul.TxEur >1e:e[:aq ,,1u1o[ lrlatu-c16ru 1e:1t1raA,, 6upspdlp nlrad e>1eru ,166u11 dnrlnc 1eq 5u1pu1p pltTf -
>tPq TelueluroITPT T S.
0. = {pq resPp PpPd uauow{eq 6u1pu1p
'{Pq reluPT ue6uap qPsrd.ral TsuPds{a uep rs-{p.r1uo>[ rlue.f,B6uau qsl Eutpulp .re6p ', ( ,PPd raqqn.r,, ) 1a]e{ uPsrd
E1 ednraq ,luTo[ 6u1p11s,, 6uesedTp e^fiuespTq '6u1pu1p .resep eped
'm'cf TP lEraq ) p[ps Te{TuaA uEqaq lPqT}te efueq uP]t
e{Ptu {eq 6u1pu1p up6uap n?P^uaul {Pq TPlueT e{Tf' ( TirTpuas lE.raq +
-EuPcuarTp TelueT
LTZ
; 242
holr^a n lat adalah beban segitLgr.
jepit
sendif. Penutuo bak
ilika dibuat dari beton, maka harus diperhitungkan sebagai berlkut:- jika menyatu dengan dinding bak, rnaka perlu dipasing"movemeht
joint" pada pelat penutup tsb.- jika pelat penutup tiilak menyatu dengan dinding bak ( memakai
"sliding joint" ) maka "movement joint" tak perlu dipasang.' - pelat penutup direncanakan memikul berat sendi.ri, beban hldlup,beban-beban peralatan yang berida diatasnya dan gaya tekan keatas pada pelat yang terjadi akibat tekanan gas.
g. Tulangan minimum.
- Tulangan minimum pada irelat lantai dasar, dinding, pelat.pengtup dalam. 2 arah yang saling tegak lurus sbb :
tebal = l0cm ----) A = 0,3t x luas penampang beton (=Or3t.b.h. )
1Ocm (tebal (ascm ----) i : E:rl .b.h. ralu berkurans menjadi
. O,2N.b.ht ( untuk t = 45 cm ).t> 45 cm ----) A = 0rZt.b.ht
h. Selimut beton minimum.
Minimum 25 mm atau diameter tulangan (anbil terbesar):
lvrr.t.ll.5lBak berbentuk silinder dinana dasar bak tidak menvatu dena
Gava garik nglinqkaf ( "h,qntensign') .
uhtuk tinggi tangki :tr I ,rn
T tr *.{.n. d
Luas trrlangan melinqkar ya-nS",.dip$,rlgkan
untuk tinggi tangki = 1
[- +* -4n'u ( cm2 )ua Categ. tarik izin baJa.tr^=
t'rubber
ue{P x ueT66uTla{ epedg'12
p':0
. ue{e1a1.rad TSTpuo}t eped 6uil1ue6ra1 rer16u.11 e,,{e6 + TpluozTroq.rasa6 qaTo Tn{TdTp 6uer{ B,{e6 resag
. rel6uTf au Br(e6 + Teluoz Tror{resa6 B^fip6 e^f,u1nquT1 qeTppe ,:resep
1e1ad ue6uap nle.duau 6u1pu1p leqTltg
nple a f? =p'10
(p Z/Ll .>[T;P?A = d
.
^ Jpsaqas ueqpqn:ed 1ue1e6u
t>16uetr Trp[-l:e[ , .rTB ueup{e1 lpqT{VyT-rPf.
= -''---Z = )tTrP1 ue6ue6a1
1 Z ---^t ----? --:-: = >{TJpl ue6ue6a1
p'*J){
bB:B[ pped
D
ft=beu
7rF'o =
ro
.re>16u1Taur e.fie6 (----.rpspp Tf,Ep x(x-r.r) = b +
b = TeluozT.roq rase6 ue6ue6a1 +
= re46urTau {Tf,p1 ue6ue6a1 r{eTo Tn{TdT1 6ued
lF{rl "
p nltTpUTP rPsrp PueurTp
f.-t{'q'gE'o = v
(ZvZ ueueTeq 1pr{Tr ) Ts{nparTp snleq uc o r(+ {nlun nTpTlq'q'tt'o = [ (--:- ruc or]+ {nlun
-T6Bq tnlun EuB66uer qrqar sn.rpr{ e6nt :e>16u1Tatu':::::;:#-:r:;6u1pu1p splp 6un[n a{ rpspp rrep 11ca6uaru ,, uoTsual dooq,, Bua]ex
re>16utTau ue6ue1n1 sEnT = V
T):-}=u *i-Y-
=ogffi==b ru [ - TTque 6u1pu1p 166ur1 '1 = 6rip1u1p Tpqafi
9'L ['J'rr^
244
q =tth-x) -%=/tn_*l _ H
Kita tahu bahwa (dari Mekanika Teknik).-2. -2 -4
M=Erp.;D=Er=;9=Er=ctx ox clx
Harga EI tsb akan berubah jika pengaruh tahanan lateral ( "lateralrestraint " ) dimasukka.n dalam perhitungan o
3
E13Jadi q[ =
EI berubah men jadi ' t--- u (A= poisson ratio )
-4cty12(1-&2 I
1z(1-/rt') a*4dimana q = %
;4d- lz(t-rfr) dx4q1" + +a(r-Az )v - tz Trn-xt (t-/a2 |
dx4 t2a2 E t3
Ct= 4 tz(1-.,4zlt2a2
&- + 4 d4 tz Y$-xltr-/tz,dx4
' =r4 Y E t3Persamaan diatas adalah persamaan differensial.Jawab persamaan tsb (lihat ANALISA II JILID 2 seri DELTA).adalah : y = "X*{c, coso(x + 9Z sin,;.xl .
vx) + Xt.-f;rfl'. .. . . .' (*)+ "-*,*,arcosXx+casin rJ
harga dai"*bil = 0,2Faktor c
1 ,czrc3,c 4 tergantung pada kondisi perretakan .
d. ;ika perietikannva iepit, atas bebas.(hubungan dinding dan pelat dasar kaku sekali).
pada x = h ----) M = 0, jadi
o,"oufl n Er $=oD=Q'"]"il=
j
,I,I.t
{I:
"J*:
-"JtI;
=-tr--n1-w.Jctx,Fjepl,t x = 0 ----) e = 0, jadi
*,'EI =s
x = 0 ----) , 0, jadi EI y=0(defleksi ) .
harga-harga diatas disubstitusikan kepersamaan ( * )
akan diperoreh ct s/d c4, sehingga 3 y = f(x)dipgroreh untukberbagai ni-Iai x.
.lJ'E rZJ, L) qaT6rad1p ( + ) s.rad eped selpq e6:equeltTsnlT?sqnsuau ue6uap e66utqas
0=i$ rs Tpe[ 'o=o
d fg Tpe[ ' 'O=l ,0=x
zra = iE rr*
TPEI 'Zg=Lg'r{=xI rE Tpe [ ' 0=r( , rl=x eppd
: ( iruo-TlTpuoc ,{repunoe,, ) sp?pq .1e.re.[g
'dn+nlPTnraq sElE , lTr^uuEte1a1:PtTf 'c(|sz p/s 9v7, NVWVTVH VOVd NVXUrdwvr rwvx rssv,t)
_;rad ue{eun66uau ue6uep ;:::Jil"1}Jr;:_,;l,lu5::; 'rp>16urrau >{TJel Br{e6 'lasa6 , uoruou e6:eq ue{nluoueuPpuE up{r{ppnuau tn?un Br(ureuaqas ,3rSrde6;Br{ TaqE[ +
'up{TaqplTp I e6:eq( z{''. ;li.;o:-;n;:::,;^"'
+
( e r{', ) g = Eulpurp eped ef;aqaq 6uefi uaruow +
up{Taqe?Tp d uaTsTJao}t(r{-.g )d = 'g,u=urTp
dooq,, 1 rel6urTau )ITJE1 er{eg* .@e66urqes qoTotadlp (x)J = { e66urQas
tJ,tJrZ1, LX qalo.radup{Tsnlrlsqnsrp qpl sp?eq e6:eq_e6reg
. ( TstaTJep)'0 = d rtr Tpef ,0=u( ,0 = x'o = iE ., TpPt 'o=g ,o = x
'0 =
'0 =' ,, PPd
I0=
xp7F
0=
7, .'J = ( ,, uoTsuo1 p'TrpcTp ledep 'J uep ,!
-Tp ue{P ,(+) upPues:ada>1
XD eB
F6 rg TPPI '0=o , q =t' XD aL-
fu rtr Tpe[ 'o=w 'r{ = x eped'raqqn-r,, 6upsedTp 'lesep 1e1ad upp Eurpurp uenuelred pped )
'q-p
^['csiE' = -rJBueuTp ,trecrp ledep :! + (x-r{)g = ,.0 e6.req
- qb7
r 246
Tabel' 1
Koefisien p untuk kondisi perletAkan jepitdan atas bebas, tangki silinder.
Gaya tarik melingkar( "hoop tension" )
T = p. f .h .d/2 kg/m'Tanda + mbnYatakan tarik.
Tambahan untuk harga n2 /at antara 20 s/d 56.
lFxJE
h2m
Koefisien punt,uk harga x = ..0 .75h 0.80h 0.85h 0.90h .95h
20
24
32
40
48
56
+0.716+0 .7 46
+0.782+0.800+0 .791+0.763
+0.654+o .7 02
+0.768+0.805+0.828+0.838
+0.520+A.577
+0.653+0.73 1
+0 .785+0 .824
+0.325+0.372+0.459+0;530+0.593+0 .536
+0.:l 15
+0. 137
+0.182+0.217+0 . 254
+0 .285
Catatan 3 Tabel ini kami ambilkan dari tabe L 17 .2A "Concrete structures " by V. N Vazirani .
.2nffi
Koefisien p untuk x aao
0.0 h 0rlh 0 .2h 0.3h 0 .4h 0.5h 0.5h 0.7h 0.8h 0 ,9h
0.40.8
1.2
1.5
2;0
3.0
4.0
5.0
5.0
8.0
10.0
12.0
14.0
15.0
+0.149+0.263
+0.283
+{.265
{{.234+0.1 34
+0.057
{{.025
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{.011{.011-0.005
-o.002
0.000
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+0.41 3
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{{.533
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io.6g7
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+0.582
+o.ool I
+0.01 i
+o.o1ol
+0.0231
+o.o3o I
+o.osl I
+0.072 I
I
+0,090 |
+0.1 12 I
l
+0.1 50
+-0.172
+0.211
+0.241
+0.265
IF$::
uP{P EpuE nTeT
, x T SPT.rpA
( 1' P I / Ztl tuoTolt eped
reEeq.raq {n?un S TETTu uerlledepT.rpc , ( 1. pl / ,r4, e6.:eq nTnp Tf,eo C
9S
It0'ztiZOZ
9s p/s 0Z = (1'pl/{q e6;eq {n1un uEqBquE;
0'gl0'?[0'z!,0' 0l0'g0'g0's0'?0'E0'z9'lZ'L8'0,'0
uZ'
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Tpe[ra1 ]tTf,p1 up{ele.,tuau @ epuel,ur/tu6>1 Er{.0
.3 - uoruow
.repurfTs r>16ue1'seqeq sP?e UPP
up{p1a1.:ed TSTpuo{ {lnJ r.lsrJeoX d=-'lTda I
tz00'-9200'-
zt00'-0r00'-
e 900'-
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t [ 00'-
8 [ 00'-
7000'+
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9000'+
1.000'+
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S000'+
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L 000'+
9000'+
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Z [ 00'+
? [ 00'+
0000'+
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2000'+
1000'+
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""' - x {nlun IUaTSTJAOX
6100'-
0600'-
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9? [ 0'-[g [ 0'-ZZZO'-
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zz00'-
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6900'+
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6600'+
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1000'+
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8000:+
9 [ 00'+
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9L00'+
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t900'+
I200'+
0000'
0000'
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[ 000'+
e 000'+
8000'+
S [ 00'+
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SE00'+
[ ?00'+
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lE00'+
7 [ 00'+
0000
0000'
0000'
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000'0+
1.000'+
2000'+
8000'+
9000'+
0 [ 00'+
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#.
fl'
li LVZ
'Z Taqe,tr
248
Gaya geser = t. t n
tanda @ menyatakan gaya geser berarah
kedalam.
Tabel 3.
Koef isien ts untuk 'kondisi perletakan iepdan atas bebas, tangki silinder
h2Ar 0 r4 0r8 1 ,2 1.6 2.0 3.0 4.0 5.0 6.0 8.0
E +0.436 +0.37 4 +0.339 l*0.317 | +0.299 +0.262 +.0235 +0.21 3 +0. I 97 +0.'17 4
ffEr 10.0 12 .0 14 .0 16.0 20.0 24.0 32.0 40.0 4g .0 s6 .0
e +0.158 +0.145 +0.135 | +O-.'.127 +0.1 1 4 +0.1 02 +0.089 +0.080 +0.072 0;067
Tabe1 4.Koefisien untuk kondisi perletakan sendi
dan atas bebas, tangki silinder.Gaya tarik melingkar (=T)
Tanda @ m"tyatakan tarik.,T = p. 6 .h.d/z kg/m' t
.2nAE
aKoefisien B untuk x,=0.0h 0.1h 0.2h 0.3h 0.4h 0.5h 0.6h 0 .7h 0.gh 0.gh
10.4
lo.tlt.zl, ..lr.oi3.0la.olu.ol..olr.olo-ol2.oln.olr.o
+0 .47 4
+a .423+0.350+o .27 1
+0 .205+0.074
-0.017-0 .008
-0.01 1
-0 .015
-0.008-0.0020 .0000+0.002
F0.440
r0.402F0 . i55r0.303r0.250{.0 , 179
+0.137|0.114+0.103+0.095
r0.095r0.097r0 .098r0. 100
+0.395+0.38 1
+0.361
+0.341+0.32 1
+0.281
+0.253+0.235+0 .223+0.208+0.200+0. 197
+0. 197
+0.198
+0
+0
+0
+0
+0
+0
+0
+0
+0
+0
+0
+0
+0
+0
35d,ud,tl,u1371
,,J35J
,u4341
,r)31|301
,rl2sl
a
a
a
+0.308+0.330+0.358+0.385+0.41 1
+0 .449+0 .469+0.469+0 .463+0.443+0 .428+0 . 411
+0.408+0.403
0.2644.2970.3430.3850.4340.5060.5450.5620 .5560.5640 .552
0.5410 .53'l0 . s21
+0.21q+a.24d+o.3od+0. reJ
I
+0 .41 9
+0. s 1JI
+0.579I
+0 .61 i+0.531+0 .65li+0. oot+o.oGd
+0.659+0 .65q
+0
+0
+0
+0
+0
+0
+0
+0
+0
+0
l+oI
l+0
l*o
[*o
. 165
.202
.256
.314
.369
.47g
.553
.605
.6 43
.697
.730
.750
.7 6T
.764
If'
0.1110. 145
0 . 185
0 .2330.2800.3750 .4470.5030.5470 .6210 .578a .'7 2A
0.7520.776
+0.05+0 .07+0 .09+0.12+0.15+0 .2'l+0.25+0 .29.
+0.32+0.38+0.43+0.47+0.51+0 .54
h2at
Koefisienpuntuk x = .,.0 .75h 0.80h 0.85h 0.90h I o. esh
202t32404856
+0.812+0.816+0.814+0 .802+a .7 91+0 .781
+0.8.l7+0.839+0 .861+0.865+0.864+0.859
+0.756+0.793+0.847+0.880+0.900+0.911
r0.603r0.647r0 .7 21r0.778r0 .820|-0.852
+0.0344+0 .0 377+0 ; 436+0.493+4.527+0.563
890'0+EL0'0+6L'0+180'0+960'0+?0'g I0'yLO'ZL0'0 L.0'8lp
zr4
0 [ ['0+LZL. O+l8 L'0+85 ['0+58 L'0+tlZ'0'+0ZZ'0+lttZ'0'+gVZ'0+3
0'g0's0'170't0'z9'LZ'L8'0,'0g
uZ'
urpT ppe>[ qp.rp.raq .rasa6 pXp6 erqqeq
-1ffi@,[
= x >[n1un3 ualsrJ:eox
uE{PlPr{uaul +
'9'? = -rASa6
PpuPJ,
edeg
00000'000'+
0000'+0000'+S000'+8000'+
9S870?ZI,VZOZ
?pu
Z'
0'9 [
0'? [
O'ZL
0'0 [
0'g
0'9
0's
0'?
0'E
0'z
9'L
Z'L
8'0
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ueuofi
..renT bulpurpTpP[:a+ ]tTre? ue{pletruaur @
. TUEJTZ
En fq rsejrnlcn-r?S a1
a.rcuoc,, g. t. L I TOqel-Ip TTqureTp TuT TaqP.t
0
0
0
0
0
0
0
0
0
0
0
0
0
0
8000'+0 [ 00'+[ [ 00'+e 100'+I [ 00'+0200'+
1000'+8000'+I L 00'+7 100'+0200'+?200'+
t000'+?000'+9000'+6000'+S L 00'+0200'+
00000'[ 000'+t000'+9000'+0 L 00'+
' ' ' '.' =X {n?.Un 3 UaTSTJaO}I
6200'+
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6t00'+
S700'+
?900'+
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8L00'+
26001+
I L [0'+g, [ 0'+
99 [ 0'+
Il[0'+L8 t 0'+
16 [ 0'+
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9ZA0'+
Zt00'+
t700'+
L900'+
BL00'+
?600'+
8 t I0'+
t9 [ 0'+
S0Z0'+
ZIZO'+
t9Z0'+
z6z0'+
Z L t0'+
8000'+
z [ 00'+
4 L 00'+
9200'+
8t00'+
2900'+
0800'+
60 [ 0'+
Z9[0'+'
617,0'+
9920'+
9620'+
6ZtA'+
lSe 0'+
,000'-00000'
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[ [ 00'+
0200'+
6800'+
1900'+
8800'+
LZLO'+
66 [ 0'+
SEZO.+
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2000'-
1000'-
00000'
2000'-
1000'+
6l 00'+
7800'+
lS00'+
2600'+
8S [ 0'+
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Ltz0'+
L LZl'.+
[ 0t0'+
L 000'-
[ 000'-
2000'-
[ 000'-
0000'
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9 [ 00'+
tt00'+8900'+
?[ [0'+
9? [ 0'+
LLLO.+
L0Z0'+
0tZ0'+
00000' I 00000'
00000'I 00000'
00000' I 00000'
00000'I 00000'
000000'I 00000'
00000'I 00000'
[000'+ I 0000'
[000'+ i I000'+
8[00'+ I ?000'+
tEoo'+ I uooo'*
??00'+lzL00'+
8500'+ I 9100'+
7900'+ I 6 [00'+
zL$a'+ I 0200'+
00000'
[000'-
1000'-
2000'-
2000'-
2000'+
9000'+
9 [ 00'+
0?00'+
tl00'+[ 600'+
[ [ [0'+
Et [ 0'+
IS[0'+
6VZ
'S faqE,f,
250
vrr .1 .11 .7 Bak berpenampa rseCI1"
penampungan
persegi.
I
nlan terletak diatas tanah.sedikit maka dapat digunakan
Momen yang bekerja adalahdalam 2 arah.Untuk mneghitung momen-mo-
men digunakan analisa pen-dekatan ( "approximate a-nalysis " ) .
Untuk L/BS 2 If,= panjang tangkig= lebar tangki
, , ji*tiTangki direneanakap' gibbagaiportal ("frafte") dimiina be
ban yang bekerja adalah be-
J ikabak
kapas j"tas
penampang
lzT.J
h=H/ 4atau1 meter.
H-h )ryH}
ban segitiga.Tegangan yang bekerja pada dindi-ng tangki 3
pada ujung atas = 0
pada 1 meter dari dasar atau H/4 (pilih terbesar) r
tegangan mencaPai maximum
Dinding yang terletak setinggi [= H/ 4 dari dasar diren-canakan sebagai kantilever -
Dinging tangki direncanakan memikul momen lentui + gaya
tarik yang diakibatkan oleh tekanan air pada dindingt.Momen lentur dapat dihit,ung dengan cara CROSS, dan gaya
tarik yang bekerja pada dinding ditentukan sbb :
Untuk teng.ki-dimana L,/B\ 2 :
Dinding yang panjangnya t:lrI airencanakan sbg kantileverI
b Dinding y9 panjangnya B dit-I rencanakan sebagai pelat'l' yang menurnpq pada dinding
l< l' fl yangpanjangnyaLt dandindi ng yang ber jarak lN/ 4 da-
(H-h)ri dagar dir.encanakan sebggai kantilever.
,+6
M1'd( s-n ) .
=H/ 4 atau 1 meter .
;El:it
i:
[,ir
iii
l1I
.rrE uEuE{al lrqT{r :fTreletre6 lnrlluragr e6n[ rpsop lETad .ndunuad 1o1eq eped ndumueu 6uBrIqErp I nele qprp Z leTad le6eqas up{EuecuarTp rpspp lpTad .qEu
-p+ spleTp rppraq 6upl{ r{pq ue6uap rutes ur{puecue.rrp 1u1 ad14 rlea( u{uE1 PeAqraAo,, )
6't1:t'rr^.1eos qoluoc lpqTf Epur up{r{ET1s eluselat >1n1ug
.1erer(s Tqnuauau snrprl {Eq qEl^pqTp qeuel ue6ue6ag, -.6uoso{ ueepeat ueTpp ,{Eq nlr lrrs , ( ,,1JTfdn" 11e>16ue ede6 depeq:a1 {pq laca6uaur ednl ue6ue[ ueg _
'(Et.ra{eq >1e1 de66ue1p qpuel .rre ueuE{a1 e6n[ uelrllurep) Et-ra{eq 6ue.fi qeuel upuErlal EpE {81 de66ue1p nf,T lpes uep (rTE qnued TsTTp :(rq) rTE upup{e1 TntTuaur uptEurcuarTp 6u1pu1p -
.6uoso:1 uepppa>{ urpTep {pq nlTlees'qeuel fTE + qpupl uEup{e? fn{Turaur up{pueouarTp 6urpulp -
: qpuel utETpp Tp {Eq ueeuecua:ed qe16ue1_qe:16ue1
urrrrnsEurTp r{pupl rrp ueuprel uEp qpupl uru,{al ,[I"i:H:JtI:-epaq,qeuel sElrTp 1eq ue6uep efes eues 1u1 ad11 {Eq uepuecug.red
'qruel upTBp Tp Eprraq Sueu{ tpg g.JJ.[.rIA'Tpos qoluoo lprITT ppuP upltr{pTrs e.f,use1a[ {nlun
x
qelo rn{TdTp 6ue.f,
-e1 B.[86 uE{T.roquau
zs( (q-H), )
',1 -W = O11aU W
'4apuad 6u1pu1p pppd (q-H)ralau [ 6ue[uedas 6u.BCued
t >{T:e1 ue6u€Tn1
= .l JPsaqaS {Tf6utpurp de66uetq
ZLI
aa
: {apuact buTpuTp {n1un snsngx'(rPseqra? qTTTd)
[ ' H' I '? nPlP Z*{'H ' L '? = xeru W
: rEspp Trpp q upr66uTlat Bped
zs((tI-H)g ) + = ue6uedel I{rpspp Trep ( q-u ) upr66ur1a>1 Eppd xpru w
splp 6un[n Trpp ( e-U ) p/s 0 ueT66ut1e>1 pped
: g eAubuPfued SueA buTpuTp epPd
g/L = t,/H'ZH'g'?./[ = .reseP wv
ea
tc7
tH' I
252
Kolom-kolom penuniang memikul te-kaqan, bngin.Pada ketinggian . ha dari tanah, te-kanan angin total = P
rr'r2'r3 ... = jarak kolom kgIam ke ttk berat(Centre of gravi
olom t'exterior ttty ) pelat dasaryg sejajar arahang]-n.
Gaya pada kolom- 1 akibat anginP.ht' f1
tn'<' f .t=. f.I=l
Da1am perencanaan kolom r gaya ge-ser horizontal yang dipikul olehkolom-kolom "interior" (dalam) a-dalah 2 kali gaya yang dipikul 0-leh kolom "exterior" (1,rrr).
Laqqlceh-la$qk3h pqfp,ncanaaL_bak vanlr ditumpu olgh kotom .- kolom :Jika bak' air mempunyai 'tutup pada bagian atasnya, maka dinding-.bak dianggap menumpu pd tutup tsb. Pada keadaan ini perbanding-
Ln/Ly antara 0,5 s/d 2.I - lebar bak ; 1-- = tingg i bak .
Harga-harga momen pd dinding (2 arah yaitu horizontal & verti-kal ) ditabelkan sbb : [= Iokasi momen max + utk ben
B-tanq IJY
lokas itanq 1.-rILoka s itanq 1.-hLokas i
/tan/ t
, o tv
max utk*1 '
utk
Q=
D-
cIV
momen max + utk ben
momen max utk ben
momen max utk ben*1-'Momen = k.
dimanac.i=
L2
k=
bentang, jika momen untuk 1-- maka I diambil=1--
f aktor yang dapat ,Cilihat pada tabel berikut:
!rl
thr-v
x *I AFaktor k untuk
Momen maximum + Momen maximum
Lbintane rv bentan; i,l[";.r* r- bentanq Ix0r51.01.52.0
0r180.330 .420 .4.5
J.40).47J.48) .49
0 ,250.400.450.48
0000
006013027032
0000
034012004001
0.0070.0260.0430.055
0 .0500 .0260.0130-002
?62' 0 =
0asoc - ?soy- ysoc
gzso) - YTsoc/ +EocoDt tsoc-o0'[ rso7 + o0 [
ovt .zsoc- o9t zsocr - o01 so-
/asoc - Trsocf* soc
- YSOC
'v6z'e =
psijc = d,
oolsoc =
?SoC ="X
3 JTSPCTurp JTlrtB rlpurl urup{el ua1slg-6@:-TffiEr
'ZEn B[eq 'gZZ>l uolaq nln6'ralaw [ - rsepuod ueupT.Epay
,vtc/64g'1=(qpuel uTzT -6a11 bc
-wc/b>1 ge , O = ) z'_w/l g'L =g ,e. = =.='.. , o'e = fr
rlruBl plEp-PlrqrTP PpB tEl
tr/4 s'L =9o?t = fr
3 uP{nrn qPuel Plep-rlrq'TeluozT.roq s1:e6 ue6uap o0 [ +op
-ns.raq splp qef aqas qeupl uesTdetTTsp r{eup1 upe{nurad splerp ralau g 166ur?as uPrtr
Eueul{n1un.raATT1uP}tadT+qPue1ueqeuad6u1pu1puP{Pupcuauffi3
3 f tos rtl Oluoc q oluoe Z' Tr A
: TTSP
'r{EuPl
'z(t{T)(i I + =Te?uozT]oU 6uelueq eped e[ra>1aq
.^_z(^t)
( -s ) 1+ =
z(^r)( s) 5+ =T TEltTl.raA 6ueluaq eped e[ralaq
TTSP
,tqt,)(-D ) + = + xPtu ueuoru
i XPIU UaUIOUI UPp
uPtP uPqaq Pnuras
g'o ntl{r {n1un
+ xeu uauou
xeu" ueuou uPp
ue{E ueqaq Pnuas
z f,t/\t {nlun: uP1r1r3'TUPrTzea.
TaqPl Tf,Ep TTqurE Ture{ sPlpTp Taqp.tr,
.t.9'
qPuPl
qeuPl
dq , sa.rnlcnrls alarcuoJ,r g. LL
254?
Iutenentukan ulruran-ukuran diBrFinq pe
nahan tanah.Teba1 pelat dasar ambil = b h
1
iZ(800 ) = 67 cm ------- 70 cm
B dicoba = 016 h = 480 cm.
ambil panianq dindinq = I meter.Pada bagian badan :
P=*.t.h2.K"l=2 ,43m = | t1 ,5 ) ( g-o ,7 12 .o ,2g4
= 11t75 tondiurai atas PH din PV
PV= 11 ,75 sin 10o = 2 ,04 ton
kemiringan dinding ambil = 1/48.tebal dindinq bacrian dasaf :
H 1,5(11,571.1000Lbu 0,9. 1oo .(9rgtb)
' (tU, (=9,5 Xg/cmtldiPeroleh tL = 22 t5 cm
D
sedangkan menurut persyaratan, tebal total sebelah atas minimum201rcm. jadi tU = t" + ( 730 iE)
ambil t- = 25 cm t- = {0r2 cmAD
Ukura4 qlsdinq spq terter
I
t-= tebal total dinding bagianD
da'sar.selimut beton = l0 t tU
ambiL = 42 cm
xo
6>g t'trlst
'Z*/brt 00EZ =
,vrs/ b4 gZ, O = ,,,
ew/bl 0og[ =
t*/1 g,L = g
xO ------- s't (9'tu. uo?
( (:q6
OZgS L= ffi =.ras<i6 {ln(08?)(92,0) Lg'o + rzE
uol tvg, Lt, =6>1 E?gtt =
n6z'r' ,( t ) ( oo9 ,4 +(r) u.6z\t (oosz) ?, =
ueuPuPa{ - JAO{
6+ g?gLt, = '.{u
s' L <s3,, z
= 9?Ett +
dlzr,{.r o'[' !u' 'f * lq'
uU cz =da eueurpd + g .c.Lg', + fr61 u: ;asa6 )tn1un NVNVWVfiX
g L' Vl c.-----^ ?
ffi = 6uTTn6 pql ueuprupo{ .ro?{Ed
gLtvv = (B?,g) f . zg,s[ =
tul LgZ, g [ [= ueqeuad
luc 8Dg = o0[
r =.{uo&xv.{
uauow
uauow 3
pTq -i) 6u1pu1p 6uBtupd u L )t1n ) V )tT1T1 depeqral: qqs uE{faqPlTp (rluauoul 6ullstserr) ueqeuad uauow
uol gLrZ = o0[ uTs d =Aduo? ZgrSL = ogl soc d =Hd
uo? Sgrg[ = (v6z,,o)r(g?,g)(srt) r =
'r' ,, r{'.[ 'f = duTs BLZ + 0L + OEe = ,e
L9Z,9LLre1()1uo? t.Vg, Lt = TpltTlren er{eg 3
Z, T,Llu g'luol gL'Z = A"E
tge '61uI ?'zro1?90, g= (l' z) ( g, v) L' O?
LS9, ZtU LL, L
$z'vzv't| +s't
uol 6gv, L =(v'zI ( E'r ) ( E Z, A-Zn'o)9LE
ZL9'8| ( gz' 0-zv, ol +9, Luolgt t V=l t Z. e, L. SZt O
.l
t
Z:
TLI, LL
tv't =7
dfr + zt'o+e' L' uo* se'ffit
( rul ) uauowv {11 pr{1g.auou ue6uat?eraquer6eg
qc7.
36U
2s6
Ir{enentukan
Thd ttk A olt=.A
eksentrisitas1a=v2
lat dasar bekeria t aaanqan se-lenl.sM IvI !.- -penahan ctuJ. Lnq
P vertikal115 251-44,15
37 ,643thd as pelat
(480) - 188
= 1r88 m
dasar = e
= 52 em.
e r(*B.\-6y
berarti terjadi tegangan sejenis yaitu tekan.ltenentukan besarnva teqanqan yanq bekeria pada pelat dasar.
e*"*= B- (1 + $r'dimana L = 1 meter37643 ,1. 5(0,52).ta;dml r rr LB t
= 12,g3g kglrn= "a2rg4 t/*2
37643 , aYmin (4,8)(1), "'= )745 kg/^Z
= 2 ,7 45 t/*2
6(0,521 ,4 r8 '
L emin
s----- < q ( =1 ,3 kg/cm'max
Perencanaan tulanctan pelat dasar :
,3m
,7m
Pa.da ba.gian "tge] ( tapak ) .Berat tanah yang menekan bagian
= 0r3 x 1 x 115 = Or45 t/^2.berat sendiri pelat = 0r7 x 2,4q total = 2 ,13 tr/mz .
)2a = 13 t/m- ) -- OK
Pada baqian 'heel" (tumit).Berat tanah yg menekan bagian "heel'selebar 1 m
(7,3+1,'l8sinloo) +(7 3+2 78sin t 0o )
2
1 . 115 ton= 'l'l ,544 t/^2Berat pelat = 0r7 x 2r4 =
q total = '13 ,222 t/m2 .
t ,58 L/m2
tt toe tt selebar' 1 meter .
= 1168 t/*2
e=0 ,52m
lq7illffii
' ( ,tucVZ,6L = g-7 lfi 1e>1ed1
gg'001'6'A000 [' (?,8 '6ll srT
uol zE'61Z
, = 0Z |Lfr ge>1ed) ,uc Srt =rV cgrLL =0L.00[.tgZ'O = urnurruTur V
a
t6'Erui
z( ruc ,Z
xo (-------
,wc / b>196' V
gerL 'uI? 6LL'8 =.
E rr't'z(s€' [ ) (sEZs' t)l =9-Qw
zr/lst.zg' L=( Lg't+LLv'orrf = ,,b
ur 6gr[ = urgl.'Z.f {E.re[:aq)
, G=) "q] >
6V'g =
q-q uebuolod Uped- uauron: ue1P1P3
(s
-
(LM=:t'4( EZt'0 gL'zl
LL?' OI
g'v
rxgzr,0 4L
08 LZ= W (99)(00t)EE0'0 =[990,0 = b qaTo-radlp Taqel Trpp
Z'o = a)
9ZZ' ( L) (g'O)Z ll
=rU ,t6,I
I
re6ffi(ruog8'[=0 1-Olf, te>1ed) .urc LL'g = UZ'O =rV (-ruc 7O'GZ = c Z
LZB'6+vz'6[ =g-olfi +
B : V Lil 1e>1ed ) aruc E6' gZ
6t.' v
(BL'rtr'sL'z 'ffiT + z(gl'z)
( aE ,rrfg?L'rffi,"l
Z
E9
rul LLgt Zt
=nJ
=XEllI Ir,I
I
Z6un^ |
z'/.a fr': ZZZ' T, L
g), r
L't1
ZZZ, Ei
,w/lsl' e=
z*/1 v9'6 =
, F2
_zb
=[b
(szz)(r)(E'0)z000['(6LL'8)9,I
ffi
258
gagM max
Cu=
,3 ) ( o ,2941t 93 E/n
+
=
*rr,s42-2,13)(r,612'?(1'6)
"a 2 ,387 tm65
12,9
= 7 r15
312193 = y z 7 13
q.,=0 i 441 y1Y, .^O. . ,i-(n, cos loo) .y.1/3 y
20 ,0 7238 y'tebal badan pada jarak Y dariujung atas ) .4z v + zsr:t,3 - vl
7 ,32,3288 y + 25.
Aminimum=0,25t.100.70=17'5"*2(pakaig14-8A'= otzA = 3r5 t*2 '(g1o
- 20 = 3193 t*2) .i
5cm2 (g1o-20 = 3,93 "*2)Sebagai tul . Pembagi= 202 '"17 ,5= 3,
Catatan : Tulangan pembagi untuk bagian "heel" '= ZOZ A. utama= 2OZ x 28 r 93cm2= 5 ,786 .*2 (g1O-10 = 7 ,8cm2 )
untuk potongan b-b s/d pertemuan badan ( "stem" ) dan pelatdasar IaIu menjadi g1O-20 sampai keujung pelat dasar.
Merencanakan bPdan ( "stem').25r--rl
qy:
t (=v'
Penplanqan : M max = 28 ,157 tm,( 42-4 I
selimut beton ambil = 4 cm.
= 2r77
1'" =
L5 = 012
f,= 0,14687 (l0O)(38, WA'= 9 ,03 .*2 (g12-12r 5=9 ,05 )
tuI. pembagi= 20t A= 9r03cm2
= 45117 cm2
( pakai 2 lapis g 16 8,5
= 4713 cm2).
q,5421,5(12 837 ) . 10002(or5) ( 1 ) ( 225',1
1,5(7= 3rZ
v l,I, ( tm ){--y=2 ,3288y+25 ( cm )
2m4m6m
7r3m
0,5794 r632
15 163 42-g , 157
29 16634 r3238 r9742
lls(zg,157).loooTto,5)(11.225
(pakai 912-12,5 ) :
!.lli
02.-0
lx9
9V'g =
'urc Zt , Vt =
03,-oLfr
( ,wc7g '
$
tJtL t
ZflL68NLil
9, ZL-2,1fi
nrE,-ZLb
'(sz ?,Jil 1e>1ed) ,uc zg, L = v *oz =t6equ:.":I:H-":*'(rtuc ZS'V = SZ ZLfi te4ed1
Z*, ZgrL =rVLL=Lt-9t fr te>1ed)rurcgE,L *(,28, 0t ) ( 00t )t gZ, O -unururru V
(SZZ) ( L) (9'0lZ 1
/tv-zt ' vt
d* t ur1 z1g, t =
=nJuauow I, l
' (gtzL -luc .Z
( atuc 9S' 6Z
?t h- gLfr
ruc EE, gZ =
ZLfiI Zs0',6 =
uc[E'S = V'EgZ = r6eqruadgtZL-ZLfi re>1ed)
aruc IErE
: sPlP 6un[n TrPp ut :1ere[ Pped
'Tn?
=rV= [6'9 + gg'EZ =
s1de1 t + Erg-glb srdel 1 re>1ed1
A8 LZ(7:fTF5.1Z ( L6' rE ) ( 00 t ) sLE60' 0
gLe60r0 = b qeloradz'0
-vTP
-sl
..1 ,,,
"rJ
i (9tzl (.D,(9'0lz ^,
7,1' t, =
ruc L6, gt
v-L6',88
=tr1 ! url ?Egrgl = uauo6
lx8 LLWZV'
O
Z-OL0r-01fr
6q7
: sP+e bun[,n TirPp ug lere[ PPrd
gambar 3
tanah
erfor
260
Soal 2 : Diketahui dinding Penahan
Jawab :Langkah perhitungan :
menentukan koefisiencos o( -= cos
.-o=cos I 0
= 0 1373
COS
tangh tipe "count€rfort" spt ter-Dafa-dara tanah r .
trtanah = 1 ,6 t-/m'
t ( ftanarr jenuh air ) - 'l ,'4t/*3g = 3oo
:=o )
mutu beton k225 , ba ja tJ24 .
Diminta merencanakan dinding pe-nahan tanah tsb ?
*lbd
tTcos + V "o"
2* "o=2
g .
"o" t so-
cos l 50+
p
cost lmmenentukan ukuran dinding :
' tebal pelat dasar ahUif = 1/14(800)
lebar pelat dasar ( =B ) dicoba = 0 ,6hJarak counter fort satu sama lain =
Tebal counterfort ambil = 40 cm.
Tebal badan sebelah atas ambil = 25
Tebal badan sebelah bawah = 25 + 74((ingat kemiringan dinding = 1/481.
Gambar :
cm.
1/481 = 40,4 cm 42 cm
= 2 ,502
= 57 11 cm
= 480 cIIl,.
0r5h = 0r5
ambil 60 cm.
(800) = 400 cm.
tekanan tanah 3
cos? cos'S
-l ,4m
il
tuo+ sL'J = (aos,zl.-,(t)(t,Ll f =
ts'z =
"#T ((s)f,noE'9 + (E'zlzt,ot + (
Ew/l?, | ={euel ,9 , tr/l g, L =qpup?,
o ralau [ = 6urpulp Eue[uedA
uo16g9' [ =Ed t uol sgg r 9 -uta r uo? etgt z
uo? gg0, L
uol gLzg, g= z(gr
g+? , vl (ELE,0 ) ( ?, Iuol L' Ll =(9,0+?' tl (zL, E) (EtE,o) (9,
uol t.L' V =Z(ZL, O+E)-(tIEr0) (9,
url L6, egt = ueqeuad uauo6]1 [9Sr0E = Tefol d
d{"!u' ,Li = da PuPruTp
a'+ E'c'Lg'o + 061 u =.,*d
=6u1Tn6 {1n ueupurpa{ rol{pd
. .u? zrzg =
(zL'tl f +s) 6g6'E = 6uTTnE
'TTcat PuarP:t UPITPqETp
Erz1, l,Z = uolaq{n?un 3 v {T1T1 clrprqralA- II- Za ! uo? ZL.OL =-ZdA. I{.
= td ! uol 6g6rE ='[d
+ = !u' *rr' ,l
f =€d
t ) =Zr{,'( t q'
Pr[ ' !-- l =Z d,
,+ = fu',=rt' g '+ =[d: qeupl ueuetal uer6egg
xo s, t (
uauow4-M 3 .Ut1t1E3)
fg
t0 [ 'g
06[ ' E I
gZ,L' S
6gs'9 t
ggs'z
8Et?'8g6E'gs
z96' 0s
ur8' ?
ur8'?
tII8, ,vtl, Z
tugl[,[ =(SZ,O-zv'ol E + s'Lutg6g,[ =gZ[r0+gz'0-z?'o)+9,I
, ut lv' €,- rins'
urL?,€ =(gL'Zt
f . zt' o+s, L
A
1Gggrt = td
A
1€Lg'z = 7a
Auo1 g90r[ = [d
zL6'g = (l'z)(g'?)g,0uo1 960s, L =
I' Z) ( tr'r ) ( E Z' O-ZV', O)1
E
E
Z
v
t
(ur+)uauow
- ( .ra1au )v pq1 uauoru
ue6ua1( uo+ ) lEraquet6eg
' roleur t - 6u1pu1p 6ue[ uedtntun 3 buTpuTp srlTTTqEE
FrP.n
262
= (501561) t,g 3oo + o + 1t75 = 30194 ton.
= P" + P^ + P.'h 'h 'h
= 31989 + 10r72 + 51305 = 21 ,014 ton
,rr#rr 30194 IFaktor keamanan untuk geser = iffii4 1 ,472 1,5 oK
uenqecek apakah qadi pelat dasar betGE.te,,.,$g$Fnqan seiqpis (tekan):
penahan ^'gul inqx=
=W=)rol3m1/2 (4r8) - 2,013 = 0,387 m
1/6 B
1/6(4,8) ----- 0,387 0,8 OK, timbul teg.
pada pelat dasar , =elenrs 'P ,-. 5e.9*.*= f;5tt*
g;) dimana L = lm
g=
e?0r397
Tegangan yg bekerja
50r561 (1+ 6(0r397)l4rg(1) "; 4rg '15r63 t/m-
)1 ,563 kg/cm' q ( =1 ,6 )
OK
q
emin= ?ffifu(t- 'gq#-I = s,43 t/n5,43(1,6) + 15,63(4rg - 1r6)
=- | " r"
5,43(2,021 + 15,63(2,781q2 =-
Catqtan,: Jika tegangan yq terjadilebih besar dr pada A ma-
ka kita harus menggunakanpondasi tiang pancang.(dalam "soal ini diketahuiE = 1,6 kg/cmz).
2-,2-= 0r5a5 kg/em-
12,23 t/mZ
'= 11,34 t/n2
Berat tanah yg menekan " heel l'
sejarak 1 meter (panjang pelattegak lurus bid gambar = 1 m )
- wr * w2
= -lll,gLs + 17,1z4gl.2178 2,78
= 111535 E/m2.
1
ma
2,78m
'ur1 zLA'ZL =
z(?)(Er9'L\ "iru? 90,0[ =
z(v)(s?s'Ll ! = dBT w
'uP{81e1:ad-up{efalred sPlPTp -resep 1e1ad d1sul-rd ue{.resppreq qs1 respp 1e1edue6un11q:ad e>1eu 1.roJ-ralunoc epe ( ,, TaeQ, ) ?]run1 uer6eq eped eua-rex
w/+ SIS' L =ur[ Zr/1 SlSt L =bffi
Zr/1 EgrS[ -xerr
vt'Lirli*='\ -lll vt'st
ffi+zw/19 L
z'/1 998 ' I
z*/1 gL6' zl
g?g'. 'Ltv,
tvn s
gL6'rrfu
vv'L + zv'ovl'L+gts't L
z*/t vv' I
= ,, oo1,, Ue16eq {n1un Tplol b
= , Toor{,, uET6eq {nlun Telo1 b ;pe[= ( ,oo1,, uETbeq ) -respp 1e1ad lerag
zr/l ztrT = ?rL x €r0 =' ,ioo?,, up{uauaul 6ue^f, qeupl lerag
( urc 09 f Eqal; 1e1ad TrTpuas le.rag
'zL=b
e9z
z*h v'lL =v'z x g'o =
i' 'heel'
,i
264
Penulan
M lap 10 rh6. ._,:
fu= 55' cm
= 6171
:.
CtI =
pakhi
l4 tump =
Cu=
6'
A
Catatan ,:
Check seser :
ambil untuk selebar 1m qt =
1.
Lg12-7 ,5=1 5 r 08 )
t-m
25
A 'minimum= 0 ,25t .b.ht=12 t072 tm
55 = 5r13
=0minimum = 15
"ro2 (pakai g12 715 = 15108
",o2)*Karena beban yang bekerja pada pelat "heeI" tsb tsb makin
kecil untuk potongan mendekati bagian pertemuan "heeI " dan
"stem" maka jarak tulangan makin diperbesar 912 15
*Tulanganpembagi= 2OZ, A=3"*2 (fl12 30=1,77 "*2).
7 ,545 (uu
7 r545 ( 1 ,28)t1 ,635 ( 1 )- 5,419t/m22 r781
; (7,545 + 5,4'19).4
25,928 ton1 ,5 ( 25 r 928 ) . 1o0o
PenulanFqn ba<tiarl rtoe' :
Karena tak ada counterfort maka lat kantilever.+ 15 t63-12 r23
.a,L
(1,6)..,.6)(1,5)'
0005)
0355.
.10225
0,0
lCal
1t2'
2316
u=
{-
ipe
I-rer
.=-b
.=
I"Ir
d
d
M.D
nakan sbg p:14t,
12,23)(1,6)z !, -
(r,6) - *(186)(,17 tm
'ffi0"2(0,5)(1)(22510
rroleh q = 01035
pe
fl8
,
0,05).1000,s)(1)[225
1r635
fl
Ii
h
.(08 Ztfr 1e>1ed1
'2uc LrZ = ,V c(rtuc It'[[ = 0t-ZLf,) c
urc Er0t = ZV.O0!.tgzrO =f.'LI'q.tEZr0 = wnuruTlu V
0t'g =8E
qa Toredtp7,'0 3
=nD
'ru+ [EDrS =
c,(r)(880'vl + = za b + =,, f oT-ra1ur,, 6uelueq {1n xeu dBT W
'w/a ggortr = [-((v' ?) ( e Lt'olv, L+( e tE,0) ( s ) 9, [ )=
l'(zq'E>{'rI + "{'1,.{' ! )=b: rpsEp rpEd
Ut [ = ueppq t66u11 {n?un. fTque
ucnUICZ? =,,tlIOlS,, -reSpp pd
rad pueuTp ) up>te1a1rad
= uolaq +nuTfesT=lol reqe+ =1r{
luf I
3 uEbuedel
V?'g = -T bC
ue6uetnnaa
+ = drunl w
7e-q'
u?' ?=
tUt =
Elt
edn.raq eduue4pleT6qs ue{EupcueJTq
. ( lfoJ-ralunocede-reqaq spfeTp 1e1ed
tq
nQr!
=(gzt. '0-g, t ) gg, I L (gzt,0- g,. t ), z
W=q,w/1 z6'zl =
- ,, b : e-€ ue6uo1o6
@ '(aulc st't = sr-gf, te:1ed) z*" vzrt = v-toz =r6equad ue6ue1ng,
'ue6uprn+ 6uesed ElT>t dpia+ uppups>1e1ed urerep rde1a1 , 0 =, v( rruc ?t' Ll = gZ'Z + g0, S [ =
. Z L frZ + g' L-Z L fr qer{Eq ue6ue1n1 edn:aq 1
z*, Z'eL - ffi (Es)(00r) se€0,0 =r
3 ( rurals"uPpEq UPTTPUTPUPDUETnuP{PUPCUaTaIxo +----
,wc/6>1 09,, =sE'00['6'00001'fg'Ei -SfT
uo1 tgrsl
Penulanqan ttunpuan :38Cu= = 5 r75,ffi0,
5' = o,Zdiperoleh
Tulangan pembagi
Catatan :
: A min.imum = 1015 "*2 (912-10 = llr3t cm2)
= 2rl "*2 (pakai gB - Zol
Tulangan utama 912-1 0 adalah untuk jarak lm dari dasar"stem", sedangkan harga 2 makin keatas (kearah puncak badan)makin kecil, sehingga jarak tul. 1 0 cm tsb dapat kita perbe-sar sbb :
y ( meter ) tulangan laP=gu*,
7 ,4m
Perlulanoan "counte,rffoqt"Direncanakan shg balok kantilever.
0 s/d 4
4 s/d 6
6 s/d 7,
Ar,=fr12_30
A, =012-30A, =9.l 2-30
Pt
Jarak caunterfort=^ 4 meter.
3m
{ ,' Attr
satu sama lain
1 r7 gA4
20,59o
rEt
P;
Pt
tj
4r088
tgO = 278/740 = 0t376 -+--- g =cosg= 01935 ; sing = Or35Zdtot.l= 278 cosg = 260 ,Z cm
deffektif= 250,Z 10 = ZSO,Z cm,1Pt= ; (1,7904)(3) = 2,6856 ron
P3= 1/2 (4r088 : 1,7904)(4,4) = 5,055 ton.li.]|,
ffii
A=fil 2-30A=fi1 2-20A=il12-tO
wi:10\
' z*" LE' I
iuol ?,s€.t9L =
'ulf = uTPT PtuPs nlPS
lro3ralunoc >1ere[ pupruTp (, ) ( 880' ll=( .ralau [ =uPpEQ 166u11 {nlun TTqurE )
( P-p ue6uolod ) utppq resEp Pp
Ed 6ueq6uas f nltTdTp 6ue[ tTrel e,{pC
( _urc L' 6l = c
_ruc e 6, VV = c
z'" 8'IL =
o8o-2.., = .. :'uO, =Te?o?s-, 6o6rTz6rer)s'r Fffi v
sr- -V = ue{nTradTp 6uBl{ ,6ue>16uas ue6ue1n1 sEnT
= z(t) )t'v/t 'Z = 6ue>16ues I senrruu OL6 6uedrueuad Z 6ue>16uas 1e>1ed
880'',,ru-
SZ,fiO I te>1ed I
0802
uo? ?Egr6E =
o6E'oz u, it? i1i : s?t ' os = uolaqq
ruc E, Lgz = W = socylTl{a}J"p =[p pueruTpv Y-v
[a, Y
o6r+ *o=uolaqo
'uo1 grt'og = (p)(o5[ soc tsgLt'g + g[gr[ + 9E8g'7,')l = TElo? d
@S?65 (---- sPlP Tf,Pp tu L'E rlereC eped er{u
-qe6ua1as Tpe[uau 16uern{Tp qsl 1:oJ.ralunoc 6ue[uetuau ue6ue1n1
0 =rV
(szz ) ( E', 013,(z'oEe)(0?)Et?0'0St?0'0 = b qalorad
=[TP
g"\
"rl
((v'r)i)sEo's+(
ru1 lg, L os I
z'z)8t8'L+((r)i +v''.Ialalll n =
0=
Z'O9Z
= ( .resep pd ) xeu uauol{
nles lroJ.ralunoc {e.rPr
soc 66'9E L
soc (?)
?) 9S89'Z,lUTPT EIUPS
(9ZZ)(tr'0)(E'alz0oor=l7t'IsL]9'L
I q7.-
-268-
juitl|h sengkang dalam 1 meter = W = 7,5 ----> 8 bh
Jrrak sengkang akan bertambah besar untuk potongan-potongan ygmendekati puncak badan tsb.
Sambungan ncounterfort' ke pelat dasar.Ambil panjang pelat dasar=1 meter
O ,7 4m [4]= berat tanah yg menekan pelatdasar selebar I meter.
[= 1 ,6 x/^3
l'= " n t/n
r4/= ( (3+0,4??l+(3+0,7a) (1 ) ( 1,61
(4,4)(1) (1,4))= 'a1,9336 tonGaya yang dipikul sengkang= W . L
= 11r9335 (4) = 47t7344 tonA=*qkrrrg
total:=
= 34r42 a*2.II#F",' kJ+lpakai sengkang 2 penhmpang g 12 ,.
Iuas I sengkangjumlah sengkang
jarak sengkang =
jadi pakai 2 panampang 912-6,5nampang 912-13 untuk potongan
= 2.1/4.dalam. 1
100w=
12 = 21262 "*21I ,42 = 15 ,z2,262 'i r
laIu mengurangyang mendekati
fi . (1,2meter =
6 15 cm
men jadibadan.
pe-
Ganbar a
912-30
fia-
fi 12-
30
potongan 1 - 1
g8-15g 12- fi9-ts
$o
ffi
OT- ZLf,
ruc00 L
F-,r/
"g-z L
EZiO L
0€-z ti
0z-efi
0[-Zt'i
O€-ZLfr
uPburTnuacl
rqNI
(rl
= 270
Soal 3 :
114
1s
--l ,5122-3 0
6
98-
912+29
g1
3 r7m5g2s
g 1 2-25
912- 1 32912-6,5
0925,2910-12,5
12-17,51 2-30
139fitz-7 ,5
tonsan 2-
berbentuk silinder yang dapat menampung a-tangki tidak menyatu dengan pelat dasar. Rencanakan tangki tsb jika mutu beton
Jawab : Untuk merencanakan penulangan tangki tsbnakan cara elastis. (bukan cara kekuatan
Ambil tinggi tangki totalketinggian effektif ( utkir) = 4m -0r2m = 3r8 m
diameter tar.rqki :
isi tangxi = 400 m3
Diketahui tangkiir 400 *3. Dasar( "f lexibrle base" )
K225, baja U24 ?
zocm]f,.
= 22900f;
= ru.t1 400
kita paka L 2
kita perggbatas ) .
=4mmenampung a
+ 7Td,2(3,8) = 4oo ----t d = rr,58mambil d = 12 m
tulanqan melinqkar :
C.V" tarit melingkar (utk tinggi tangki = lm dari dasar)
= *.{.n. d = *rr) (3,8)il2) = 22,s ron
luas tul. m€lingkar (utk tinggi tangki 1 m dari dasar)i'; 2dimana O- untuk V24 = 1 400 kg/cm-a
16,2g .*2 (pakai 12g14 = 18r48cm2, ataufilq I cm).
lapis (Iuar dan dalam I g14 16.
912- 3 0
( uc g' LTeqal l e[.ra>1
0z-8i
(02, Vtfi 1e1ed 'Z*, LrL
'uol 9,0 [ =
ledse,, pedraeenr.
92,-Z L
'0v vLfr
VIVLfr
: rPqurPgs1de1 Z re>1ed
L'L = 8E%+ =r
= :er16u1Taur tTrPle.(eg
L
VLfl
tiZIV L fi
Z
= Vlfigl Zr"(zL)(8'r)f
w/19,8=(grt) I
' SZ n lfr srdel Z te>1edr g ,w/;;g, Z=
(S,ZJ-VLfl[ nele , F. grEzt,, zt = n 1fr8 )rruc71 = fft_h
_ y B'E' ,=
re>16u1Tau ue6ue1n1 sEnT'uo1 grg[ =
(ZL I ( 8' ,+ = :e>16ugTaur' tTnpl e.f,eg
'ruc 0? = 6u1pu1p Tpqal
z,*r/6.\ E'E = =o12
LZ = u (---- ?Zngzz>t {n?un
PpPCt rerlbulleu ueTnuad
=\'.Q
ulz't
'({PCTrPp ) trutItrPrPpat
. ue?ElPo. OZ gfr ue6ue1n1 srdel Z re>1ed ElTlt
(0[ g/ nele 'z*rg = gfiol Te{edlrurcEry = E[.00[.tgro = urntururru vruc S[ = fTque 1e1ed Teqal
(autc S0,6 = S'ZL - ZLfr te>Jed;
z*. 9E'g = 0? x 00'[ x #*:16' = Te{T1,:raA Tn1 senT
B ILZ'T = 3+# 't['0 - t6'0 = 'Tn1 aselr.raso.rd3 TP{T+XaA qErE ue6uelnua6
TTqUIP (----. urc Z' gt = 1
00gzz = (8'EzE+00t)E,E
6Z' gl' ( [ :LZ)+100 [
: buTpuTp TEqa,f,
='=b OOSZZ
272
Soal 4z Diketahui tangki air berbentuk silinder, dapat menampung air2
500 *'. Dinding tangki pada bagian dasar berupa jepit (menya-
tu dengan pelat dasar) dan bagian atas bebas.Dlutu K225rbajaU24
Coba tinggi tangki = 4m0cm
Permukaan air= 20 cm dari bibir a-tas :tangki.D*a$e9gr .t+pqkt I
*o'u'(3,8) = soo
fl= 12194 m
ambil d = 13 m t
tebal dinding ambil = 20 cm.
h2 (4r2d.t (13)(0,2)
Pada tabel 1 ,2 13 { lihat halaman 246,247 | dapat
P, - koef gaya tarik melingkar.P, = f 1.t.h.d,/2f, = maksimum Pada x = 0r5 h
secara interpolasi 3
f, = ( X*8 (o's7s-o'514) + o'sl4) =
Pr maksimum = 0r5186( 1) (41(t3/Zl = 13,4836
harga-hargq koefisienp aan f, ait.belkan berikut
n2/(d.t)=6r15
= 5r15
anda peroleh 3
0 r 5186
tonini 3
kedalamandr ujuBg gtaS
koefisien untukmomen 3uf,t"itnn
koefisten untuk,.r? :a;i* T"rinkar
0
0rlh0,2h0r3h0 ,4!0 ,5h0 ,6h0r7h0 ,8h0 ,9h
h
0 ,000 1
0 r0002850r0007550,00181750,00308o' ,0044650,00500250 ,0 a29
-0r0039575-0r0183925
0
0, 1 17975
0 ,23280r34330r441150 r 50625
0,5185750,453230 r 307
0r113
il1
"4
: .(urpTPp ueprpnT sqdel eped 6uTsBtu-6uTseu gl_gb 1e>1ed 1
,uc iVO'g = t,?, -00[ . t Egzro = c
]u'0" {8i+* (zz' o tt'o ) - re 'o} - unuruTur v pr{?
Z*, L6A'L = V gr0
z*" Lt'l = (gt ) (00[ l zgt000r0 = q.q.(y
zsLooo,o = lrf+f* =0?
9'L =(yru 00t qaTo.radlpg'o = fr
(og?r)(r)/'I (9t'oztltzl I r, L, f =cJJ J Lle
uoloq lnurrTgs(---- tuc gg'Z,Z = 1 qaToradlp
'(fgd lprITI) ,ruc,71>1 g,E ="k1PI{TT , VZn e[Bq '1ZZX {1n) LZ =u-urnuTxeru :er16u1Tau {Tf,e1 e^,{e6 =rd pupruTp
v( t_ul + +00t = "0, d-
rul zLLLL'L_ = .(v)(t) sz6Eg[0r0_ = L ,vlv v
splp 6un[n Tirep ( rpspp eped 1 r] >1ere[ eped Tpe[:aa xeu uauol{ru? gtoze'o = "(l)(t)szoogooro =
sp?E 6un[n Trpp q L, A [=rrE eped Tpegr"+ + xpru uauoH
.02ue>16uepas , sele 6un[n Trpp q g, 09 L - A L frZ .re>16ur Taur tTrel ue6ueln;
3r reqbulTau Qe-r@,u/ru6>1 (tq'.1)g = urEl
{3aqc=rV=V
'ruc eZ -1 ,urcE = fTqwe'luc t.Z = TTqure
'(rgduol gtgv, t L
-01fi2 6uesed-tp q p/i q g,0 Trppp/s 0 uBr66uT1o{ tln 6uesedlp
(atuc 8?'0[ = vz's x z = sl-olfr stdel z re>1ed1
luc t,9, 6 = Z,
#8*h = -:e>16utrorx v' . r>16ue1 selp 6unEn
TrBp q9',0 >1ere[ eped uo? gtg v' t,[ = ,HnuTs{pru -re>16utrau {Tirpl ereg
ll7 F
fo 274it
Penulenqan, tlomgn max - :
' ,ca = 23-'5
21 ( 1 "a77 12',) .1 000( 1400 )
= 4 13
t f = 0,8diperoleh 1 00n&) = 5,823
unt,uk
0 ,19
0,195275 (
D0r875.b.h
= 6 ,5 t<g /cnz I
7
1
A minimum total = 0r3t.100.15(pakai 2 lapis 98-ZO = 5
Ganbar penulanqan :
= 4r5 cm22.cm,
00)= ffi '002773A = 0,002773(100)(18) = 4,gg.,02 (pakai g8-7,5
A'=018(4,ggl = 3igg.*2 (pakai l8-7,5Penulanqan qeger :
Geser maximum terjadi pada dasar dinding.Koefisien geser =L
h2 (d.t) = 6 t15 adalah
9'15;5 (-0,174 + o,1g7r = 0,r9527586I (4f = 3 ,1244 ton
% = 1,984 *g/emz01875. 1O0 q(Zu ( utk k225
Pgnulanq_en pelat dasar :
Ambil tebal pelat = 15 cm.
910-20
88E', 0
Jzgt.' a---E-
-
tqsr I q.
3ogVV'0+
.atIE'0- _otLE'0+ogLE'o+ .oslE,0-
L?V ' 0+svv ' o-
IoLt' -TEE
U-t-
-r089'A- o0t9'0+-gt+ r t-
rcffi I -rr"\6g' 0+Lb, L+
I
V
.l: ssorS
ggs'o = oo7y'
zs6'o = * = ovx
L,gg,o = IE'h = EVx
z*/1 | - z'L =
rrqrq rf,Ep .raleu z ueruprrn"r'nl' .raAaT Tlup{ re6eqas
ueteupcua;rp ,re1au [ = TT.que q
'(r,yeuB? eped ndrunuau; qpupl splprp ppe-raq {pq ,;rI:::tEesrad {nlueqraq 6uBdueuad e vzn p[eq , szzr uoaeq nlnIil
' jralT T 0 0 0 ' 0l rTe 6undtueuou ledep 6fi, '6ue1n1f,aq uo1-aq Trep lpnqra+ (dn1n1 eduel ) f rp uedrul^f,ued {pq up{pupcuau
SLZ
fiq'9=J
OJZ, I=&
oo-o 3 qeTaqas -reque6 leqTTZ >(---- Et' L =
s.T
ru Z'l = 11>16ue1ru9 = 1 r>16ue1
,
a'
?,w 9z =tu og,z =
'luc 0z = {eq f,TqTq Tf,ep-ps ede:aq TTqure qaToq epue ) jralau E
.reqaT) g Tpp[6uB[ued1 1 TTqup
lJte, z
;rO, = t>16ue1 senT
{Eq JT1{aJJa 166u11f,TP uet66uTlay .(E[
= TTqUE {pq 166ut1
'ssojtSeJPS : UaUIOUFUAUOUTuEnluauad
e=H
: s TPos
E 276
UenentuFan tebal dindinq (=t) :
M max =, 41744 tm
2 t3722,372(21 = 4,744 tm
*. (6)2 2,372
2 t12g2 ,1 28 (2) = 4 ,256 tm
ft l(4,2')2 2,372
M tumpuan
M lap max AB=
=
M lap max AD=
K 225 ----+v24 ----+
= -0,167 = 01334 mt
= 75 kg/"*2= 1400 kg/cm
= 21
75
,*. .@'t. 21
terr=%GD'bd
cL
n
= a 1247
= 17 cm
selimut betonjadi t total
4,744.105100
=5em
0'u
--
ol , 'Oa0b- n
0 ,529
{o=
Gava tabik vqnrq timbul :
Pada dinding yang Panjangnyar=+(H-h).8
= + (1)(3-t).4,2 = 4,2Z
Pada dinding yang panjangnYa
T = * (1)(2)(6) = 6 tonz
A - 0,01231(100)(17)= 20193 "*2
Af = 16t77.*2tulangan tarik tambahan
jadi untuk tumpuan 3
Ab.fi{t- *Y",0,0612 ----)do = 0,247
m
ton= B = 4r2 m
Akibat gaya tarik T + momen lenturM netto = M T. x
dimana x = jarak tul. ke garis be
rar (srs "".:;:]==,;::Penulanqan tu$pgran, :
M = 41744 tm -
M netto = 4 i7 44-4 ,2 ( 0 ,06 ) =4 ,492tm.
2 ,07
' diperoleh 100 noJ = 25,86
h)= 25 rg6 = n1'illZTI = u,01231
=-L = 42oo kcr - = 3 "*2t u 1400 kg/crn?+ 3 = 23 tg3 cm2 1616-7 ,5= 26 r8c*21. 2.Ar= 16t77 em'(914-715 = 20t56 cm-)
[- -
e e/
'7 2tr 2 ,317T 2,37
*"=
Penula
Iit
t
['
I[,lit'
Ii
t
iil,il
[.I[.
I
tj
tr
lit,
iit1,,il
ijFl'l,iiItI
it;
Ir
r
1
:
I
i
'uplpnrlred n1.zad-Tlrer rwc/b4 gV'O = qeusl uTzT
tepTl r{pue1 p^f,u
6a1 rnqplo{Tp p}tT I
,wc/b>1 99e '0 = Zr/t 69grt = l++t =9 uo1 tgr6g = Telo? d . uol tg, 6l =
uol I'Z (g[r0 x Zrl x 9 +Z x Z,Z'0 x Z'V + Z xZZrO x 9) = {pq TrTpuas 1p-roq
uo1 OL = rTp lBJIaq ,;Te qnuad rsr-raq {pq p{T[ ' 2 ,rE rttr-r *rnal -r -...
. ,.,! r{pup? eped efra1aq 6ue^,( ue6ue6a1 {caqc . Z,
(ZmtO's=ol-gil)rruc g'l = t6rtZ.zoz =pru'ln v 'zoz = roeqruad ue6ue1n1 -[ :-EEffi (dry €0's = oz-gf, s1de1 z) z*r sL,t = El.00!.tsz,o = uTru v
' ( atuc L' g =9, L-gfr )
aruc E r g
urc S [ = TTqup jresep 1e1ad Teqal
= Z,Z'00 ['tgzt A = unuruTru V teledrul Eggr0 = [.(t.[) P = I
.@Zm L' L = uZ'O = TTque rV
' z*" 9'S = ZZ'O0[.8SZr0 = unuruTru v te>1ed
rul ?tt.rg_ = deT I,[
(-urc Sg'g[ = 0E Vtfr + S[-glfi) ,urc tlrgl = upta? .Tn1(rutc L't.Z =S[-nlfr + st-glfr) ,ulc 6, lZ =E+5rgl = ]tTrE1 .Tn1 Tpp[ c
_ruc E =LV Z: upqpquel tTf,p1 ue6ue1n1
z*" 6,Br = (LLr,rf;r;l;:; :';t t 10 30==(fr(---- gt , EZ = (Uu00 L r{aTo-radtp
6L,Z ==EJ( g0' al z, t-gsz' v =
LL
tu?
x'fi-wggz'l
o11eu[4
ue6uedel uauor{
o0rr I tq00['(?00
LLZ i
:ugPTPUTP{nlunT uebuelnEa6 ut
r0
@
? 278
1m
914-1s
1m
9147 ,5
914-7 ,5
g 1 4-,7 ,5 15+fi1 4-1i5
uo@
f
$s
1 6-301Ig8-10 I
ref zo
=1r5m
\0
ts.t
oon
I\o
ro
rsJ\o
rn
r\I
-\ob.
r6-tJs 1 6-1 s+g 1 q-30 fl16_i
916- 1 s
potongan a-a
uZ=
tll[=
Zw/12=
,4
El '{pq urErPp Epe>{ppTl rTE clEbbuE uep ELra{aq qPuEl uPup{at' :E
: qBliPft
'vtzL pdu6uetuwd b^ 6u1pu1p epBd nd
-unuou 1e1e.d re6eqes up>[Euecua-rrp
)teq 6un[n p/ s -resep T-rep u [ {eret eped ue>16uepas ':ana1T.luB{ Te6eqes ue{euPcua-rTp. resep -rp u [ {PfEt eped 'ug pdu6ue[ued 6^ 6ulpuTp-
' ralafTlue{ le6eqas ue{euEcue:rp wzt Bz{u6uB[ued 61 6ulputp- z Z ( e---- t = a/l
*r=t
'zf o e[Bq
t*/ 1 t'
t*/1 9, L
q nfrl!{
*fi
>{
>{ uolaqeuEl
ooe =qeue
, sZZ
[-
r.lpu
-P1 uee{nu:ad
I
wz L ue.rn{nf eq
-rp uL- epe-roq qpuef Tfe 3 qPuel elep-etrE(I'6ue1n1roq uolaq Trep lenq-ra1 'urg x u? x
'qeuel ureTep pperaq 6uer{ -rTe {eq ue{eueruau 3 T*TeoSq-q ue6uolod
0z-8s' L'gfr
I
,L-9 r0 L-gfr
-vtfi+SL
0e-vLfi
tL-t Lfr+t-9 Jt
t-vtlt:9Ll
lwr
+0t- 9 t
EYEFt."!rI
Ii
280
Ka= 1 -sin 30o 1 t)1+sE_3-0o - '/r1 -s i.n0j+
h
1/3)+2= 3,399 t/^2 = 3,4 t/nz
Pada dasar dinding :
Ir,1 max= Pl .y1* PZ.y2*P3.y3.*P4.y4
= 1/2(0,533)( 1l(2+1 /3)+d,533(z) (1 )+ *,0,s67) (2)
.1 /3 (2)+ 1/2 (2)(21 .1/3 (21
= 31599 tm
o'ua"
n
75 = 0r460
= 0 ,068 ----) 0 ,262
= 15,72 cm ----) h = 20 cm
,3.1.1/3
tJ= 7s+#N='oil2=.no
R = 1/2 .
Gaya tekangi ( pada
a=1,5 ( 1 ) (1/3')+1= rfiT Vmt
t.4 = 3 ,93 t/n tinggi dinding.R pada dinding panjang dipikul oleh tulangan pembg
dinding panjang tersebut ) .
pada dasar bak bekerja tekanan sebesar = t{n 1 .k"* Ynr.k") + te-
kanan air.= 1i6(1)(1/31+1,3(2)(
Untuk 1 meter Paniano dindino :
-
Lerkiraln F,ebal dindino :
mutu beton k225 ----tbaja U 32 ----)
-O'b
Try2
75 kg/cm2
1850 kg/cm21
r-l
d6.n(1- *tr",h =%lE
= 0 1262
ht = 25 cm (selimut beton = 5 cm)'Karena tanah menenkan dinding yang pendek (B=4m), sedangkan. dindingtsb menumpu pada dinding yang panjang maka pada ctintling yang panjag a
kan bekerja gaya tekan sebesar R :A=1 ,967t/m2
. Zr, 98, g = St_Af, geled r6eqruad . TFl >t1n Tpe[
Zw gL' Z =8'E [ 'tOZ =purEJo v : tg7 -r6eqrued v, 1e:eds
r6equ,ad .Tn1 .#J:':"; P=;,:J=.:=^ >rrf,e1 etres- selpTp urapT ue6ue1n1
(_ruc L0rE[ = grL Ztfr 1e>1ed) _uc grEL =rV ,Z r L L,bP C U L! =rV
lzrrl0'Et=s't-ztfrl z*c 8'E[ = (02)(0ot)t6900,0 =[t 6900'0 =
( LZI ( 00 r )ffi =
zs'v t = (vt00l
s-92: ur+ sr,
j] {.,0,. =
luc 9Z = TPlo1=' XPIII H
6ulputp{nlunTPqO?
tul gt7 = xeu nuailg=PIra{q 6i {TrE1 e.fi86
1efrarlaq rTp u?up{a1, epe {p1 rlpuel
.muPuP{a.1) g uEPpEa{ {nlun
ru1 669, t = xeur W
uol E6rt = raTau I.6u1purp
166u1q {n1un up{e1 er{e6
1e[.ra>1aq {e1 rTp ,B[ra{aqrtrpuP1 uPue{o1 ) v uPeppa{ {nlun
lEequad ue6ue1n1 q€To
,uo1 9 = (?)(tla/[ =u
'burtued 5U
Tn{TdTp 19 }tTf,E1 e,ts6
,kz*/1€:0
'U -resoqss >tTf,p1 edB6 1ue1e6uau up{E 6upt ued 6u1pu1p B}teu ' bulptllp E{Bu ' :lepued buBA butpurp Eppd up{oueru -rTE - eue rex
ur? S'l = t/t-(E)(tlZ/[ =xerr,D{w/l t= ( t ) l. = {pq jrqsep EpBd'rLra{eq
tffi
rlBurl uEus{el BBE {81
282
Dtbrencanakan dindinq vang]rendeL :AIR DALAU BAK KOSONG.A. KEADAA
Pada jarak 2 meter dari atas, :
; ;;*;r"" =
,",rrrrrl
I ,l Zlrr, rrn,, = t,623 rm
M lapangan= 1/16 { 1 ,967 ) ( 412= 1 ,967 tmpada dinding pendek dianggap bekerja gaya tekan yang berasal
dari dinding pan jagg ( 1 meter ) sebesar 1 ,967 t/n( dapat diabaikan karena nilainya kecil ) .
pada jarak 1 meter dari dasar s/d dasar, dinding dianggap sebagai
kantilever:M max -: ( 2+O ,867+0 r 533 ) .h/2'. h/3 =' 1/6 ( 3,4 ) ( 1 )z - 0,567 tm.
KEADAAN DIMAN BEKERJA DAN AIR ITTEITTENUIIT BAKI
,6\,
B.
Pada jarak 2 meter dari ujung atas :
. M max tumPuan' = 1/2' '82= 1/12 (21(4)2 = 21667 tm
MmaxlaPangan= 1/16 (2)(4)2 = 2 tmpada dinding pendek beker ja glya tarik sebesar (''untuk 1 m
dinding yang panjang) = 2 t/^2 ; 1m - 2 t/n'Momen netto = M- T. x
M tumpuan netto= 2 ,667 -2. ( 0 ,075 )
= 21517 tm
M lapangan netto= 2-2 ( 0 ,075 )
= 1r85 tm.
pada jarak 1 meter dari dasar
sebagai kantilever : M max =
s/d dasar bak, dinding dihitung3. 1/6.h2 = 3.1/5 (lf = 0,5 tm-
?l
. .6uoso>1 ueepee{ uETEp {Eg'ruc 0 t = Jpsep 1e1ed Teqal de66ug
' ( "'IJTTn" ) 1e>16ue( atuc 9€'E=S[-80)z" 99L'L =
(aurc9o'6 = g'ZL-zlfr te>1ed) atuc gL'g =
eueln v. toz
'uendunl
Epeqral rTE {rq ueqEca5ua6gL, g 'ZOZ, =r6equad ue6ue1n1
,9,21 ZLfr = TV
B0'[ + L,L = Tp1ol V 'flpef'
-uc 80,[ = gfgf = rv z 0002
= l6eqruad . Tnl 1e:eds
l6eqtuad ue6ue1ng,uebuep erues ?enqTp ue6ue1n1
'url Sg, I = uebuiadel 11
V = ,V'
( r z ) 00 r s8e 00'0 = SEd5-gg0'g
Z"L' L=(0Zl (00t )geStoo'o =
[=V
I=(I
=(Tu00 t0Egt'L y'
000 t' ( LL9' Zl tZ/I Lt t =9-EZ
Itrl LJS, ?, = uEndurnl n'TUT ueepea{ Z {n1un
g uPepeax
ueepBax
=PC
ru1
dtunl W
'g uepppa{E'0 =(-rpspp .Tp url
u1
tu?
{nlun {caqc BlT)t
) ranarTluplt r{
uo1 Z ={T.rp1 dgBr[ = deT I/t1
LLS, Z. = dunl W :
rul
ru1
L96, L =tzg'z =
dBT
dtunl,{
W:V
rffi
eBZ =
3 rsepuod 6uTp-uTFE-e6-uETnuad
284
Berat' dinding vertikaldinding panjang =dinding pendek =
berat pelat dasar = 1 3,
3
2(12,5 ) ( o,2s ) (3 I t24OO = 450002(4)(3)(0,251(2400) = 144001(5,1)(0,3)(2400) = 48103 kg
kg
k9
beban yg bekerja pd pelat dasarkelebihan ( tanah+air yang beker
lm ja) = (13,1 . 5,,|-4;5 .lZr5l(116 .1+ l13 .2 +.1 . zl
-
L<tanE-h--- -Eff= 65,472 ton= 65472 kg.beban vertikal total (arah ke-bawah )=45000 +1 4400+48 I 03+ 65472
=172.97s kg.Gaya tekan keatas oleh air = (13,1)(5,1) .1.(2 + 0,3)
= 153.6G3 ton (arah keatas).Karena gaya tekan keatas rebih'kecil daripada gaya vertikal kebawahmaka bak tidak terangkat keatas pada saat kosong.catatan : 0 sebenarnya antara dinding bak terruar dan tanah ada geae
kan sebesar Iuas dihding total dikalikan tegangan.(cara menghitungnya idem dengan mencari n"r"*.r, antarapelat dasar dan tanah pada dinding penahan tanah,. dinanakoefisien gesekan diambil = 0, l5 ) .
€ Jika gaya tekan keatas yang diperoleh melebihl (gaya ge_sekan d'indirng dah tanah + gaya vertikdt kebawah)maka bakakin terangkat keatas pada 6aat koEong. Untuk mengitasikasus injir rlaka ukuran pelat dasar diperbesar Iagi.
Perhitunqan pelat dasar :
{'
4
' LZeo0, o(v
Pu oot
llrl
' XO , ruc 0 E TE?o1 1e1ad
'ruc Arg[ =
SZ
= ( unurT steu ) ue6uedel D{ }tn1un
[-osg[ 'I
000 ['( 6gt m
: UPDUETnUad
uee.rT>t.rad Tpp[
000T-36877.92t 0 =
raAeTT?upt lTda[ W
eueul[r
:.
lTdo[ w
TN{TdTP
I6gv' E
1l L6'E =
TBqAI
t
qAr
TJi : 1e1ad r
" 'uI1 68?'E =' ( s [ ,0+E z, o+zl (8, o\zr g- ( gzL, o+z) og.e ,0*0 Lgg
' Arz+ (L99' ol L9g'0+( t ) (990, [ )+(s€t, z) (g992, o)+,( gg, z) (gs, t ) LZ =: qe6ua1 qe6ue1 Tpe[:a+ 1fir,*r*=* 1 ue6uedel n
( Lgg' o')Z + ( Lgg' o) t9g'0 + ( t )
(990'[ }+(ttE'ZlSggZ'O*r(8,0) (8S, [ ) i =
ru Lgg'O = (Zl t./L = ?^
rul z = (z)(2,)z/[ = 7d
ru Lgg'O = (Zl E/L = E6
uo1 tgg,0 = (zl(Lgg'o)r, = Ed
ru1 = Z5
1 gg0'[ = (z)tEEr0 = zd
ru ttt,Z = e/L + Z = t6
s99Z,O = (r)(EtE'otr = !d
. z(t'o ) ( z' sl r ?,{'?d + q
Ed' €d+z^'zd+[r{' I d*2,( 8, 0) ( gs',)l = iraAarrlue{
6>1 69t = 6utseu-6urspru up{e1a1rad'6>1 gt,L =
(E'0)(0029)Z - OOBt - 0O8t - (t,E) 08St = TE?o1 rr
..6:t OOA! = OO,Z .e. gZ,O - ,ur red 6urpurp lpraq'2./a z'9 = (z't + z'e't + l'9't) = .b
' .b = uEr{IqaTa{ 6uer[ resep 1e1ad uelauaur 6uptr qeurl ueqaq
z,/6,t ozr= oolz. r.€,0 =,,5Y:'=:I;"=.I:,; ::::"=r:il:""-::::z./l e'z = t;(€'o+z) = rEsep 1e1ed eped splre{ ueut}{e1'(rq6 6uep1q sn:n1 >1e6a1) ut = 1e1ad 6ue[ued ]tnlun TTqup
o,o= tI
ffitri
: buoso{ {Pq uEepEax
286
A = 0,00327(100)(25)At= A
untuk tulangan tumpuan, idem.tulangan pembagi = 208. A utama =
Check qeser :
n6 ,2tt/mt ,.--->
2 /-'-r-'- 12,5 = 17,5
(g12 '12,5 = 9105 cmz l
.
)1,635 em-(pakai frB-ZO =
= I t'175 "*2
2 ,52 emz )
1,58t/m
n3
912-25
B-1s
D =(6,2 1,58)(0,34 0,8085 . 1000L6 =--
CM
0,37 kg/cm
0r8085 ton
( fn --,--)2OK
IO
lo
12-12 ,5912-12,5
potongan a-a
912-7 ,5
,31 fiBZo-20 12m
potongan
n[]
L- L .1od
3 qprrlrfe e66ue^[uad uoro{-uroro>t 1n{Tf,eq qs1 {eq ue{eupcuau
. vtc/b4 gr0 = qpuel uTzT ue6ue6a; z ' 1 v v -l---'7 - i- i. ZEn e[eq , ;ZZN uolaq nln14
.6upf n1;eq uofaq dn1n1TreqTp qs1 )teq . ( ul? ={Eg t66u11) ur? x lus x rus I qeTppp qs+
{eq uprn{n 'qeup? uep{nru,ed T-rBp Jaleu 0l ueT66uT+e{ Bp-ed epE-req '6uetuBd r6es-rad 6uBdrueued:req rB )teq rnqela{To :-Z-rpos
. ( Jesep Tpdues >tPq 6unIn TfPp Jesequau ur{Eru uoruou uE{Trseq6ueru 6ue6B6T1!6es ueqeq edn:aq qsf 6uTpuTp eped e[;ra>1aq 6uBd ueqaq
Puare{) st-zLfr uE>[sn-re1'rp nrpT ,-resep Tfep -ra1au I efT{-B-rT{ ;edures ue{sn1nd1p s' L-zJfr .
rn1 6uB[ued 6uTpurp eppdq-q '+od
tuI
-1o
3 uBlPlPJ
,;r-'l,{
.I
ZI-ZLfi
s r-ztg
LgZ
Mergncanakan pglat penutup bak.
l(rr=q*)l ^Mt__ = Or',25 q ti
xcheck tsb_el_pels! _i
h -o( /14'o / r= 0,262 @= 10 r 36 cm .- ---)
=( selimut beton
288
Jawab 3 A.
T
#=,), ----) "one,xPBI tabel 'l3 ,3 .2 untuk
diperolehMi = 0r'125 qx
Mt-- = 0,025 qv
beban vang bekeria _:
way
t /Lv't2x'
L2x
K2Z5 --;-) d ,nu 32 ----) F u
n = 21-l\,, rb
0o- -l---tob' h
= 0r460
slab"
x) 2r5
tebal pelat ambil = 15 cm
berat sendiri perat - 0 , r 5x 2400:=3 60kg /n2
-::::, :i:T,,,:'l*l'l::,izl;:( liha, ppr )
totalq=500Xg/*Z
= 0,125(500)(5)2= 15G2,5 kgm
= 75 kg/nZ
=1850k9,/cr'
75
ha diambil1 5cm --) oK
= 3 em)
= 2 rB5
75+ #
o'b - folr- *f.,= 0 ,068 ---) do =0 ,262
9r06-e*2)
= 2r52 "rn2).
lf, = 15 m
1; = tinggi bak = 4m
t.+= 3,75 )2vdinding direncanakan thd beban yg be-kerja pada bentang vertikal.
o(2o
61 zo
15-3
{;di
21 (1562't .
0 12
role=
pe
1 850
h 100 nC[ = 14 ,29
uJ= !!+?e-=, :TffiTZTI = 0,00680
A = 0r0068( 100 I (121= 8,17 "*2(pakai g1z-12t5 =
Ar= 012 A = 11634 "^Z(pakai ilB-tS = 3r35 "*2)turangan pembagl = 20?.A utama = 1 163 4 "^?t (ga- 20
B- Ir{erencanakan dindinq vanq paniq4g.
{
1=V
=4L/m
ELt,0 = ttr0 +
-oracl"rp T selodrelul e-rpc ue6uap
(€E,a-zu'o) ##+ = x
qar(z9z upueTpq IEqTT) .Toqpl Trep
. A- ? Az )sz,t ={ {,,1=;l( ue{TeqpTp p{pru I Tce>{ Euo-re>[ )
,w/b>t 0SZt = (uS)(ruty64 00S,, f = (6ulp-uTp , ru :ed ) 6uTpuTp eped ue{rn{ldtp 6uBd dnqnuad lprad Ti=p ueqag
(z*, 9e'e = S[-Bg) Zrc Z6'Z = eueln V.g0Z =q Z/ I p-rT>t-erT{ rse>[oT eped Tpe[.ral + xeru I^l 3 uplE+EJ
e E€ 00'0(tz)00r--Tll.,_t,-L
(rtuc 88'E :0e -Zlfr)aurc LZ'L - VZ,O =rV(atuc Sg'L = Et-ZLfi te>1ed1
ruo 9t'g = (gt)(00t)€9t00,0 = v Z
Z, O
osg['Im
'3
gLV'L = t), 00[Lg' t. =
'rEnT 6uTpuTp eped Tpe[:rel( Zr, 88't = 0e ZLfrl
(auc l0'Et = g,L-ZLf,l ,wcg,rt
zJg00'0 =( [ z ) 00 r-mrzt =
[Y f \rl, L-t y r
60, LI = (y]r 00 t It 69'z =
l,l N-ZZup{T re1
z*" z6z+ ( B [ ] ( 00 r
3 rr+ 16
= V Z'O1-+xEllrt{=rV
=[ )zl'B00'0
z'0oEBt't
0001'(Lgz'tltzV-ZZ,
=3
=PC
-uTp eped fDe@lrPsPpepedTpe[:a+ r@LgZ'?= XEIII n'([rc 27, =TPlO1TPqAIPLUuTpuTpUEDUETnUad
tg'tE=
-(vlv c
ru1 [6 'L =
( luc? = TTqueuo?oq 4nurT r as I utJzzgt =r{ fTqurpTp ,ucL'LI
Z9Z'O
=tl: buTpuTp Trqa+
.resep eped xeu hl
.s'eE
2r'-0
,p
=+XelllW
=1,{ (--- ruc
EI
z(vl't9t
A,1'J)
oooi=rEm
tu1 LgZ, V =i
1.'
-. 290
max
max
mas
max
1v1r,
I v1n
+
+
:
+ 0 14
bentangbentangbentangbentang
= 0 1475
0 ,425
= 0 ,020= 0 ,008= 0r0345
=r0r0195
, ?,,T,
[= lokasil= lokasiC* Iokasi
5rIItP= lokasi
Ir{ max +Mma3+M max
It{ max
utk bentangutk bentangutk bentangutk bentang
I vth1r,
I v
TULANGAN ARAII VERTIKAL.
f;
{";
'll
@/m= 0r2
0,004155(Or2 A = 1
100)(
,496tm3+ = 1,28
A
Ar
22-4
= 0r2
0,00238(100)(ArZA = 0r858
18)2
cm
= 4 r29(910 -
1 '25-1 (0r48-o ,47) + o r47*1 = ffi [ur+u-ur
^ = W (0,4s-0,41
-b
max + utk bentang lu= 0r02l.t.LZ = Q,O2O (4)(4t2 - 1r28 tm
M max + utk bentang fh= 0,004. fl.ffr = 0,008'( 4l(5)Z = 0,8 tm
M max - utk bentang Iu= 0,0345.[.ri = 0,0345 (4)( 412= 2,208tm
M max utk bentang th= Or0lSS.O.ffr = 010195 (4)(512= 1,95 tm
tebal dinding total = 22cm, selimut beton ambil = 4cm.
Penulansan dindinq Ben{ekj tebal total = 22 em.
Irt max - = 21208 tS,,:
22-4
A=Al =
It{ max
harga k untuk M
M
k untuk M
M
I v
tarikan terjadi pada dinding sebelah da
lam, terjadi pada dasar.
= 3r595 L
J 1oo nt,) = 1,r_r:
h)= ffi'= o,oo4lss
18 )= 7 ,48 "*2 (pakai g1o-10 = 7 087 "*? )
.
"y2 (gto-30 = 1 ,967 "*2 )
.
tarikan terjadi pada dinding sebelah lu-ar.
= 4 r72100 nh) = 5',002
h) = ?r99?,,100 ( 21 )
"*2 (sro=-';:o:'X,r, .,02 ) .
30 = 1tg67 "*2).
(urc e -'ruc 0z
uolaq
Ln wc
ru6>{
ul6t
w/b>t ooo? =vt/6>t 0g? =
lnuTTas ) urc t,z
8'6r = *_l
= TTque ?.r rpp[
zsz,o = fr/"o=
r{
I@rsTlsPTa
+lde Iaq
sT+sETa?Tda I
-
Z
Z
xT'b'Z-xT'b'Z'
V9VZ =
Igo'o
t s0 ' 0 =
3 1e1ed TEqol-@Ii
z(9)(o'tt) zzo,o - ot^
zLLs = .(s)(ognnrLEoro =*t^ = *t^
c'
zr/brl 0gr? = Telo1b61OOOt x ? =(ru? 166ur1e$)-rTp 1e-raq-
OOVZ x Z, O = 1e1ed r.rrpuas 1e-raq-. ruc OZ = TTqupTp 1e1ad Teqal
1w
Tr{
\')ttI
xZT'o' ZZ0, 0rgd z'E't.l
T
= -I,ul
TaqP? urg=xT
[=rT
pqaq
rusxlus=1eued I up]n{n'e.,{u6unlnpuad {orpq ue6uap aTTouoru -roc rp lprad
'resrp +eT . (I(atuc L96'L = 0t Olfr) aruc ZS'O = v ZrO =rV
(rtuc V5't = OZ Otfi) ,tuc L'Z = (gt)(00tlVVtgg,0 = V czvn Loo'0 =
*tsS'= tn
Ee 0,e = (rJu 00t
' .resPp Tf,Ep rut
1l
( atuc L8' L =0r-0tfr
ggE00'0
( [ z ) 00 [T?W
8ZL'L
( atuc L96' L
re>1ed ) auc
=(u
= fl), oor
L6'g =v-zz6'L lere[ eped Tppp[ra+0+xPun= 0t = 1Li lerledl SZe'L
vzg' g = (8t ) (00[ )89800,0
eped Tpe [:a+ ul1E5',[ =--. xBttr n
ur} g
=rV=V
gt
[ .,,e =
1 >1e:e I '-resPp T.;rep u
z'o =
L5Zet;V.,4r
'TV,f,NOZTUOH HVElrNVONVTN.I
;292-
total = 4000
Mt=Mtxx
Mt
+ 0,23 x 24OA = 4552 Y,g/m2
5 0,051(45521 (512 = g8o4 kgm.
M, = 0,022(4552)(5)2 = 2504 kgmIv
Penulanqan pelgt :
Y
(zz'9[=)-aq 6uBfr
,utc/b>1 e t 6Z
EE '92
'ue6uedBTuendulnl
leledrp
{1n oz-gfit?n 9L-gfr
6uer16uas
e I ea>1
oz(
(s)(svstr
( rwc LL'sg =
v'0 =
:) -r, 09ru6t SZ, eg 6 ? g
,w/ b>t 9, s
rlrgnTp I1OTEE
t0s0'0( r z ) 00 rmg ' E0 [
'u6t zgtg? =: PlEr
9n
+ (E,EL,EZIZuep e61116as
lt 0 =
= XPIU W
uEqoq z lPqT{n
=(n
=Mu
:l
u oss t-:Zdreq,'. ut _uz =J
-urcZ,g, 6 = Z(g'Zl 2l V/ L. Z =rv C C'- rr-tuc sz =
TTqurp ,6u1.r1ur . Tn1 qerefi =r" . ue{{or16uaq1p SZ,i . f nl Z,
.6uele{as l 6ulrTu ue6u-pTn1 1e1ed1p rasa6 .
fnl leEeqas: :esa6 ue6ue1n1
( ,vtc I 6>1 gZ= ) cErP?uauas . qaq ) rrrTSa ZZ, gl
6>1 E t 86z
+ (9'Z) ( 09 LZZ) f- ( _uc vs' lz, .Z
szfrLt 1e>1ed) ,urcvnzO'0 =
( r z ) 00 [
z( s) ( e?sl 8/ L
00nz x Egro x
sTfi L
a{ uet1ntTdTp resa6 qnrnTas
0g'EE' gLgr0 Q2gLg6Z .L
= up{p1a1rad Ts{EaU @= SZfrg 1e>1ed ) ,urc S, OZ = r Vvz'Ls = (09)(5t)nvz0,0 = v
szfi t
,rTr = (v (Vt., lS = fnu 00 t ]L [Er[ =
ueJn{n , tesaq neduelral . Tnl spnT
,uc 6'Lg = (sr)(0t)tosor0 = v c
00 [
z(s) (oecr8/L16eqre1 ueqaq
E0'[ =
+ (g'ELgeZlZ = xeu W
gEr0 = TrTpuas lpraq
e6Z
294
Balok as | =Balok tsb
-berat-beratbeban
4.Coba 35/65
memikul dinding bak yang pendek :
sendiri dinding ( tebal 22 cm )' = 0 ,22 .4 .2400 =
21 1 2kg /nsendiri balok = 0135 x 0165 x 24OO = 546 kg/n= 4552 kg /^2
11380k9/nM lap max 2361 3,5 + * (546+21, 121 .52
31919,75 kgm
= 1r88
r 00 nA) = 32 t58
cr\ = 32 ,58- 100(21)
= 0r01552 (pakai 7g25 = 34,362l
s45 ) (5 )
20870 .. ^? a , 2rE ,Tb = ffi = 11,35 kg/cm')ZU beban sementara =1Okg /.*2 )
geser.
tumpuan.lapangan.
kg /nke dindi-
{";
A' = 13r02.*2 (3fr25= 14,73 cm
Check geser :1
Reaksi perletakan =i (11380)(2,5)- )0870 kg.
cm2l
+ t (2112 +
jadi selrrruh geser dipikulkan ke tulanganpakai tulangan miring 2925 dibengkokkan.dan sengkang dipasang praktis 08-lS untuk
BalokasA=asB:g8-20 untuk
ukuran balok dicoba 35/65.
1 3 8 Okglm
berat sgndiri balok = 0135 x 0,65 x 2400 = 546 kg/mfberat dinding yang panjang=,0,22 x 4 x 2400 = 2112 kg/n
. berat atap (dipikul oleh dinding panjang).( = tebal 15cm) = ( o'15 '5' 15'2400 ' 1
Z : ). E = 900
berat hidup yang bekerja pada atap dan dilimpahkan.1 1ng panjang = i (100 + 40)(s)(1s). iT 350 kg/m
9tot.I = 3908 kg/n
beban segitiga : q = 11380 kg/nanggap balok as A terletak diatas 4 perletakan arb ,crdl
.:
{.
I
e=2658kg /n
21 (31 338,5 )0 r35.1 850
1 380kgp . l
O
,( -lllc lit lZ = Sz,igl - cct80t0'0 =( 1Z)go1 = E8'Z,Z
Eg,ZZ =lylu 00tt
ul6t
-(s) (zss?) ( s?,' o)99t'0 + s'zlzzt L
i+ gT (b)(92'O)99t'O +r(9)(8068) I
(rwc Z,g'G = SZIZ) ,wc t,L'6 =,V cruc Lzg'zz = (0g)(st)tg0[0'0 = v
'. l
?10 =
09
qEt l' l6lJZ = uEbutdu[ II3 g = [ st )toTtq utbutTnuad
TI,r-r {,,,,=
6>1
61 ?,6' gt,d6? = 'Qd ! 6l gq',
v''16?-lz *. =+ g'vet?t- =
Zg'69982ue6uedel W
g66ez = 'tQuOE
L9?,8[ = ' U
" - g ' o9g8z g', 69c)82
ul6tg 0 6 g
co Eq 'utr i* l.qu5998
url6ttOBt r Iu/6to e g t
: uauou TTseHI| 9'699gz+9'69982- o
4qe ur6)t L' ge GZZ =, TP1ol = .EW
L't?t8 !E Z(g)(B06elLZ/L = sW
: tler 16eq:a1 ueqaq lPqT{e
rft .T
-
v
/ 6>t
058uIcrp
't
0 6' LgV[ t- 000L'Se6Z
19€6zz- L'se6zz+ LtseGZr.- . Ltgt6zz+ L' 9i6zz- L'9e6zz+
- f'-rl -rTrloF-u- lrl- pq
ru6{
61 osr8z =-(Et(ZSSr)EZ'O = c
Za'b 9Z'0. =ttl PuPtuTP
T.Irl .?0110 = dI{e61116as uEqoq tPqT{B
'BrIIr g,n ue6unllqradttnlun ssorS eres ut{TeseTasrc
0Eg t 'Et'0( r' [ 6t LZI L?,
962
296
, '611 vg?z =
( OE/ Oe I uoTot TrTpuas le.rag61 gLEg =
z(9'Zlg( = tuoTo{
6uTsBur-6uTseu eped Brte6 Tppftu g,Z = I qs
a{ uoTo{ 6uTseu-EuTseur'>1e:e[ = r,'7 = d'lil
ruofo{ buTsetu-6uTseur eped BfreD
r urlPlsct,9 L9 =
lv'i [ =d' 1 116ue1
6ueplq pd e[;e>1aq 6I ) ui6ue ef,eg( ,w1S>1
gZ = unu-TuTur u16ue ueue{a1 ,tg fgdd pd)
Z*/6tl 00[ = TTqup u16ue upup{gJ,\tcZZ = 6u1pu1p Teqal pueurp
ur?trEt = (trr1,ZrO)Z + urg[ =f
-uBd 6utpurp 6ueptq eped et re>1aq b^
*@r +
)r^
r'(u? x rxg[1 6ue[ur6uB e.t(e6 ue46un1rq.zad plTN
3 UTbUE PATL PLTa{aq {rq rprilt
-o{ upp selpar {rr=lralfse ruoror-ruoro{ "r;:ilil}r:;-H:*rlr=;J:l ur1 ?,9, L0[ = u6t VVrO?,1LOL =
(E'[ + E)006 + (E',[ + s)002[ +. (E'L + 8E'|,LltgLg =(uIE't = lsepuod ueueTepat de66ue) Tsepuod fTda[ pd e[:a1aq 6^ uauon
. r{puE1 ueetnurad Trpp luE =(ur0[lz/t >1ere[ pd e[;e>1aq 61 006 = (00t)(Et)(E,oli-, =
' ,, 6uTcEJq r Z, Bpe , ( Ot/ OZ de66uB ,,6u1cere,, . {n )
' ( uoro{ n>1e6uad ) ,6u1cere,, eped e[raqaq 6ued u16ue e.f,egr.{puEl uee{nurad Trpp rus = ( u0 [ ,tz/ I lere[ eped e[:a>1aqr Qe/OE dB66ue ruoTot .tn 61 00Zt=00[.(0[)(0t,0)? =
( urorolt v Epe ) uoro{-ruoTo}t eped e[.ra>1aq 6r d16ue e^egqeup? uee{nu.rad Trpp
rrr g e , Z,L = ( gt, ,l ,, +t+s, t+S, t ='>1e.re! eped e[.rargaq d
urcEz = -rPsPp 1e1ad -Teqalruc Et = dn1n1 Teqal
6>1
00['(EZ'O,+ sIr0+?)
rus
ut
tu
'rl =
,'{
n"l
,{
f-
t
-&
FI
L5Z
298 -
Kombinasi akibat angin + tekanan air dalam bak.Keadaan bak kesong 3
Kolom-kolom as A- 1 = 'l 4131 , 0 8 -537 6
as A-2 = 33538,92 5376 +
kolom-kolom tidak tertarik keatas.Keadaan bak berisi penuh air.
Z Kolom-kolom as A- 1 - 39 1 3 1 ,08 537 6
B-1 = 39131,08 + 5376
B-2 - 83538,92 + 5376
+2484 = 11239,08 kg2484 = 306 46,92 kg
Penulangan kolom :
Semua kolom diambil ukurannya 30/10 (akan dieheck apakah cukup)
thec,\ kolom as A-2, B-2 r. A-3, B-3 :
Ukuran 30/30:.mutu beton K225
baja U 32
P_embebanan sgmentarS .
---) O'b= 125ko /"*2---) Fr= 2650kq,/cm
+ 2484 =
+ 2484 =
+ 2484 =
+ 2484 =
36239 ,08 kg4699 1 ,08 kg
906 46r,92 kq.
91398,92 kg
n= 14
. '. + A.DA+ 2650 A
artinya be-memikul P te
Gaya kolom as B*2 = 83538,192
86022t92 = (900 21 A).A - $7 ,36 "*2
A maximum = 6Z . 30 .30
Karena A yg diperoleh A
= B-2 = A-3 = B-3 diubahBerat sendiri kolom - 0,4 x
Gaya kolom ( tanpa angirr ) as
= 83538,92 + 4416
87954t92 = (1600-21A).A diperoleh negatif
91398,92 = (30x30- n' A)
91398,92 = (900-14A) .125' A diperoleh negatif,
ton masih cukup kuatkan tsb.
Check th_d pemb,_ t-et_ap ( t_elrpp apqin ) ,+ 2484 = 860 22,92 kq
75,+ 1950 . A
= 54 "^2
maximum, maka ukuran kolom untuk as A-2
----> coba 40 / 40 .
0,4 x (10+1,5).2400 = 4416 kg
A-2 = B-2: A-3 = B-3.= 87954 ,92 kg
75 + 1850 Aartinya beton eukup kuat memikul P te
kan tersebut.untuk tulangan kita pakai.
A- 1,5*.40.40 =
Sebagai sengkang pakai glo 1 5.
24 cm? ( pakai 4g22 + 4fi19= 26,61 cm2 )
e=zsPTsepuod
l ucOzt
tug' L= g' Z
g'0
tug' 9=E ' I
ITq
g'0
ue (-- ruc Stt.4a\. g' 00 [ ' S' L)(ffi
+S=rn1e[6ue[uedruc 6'E9E Qa
eqoc
+E
00['g'9'9(26' ZZO98 ) ( Z) L' L
=;n1e[ 6ue[ uBd TTqure
( lede: nTEf ral ) uro 0E [ =
Z-A sE uep 'Z-U
1u16ue edue1.1 ?-g = ?-v = [ -E
( .,uc LZ' g l = ZZf,? 1e>1ed; c
d Tn{ruaur {1n lpn{ dn4nc uo4aq etru11re
' v'oE8t +
6>1 Bo'ElgLv = tgvz + go'Itt6E
'-rnTEf TsEpuocl TErlECl ElT>{ {Teq qTqaT
0E€ - 009 = >1ede1 a{ 4ede1 qTs-raq'>1e.re[
se l sepuod >1ede1 . {n ue{T1eq;ed E1T>{ e{T.C
ruc ov1 TTque urc L' 6EZ I a
Zg8'0)mEr9[7) I[=o
'ruo ,EE = TTqrue 'tuc 6'tvr<;-o se rsepuod
' B'o> nery#tff,r =-Oz z-v sP Tsepuod
' 1xe16ues irn[nq ) 1ede1 lsepuod le4ed eqoJb4 zG'zzog8 =vT?z + z6'gtst8 =
t:B = € -V = Z-g, = Z -V sB l sepuod eped e^deg
64 Bo'g rglv = vgvz + 80' I E [ 6e =
(20'V - glfiZ, Te>1ed)rurcgL,Z =(szl (0zlzL qqz r08 LZ
= uTuI V
= [-V sE lsePuod ePed edep
'1gLfiZ te>1ed rV
ne-0
unuTuTru . Tn1 1e>1ed1p
' Oe/ OZ' {o i;5ETcprq"- ueeue5uardE
S[ - OLfr 1e>1ed bue>16uas {n?un
-ulc S't[ = 0e '0e 'tg'[ = TTqUIE Vc
'uB{41
'JTlB6au qaToradlp vsL'(vLz -006) = 80'sL9LV
= 1 u16ue eduel ) aloTolt edeD
u
- 662
'0t/0t"'{n: ,-g = ?-v = l-g = l-v sE uoTo{ {caqc
300
Ch-eck gkrlggrn belok poqdasi s
gnetto=ffi= 7,169.-/n2pada balok pondasi
q= 7,159(3r2) = 22,9408 t/nM jepit = *rr2,e4o8)(r,25f
= 17 ,9225 ton m
M rap = *tr2,94og)(2,5 + 1,2512g5 ,023 (2 ,51
= -53 ,755 tm
Coba lebar balok = 45 em
= 9016
beton
h =VOA- o ,262 /53 755.1 000
ambil h. = 95t, uk. balok pondasi
Penglanpan bq.lok -:M_ lap = 531755 tm
em
a5
{:;
0 ,45cm ( sel imut- 4s/e5
cm).
100 nLrJ = 1g ,51
90 + 2,44
/0 r4 {ll = 19r51_- 100(21)
= 0r00929A -- 0,0 0g2g (45 ) ( 90 ) = 37 ,62 "*2
(gg25 - 39 ,27 em2 )1A'= 0r4 A = 15105 "*'
(pakai 4g25 = 19164 "*2)I[ tump = 1719225 t* .
A 0,00288(45)(90)A' - 4,67 "^2
(2g25
theqk geser :
Iintang max = 86r023
4= = =2!31!0 ,87 5. 4 5;06 = 16 ,'
jadi seluruh:. Legangan
= 11 ,664 ema
= 9 r82 em')
] roo .tJ= 6,04sI il= L,inr-100 ( 21 )
= 0 r 00288
' (tg25 = 14 ,7 3 "*2 )
22,9408 (1,251 = 57,347 ton
18 kg/cm )7, beb. sementara (=10)
geser dipikulkan pada tu1. geser.
Ca= 90
9225).10001
0,45.1950
= 4 r2321 ( 7
tr= o r4
'llIC 09t = ITTqUP
It gtL
ruc g, zs.< Ig'00['s'[ _f {
= rnle[ 6ue[ued TTqUIP80'E[91.?=dlrnle[ Teted:
'PUPIn 'Tnl .IalaUE .
TP
1e1ad Teqel P{Pu
xo (---- (
,wc/6>4 ZZ, 9
epedlirep TTca{ qTqaI(aruc €t'[ = S?,-gfit
er{ueselq l6equad 'Tn1 :a?EUTBTCI :
Zr, gg7S'g = egz't x tg7 = l6equred
urlrlEc'Tn1
ffi'
I
c
g0'sIg[?=
( ,tuc Lg'Z = S?, - gil Z*, 99?8 ' 0 = r V
(Z*, E0'E = OL-gi r"rlef )ruc E€,2'V = (LL) (00 L')GVZ00'0 = v(lflqgr =r'r4 z'o =Q 6V200'0 =
ffii',oor J -=g!?r,:i :;3
L ts'? = = EJ :;3 l )Jg I-:- = el)
EOZru? LLo'g = z(ilLltL)(t)(69 1'Ll r = lTda[ H
T!&
1
: TsPPuod ltiladue6uedel qPraep {nlun E t- OLfrZ
upp uendurnl qe.raep {n?un S' L-OLgn 6ue>16uas 1e4ed
uc s6'L= =9ll?l;tI = 0E8 ['? ['E
,Q .SV
ly 'v ="v'OLf, Oueanr8uad V 6ue16uas 1e>1ed elT{ TuTsTp
'6ue>16uas nelp Eulrgrl 'Tnl ue{eun66uaur ledep e1T}
' ( resaq:ad1p /
q:, (qz BltTr I ::ueiETE5
,wc/b>1 g'9= ) q3>(.
l['00['E[8'0 i*t :=J[0'8v26
uo? tog?z'6 =(g80'0 EtE't)( Ll6gL'L =[-LO
ru L = . lgf ad 6uB I ued tln
: raseD Pq+ 4e1ed TPqax {caqc:tltc E = uo1
-aq lnurTTas , urc 0Z = TTqurE +pTad Tpqa&
E,dffiI
w/a6g L' L=b g ?v
I nc
416 1 5 ,08k
netto
: 3024161 5,08k9
= 77 ,1 cm ----)uk. balok Pondasi =
Penulanqan balokiondasi :
S laPaqan = 251008 tm
2.@= 6935,8 kg/^2gnetto = ffi= 6 1936 x/^2
beker ja pada balok = 6 ,93691 ,6 ) = 11 ,
097 6 E/m
M jepit. = + (11,0975)(1,2s12 = 8'67 tm' f -4,G1 5(2,5) = -261008 tmM rap = + (11,0975)(3,7s
-Ch.eck uk. balok Pondagi -:cobab=30cm.
h =do /T= 0 r262
ambil ht30 /80
= 80 cm (selimut beton = 5 cm).
75 = 2 r39100 nti= 19r51
r^)= +tdh= 0'00929
. A = 0,00929(30)(75): zo,go.*2 (5g25 = 24r54 "*2)Ar = 0r4 A = gr35 .*2 (2g25 = gr82 t*2)
I{ iepit = 8157 tm ----) pakai A = 2g25 i A' = 2925
Ghegbqeser:Lintang=41615,0811,0976(1,25).1000= 27743r08 kg
ta=ffi= 14,0e *s/cm}
)70 beb sementara ( =10 )
semua teg. geser dipikulkan pada tul . greser.
pakai sengka ng 3010, A" = 3 '1/4 ( 1 )2 = 2 '356 cm
5 -A"'4-= (2r356)(1850) = ror3 cm
pakai sengkang 3gto-1 0 untuk daerah tumpuan
dan g1O 1 5 untuk daerah laPangan 'Untuk pelat Pondasi :
I*gi ide* untuk pondasi as 2 yaitu l=08 - 10 ; A'=80-
25 ; tul ' Pembagi = flA 25 '
[;
zt (76,008 ) . t ooo'0 r 3. 1850
= 014}
ul
c
,
26 ,008. 1 0000 ,30
sg t 'sZJL, O
!!!---
-
Iul=r :gB ['g+ gg L'g- ?,lL'r. zL t'gi
EEg'z- vtg'ygtg'z+ E L'o-
?89'o- E?E'[- ?,ve'[- g Lv! L-BIt'[+ EEg'z+ E['o- g[?rt+
t'0+ Lg'E+E'o- t'o+
L9's+ ggg'e+vg'I l,- to'I L-
EEg'e+Lt'9
'tulE'0=Z( E ) ( ?? t ,0 )
rul L9, sruc64 000 LgS
zL cp--T = dl^r-
tu/l vv L
ut/b>t VVL = QO7_?, x Erg
pc qc co= -dW = 'dW- = jlrl
,O = b Tf,Tpuas le.raq lpqT{B
qPO
= jt^lop
= '41!
.raut.rd ua'urory
za
z(ooE) )
wi@=
qc oo= gW= jhl
z1 P
a.IgE= ;^ ) ' tuc E = leqT {P ueuofix Z'O = ureoe aT+ TrTpuas 4e-raq
0t / 0Z up{p.rr4;a{1p ueaq eT+ up-rn{n'tuc E = TTqUP PlT{
' 1( :es-aqas rnlef lsepuod EuTseur-6i1""*@-pd upun.rnuad TpE [.re+ uB{T sunse p1 - |
- Tt ' ,, rlpoq aT1,, ue{pupcua.rou {n1un I
:
L p 1-.--)'t$r{;l'*-r* :
p b\ q) pl
o:8=VSe
r ueaq aT1,;
' 1 .:n1e[ lsepuod.
E'0-e '0+
r tuPaaTl "
cnc
6uTseu-6urseru e.rplue 6unqnq6uad {oTpg )rurpaq aT,f,.
r:lII
l
l
ii
;
304
Penul-anqan tie :bgglLj uk. 2O/3O--#
N tumP = 61712 tm (di b)
[=
25
z1-G,712).100!(a,2).1850
0r4
100 ,nu)
"*2 ( pakai
8 ,55 cm2
= 1128
A = 0,0335(20)(25) = 16,75
A' = 6 ,,69 "*2 ( Pakai 3g1g =
Itt turrlp-= 5r185 (di-c)25
= J0r3570 ,35
=-1 100(21)
= 0 r 0335
efi9 = 17 ,1 "*?
100 nU)= 53r61
IZTTI lBs).1ooo
(a,2 ) .1850
cr.
(_0-A=Ar=
Catatan
0r4
= 1t46
denhh
o', ilt-15
53,51= ffiiffi = 0,0255
5g 1e
5g 1e
0,0 255(20) (25) = 1'2,8 .*2 (pakai 5g1g = 14,25 cm2
5 , 1 .*2 (pakai 2g1g = 5 ,-l cm2 -)
: Momen bertanda + untuk tumpuan B dan untuk tumpuan e
lah j ika perletakan b t,urun sedangkan yang lain tidak .
Dan jika perletakan c turun maka *0men di B= -5, 1 85 tm dan
di c = + 61712 tm
ada
sehingga : tumpuan B perlu [ = 6919 dan Ar
tumpuan c perlu A = 6919 dan A'
sebagai sengkang pakai fr8-10 untuk tumpuan dan 98-15 untuk lapa-an.SKETSA PENULANSAI'I.
gl?-30
910-30
i
l
[,&
1t,,
22 150
dagar.
I
I
a?.--sfr
-T ,lI
.t T7V
t).TaNI
l\)
il-J t
Ia6l rl llJrlI
l,_
:T
,*
-Z Lfr
L-Z L6L- et
0€-016E-OL6
az-016
s r- ,sil-,@
9'L-?,1fi
, ZL_ZLHOT-ZLfr
0r-01fi
OE-OLfr
OE- OLf,:
g'zl-zl
z0 Ll
305
r--- - ---]
tulangan susut z gB 1 5
35/5s
35/65
35/65
Penulanqa+ pelat da
o35/6s
33/55
35/65
Balok as =asA
Balok as 2 = as 4
3s /65
{
ZRbt.g
I€5--,pot 1 -11p
8-2 0
3s/6s
35/65
35 /55
)d24a
P-e .1Od
. Szfrt. 6uBsed ElT>{ p{pru 6uedruBuadt 6uer16uas ue6uesBuad {nlun eue.re>[
-rde1a1 gzfrZ= tV, SZfrZ=V : uendunl {nlun 9Z-
l*foz
a
B Se = V se )tofPg.
@l
Orw
Fat
t
rIJ.t --
ffi"rls
il"1
(^cqC
ot/02
I
Aul
\0Ut
vL/l"
I
I
t\'oF(Jl
\oaJl
,A
UL/ OE
/ovottay'
ot,/ot./ov \
/
c /nzlt:t0t/0z
Ezfrs \
308
l,
iilly
pondasi as 2 = as 4
40/4A
7rlgs
-v_
T1s lru
sil
pot. c-ctumpuan diperoleh [= 3925 , tr I =
untuk tumpuan dipakai 4 penam-
4925 .
.tafrZ1 -lt\-/4910-7 11
;4g25 fia-zl es
ffiJ-nLg6-
320 h,0lc-4+t
rffi]Catatag : Dari hasil perhitungan untuk
2925, tetapi karena sengkang
pang maka kita harus memasang
t., 30 .r
kolom as A- t = A- 1
Fi[rA-4 = B-4
10-15
=B-2= A-3 = B-3
T2fi10g8-15
30 2g 1e
"tie beam"20'/ 3o
senEk.ang g8-15 untukg8-20 untuk
tumpuan
lapangan
\IAfi22+4g 1e
T
tt bracing "
291 6
Ij'Eii'I
I,. uP{TEqPTp
1dp glsed qeuel ueup{o1VtA = qEuel upp uolaq
pjre?ue up{asa6 uelsTJaoX'z'" /
6>1 gr0 = .respp 1e1ad qpaEqfp qeupl uTzT ue6ue6a4
:.reeu{ra1 rds raAaTTlue}t
?nqasral
e ue6uelnuad Esla{s uEp ueEuelnuad der16ua1 qsl 6u1pu1p up)tpugcua*
adT+ qeuel ueqeuad 6u1pu1p Tnr4ela{To %. r{Eupl ue6uelnuad 6u1pu1p ue6uelnuad eEla)ts - a
e er{uue6ueTn? upTluaqraqurad up{eupouar n1e1 1e6e+ 6u1pu1p TrEp uaurou ruer6elp requrE9
.e_E uPolod eped ue6ue1n1 ue6uep uBr(ButpsTp nfred tp1 qs1 6ulpuyp 6uefuedasue6uelnuad - ue6uelnuad ,Ip6al 6u1pu1p eped .1n1 ueleuaq6uad {n?un .pe-p uE6uolod eped ( rualsr ) tp6o1 6u1pu1p eped ue6uelnued. ue>Jeue"u.*
....qs4 .rE6pp 1u1ed .1n1 up{purcuag .q
z qsl buTpuTp rrsEp 1e1ad eped 1pe[ra1 6uez( qeuel ub6ue6a1 6un11g .e. : eluTutTc
'seleq uplen{et erec ue6uap uepupcua:ed ,li:n elcq,S4!X uolaq nlnH.uE{TEqETp g:tsed qeuel uEuqtai
'62'O = rTltr'.I'uel u.uE{al uersTraox .s.lprp :eqrue6 eped unluec:a1EIuuP.h{n-uernqn 6uezt raAafTluc:( ad11 qelepe qsl qpurl ueqpuad 6u1pu;g
Gyo
Eul6rtO OLL -r4euee^Xvlr-'
0dFdl
-uTTau uebuolod: TUT
PuPruTp r{euPlqPltlPqTp requp6Ja? Bz(u6ue1uPqpuad 6u1pu1p Tnr{e1o}tTc3 [ tEos
w/19' [ =,QBu'l/C
qeuPl
,m/l1u=b
608
uPrITlPr TPos - Teos E. rla
310:
SoaI 3 : Rencanakan'r.J
ti r*'
Soal 4 :
SoaI 5 :
Soal 6 :
SoaI 7 :
dinding penahan tanah tiPe ucounterf ort tt .
Tr^nuh= 1 ' St'/n
g = 3oo
teg. tanahizLn= 1 ,2t/m2
Rencanakan bak penampung air berpenampang persegi panjang ya
ng menumpang diatas permukaan tanah. kapasitas bak = 1 00 *3,dalamnya bak = 3m, perbandingan panjang dan lebar = 1,5
Ferhitungan secara elastis ?
Rencanakan tangki berbentuk silinder yang menumpang diatas2
permukaan tanah. KaPasitas = 250 m' ?
Rencanakan bak penyimpan air yang berada dalam tanah. berben
tuk persegi panjang dengan ukuran panjang = 7m, lebar = 4il,tinggi = 3m. Bak tsb diberik tutup (pelat beton).Rencanakan kolom renang yang penampangnya berbentuk spt ter-gambar.
40m
t
0m
,2m
20m Petuniuk :
Gunakan eounterfort u-
-1
ntuk dinding.Soal-soa1 1 s/d7 tdkami bertkan jawhban-
nya, anda dpt melihatcontoh-contoh soal utkenyef esaiiiannya .
1 ,6m
H
[1
[:!,;i:i
- tlt -
.uTzT ue6ue6e1 ppEdTrep TTcat qTqaT snr-Er{ p[pq upp uo?aq pppd TpEFeaa 6uert up6uE6al , TuT apoxau pppd
'sElEq uElpn{a{ pJpc eped up{rpsppTp upxpn{a{ uB-6un11qrad 896 t )tp[as upp .'096 u p/s 006 t rtl TBTnur uptpun6radrg
'v
( r TaalsPTTu,, ) esPTq
.(uu.rsep :f;;;l:,
("g=)
,wc/b>4 t0,0 = Tsd 1 1e6u1 F-iiETEiE,
,wc/6>t 901 . €,0,2 = "g
,
. : sxw uBnles UPTEp
ge6ue1n1 eteq tln TSd. 000.000.62 ="g
a?eurTlTn,, ) sPlEq UPXPn{a{ srPS . gssa.r?s 6uT{rorril ) sTlspTe p.rpD .
V! iffi t.rrrr
?' Z 'E'8 t sd IDVs
r-E = r - Z-rrrAell/\T I s 1e6u1*
(siz) EB'o = ?ltw/61
g81o'91
,utc/b>1 E[,,9g !
.uo+aq ,, lq6Ta4 1!un,,
'a
It, t ="^t 3 sxn uen?es
eu/qT Es r p/s #/qTTsd ::/
p{eu 'gzzsl uo?aq n?nu,uEtTpsTlilr ffi+-iEffi.(
rwc1o47 3y 00 [ '9 [ =ts =sxw uenlps rueTpp( Tsd I 3:Jr ooo'ts ="s e6:eq
3 ( tr/l ?,Tt.'z neaP
,lltqit stl =-M r
leiulou uolgq l'(ruc 0t 166u11 ,ucgt ;aleupTp) rap
tftfs ednraq ?sa1 qoluoc tnlun ( Tsd))tT1srral{eret uo+aq uP{a1 uplpn{atc
t*c
= !l EuEluTp
( e*/+
rupTEp )
oG ='u {nlunc.c
E, [M 'EE =-f,: [.E.g.sd.f3v
gv'z p/s e*/1
' [ 'rrrABr(urel
-uauot lnltTreq E 6-g [€ goOD rJV Trep TTquE Ture]t ue>1o1ed le6eqag
I i r'L
tf,ro gqoJ rcv
-
rrrA gvs
.' ,1,:
'
J . IIIA
- 3.12 -
ACI appendix B.3 dapat anda llhat tegangan izin, sbb !ACr B-3-1 Teqanqa tekan beton tak boleh nerebihi harqa -berilut :
". :::::_l:ntur (tesansan yane terjadi pada 6erat tertekan)
---) = 0145 fr"b. Geser :
untuk*
*
balok, p€latgeser dipikulgeser dipikul
1 arah, pondasi :
oleh beton ----, ,"oleh beton +
= lrl4
, '" = |r|rfl
= n+ hrG;
20 .000 psi2/t .000 psipetat 1 arah(bendan harus fu""*
baja tul.untuk pelat Z arah !
* geser dipikul oleh beton
dimana lg =/ c s:-sr yg pendekB -3 -2 Tegangan tarik baja f= tak boleh melebihi :
a. Baja tul grade 40 atau grade 50 __*_,b; Baja tul utk srade 60 j-- --__)e. untuk penulangan lentur, frtuL 3/g", pd
tang kurang dari 12 ft ) ____) 0 r 50. f__
ng dari 30 .000 psi. Y
diperoleh dari mengalikan,servieepembebanan.
ACI ps.9.2 mengatur tentang "faetored9.2.1 Untuk beban mati + hidup :
.U= 1t4D + 117 LU * "factored load,,D = beban mati ( "dead load " )
,L = beban hidup ( "live load" )
__-) u"
f, = tegangan le1eh baja.B. 'Ultimate s th methgd." ( kekuatan batas I .
Pada metode ini perlu anda ketahui beberapa istilah sbb ra.
:
adalah beban yang bekerja p,d suatu konstruksi tanpakalikan faktor pembebanan.
"serviee load" terdiri atas "- berat sendiri ( "dead load " )
beban hidup ( " I ive load,' )
beban angin, gempa, d. l .1b. "Faet-ored lgad "
di
f oad'* dengan f aktor
load" sbb 3
i;' 0L' 0 = i 1-- 6ue>16ues ue6uBf n,f, Z. qSL'0 = f, (---- TErTds ue6ueTn,f, !.q
-rnl,er uauou + up{e1 TEU-rou nEJe 'ue>1a1 Teu-rou . q06'0 = fr(--- JntruaT uauou + >{TJpf T'u-rou nele , >{T.:re? Teu,ou . e
, '-rnl,Ar + reurou nple TEU-rOu >tnf un z. z. t . 6
0 6' O=f, (---- ( Tpru.rou eIe6 eduel ) Tu-rnru rnluaf {nf un ;.;.r. u
3 i 6ue1ue1 rnf ebuaru f jV -2zt=6 rsd
:@en>{p{ Ts{npo-r -rof>[pJ up)tTTE{Tp,,q16uar1s TBuTrr";j';:r;]
'a .ue>16unlTqred1p unTeq (fr=) ue+pn>[o>[ rs{npa-rfol>t'J puerurp 6uedueued uelen{a{ ue6un11q:ed ueeun66ued
ue6uap eduue6unqnq uefep 6uedrueued n+ens uelen>[a>[ qpTepp
'p. ,. pEoT
peJoJcefl r Tn{rueu {nf un6uedueued nlens ue+pn{e{ qeTepe' 1 uelnf-Tp Ouer[r.r 'c
. ' 1nsns nelp dee-rc
nPle +uauorlles rpT+ua-raJJTp,, lpqT{p rnqurT? b^ ueqeq = .r
, '(.L + O ) Sg,t'(T LrL + l, ?rL + a vrL) gLro = n
qaTo ue{leqr;;ir';i}=,;:;': ;llr':'lilu':'l;;"lJ;l; L.z. 6
i L'| + o 610 = n. qeuel
irTB uEue{af -Ht + T L'L + c V, L = fl+ dnpTq + Tleru upqeg V.2.6
ff t?,1 + o 6,0 = nnPJP
, peof a>1enbq1-rea, tf,L8,L +T L'L +C V'L) gL'O =N
edua6 + dnpTq + Tleru ueqeg t . Z. 6M trl + o 610 = n
'-rpsoqrol r{TITd
'qeuel UeuP{A1I u.ue{a*
";*;,nple rTP r,rPue{o1
. reseqre1 qr r Td
{, "
{,^^
a
t; t + T
utbup
,, PPOr PUTS,,
L'L + o V'L
nele=[
) El'0 = n
tlt
resaqral qTITd
+- dnptq TT+eU ueqag z- z- 6
314
Harga jika t, (il)
60.000 psi (=A2OAkg/cmo ) , tulangan simetris
- h-d'-d_dan = ? o'70--TUhtuk penul-anqaq spiral.
fi=o r9,A-0 r20 T-
mana f= 0rl. f; . og
A = b. hg
fl= kekuatan beton karakteristik (utk contoh silinder)prr= normal nominal yg beker j a pd suatu eksentris itas .
Harqa I dpt dioLbarkan denqan qrafik sbb :
fiP',fi=or9-1,5 Frlorzs (tur.spirar)
6P0,9 o-2 #> 0,75 ( tut. sengkang )
9 ti .2.3 Untuk geser + torsi ----) g = 0 r85
9 .3. 2.4 perletakan pd beton ( "bearing on concrete" ) ----- fi=O ,70
vrrr. 1 .4
"Required strenqth"." Ir{--= " Factored moment " pada suatu penampang - .
up = !'Factored axial load " yg beker ja pd suatu eksentrisi-
utas.
\f = "Factored shear foree" (geser) yg bekerja pd penampangu
T = "E'actored tOrsional moment'r (momen torSi ) .u
.
[r{-= "Nom'inaI moment " pada suatu penampan(l .n
!{. = "Nomina1 momentr dalam keadaan seimbang ( t bala[cc" ) .Dp = "Nominal axial load" pd suatu eksentrisitas.n
p^= "Nominal axial load " pd eksentr'isitas = 0o
P.= "Nominal" axial load " pd keadaan seimbang .
V:= Gaya geser nominal Vang dipikul beton.g-= Gaya geser nominal yg dipikul baja tulangans--T = momen tsrsi nominal.
nf-= momen torsi yang diPikul beton
?,\,f-= momen torsi yang dipikul baja tulangan.
s
i
' E, Z, J . oN Teos qoXuoc +.pr{TT uEtr.{ET;s , efiueulefi tn1un
(69'0 ' fr# -p ) rr.sv = J'-v E
l a/ L-pl dJ
' "v =9tt
= P (r---,f * )IsI'V=.1
sJ' V
',. ie:q: ?l SBrl0, = ) l
., "; ..: .,,, .,,.:,
3 upf Eosracrat TTEqTI5X
I
( 000?
sg'0 = t/
..'
: qqs
E9, O < snrpq uppc'. 000 [
f,::LLdqure6lp
',' '
x1n (---- (.rapurTTs .L.Z
11gu.r6- llnlueq rueTpp. : ':, i'r ,' ..
* t1 (,;,-;.. Tsd..i,:.
ruc/bt1 0,gZ= ) Tsd
LA'a*-1
x'
ccr f ' '{
'sElEq uelen{a{ Erec uPtTpsaTesTq
' s' J'rrra:-
' ,g16uar1s f puTruou$ uetTIBtTp , = ,,q16uer1s g6TSop,
,,( s,x
+ ',1 ) i =puESuor T srol uauolg = u,r'
,;"; * 'n'l fi = puecuar'rr".u =;;; = ";;
)r TETxP eFBg = ua
/'6uedueuad nlens eped Euecuar uaruol^l =
tW fi
ue{JP
-,1
(?r)?r
,l
000,
0007
: qgs E' L'Z'0 J
IHM r{aTo,LUP{TESTTE apTTp
I sd rJE uBtep rntrrp ?,"rreu
-?,
'rrllbUArls UbTSApr
316 -vrrr. I .5. 1 Xgadaan geimbaqq ( balance ) paJla Iiekllitaq ',bagPs .
Pada keadaan seimbang 3
t drtc
e=c9r=e)c-*b
*b-il
_ 8700087000 + f
v
lrofv
t*b( r'a-
ASD
= b.d
Catatan 3
vrrr.l .5.2
dalam satuan MKS
*b oroo3'-=-d,\'2,6tm5 + o'oo3)
fo= += -f,' , {ry+6ffi '}
l{erencanakan ukuran penampanq perseqi seeara kekuatan batas-
Jika diketahui momen yang bekerja, tegangan baja dan beton.6c=0 , 00 3
w$l"=orooir-qo,8s' f;
A.. fsDy
0r003,, ( regangan pd keadaan Ieleh ) .
(2s+
f(ffi + 0,0c3)
nmtb 'b0r85, f'r.ar.b ='Asb. t,
orSs f; .fr.xo.b = Fo.b.d.
diperoleh fo !E :-t' .v
0r85. f;==-r'fr' ${$f;ffiACI ps. 1 0.3.3 membatasi untuk tulangan tunggal :
f^u*= a'7'fo
A hg4s. r;
.: ;
,'g uup. , . oN Tpos qglt{oc XE1{TT EpuE up{qetrTs I.
I ' ' ',u")tnruaiTp rpd=q p ', ol = svE irB6uetna , ,, ,
"u upp a/ t"ttu T6ET rreca{ qer16ue1 TrBp rlaToradTp Euer{ p uep q TETTu ue{xEsppraE
.tococ 6uer{ p uep q e6req qTTTd ntpt
's
PlT{a? ,t
unurrurru I - , d PpPoTrPp
t.'t'0['sd rJv
nu
' . :'' ' li-r3 ,P Q i :.,' U-- iry
3 D q e6:eq up{nlua& .gi-Zt
(l's.01.8,d lcv) # - umnsuTurv.vv
rPsaq qrqer snr'q upp ,/ ,r'o> )?p:Bfis Tqnueurau 6upr{ e6*eq nxens TTqurV .
Z
( I 'S' l,' rrrA lErtrTT:
n
f,r t -tf -r)t =) np?E tn
_u -ruz _J,gi- r) ^r.r/=#='u
( racuP?sTsar lro luaTsTlaoc$ ) uEuBqel uaTsTleox" U=
lU=: uEtprupu
(p'c,
J'E8'0lrt? -P)f,r'p'q')=
'p'q ';l = uw ?l'(sfr
-P) Ia
(p
'Eg'0?i ]lf L,
:)=U!t
, aF
318
rr.1.5.3dihitung dengan cara kekuatan
vqec
batas.
0r85.f:
,i\la
-L
Diketahui nilai momen yg bekerJr, 'akan ditentukan penulang
an.Lanqkah perhitunqan :
El. Tentukan dulu tul. tarik maksimum jika balok ditulangitunggal .
x.=D
x*ar, = o'75 *b
C- = 0185.c maxar*c max
T;i =* =-'s max ty
f t. b.c amax
;,1t.
klIrII
i
1
M = T(d ; "*"*)- -n max z, max
b. Lalu dibandingkan dengan momen yang bekerja.jika momen yang bekerja kecil daripada momen max
yang dapat dipikul oleh tul. tarik tunggal rmaka ba
lok tsb ditulangi tanik tunggal.jika momen yang bekerja lebih besar daripada momen
yg dpt dipikul oleh tuI. tarik tunggal, maka baloktsb ditulangi rangkaP.u url. 9$*e..:,
-
c. Untuk kasus momen yg bekerja lebih besar daripada momen
yang dapat diPikul tuI. tunggal :
e'
j ika le-c
s
S
c'1'Menentukan Msisa = MbekerSa Mn max tuI. tunggal.
s
check
Ms 1sa
a-':t-dulu apakah tuI.
s 'Zc = i(x-d') :
8S- " " 't'" ( tv --l
tekan sdh le1eh :
x
tu}. tekan belumleh tegangan baja
t'"'E"EC. Lg - "yjika 2' ,7 ty --?
^ " tt-*-r,J'EB'0
(p'9',) lt?r
^, J-0609, tl
' 0609 '
oa <',7
),) -) t
_.d 98',0- J )
u5'PI J.*lt
=-J
s,3
: qalo;ed1p
[^a _ _L_. _ -' \ (,p*x)cZ
qCITaT qps ue{otr . Tnl trere{s
( , P-x l'jx 3 (,P-x) =
t ,/ '?r's8'o-__-7.€:^ p' J
",?,3
(+)'
<*-l--)l
: sp+ETp upBruesrad- r{aToradru5irr-rlreJ' SXI^I uenJeS (----
^ ss'o ( t"--= -Lt ,) -) , nele, ]'98'0
: qaTaf qepns ue>[o1 - Tn1 1e-reFg
3 UeqpqulP.f, . p
'-?r;tl l.Lw[
)roar{r@
r =vO
2g
p,P
?r'EB'o
,/
dJ-000t8
000r I
s r e66un1 - Tn1 xpru s rV + -V gL'O =
: apoc rJV xeu
c3+Xetu J=
V
V P}tT IXEU S
V
Sv pr{1
"':'
^
' rnl snse{ >{1n ) - -
J eu'urp J,tr,
c,{ ( irsS'o-"J ) ( qaTar p!{s ue{a1
6rt
=?oO
- 320
€.
ry2€,f--
x 7 2ro3.106
f
atau x ) 9%-.6' ..... (**)4 5o90-tr'* " "pers ( ** ) disubstitupikal I: E?t= ( * ) diperoleh :
&(f-('t,-=il-))) ffiatau ' ( (- f 't'- -t+li) ) ) o'8
'' f ,'\ ZACI Code ps.10.3.3 menentukan b untuk tul. rangkaP. ,
. f t=b . '
=fa + '(==-)= f A^lance untuk tul ' rangkaP'
= (A^Iance untuk tul . tarik tunggal .
, ,6090 .'= o,8s.tL/ty.fl .t*mf;
15-090-- q
'6.090-f ''vr'r r r-* r',:,.. -'t' a):
(a
(,
Po
tr'caIt
SD
Ars5A-87000- $t 87000+f , ) ( t, --, satuan "us cusro!{ARY"
6090 - # (6090 + fy)( fy --) satuan MKs
Balok penampanq perseqi memikul lentur nurni.
Dihitung dng cara "working stress design" (cara elastis)Pada metode ini ada beberapa asumsi sbb 3
€1. Bidang potongan tetap rata setelah mengalami lentur dan
tetap tegak lurus sumbu konstruksi.b. Regangan sebanding dengan jaraknya terhadap garis netral.c . Bcton tidek trelnikul tarik , ba ja tul . memikul tarlk tsb .
d. Ikatan yang kuat antara beton dan baJa r sehingga tak ter-jadi slip pada baja tulangan.
Di dalam ACI code 318-83, metode ini ditempatkan pada appen-
dix B dengan "ALTERNATE DESIGN !!ETHOD'..
tegangan J"eleh ba j a tulangan .
(^^; utk tul. rangkaP = 0,75. Fo * ('Gl-LsbT
v
wc/b>{ 00gZ
TSd ooo'o?.d
=Jedupns{eu (---- 0 ? epe.z6 >{1n'ope:6 6uTseu-6.rTseu {n1un
. c.,J gl'0 =
: Z- t - EI xrpuadde
,wc/64 3:J oorsl = c
,wc/b>t 90[. tO'Z = s
IesTUJ
cJ
IJV PPEd
g
u eupuTp
c
,3:
x_p =3
x 3 (x-p) = '3 , '3S
Vw=
'e
e uolaq u,p 'Tn1 p[pq pd Tpp[;a+ 6^ uB6ue6el ! uBeduelrad
: ue6ue6e:r rue;6e1p Tf,pp c
# qaro.radlp [ "J. sv = ,t
J fi =J t q.x.rr'g = D
-eJ'f nf uep , q/q >{of Eq up-rn>[n , (.W= ) e[-ra{aq 6uBd
.S'
ueuow ,";'l"lr;O=f,'=3 =sJ.,
f\
cE
B_g
cJ'U
=U
sI
c3fif,'3 =
L ZT,
,J' l'lr' t 'rrr,a l
322
B. Diketahui tegangan izin beton,rik tunggal (=A=).
Pertanyaan M, Yang daPat diPikultangkah perhitunqan :
€r. Tentukan lokasi grs netral :
b.x. + x = n.A- (d-x)zs=aaaaa
b. Check utk kead.aan teqanqan beton Eimanfaatkan penuh
,F :F!! CC
I_
.Jba ja tul', uku?an bhlok, tul-angan ta
pena*iang tsb ?
4
f e .ES 'S S=:-t n n
d-xf=f!.4! t,xc
a1
apakah , ft lebih kecillebih besar daripada ftf.){ artinya tegangalrbaja yg terjadi melebihitegangan izinnya -) tidakboleh.
j adi
criecratauj ika
Jadi M, = T.z dimana z = lengan momen = d-x/3c. Check untuk keadaa teqanqan ba'ia dimanfaatkan penuh.
sLFr- = ft ----) f. = a:i .ftt, (J Ll-x L
check apakah f" lebih kecil atau lebih besar drpd %Jika f. ( { ----) oK
d. Dipiluh harga terkecil dari langkah b dan c.
vrrr. 1 .5. Balo\ perseqi tuI. tunqqal dalam keadaa{r 'balqncer. 'underreinforced' r'over reinforced' .
Keadaan 'balance" (seimbanq) :
Terjadi jika tegangan beton yg bekerja = tegangan beton
izin, dan tegangan baja yg bekerja = teganoan izLn b.ji.Keadaan iunder reinforcedi (tulanqan lemah).
Terjadi jika tegangan baja yang bekerja = tegangan izinbaja, sedangkan beton yg bekeria lebih kecil dari pada
i zLn beton.
i
Slt
e . .-J*'{ *l
,l 'zcJ '{
G
r"J '"v
UE
l+=J
-, f -J, 'p.q. = oJ .=v = ,r l-v 1 p'g'{''J S= q.x. cr E= l
{ S *t = f,
P' {=x
=[
'. p
'c J
p'[ = z = uauouue6ua1
uauouruebuaT ualsfgeo:gurtnlueual{ . q
TPEI
1l+'J'{=p-E--=X
-Ib+31 3': Bsx
1l r cJ = (x-p) ! ;
.P
P'/ =2
ao
J'u
uPp{ofpq ue}n1nuPlIpqElbtTE-ffiUP
4ad {oTE8€'9' [ 'rrrAledecral m pocroJuTar .:raAor upeppa{ ----- qx
l edecral r pac.roJuTar .rapun r uppppa{ qx
'pfeq uTz'T ue6ue6a1 epedlr.ep TTca{ qrqar e[ra>1aq 6d F[eq ue6ue6a1 uppuolaq urzr ue6ue6a1 = e[-re>1aq 6.d uolaq ue6ue6a1 p]tr[ Tpp[:e,l
rat
(x)x=X
\
e{Tf
EEE
'(lpn{ ue6ue1n1)
- 324
d - Meneffi.ukan ukuran balok.,4L I
F1-- = T.z = C.zw = .b.d t (j
D
€. TulanqdA-f | .z
S
d)
bd2
1-=lG2-c .b(k d) ( j d)
.) k.f1
z.
(A"
f^.jS
Mw
M-
lvrJl.1.6.J Balok penampanq peresi t]rl. 'ranqkap dihitunq secara 'wor-@''-
Jika ukulan balok ( ut,k keadaan 'balok dengan tul . tungrgalleh diperbesar lagi, maka tulangan rangkap dipergunakanganqan beton dan baja yang bekerja lebih kecil dari'oadaizinkan ) .
) tar( agaryang
bo
be
di
|li.
!
tapqkah perhitunqan :
a. Tentukan M = R.b d,2
ul*.". R = +2b. Tentukan A =P.b.d
"lCheck terhadap A-sl
C. ItI . E M = M ltlsrsa *2 w *1
M* = momen yang
M ditulanqi A dan*z r s^MAJd.c2=TZ=ffi
T)€.A=.,,5(t*2 f"'8. A = A + As =1 "2
g. f,. .: f" = (x-dt) ! x-1
----) utk
fc . j.k
Ir{wl
fs . ( jd)
beker j aArs
t. \c----7 r
tul. tarik tunggal
ISd-d'kd .f c
x_df -=-f =c. x c
I
rfr
ii
i;
t ' (t'l-ar 6uedulBuad ) u uoTlcas pa{Dg.rcm eped pTs.rauT uauou =r"r
JD xPu xPuI_ _
'or'( ,(f=fi ) -r) + ur.r,q,=
acrgF
: e1er 16eqra1 upqaq lpqT{V
d- I buelual :n1e6uaru t. 2,. E. 6. sd IDV
_xeu ,)
Ia
Itt
-uoe qe6ualrp 1esnd.ral ueqaq ;#;"r ''g ?gE -x'ruo# E -E- = J
1:
ruoT+caT Iep auryffiPtbururlnpuaTl' L' [ 'rrrA
LZ
- = T
s'gl
- = T
Iculru't{
UTUI' I{ v--
uTru'r{
( TP1ol 166u11= 1 uTur{
l, .l
_tr i r +
11 = )l
*
HI
9t= T
rrBrt-
3 qq! q"Tupe uer16un11qrad-Tp t{Esn {ppTl uelnpuaT pueurp }torpq unururru t66u11, L. Z. S . 6 . sd
{oTpq uelnpueT 6ue1ua1 rnle6uaur Z. g.6. sd IOV
(r'+ - irZa
lcJ'( [-uz)
Za
(E's'g
tl' [ 'rrrnf'v L 'oN r,tro1uoc lBr{TT epup up{qeTTs
= ?v E{pru "J ( t,r uz BITF
s s [-,
' sd rDV ) st upp t',
uz : ue>16urpueg . r{
szt
- 326
"Crackdd-section" :
irt>t Menentukan
Stat i sb.x.
x:momen
1/2 x=
x=
b *3 +
thd grs netral
. :l:n A" (d-x)2 =
=0
"l
crJ aaaaa.
As
f rcr yt
f, = 7,5q psi a
atau = 1 ,984 trL kg/emz
?1?r ='a bh"g .z
f; = 0,83 . * bk
IgM
Yt= 1/2 h
B.Frs.vqrglF,qn !,endutan iAln- ( AQi tabel -9.5 . 1 b )
f,= # (akibat beban hidup, pemb. jangka pendek) .
f= # (pembebanan jangka panjang).
vrrr 1 .7 .2 Lendutan ianqka paniapq ( "lonq time deflection" ) .
f,i.r,gL. panjang = fi"r,gL, pendek + f,.nro"t creep+ f"oibat susut.
ACI ps.9.5 .2.5 :
Gxiurt ereep +
dimana { =
' f,:irngka pendek akibat
beban mati + hidup) (a )
shrinkage -
f
r'=
1 + 50
Ar
i'A;= tul . t.ekan pd bentang tengah un
tuk balok menerug atau 2
perletakantul. tekan pada tumpuan utk ba-Iok kantilever .
A = tuI. tarik.harga
I ^iambil dari"t"b.l berikut :
jangka waktu
)s thn1 thn
1 /2 thn3 bln
2
114
1 ,21r0
t{
x TB-r1AU SJ6%
FiT'i.
'sp?Pq uE1€n{e{q
e.recss up{TEsaTasTCI;-TEE
unl upTnlraqTurnu rnluaTuoluotuTn{Tuau .l
q Z/L19
T ZL/ I
T Tco{-re1q
T V/ L
1 9t + ^qT T ca{-ra1
adz'8 ' [ 'rrrA
T >toTeg
.tr {oTe8
{OTPq
, : e-re1ue
'{otpq 6ueluaq =J
3 e-rplUe
i :;lqlrtd:
; ";i qtrtd :
lr ( {oTBq uelen{e>1 ue6unl1q-rad }tnlun )
: suar} grlq z. ol . g. sd rJv'JTl{oJJo .reqaT a "q
. suaTJ 6ue[ uedesue{e1 ue6ue6a1 TsnqTjrlsTp
..rTS_tP
lp 6d +ds ue6ue6e1 ue:6etp{n1un uaTe4T}te suoT } -rEqaTffi
: suaTJ reqaT[ '8' [ 'rrrA
{oreq
LZT.
: & {oTBg
- 328
Jika
utk
x ( t --) balok persegi dng lebar = b", tinggi = x
flens: (x=t)keadaan garis netral meqelg rbatasan badan danf,.eooot o.AJ fL
-il
'iFri H
mft_ rlt ,il I le.rd
-a--l-:' -=L:-!u>2y
C= 0r85'f; .Er. b r A -f
T=A - f ) c eT- A= t,
Mn= T(d-1/2 a)
= A- . f -- (d-a/2)syA.f
= o".ry (d-ffitUtk keadaan aris netral memot badan
tinggi blok teganganb.eton=a= tebal( "web" ) sedemikian sehinflens :
!d-L
Q= 0r85
f=A .-S
Mrr=T(d-a/21 ="Untuk keadaan aris netral
F6-^s5fi'
es ) tv
----) t =
(b -b )c' e w'
harga a
CL uaaaaa
ba aa
GtL
fr'*.f t.tc
fv
.b \elIc=T)
A .fS
,l'= +
fv
c1
AsT
-2 (or85.ff
I
aaaaa
I,i:,s
Ii
r . diperoleh
cIJsg'0
,3
'tEpTl nEXBTurnu ,IltoTpq
,q1= p de66uV
-au qsluEfratequauoutIB)tr {acaz1
or*
zc = zr, eu'urp j= q '"o ' L zr
[c = [,r, EuEruTp ]= q t'o
r 5 [.I,
l6asred {oTpgTunnu ,[
s
(-- ?'
(----
(----(
v
(^
o'p
1
1
9"-- -.,- xp.tu sv SL'A = -vor= qs
+ V= V
q-qt?t s8'o = zc
'^q'?J Eg'o = tc
>ex(, q* : nTnp {ceqc.r * r*.r/= qp
grI
800,0
,, acuBTEQ il
c= ? euPuTp p-
Buuqurtos u€epea1
=? + '3o?
urET Ep
?
o
of,E I t-tJ. t.-'1
ig Eg'o
E. t.0 [ . sd IDv3 upturzT-TP b^ unurxeu 1 1e66un1; )tTrel .
Tnl senr nTnp ue{n+uafi i 1
: ue6un11q;rad qerlbue'J4 1e66un1 {TJE1'Tnl upe,{uBlrad , eE:ar1aq 6/t ueuow Tnr{p1a{To
a
Y'gI
,) nr o))
-Tp Turnu rnunl uPbuErn1jtaq & 6uadt'8' [ 'rrra.LL'oN Tpos qo?uoc leqTT Ppue
" = (Z/l-p) tc + (Z/P-plue{qeTrs
7U"3 = I,{aaa
679
-330-C=0,85fl.b".aM' ys dlapai allixur penampang = cld'a/2) = ... "
. heck apakah Mn yg,diketahui lebih besar dari padla M yg dapatdiPikul PenamPang utk keadaan a= t
Jika YA ----, artinya a) t ----- T murni '
' TIDA( -) artinya a ( t ----- balok persegi
3 : ltenenlukan tglanqan :
Mr, = 0,85 fi b, .a(d-a) + 0,85 f I . (be-bw) .t.ld-t/2|. dimana Mrrr f:, be, b*r dr t diketahui
sehingga a daPat dicari -
Ct = 0185 f;' b, ' E dPt dicari -l o",Ttil
vrztv
dlmana T. {-"'-'- -1 -l
dimana T, =C2cz = $
A = As
eheck
,85 f+
=1 '2rhdA :
S max
) . dpt dicari -)A- =o2
A_ ----)smaxA-'s---- ----)max
ok
ukuran balokdiperbesar.
| (b -bcewA = ..
jika A" (
vrrr.1.8.4Silahkan anda lihat contoh soal No - 1 I -
Penampariq T bertulanqan tunqqal memikul lPntur murni.Diselesaikan secara elastis ( "workinq stress desiqn" ) .
M-- yanq dapat diPikul ?wtJ
f'c
r,anokah perhitungan : fsLanqkah perhltungan :
Check apakah balok tsb r Turni atau balok persegi. Anggap
grs netral memotong badan.
A. Untuk keadaan beton daerah tekan diperhitunqkan :
bru. x. * x + (be-bw) .t. (x-1 /2 t ) = t.A= (d-sx )
:I= ..... .rika x ) t -- -, T murnix ( t ----) balok persegi.
. TEUTTUOU
'lA Lt L + cn ?, L = ,r oc.roJ -rpeqs
: qqs resa
;rase6 ede6
pa-r01cPJ,,
Afi
irilr
l.'
,ili
i
n
A
nA
uplpntat EdqlualrnEolu I['srJv>tor eq upppq _r*qa r = ^q
-rasa6 pr{p6 = A-resa6 ue6ue6e1 = A
/vlI)' o
-4
: .raso6 ue6ue6e;
rutnu J' {oTPq16es:red {oTeq
'6t .oN Teos qo+uoc leqTT uE{qeTTS'V lds uepT e.B ! Z.g j [.g qe46ue16---- 1 (, x(---- 1> x pltT[ ..... = x
(x-p)=v:u = (+ Z/L*x)+.^q3 uE{Teqetp
Z*ZJ * [r.[c =zz -p =
(+-x) + +
tz-p=["- c
= t l! I r'["J
z + tJ ttc
zc upp tc r""rtor
ze TpEI
1 = Zz
I E TpE[
? f =!,
zc
I, 1-x)
IcI
__t
qaroq)tp1
yo
3 rrPcTp
cJ
cJ E>tT I-
ldp tJ e{eusI=
+-- "; (+-- =r) "'x 3 (x-p) =
PtT [ {car{Ds
J
I{eToq {81
xo
/vtq' ( f -x )
.L. r...
=-DrJ
xr(1-x) = "J :
=zJ
J Pueurp
= [c
IcJ
Ic
Z
Lt
" =c
J
J
rI*sJ
C_CJ= J
cJ
q'1 'P)tT I tcaqD
s
Ic'J
zc
.c+Juep [c
f-ir-rl q .Ii,-+ua-i-T56m_inl
F'r.rfil-
"^o
T
t tt
nplp upp -JuP{-resepraqr-rPc@
- 332
6v'nvn
gaya geser yang direncanakan ( "design strength" ).Gaya geser akibat beban mati.+ V" ) diPikul beton+ v^ ----) dipikul baja
5
akibat beban hiduP.Gaya geserVcVs
1 1 mengatur tentang penulangan geser, tegangan geser sbb:GESER AKIBAT MOIIIEN LENTUR
vr=
ACI bab
Gaya geser yg dpt diPikul o-leh beton(akibat momen*Urrtang).
lv "=2 q;b, .d ( satuan: Ib )
t v.=0 ,53 1/ t L . b\^r. d ( satuan : kg )
Rumus vs lebih mendetail 3
rumus 11.3
,V =IC
tva=
r v--'du,t /r; + 2soof* t).b*d(satuan : fb) V .d@,5A + n6 f*g;.or.u( satuan : kg )
AS
rumus
v .duMu
M=u
11.6
(r"factored momen" ygteqjadi pd pot . dimanaV,., diperhitungkan.
f,'A"b
wfvlc{ (s
Lu"SA
.das tubar bsnc
:Lb927 5
: kg)
=E:w
1u
=le( 3,
atuan
tuan
1. tarikadan
.br.d).,tr.bw.d
Cara kekuatan batas
"strength design" (U=0 r I5 )
V =V +VncsV =V +Vncs
Vdimana v = ns-"rs"E 'n br.d
b = lebar badanwd = jarak serat tertekan ke-
sb tul. tarikV.,= gaya geser maximum terjatr di pada jarak d dari mu-
ka kolom o
Pada ACI code bab 1 1
rumus 1 1 .1
rumus 1 1 .2
No.
1.
il
':
ccc
?['l Isnun-r
LL'LJsnunr
8['[[snunr
: rPquPg'6u1:1ru ue6ue1n1
(Snc7S4 : uenles)P''q' A zLtz ) tn.l
( Tsd : upn?p6 ) tp'nq . ?J! B>_tnJ
wc/b>1g027=) Tsd ooo. Og ) ^t,
( ,ulc
3 upnles ) .A
{ E' E -unururtu nVs.^q "
l ( ,Qcu1 3 uenles ) I6C'l J ,_ o E _unur uTrx no J
/Y\.s' q
. 6ue>16uas . Tnl spnT =nv bue>16ues >1e.re[ =s
Tn{TdTp 6uer ,;3$'*:; ==ASS
;.Ff, A
:
(6t) p'nq'V sLG,o>nPle,(qT) p'Aq' A e>
)OuTs nJ'nV
= "AuP{
-Elo1rad rp pues bi >1e;e[ pduelt)to)t6ueqrp 6I Eueleq , dni6nE?E 6ue[ueuaul 1.1e86un11 6ue1Pq 1e>1edtp .resa6 - n1 6qs pltT I
'q
e{ 6u1r1rl . Tn1 >1e;e [ =
'resa6 ue6ue1n1 qe:lT =
s
V
J
A
T 'p[Bq LIaTaT ue6uuhffi =^JTpluoz r_r.o{
sr6 6up 6u1:1tu - Tnl 1"pd #
( Tbsoc +,X uTs hi- r =sh,.',
- p' J' V'e
i rasa6't
'tt I qpq apoc IDV eped"( EB' g=fi)
"u6Tsap q16ue;1S,
334
*"1 "Strength design " (fi=O.,85 ) Pada ACf CODE Bab 1 1
;l
tr.)o
Tulancran o.eser diDerlukan iika :
Vv >s;uv
atau ,., ) ;kecuali untuk :
-peIat.-pondas i-balok Yg tingginYa.(, ?
S -.7
ps.11.5.5.1
Untuk .,sengkanq berlaku ,:Jaraksenqkanq=s:
s---- = *S.o cmmax
jika ,"( 4 4 -br.d (satuan:Ib)v= ( 1, 06 ltL'biu'd(satuan :kg
=*"*= t <3ocm jika v=4] 4{E .bw.d
ps.11.5.4.1
ps.11.5.4.1
( cara "workingACI APPENDIX B.
GAYA GESER AKIBAT LENTUR
stress" = elastis )
No "Working stress design" ACI APPENDIX B
't . Tecrangan geser :vo+vl
v= b .dw
rumus B-1
rumus B-3
ps.B .-l .4.1
2. Tqgangan geser yg &t -dipiluL oileh bebn-
Rumuq yanq leFih merldelaiI
'"= JF * t3oo (* H-<t,g f (satuan : Psi)
u"= 0,083 ,E + 9 f- ifSo,5035,Q (satuan z kg/c*2)
Rrurnus vancr disglerhanakan :
," 's1r1 ,/tL (satuan: psi)
," {: o ,2915 .tr ( satuan: Yg/cm2 |
'i:f
-alu 'Tn1 e[eq' ( rapuTTTs
{T1sr.ro1{Pf,p{
' 6uE I uetu
qaTaT ue6ue6a1ednraq qoluoc )
uolaq uelPn{etsgr0
; 1ulo1or1
r=,
-ct -1,c
=)t
----.-:i 'TErTds rdnraq ( saT4Teral
TrPp I{eTaT uplpnta{= ) 1er1dsednraq ez(u:ase6 . Tnl E{ uroT
-olt {nlun feuTurou uetal ele6 ="dd"O .
^,' "r[ + l"v ' ^, + "v . ?l .c{ =ud
-ETPAUTaSOrnxuP{el Tatrrou Tueqeqlp- uroloX[
'uE{e1 Teur+ rnluaT uauou Tn{Turaur
['0['L'rrr^0 l ' J'rrrA
( Tsd
l7wa1s>4
?Jt e s jo
( Tsdc
3 UEnIPS )
('n-n) P{TI tucot]* =*"'"3 uEnlEs )
)(tn-n) nElE: uEnlPS )
( "n-A ) r EltT E uco r)6 =*'*s f
a'l
g-g snulnr.
L-A Snunr
'Tn? e[eqsJ
Av
"q.
A Jr,v'o '
( cn-" )
( Tsd 3 UEn?Ps )
?Jl 9't >1cn*n) pu'rxTp
lours '=r AA.L:: p'q'( A-A)
: uendunl rp pups 6fr, >1ere[' pd tiBBru-psreq uet{o}t6ueqrp 61 1dnr6 nplp
1e65un1 ) 6ue$ ueureru ue6ue1n1 6uB1-eq ue{eun6lp rasa6 'Tn1 6qs e{T1'
'r""3:*ff5i1;i = nv
r tE
...^q "(i?^ia
)
.E 'rasab 'rn+ qaro Tn{TdTp xasab EArg
I XrquNddv IDv,u6Tsap ssarls 6u1>1roat,,:oN
qrtr
: 336
luas tul. memal;angkonstanta = 115 s/d 2 r.5
biasanya diambil = 1,95tegangan leleh baja untuk tul.
spiral.O"p= luas tul. spiral.
Mehentukan luas tulangan spiral (=A=o)
tt r Gaya normal yg dpt dipikul oleh tul .
spiral = P_ = l<-.f- . A .- ....... (a)-! n s s__ spYcg
A .=st'k^-
D
.cL s
v
dimana
/, '=
A=sp=
iadi P = T(" - uol'"= ..... o.... (d)JL'\'4 (s I TC D"2.s
persamaan (b) disubstitusikan ke (a) di.peroleh :
Prr= 2 r",, . fr. Ac ................ (e)
Dianggap bahwa kekuatiX yg dpt dipikul oleh tuI. spiral sama
90t dari kekuatan yg dpt dipikul oleh beton yg berada aito"t2 ,"., . f, . A. = 0175 f; (o, A")
og = Iuas PenamPang kolom
A: = luas "core" = * rcO|"c-l
dipe roleh /, = 01375 ,* 1) *r.................
(f )-(
jika faktor keamanan diambil = 1 ,2 maka pers ( f ) menjadi
/, = 1,2 (0,375)(on/o" 1). f; / t",
A f I -
/o = 0,4s ( # -r) #f D c -s
berlaku untuk ty ( 60.000j
k diambil = 2s
A- /e- .................... (b)sp' cvolume tul. spiral utk 1 lilitanE(o" du) . ds ......... tfl)
D = diameter tt core tt
ed. = diameter tul. spiral
D
a^ = luas tul. spiralD
Volume "core" utk sejarak s
( s= jarak lilitan spiral yg sa-tu ke lilitan yang lain )
= 1 TC o2 . s ...............4 "c
dengantt corg tt .
.... r..... o.... (g)
rumus ACI 1 0-5
Ac
a
(c)
p"i ( ( a2oo *g/cnz I
qq9T ErSds
TITTTTTp 6uetr 'duleuade6t1r6as
ue.re>g6uT T
:e>16ues rn [nqnElP t6asrb4
uTlllueuau 'Tn1 qeTun6uedueuad tnluag
?, ' 6'0 [ 'sd r3va
6 unuTuTru 6ue[ueuatu sEnT = "V aa
6o'* 8) 6uefueuraruv]6v' t[: 6ue[ueuau ue6ueTnl t Tse?Equaul I G.0 [ . sd IOV
' buEf uEureul uPburTnfi,
sa!1 Te.relET .ralarueTp g?>6uB[ ueuaru ' Tn1 Ja?arueTp 9 t >
"seTl trralrrr uEburTnl {erPft
t0
v
TTca{raf rTqup
(ruruE'6=) E l* (
7( uurE ' Lg=l
Jt,rro'E?=)It**r's€=)
( rJrx€ ' ZC=l
8t ltFnL*LL*0r # >
' ,, sa!1 TP.raleT,,' .ralauPTpueueu 'Tn1 .ra?aueTq
: qqs ,, sa1l tB:5'18T,, 6+f ue?eJe^f,srad lenuau ,,r(.re1uauruoD, gt0[.l.sd f3v
'usoTl TPralPTr
'buPluPulau upbutTnl e . sall TErelEf r tnlUn IJV -piepupls
'ue>16unlTqradtp )te1 ud
TeTTu depeqral 6ue>16uas qnre6ue6
urle+rJn1 SEnT =1=V
q' q ; uo?eq '6ueduBuad senT = 6O
1s---r{- .1s - b c uJ + ('-V -V).;J S8'0 = -d
e'0 1 ' ['rrr
lc1, -rt
mlb-jl
= , g/€, untuTuTur T 1e;1ds 'Tn1 raloueTo '
( ruc? g' Z= ) uT I unurTuTru Tsplequau, ' O L' L'sd tCV ln.rnuau
nles 6ue,{ 1ee1ds uB1TTTT I{TSxaq {e.ref
E'0 1 'rrrA
lutu E' 6(urcz9,L=) uT t unuTxeu t
: uTeT 6ue^f, alt
338
vrrr.1.10.4 Diagram interaksi M' Po :
Hubungan M
dinamakan
Kurva dari
P^ dapat digambarkan pada suatu grafik ygnodiagram interaksi.
P'nPo
Ptnl= gaya normal nominal
= gaya normal tanPa ada-
o"= o Ma __4"
nya normal Yg bekerja -
[{^= Momen yg bekerja jikao
tak ada normal -
P. = normal dalam keadaanD
seimbang.
e = Mo,/P'
ke A disebut "compression control" (tekan).(disebut juga bertulangan kuat =
' r'Over reinforced").Pada keadaan ini : 2c =
:10032" ( ty
^ 'iadi ba ja tul - tekan belum lelehc=0r003 J
ke C disebut "tension control" (tarik).(disebut juga bertulangan lemah=
"under reinforced" ) -
Pada keadaan ini , Z" = 01003
,,-fc=0r003 zr2€yjadi tul. tekan sdh leleh
,003
=f /ny's
's ( Ly
Kurva dari A
ririk A disebut titik "balance"padakeadaaninit t =0
e= = t,Sumbu vertikal s e = 0 (ingat : e=Mn/Pn,
Sumbu hirizontal : e =t4Garis OA : untuk e = "b( e dalam
jika M.r= Ormk : e = 0 )
keadaan seimbang ) .
vrrr. 1 . 10.5 GA a normal tekan mak€ilng4-dan eksentrisitas minimqn,
ACI ps.,l0.3.5 :
Eksentrisitas minimum==0;03h utk kotoh yg dit\il spiitrt
sebagai, rt lateral ties"nya atau kolom kom
posit (kolom baja di-bungkus beton ) .
TUT uPepPo{ redPc.rol
' 0 ? t uPurP T Er{ epPd Epe .rPquPgu : -d up{nluauau :tnlun ue6unllqrad qer16ue.1
(P)^ 'o""' frJ'=o - t ?r Egro Is"J ) iv + q.ex.L/. ?; ggro =gd
^sl ' V=il,t?r sg,o - ^r) ?o =
qc
q.ex.r/. f, s8,o =q'P' ?g ggro = "p
+,$p4 ueeppax : ue6usgluT:
.( t\J + 060ep0609
'oTOrluoc uoTssar,r rlrraep uelelflffigad9'0['['rr. Teos qoluoc 1P[{TT uP{qeT Ts
qaToradlp (q) upp (e) s:ad pnpa)t TrEq c,,P-,P-P)-3 + (P Z/l-,,p-p) "c = Q".9d
: 6updrueuad le:oq s:6 pr{1 uauo14
Qa:
+(
c,J
Qd=
,rr 98
T
"l1
'q= t00
+ t00'a't'|,*;
p t00'0.6uequTas
ii,
uP,peat urp ue{a1 Teurou q{a
.ouE-qups
' ,, So]1f P.raf pT ,, 6qs 6ue>1&ps r6ue1n.q1p b^ uoTolt t1n od 0 g , 0 =
t rrSO-Tl fera?eTi 6qs 1er1ds 16ue1n1Tp b^ uroTolt {1n od gg,0 =
xeu
3 SPr{equau g. E. 0 [
Ur.d
" sd r3vr saTl TBraleT r Te6
-Egas 6ue>16uas t6ue1nlTp 6^ ruoro{ {1n r{ [ , 0 =urourrurru se?TSr.rluastg
5ti
I
[ 'rrra
oa
340
t!!,
I
rrr.1.10.7
'1 .
2.
Check apakah " ( eU atau e) eb
( e diketahui, €b dihitung )
TentqkPn lokasi qrs netral.Jika e ( "U --) maka x) *b
=A fSSdimana f" = s.Ests , 2" = (d-x):x
D
P =0nh-e-d')+C^(1/2(-
1/2 h-e ) )
c
S
T
To,rut ftc f, momen t,hd
o- cs(1/2r (d-(
diperoleh
h-e-l/2 a)+t
x=
3 . Ivlenentukan P- 3n' P_ = C^ + C^ - T dapat dicari.-n -c -sCatatan : penampang persegi dalam daerah "tension control" ("under
reinforced" ) .
Jika Pn(Pu atau .)"u --) tercapai keadaan "under re-inforc"a, "it"nf"tt
lihat contoh soal.
Lentur dalam 2 arah * normal tekan.( rBi-axial bendinq & Compression') .
Kita lihat denah balok, kolo* d1
sebelah ini.Kolom Pojok ( "corner colomn').memikul momen lentur dalam 2 arahyang harus .kita perhitungkan, t€tapi pada kolom dalam ("interiorcolomn" ) besarnya momen thd sa-lah satu sumbu utama adalah ke-cil dibandingkan dengan momen
thd sb utama lainnYa, sehingga ko
lom dalam tidak kita Perhitungkanefek lentur 2 arah ( kita hitunguniaxial bending=lentur I arah ).
kolom dalarn
kolom pojok
{r
,dt1 !
xd E'
(t! ^ax(+ts!
J, g, v )tT1T1 Tp I s
Eueprq 6uolouaru Zs 6uep1g
i TUT qeTaqas:eqrue6ral +ds ueqnlunre{
6ueptq ltque6uau raTssa:g
x'wt uwt'd : erunq*rs qTTes [. I xa
'0acpJrns arnTTpJ0{nlun: TUT r{PTaqas
reque6-ra1 rl.rades aceJ-rns e-rnTTp,{,,
udx^
7E* = r'
uw.+ =A' lrl.
A6_ __. : selTsT.rluas
-)te eped . e[:r1aq 6,{ ud uTp ue{ElprtuTpI x
LL --'-
'W , --uW , ud eped 6un1ue6.ra1 up{a1
Teu.rou + r{Ere Z rnluoT depeqral 6ued-dueued
Vxa
'reTssalg SnunrruPun-tnuad
0 ! o)c
! o)evx
?)v
ul{
t: e^f,unquns hrr=,
raqag
nlpns selPq rEurruou uelen{ax T ----1-'t
1'[TotlIPd upp .;raTsserg qaTo up{p{nua{Tp ] Tur dasuoy ", \t?.
\,
a
x_ -u tru
a' d = -W
tr". ,u _ *uro
: qpTepp e[-re>1eq 6ue[ uauo14a
4..
&
v
u,{- d
BAJn{
u
342
' normalPo = gaya
P* = gaya normal
p = cJaya normalv
nominal batas dimana e = 0
nominal dimana eksentrisitas
nominal dimana eksentrisitas
=g
=g
yB
*A
ltenentukan Persamaan bidang 32 '(a)
At x + AZ Y + A3 z + Aa = 0 "..""'Syarat batas : -
Er. t,itik o,**o ; 0 , t)
terletak pada bidang tsb
----) Al- "*o 0 + t ' A3 * A4 = 0 .."'...o"".. (b)
!f terletak Pada bidang tsbb. titik g(0 ; "r, , _* Ea'lr Pcrsa L)
--) 0 + "r"-o, . i' A3 + Aa = 0 "....""""" (2t
c. titik c ( O ,0 , fol terletak' padl bidang tsb '
Dari pers (1) r (zl ,;,"ir:;."Eq'o:+Aa=Q"D"""' (3)
r l- ( b -1). Aaal= %o y
. P_A2 = | ( # -1)- Aa --..."....' (4)zex
yB
Al = Po'A4
Harga-harga A1.AzrA3 disubstitusikan ke persamaan (a) dipe-jroleh :
pP-x ri-1)+Ja-(f -1) Po.z+1=0"*o ', -YB - x
i,+-t, . t (+ +. ,.+"= o
Tarikgarisvertikalkeatasygmenenbusdan menembus bidang S, di F 'E "*o'trr't)Et"*o'"r"'t)Jarak E dan F dekat sandiri, sehingga dapat dianggap
..... (b)
ev
bidang S" di E
*-=Qo
Sehingga Persamaan (b) menjadi :
/-L: ( t 1-r ..,',!}I (| -t A_ I ' n, Po' 'r lll- , '*"'. |1, r
1 ) -(ir+Po' '1r
1
P.1.
xoG,
f nE?Et
0=
x-u pct w
xud
^ua' d
do
Aol{
Ini:l{
*1
a-nPlE
g'0
a.g'0
xoH
rH
,L
-tMqaTo' ,, tt'61sap alarcuoc pacroJUTog,,
t Tol senT , 6uedureuad uern{n ppgd11 'P*u
anP1e0= W
ez(u:esaq ue{nlualTp ledep +FquE$ds r s{praluT eA jrn{ EpPd
' lnuTTas Teqal' e[ eq' uolaq ,itiia"l')o2 V = 'pu=p Lro uar sr Jao{
^^ -tro PpeO W= n E{TBP6:eq
x0= o nelP g=
E{T [ 'd e6reqt{;{
^ u__
"uWW
xuW eueuTp u
xo
xuhI
xu
P?T{!t T ste.raluT .rnoluo{ ue>11edep
ue{E , uelsuo>1 udaa
e6:eq eped
qqs qeTepe
'qctordde raTssarg ',.poq1au .rnoluoc peoTr
'Teos qoluoc IeLITT ue{I{eTTS
d **n n>1e1r€r{ ertueq ' laT s se.rg ueeues-rad
@6v ?t r'o
u o ^
x- -d =
d d _L dI t Tf T
o r c x c- {^ -d 'd -d --d, d _g L'['ttL'tt
r = ,pr d ) + ,p( 4 )
il.rnoluoc PeoT uuPEUeS.rAd
^u
' Ir{
g'0
.rnoluoc
uuw-dEArnx
d
ne I uTlTp 6uef, upqnlun.ra{ 6ueP1g
' ( ,, uourT ES I 6u
'glfrf ' E [ '61g TrBp TTqurP Ture{ )
uuI,{-EArn{
rbF
.D'
344 #
ir,
l;,1
l$r
tHl:l]i
iI
,il'
,lii
,lii,
Ir
jii
L
$,1
Harga berkisar antara 1 r15 s/d 1 r55 utk penampang persdgi'I dan diambil = 'l ,5 .
' ,5 s/d 2 untuk penapmpang bujur sangkar 'I
Mn
vse-
penampang dan jenis bahan dan juga
, utk perencanaan (ai^^biI = 0165
PnPo
narSa f tergantung Pada
f = o'55 s/d o'70Hargaf daPat andE lihat(kami ambilkan dari Fig -
Ieh Wang & Salmon")-
pada grafik berikut :
13.23 .12 "Reinfor'ced concrete design" 0
M ,TT
nvlMIoyl
l r00r9
0r8
0,7
0r5
0r5
0r4
0r3
0 r2
0rl
0 oc\ca<rtnt\\\,\
C'O()OO
\()f-@Ol\t\\
o oquo\nX
-pr
ox
qgPunqan mom.en lglrtuE-bi-axial
@ nLo.d
"ontour methodn, Parme approach'
,, Lgad contour " dari B-ressler digambarkan pada b idang M'o Bu xM +, f
--%rt Pd cara ini, titik B ditetukan--n t\ 'l trload contourtt a _yl -/ r(,,ctr!t rv\''rr ev s- demikan sehingga :
M -i- -- (--orJ Yn M M--
," M)'Nr Y
\ \\\\
\..l
5\\
\, \ (r\N \
\\
\ \ \ \ \ t\ \\
\ \
*
{
\
x\
\' \\\,
\ N\
uP{sr rnlTp -:#., Sr$,lil",,::i ;#*,:fi:,; {,,ffi)*',,xo,(
: qqsaa
(+) '
,do ,dr.W t74t + q
E uEueTPq ep TP
+ ^urn
epe )
IoW
m!'l
( ,{ow
L --1
1n{rraq >11ger6 pd ueq.;reguB6rp raTuTTTq uele{opued{1n e6:e. Ioxo"'r'f,=dtffil+"#
,xo vr.w T-t'+:-!tWDrI
xo,t
xu!t
xo troW.. W
t-)^, Fo
ffi4/xo_ _ xo_ r ,{o ^fio try- w/_ --vt--'-vty'
-T1ry = q{"[:Is
Ss,snrnT s1.re6 uptTpp[fp B.f,ust1>1erd
#)rr- WI/{
do-- xo(xu-r'
:#-frLt4'
z{oIr,I
mr=Wxo
W=xu
hI
,xo /ruIfnr%
xo0 = W eupurp
xo/,#( ,{
II.ri ' I
,i ,.r,,. ,
: ln{Tr&fu'&equre6f,al 11.rai{n1un V?t, uBu&tettr pd eAjrny
. TE{TlraA uep Te?uoz
,t,
-Troq nquns rE[ e[as 6uTseu_6uTspur uep ,,ih,snrnr r1e6e1 6uT1es 6r{ sn.rnT s;6 z edn:aq c-- , = *6req {?n
' xo-- fro
- r, :_=du'pr=-z# , ue>16unqnq6uaru 6^ =r,rn1 sr6 ean:aq (-- g, o =1) earcu {1n
f=
: upeuBs;ed qaTo.radTp
.y' uoLs'0 boT
E, O 6OT
g'0
-p=/
I-bl
60T F.---- I
- rL
Ior-1ffir.
)o/ ''xo
t wt)o*ow .d '
- s?t
z rl
uepy2 ueEtrnqnf
346
)utk M
_nvI{oy
MnxMox
M*lfMoy
1r0
0r9
0r8
0r7
or6
9,50r4
0r3
0,
0rl
M
unruk #oy
M,nX(M
ox
@o\F\\OO
FNOOsftn::\\\oc)ooo
t/r ht/r
h
\o r-\\oMo,-rnx
+M )=MM( utk #"x nxMn x
Khusus
nv
utk rs i van ditulansi t sisinMovMox
M
untuk til > *, )
M
untuk #"< * )nx
D.
pendekat?n bi${UierMoxM-oy
sehingga Persamaan
M-"+Mrr*(*ny
Mn*+*ny(*P
Ilubung ^"F dan #o
tersebut menjadi :
)( +ts*or() *tveMo*(digambarkan pd grafik spt tergambar befil<ut 3
(Kami ambilkan dari grafik Fig 13,23' 1 1
Wang & Salmon" ).Pada halaman. 347 dan 348.
Penielastan qrafill:
"Concrete design "oleh
atau 210 kg/
berupa silin
- $emua grafik berlaku untuk
"*2( rl( azo kg/cmz -
3000 psi (f :< 5000 Psi"z
f ' = kekuatan beton karakteristikc
der = 0r83 fr'Uf.kubus.
untuk contoh
ir
i'
( ,-o
drd
od
rd(:--
9E'o
9'0
[,e
d rsP{o
I,luoTo{ Trpp f elo+ ue6ue1n1 spnT =
?=v
q'q 6) eueurp F u/ = b : = d euPuT' f.sv --l
1>11ge:r6 Enues >[n1un n]teTraq) 0rt>r>9r0 :
0'tlq/qlO'[ {n1un n{pTraq >1t;e:6
TiSPIOU
B6:eHPNWAS
LVt
iumlah tulangan = 8 diseba
a20kg /"*2'pd keemPat sisi
21aks /"*2 ( r; t0.9
0.5
0.5
1r0
019
0r5
f =42O0kq/cv2
9.4 .5 .6 .7 .8Pn1Po
.3rP
.5 .6 .7 .8
---.>nPo
0,
6 atau I atau 1 0 tul disebar hanya pada kedua sisi
=2800k9/cmtno
\n.-< ,Po
P-n
----?p.- o
FN(n<f[n\gri\t\\ooooooF@\\()r@
P
\or\@o\\\\\ooooO\ !F- N Cn <f\\\\\c)oo@)o
a
1'-.l m m;ulf#tli till tLJ6 tul 8 tul 10,tuI
i1l
tii'
ri
if
:il
( TSlt)
)r
6qV IJ: .l
UjAt{fi
JJ
L1] lri []&r= r/': ,tl
dfrJ
,L
Efi:il
'2. I
I
'tJ''ol E q l;r
F ,E EItr 51.'000Ztr))l0tI
lt -r*:ll: | :lt: -_,1 _ ti , 16-- t.{-al.F- q-l
-. rY.lt|
;i?ruf St's-enp5>1 erperl 'eFugq-'f 6uefnl}p*T6dsrod' uoto, (G' ,,v3d ,..)poJ t B B t t IJV UO sAlON.r \/
I E-o [ 6T,it ! 6r- t t 6lc Trrip ur]r I Fqurtr trxe>t ) tw sa r"{;;;-;"T *"16""$o
,bIl*
/
,t
,vtc/6402?)? r)rwcTuooll [. . I
1s ledua pd e?E-reu ,;; it :l -asTp Zt -- Tntr qelun[ lf ?'.!J= 6?t
g 6'0
tul c*lo* 5'o
I
{
I
3s0
@Korompersegi ditulangi pada keempat sisinya.
4.0
oO:-
*
o<)N<Srr
67n ?*--, -^gh
c)C)
I
N@r
o\oOO\o@\\
OO<ilo
oNO
=# tk';J
vrrr. l .1 1 lsmgtgsan tulangp$ ( "Bar cutoffsi ) =
Tentang pemutusan tulangan telah kami jelaskan pada FS. II. 1 . 1 4
Konstruksi Beton I jilid 'l .
pada pasal ini perlu kami tambahkan ACI code tentang panjang
penyaluran Ia sbb :
ACI ps. 12. '! 0 dafr 1?r' 1 1 .
A-. titik belok( "point of inf leg-
tion" ) .
l,ii
tultuI
IJr:l::5o
{rofrM
tu
untuk tul, @
ampang kritis utk tul. @
lnzlgn| >*atn
I
I
I
Idn
utk
Vt
pe
kritis utk tuf@
8 L**v L#*L L#*0 L#*6*#8**L#*g *,*E/#v**tl*
: 1p.raq ,(uu e{ ur) re?aueTp rsraAuor rol{EJ ,"ri"llr;ilr-i"f"$ *l z 6qs uel6uequrelrp uri6ueTnt uern{n €poc IJV epe6 :_TEfEffizt.sd rcv eped 1eq111p ledep 1p1=1 uer.,1"^Lu u"rir"l
-ad T,.p pr rlererraq u"n@u,uuPn*"Tfi]:t"rtlil ;jlHl,*.:::;rpsaqror u*rru [
'P zt (
@.r.r rn*,"'"rlr,, ffi:::: IHT;'T":H:rq:::;'j:i'''o u.6uep
"*". rur 4( b,r JTlE6au uaruoru 1seio, ,, ,l on a, <t p(: qne[ashup{snralTn@uu&eto,l .c .q.Tn1 Inlun sr?T.r:J 6uedueuad Trep ,T ,l.r.[raq
""r;.il; '.rEseqrel qf tf d { p Zt ?_e -'Tnl ->{nlun
sT?Tr{ 6updurEuad fra,p r p (1e.ret eped ue>111uaqlp@up6urrni .o
D L .. L L. z L. sd 13g 1e6u16uaru
(.I < ) ue:1e1e1:ad ledues urrl=nre1_p @up6upTn;, .e
tsE -
VZ, OZ
gE'[ [
906' L
t0?'9690'Et L6' t.I V0' t.
SI,Z, Z
ZS9'Lv66' o
6Sg'0
g'Elgg' L
tlE'st0e
" 00?'rOL9, Z
vv|'zZAS'L
E?0', tggg'0gLe '0
t.' LS
0't ?
g'gtt.' zt.
L', 8Z
t' sz
Z, ZE
L,6L
6'S tL, ZL
g'6
LgZ, Z
t.69'LOL7'L
O LZ,, L
8ZL, L
000' t
sL g'00sL'0szg, a
00s'0gLt, o
ue6ue1n1-TOuroN .ralauprp
352
TT.1 .12 Konso1 pendek ('lBrackeFs, Corbels') -
Nllrr tulangan tarik ( "tension tie" ).
fsy
'!Compression strut."( batang pemikul gaya tekan )
( "shear plane" )
"Brack'ets, Corbels" diatur dalam ACI ps. 1 1 .9.Konsol merupakan kantilever dimana perbandingan bentangnya dan tinggi konsol tersebut 1 .
Kemungkinan kgqaqalah konsol:?dalqh sbb :
1 " Kegagalan geser yang terjadi pd bidang pertemuan konsol dan ko
Iom.2 . Tulangan tarik ( =A= ) meleleh akibat momen + tarikan yang bekeE
ja pada konsol tsb.3. Pecahnya atau terpisahnya "Compression Strut".4. Kegagalan geser pada daerah pembebanan
Peraturan ACI ps. 1 1 .9 hanya berlaku untuk a/d 1
Kami kutipkan svaraq - sl,arat perencanaan konsol s
Er. ACI ps. 1 1 .9.3. 1 Harga g utk perencanaan konsol = 0 ,85
b. ACI ps.1 1 .9 .3.2.1 *"n:"::r geser. ,r:a,rn beton normal ( "normalweight concrete". ---) berat jenis beton =
22r4 E/m' ) = vn
= lebar konsolw
( arah t bidanggambar ) .
Penulangan diperlukan untuk memikul momen len-tur pada bidang pertemuan konsol dan kolom.
Gaya horizontal yang bekerja I N,r^.dalah akibatbeban hidup ( Live load = tL ) c
*,r. ( fr.An f y dan harus )a,z v,,
Luas tulangan total =.r:'.s + Ah
dimana A" = luas tulangan untuk memikul tarikanyang terjadi pada konsol ) .
ur,(0,2 f;.bw.d \"<roo oi.u w
' I (satuan : lb)
atau vn( 56.bw.d ) ( satuan : kg ) .
ACI ps. 1 I .9 .3 .3
d. ACI ps.11.9.3.4
e. ACI ps.11.9.3.5
rliiI
II
L1
frIH
t
r
JvE'o = (uv uv + Jv) Ero = tlv
uv + Jv ='s s ' ' v {.rpbuETnl rsP{oTelexaur ue{TsnqTrlsTpTp
v p{Eur 'nosV E{Tf
Tf,Pp p e/Z Eue[uedas(uv sv) g'o (qvqv
= nlrer{ =v ue6ue1n1 uEEuap :e[e[as 6upI 1,,sdnrrgr1s pasolc,r) 6up{6uas sEnT ,:6.tt."a fCV 6\-/
Qv 6ue16uas
(tv+Jv)uep(uv+J^
s : Trpp rcsaqral qglillp
( -V ' f n1 Tse{oT Trep p t/Z 6ue[uedas p?p-rarus -- ue1TsnqrirlsTpTp upp -V . Tnl ue6uap f,e[e[as
'' Eeedgp 6,{ 6ue>16ues 1 t{v q=1=p= Jg Z/ L er{ues
'>{Trel ue6ue1n1 le6eqas up{nTradTp 'v + Jv
N {Trel e.f,86 TntTruau {n1un ue6ue1n1 spnT =c
rpsoqas .rnluar ruauou, Tn{Turaru {n1::-::rr"T"l :":l
( plP-reu PrPcas TsPto1 Trep p t/Z 6uB[uedas ue{TsnqTrlsTp
s-Tp uep -V .
Tnl 6up refie[as buesedlp6^ 6ue>16uas ) Qv qeTppp -nvg nlre^t( ez(u
[.-psrs p{pru , >{Trel . . Tnl 6qs uE{nT.radTp ('v + '^o
cn"N ]tTf,p1 er{e6 Tn{Turau {ln uu6ue1n1 sEnTn-'A .rasa6 ef,e6 Tn{rueu }t1n ue6ue1n1 senT
v : uElElEJ
s= V PtTfuv
T'V Pueurp
T+ 'v Z/E : (q)
(:V P{TC
v
v PuPluTp
I , JA
v+ --v: (P)
vlts
Ts E{eucn
=u
V
€.-t
:Z,
u
J^
: PpPClTrpp rpsaq rITqoT sn.rpq r.lv + "v
tsE
: ( 6uet6uas ) rasa6 uB6ueTnl senT = I{v
3s4 ?
Jika Asmakav-f
?AJ v-f0r5 (
iA +A
iA
A
A. .=n
n
uf +A.n A)n' =1o3 tr
:':' br3 0,p4 | .1,r'v
b=d=
lebar konsoltinggi effektif konsol.
an'gkar ) dengan salah satu cara
,i.;il
'il
'ii.,ti
,i,i'liri
. '(2 ) dengan dibengkokkan, seperti
tegak lurusA tsb.
S
mempunyai kekuatanf dari tuI. A .ys
Alergfoffan
tergambarbatas bag
- 3.r*= flu
1
2
dipasangtulanganLas harussebesarr
db
cara
O Acr 11 .g .7 yang lurus.tidak boleh melewati batas bagi-
horizontal maka
s'
Nu/ct-
( loaded 0rea ) maka
@ ACt [,H. f f .'|l .li
C ACr ps. 1 1 .9.5Pada bagian dePan konsol
tuI . As harus diangkur ( =disebagai berikut :
( 1 ) dilas pada baja tuI. Yang
- Daerah "bearing" (tumPuan)
an lurus dari tulangan As.
Jika konsol direncanakan untuk memd-kul gaya
"bearing plate" harus dilas pada tulangan A
bearing< 0 '85 f; . At . g
lebih luas daripada daerah yang dibebanikekuatan bearing < 0 ,85 .f ;, 0 rA1. @,
DIMANA Af = luaS daerah yg dibebani( loaded area ) ,
An = luas daerah tumpuanz
( "bearing area" ) .1
A2
n tul. I
tulangan
UT qT
ttt-F-Z
: uPnlPScgV) fr 3 Tpefueru
: e{Elu rs{npafTp
>{1n
rs{nparTp ledep xPtu n&
Xedep Tsroq pueump _rr,l1{nf,fs pped
0L/x>LI eltTf
'xog 6updureuad )tnlunz'l'9.tttm
ld=
ii'
Ili
i,'
i,'(uc 0[ -) ur t(snteq upp ZL/x Tegel .-es . upTeqel-rad TraqTp sn-rprl xoq lnpns fnp frs'uerlEuna;q.redrp snreq xoq 6uTpuTp uen{e{a)t (____x/rl, : ro1{eJ
uP{TTp{Tp qsl qoTo:ed1p 6^ (i.-x)Z ueyfese loasrad oupdruBued 6qs d"6E;;; ;;; xoq 6uedruBuad , Of /x 4q uep V/x > r{ {nlun
V/x<q 6uTpuTp Tpqal up{Tpsp '' r6asred {oTeq 6qs p.cuaeoor=io"
-au ueEuap up{n{eTTp ledep lZ*Tue6unlTe:ed
Qt=
z *-JIz*3
8Z* e6:eq , t6as-red
qn:e6ued p{pru (^ x' 3:/ rr,o tg<Ar;
J^V,
l
i&,
iir
l
,ii.,
,i',',
iiil:.ii'l
i|,|,,t, iiti 1.;.
;l,ti
riil,
tlllrt,iil
tii
,rill
i,iii
iil:ii,
]il
%G
ilfrPl
6uedureued {nlun
futs'oll ) t,r elrrr' sxw uPnlPS ueTPp
uT . qT UenfPSTPuoTsJol pa-To1cpJ,, )
n,tr e{T I uauoru +
ue{{nseuTp sn-rpr{ r s-roJ qn-re6ue4
{iz*l *@ >r1n ( iz*) .6 {+n , ^r*, . : LIeTepP
'Jesaqral TsTs =dl11ca>1roJ TsTs =x >{1n qprepe
' uP{TPqeTp T s-zo1urc6r; f---- (trr*
+-- (^ x c
Jn4uaT :asa6 ue6unltq:ed urpfepa1
fou,, {nlun upp
.(rnlueT:asa6 + TsJol TseuTquo>t) g.[[.sdre epoc rcv ruPTpp Jnlero
@ . T eos qoluoc lpr{T T up}tr{pT r s ez{use1e [ {n1unv, L = /leel
L =VyV'L =/ I ,,oJo.rcuoc 1q6rem TpuI
TDV t' v ' L' L t Teqpl eped lpqT T Tp ledep puprurp
/-^
sEt
z'i'L'LI uep L'r'L'LL'sd rcv O
]I,ti]
xi
,i
,,i,
"il
ri.3s6
atau To max
2
= fl{l, x v.---4
I^,,J
kan.ACI -s. I 1 .6 .5
-l
Balok ini mengalami torsi dan torsi tdk bolehdireduksi krn redistribusi momen tak dPt dilaku-
sistem "SPandrel" -
Balok ini menglami torsi dan
torsi dpt direduksi karena
redir stribus i momen daPat di-Iakukan..
lebar balok ( badan balok ) .tinggi efektif balok
nal moment strength" ( =momen torsi ) .
manaT =T +T_-U C S
torsi nominal ( "nominal torsional strength'}yang dapat diPikul oleh beton .
yang dapat dipikul oleh baja tulangan
"torsio6 T di,n
momen
tors itors i
-
(
=
=
u
n
c
S
T
T
T
T
fil AcI ps . 1 1 .6 .6 : Besarnya Tc\-/ -
or8 n. -
Z*zyT"=
t1+rffir2( pers 11'-22 ACI ) --, satuan: Ib in
dimana ct= faktor yq berhubungan dengan geser dan
tors i .
b .d b =-ww= F2d-?-x Y
atau dalam satuan MKS :
0,216 lT; E*2y--) satuan : kg emT
ACI Ds.I1.6.6.2 :
, CATATAN :
Untuk gtruktur yg memikul normal tarik, penulangan
torsi direncanakan memikul momen torsi total dimana
harqaTdiatas,dika1ikandenganfaktor(1+L,*c 500 A
v /0,4 v-_ ,)
l-- ulz + 1
U. .Itu
i Nu= normal tarik diberi t.tauOdan v.
( 1* N' ',
500 A 'g
"zfr.b.d'cw
gdika I ikan
----, satuan : Ib
1+ (2 ,5
untuk T
atau dalam satuan MKS
T2, ll .
utu0,5 /rL. *2y
ct
1v = 6uB I ueuraur ue6ue1n1( Tsrol Tn{Tureu {1n 1
. 6uer16uas 6uBdrueued L sEnT
's >1ere[ ruETsrB ue{nT.rad1p b^ -rasa6 . Tnl sEnT6,
' Tssrol uebueTnl ueeuEc-u-aia6. erlxa 6u
t epe snreq 6updrueuad lnpns-1npns. TO
Z t .) uT ET Blups nlps {ple [:aq 6uedruBu
upltTsnqTnJsTpTp sn-rpq 6ue[ ueueu - Tn1
sEnT E. 6. 9. I I - sa[ IDV1
=-vn
=Vnelp
e{Pu
z'6'9'LL'Sd rJV
iLre
[ '6'9' [ [ 'Sd r3v
tucZ
ucu TZ'
uEnlss
uenlES
uen?es
a'v z +
f..VZ +
1'-* J- E'E = /uls "-'g
^- (---- *n* =q 0E
v
qI :
'uTpT .6uer{
6up n+ps 6ue16uas qe.:e[ = s
' s >1e:reI eped ( rs-ro1 'Tn>{
lurau >{1n ) 6ue46uas 6uBdtuBuad L senf = 1v
1:ad) = -s ( rcv t'z-LL ,rad ) ffi
= "'l
3 pup*rp =,L
Tsrol Tn{Tuaru {1n , ue>1n1:edrp rs.rol - Tn1 p{plu
" t' b ( ", {n1un
iie[upuau 'f n1
(uc 0e=) ul5d 6uTITTe{asruru S'6 < <
TTcetral TTqueurc0e = ) uT z t ;rr
ry Id*[* {'uTer euPs nles {PtefbuP{buas {n1uo [ ' g' g'[ [ 'sd rcv
6-9-[[-sd r3v @'6ue[
-uPua'Tn
'Erlxa buplupuaur uebuelnl )tnlun 2.9.9. [ [rJV
'uTpT pues nles rs.rol ue6ueln{ :teieC: 9.9.[ ['sd r3v'JTl{aJa t66ut1 -p 'se:a1 f,PqaT =
1q
'T6BT up{nTiradTp }te1
Ts-ro1 ue6ue1n1 pupurTp ue6uolod -rp ( p + 1q ) er{u6upf,n{
6ue-rn>1as 4e.reI eped 6uesedtp sn]pq rs-ro? >{1n Eue>16uagI '' -luc /6>1 00zr) < - Tsd 000'09 > rs:o1'Tnl >t1n "J
z . prlxe 6ue f ueueur
'Tn1 ,1e:rtds ,6ue46uas splp TITpral TsJol ue6uelnuad' uoruoru, 1eu:ou , :n1ua1 :rasaF Tn>[Tuau >{Tn ue{nT-rad
-Tp 6^ 'Tn1 qequeuaul Buesedtp sn.rer{ rs-ro1 ue6uelnuad' r s-ro1 r1n{un uC6uelnua,C 1e;e.dps-1e.re^fig
'g' L ' g' sd rcvv' L'g'sd r3v
t. L. g. sq r3v
L' L'g'Sd rDV
L'9'IL'Sd r)Vn
( # 1c stz) +t7
Zl,t3
"p.^q -g EE'0A
,,:
358t
Da1am satuan- "British unit"-.pilih terbesar antara
( 1 ) : A1 1^1f I 1#) ----) satuan*: inch2
(2t, o, = -H- , fu,-,
At)( *1 ; vl)
dimana: Ct= Hdan At trarus(( +tr (
Per
TUr
ru+h'50.br.s.
T' ,3 I, Il r
Ar-(aY,fu,-re.)( xl + Yr.--.ET, rr'----F---) satuan : cm
Ar =(ff , -J$, r - -$,, 1r I vr) --- ---) "*2v ru+3E;Tabel -Lrlqt
il
{i$i
ii,
i|i,
Kondisi perencanaan ICI code Tulangan
1. &. T,-r(fl(0,5lf|,.lx"yl-- lb in- torsi diabaikan
b. vu(/ rysatg.qqJfS-i ; '- ."ffi,tilz*zy) -, Ig_sg
1{.6.1
1 1 .5.5.1
tul . greser takkan --, pakai
diperlgpraktis.
2. * i"(g (fl,s q Zxzyltorsi diabaikan
l1
11
.6.1
,5.5.3
tul. geser minimum
Av= to=o*'= (pers 11-tY. -) lal inch2Av= ry ---r "*2
3 ". r" <il(o,s J4 Zxzy)torsi diabaikan
11.6.1
1 1.5.6.1
tulangan geser :
r,-= ( v"-9-1ql rm I 1-1"r" - d f.,-d 'iY:*",
4 0r5vcz*
,fr L ZxzY IcL.
b.
g(gvr, (
11.6.1
1 1.5.6.1
, tuI. tor3i minimum
2At= lo=o*'" (incn2).ty
3r5 br.s , 2,= "-- (cm IryA1 pakai pen:s 11-24 & 11-25
ACI codeps 1 1.6.9.3
5. ;. ru ) fi(o,s ,lrt E*2y)b. g v")vrr)0 Yc
2
11.6.1
1 1.5.5.5
tul . "geser + torsi mi.-mum.
Arr*2Aa= S Olw's ( inch2 )
' tY pers 11-6
= 3'5-!-r's
(cm2 1ry
A1 pakai pens 11-24 & 11-25ACI eode 11-5.9.3
6. E. Tu Tc
b. rencanakan penulangan untukTu.
il) 1 1 .6.9.1 tul. geser lentur + torsi: (Tu-g T"):sAt= g l* "l ,;ilA1 pakai per-s 11-24 & 11-25
nn A'i rnarlrac - r
l?'[;i
'utoTot etnu Tf ep
Tq t uoTlce trean orql ) qerpz/p
Z/p 1e:e[ eped ppe-raq sTlTr{ -rase6 6uep6updlsepuod 1e1ad {n1un ueq
,'p
A.'9'9['sd rDV! s'' s [ 'sd -i3V
. ruoTo{ e{nu Tf ppp tere[req sTlTr{ rase6 6uep1q. _., (-- (r!o11ce
,t.it, ..,r,. ) !lere:-f,, eF'ra{ert. 6d-*:":ladi Teif-e!q.:'rtorun .,;.,: sT1Tf,_>t -rasa6 6uep1g
. Z qp-raep epedPl*raw ue>116eqTp ro-ri'lffin
.2.
rBlolvIv
= ( lrt qP-rp ) 6u
Tnf TP1ol sPnT
1 qe.raep uPrPp3 upT6eq enp
plBq6urpulp
qeJE upTep ue6upTnued
euPs
l v'v'91 rDV
{I
E
J-l '?od uPuP)tTp
6ueI ueqaq-ueqaqeperaq
lEqTlte (----'tsepuod eped
: TUT 1n{r-raq
[ - t.l,t
e [:a>1equP{p?-ras
' f,J'tt,rTd
rc
.Z
'Tplo1v
Blu TTeu qerP urTp .
tv = . Tn1 senT :
se?p T6eqTp 6ue1uT f eur
e[ eq uoTo)tuauouedu.resaq
. M t sepuod .reqaT 6ueI uedas elp.rau
up{TsnqrrlsTprp 6ue[ ueruaru qp-re urprep ue6uelnued
eIt4leseq'6uB[ued . lbesrad edn;eq rqPpuod leTedc prrErs otr1 e.,{u-resaq r^1'Tl PIUESe^dulesaq 77'
eur== W)
uenluauad >r1n np[uTlTp r:- :::uo1od z'v 's [. sd r3v
q
T
L'V'V'g['sd rJV. v'l 's [ 'sd riv
ruoTo
L' V' E [ 'sd r3v6ue^[ ueuory ffi@Truelt 6urlued Tpq ede-reqeg
' I L qpq rDV ueTep -rnlerc
6ge
360
Untuk kolom baja 3
ffiHltni.,fl
ftrii
ffi
bidanga/2
geser kritis berjarak d/2 dari tengahtengah antara'muka kolom dan uju-ng pelat perletakan.
@) ACr Ps.15.7 Tebal pelat pondasi minimum.\/Untuk pondasi dangkal i dmin = 6 in (= 15 cm). .
Untuk pelat pond4si yang menumpang pada tiang pancang
d ,- ='72 in (= 30 cm).mtn
i- lnqasfaf 6uedueu5d Tn{ldlp fdp 6d uerue 6^d :n1ueT ueruou ueeduelred : dnp-Tr{ ueqoq +pqT{e g0? upp Tleu.ueqeq lpqT{e qeTepe qsl
Telo+ uouoru Trpp t09 emqeq Tnqpla>tTp , [ . oN Tpos Eped3 Z Teosrulggv'tL =
urc6>1 v'vLEEte [ = (z/i '9-ES ) gzLsz =(z/e-p)J ='w'o, qeTepe sefe t
-Tp re>1ed plT>t b^ eIeq .6e1
Tpe[ ,t{eTaT qps -TnJ B[eq (--- o, ( "3run-rp>{
E0t 'to'zstrA =lr---? g9 t 00'0 =
6>1 gz Lsz =
e szle'(0t)(sL'ggl')
(-ropuTTTS
,wc1b4 SL, ggl =(snqn>t
: qol uoc >t1n ) ,wc 7 b4 SZZ
,J
c, J {nlun :
-tc
=EgzL'ZgL?J
.^s ) = l 'V
00zt
€.zo,o = =3 e---- gtrg : ggrg, = t00r0 , =3TaT56uaI^I
ruc se ,e= tf,{ = x e---- x t/ =p
s8,0 = 'd (---- ,vtc/b4 082>ruc 0V, S
8Z L;Z,l
(002r)(?0'8'zglv =gg'0 = e'q?r 'sB'o =)
_T-rL
: qa1e1-qwc/64 00Zt(---' Ztn e[eqt{ofuoc {1n
(szz)e8'0ctt
*'I
- TIIIOU
'gLbv'09/'SXI,{
: qertref- fJV qefepp ue4oled le6eqag
e qs1 {of eq In{ldlp ledep 6.d ., o1euTlTn,, TEUrnluaT uouou ue{n1ueJ, .ZEn e[Bq ,;ZZX uoleq nln6= 1e66un1 ( >tTJp1 ) t6ue1nlTp , urcs = uoloq lnurTTas
0 e ue-rn{n T6as:red Eupdtupued uoleq >{oTFq rnqp?o{Tc . -t TpoSqeTppE 1e>1ed ture>1 6d upnles , TuT Tpos r.loluoc qoluoc ueTEC
t 9t
362
Jawab :
SoaI 3 :
SoaI 4 :
M=nM=u
momen nominal (dari soal No. 1 ) = 13 1456 tm
momen yang diperlukan ("required moment").
Mu-rt
= 1r52 M
= 7 1972 tm
penampang = 7rg72 tmtul. tariknya memenuhi
syarat ACI ps.Jawab :
10.3.3 ?
A dalam keadaan balance 3D
5090
Menentukan
^b Gogo + f .d=ffi(55)=36,05cms
I
fl
I
ri
fl
I
;i.ni
ir.
I
iii
l
u 32 ----)f,
untuk f;
(225)=186,75kg/cn2.12Kgl cm2 -----) = or85
i"
riii
ilil'
'iiii
iiiilii
tlIl
lr;l
iiiil
iti,illli
i;,1i:l.lhrl
llril
iflii
ii;i;
ffiliiiiil
t-ccv
280
0r833200
kg/cn
' = 1 14 Mn + 1 r7 Mlg untuk lentur = 0 ,90agar penampang tsb aman maka *n )
dimana Mo = 0'60 Iu (diketahui)Ml, = 0140 M (diketahui)
. M-_ = 1r4(016 IvI) + 1r7 (0r4 M)'uM--u
= l,:? M = 1,688 MT - 0;90-1,68g rtt = Mn ----) !t= %8
Jadi momen yg aman yg dpt dipikulPada soal No. t harap dicheck apakah
/, = o,8s ,nt[r f : 280 xs/cnzjadi .b = 4.*o: 0,85 i ru, os ) = 30 ,64 cm
Cn = 0,85. f;. .b . b
= 0,85 (186,75)(30,64)(30) = 14591 1,5 kg
- 145911-5 2o=O= -ffi
= 45,6 cm-
Sayarat ACI ps.10.3.3 , A= max = Ot75 O=O
a
A- = 0,75 (4516) = 34,2 "*2S
2 (A" *.*b.rartrKarena A^ yang diketahui (4916 = 8104 cm'SSMAX
syarat AqI ps. 1 0.3 .3 terpenuhi.Tentukan ukuran balok dan tulangan tarik tunggal yg di-perlukan untuk memikul momen lentur = 60 tm (faktor pem
bebanan sdh dimasukkan untuk nilai memen ini ) .
Mutu beton K225 , ba ja 1132.
Jawab :
I,angkah perhitunqan :
a. Data-data : K225 ----) f
il.tiLil
r$
.!
[i:t
Ii
-Tp T6et n1-red te1 ue?npuaT e66u1qas upT)tTruapas ,"r"*]ffiu6leq gde1a1 , rlcaqc Tp nlrad {EA uplnpuaT . 1 1e66unl; efiu
-ue6uBTnl up{n1ua1 uep srruouo{a 6uBr{ {oTEq upJn{n up{n1uatr,'Z'tn B[eq , EZZ>I uolaq n?n14 .w/1 Z =dnpTq upqaq , ( to;
-Pq rf rpues leraq {nspure? urnraq ) w/l z, L -Tleru upqag . urz L
=6ue1uaq,seqaqde>1e1e1:adzqaTonduln1tp{oTPqTnI{P1a{Tpffi
SZfr6
( aurc6 L' V? =g 2fr6 1e>1ed ) aulc
gg' Ll
Z6LO'O = (
g5'6tr =
0509T
: uBbuelnuacl-re@ . J=(29)(StlZ6[0'0 = p'q' =sv .a
) 6s-LiLE = -t -t
I-
J -l
neu llrz
z(ze I 1.!q )
E,efgETT7Pq
=*- :1: u.:W
lu-t) T =
n ::u
'luc E = uo?aq lnuTTas'luc Zg = TTqure <-r-- ruc Ir[9=p f----
tuc 6,99=p (----
: upp ou T6Bt {carlJ
Lg/ge = {oTpq .>{n Tppp
9t=9.0t=Q : eqoc
'p
960' t s
ruc
tucn
u=u
W E'" 89?ot t =999' ggg' g
urc6>1 ggg. ggg. 9 = 99:g
frT 860'IE = (6Et'Oz. (oz0'olz/t-t)(0028) 0?0,0
(r./.2/L_Ll ^r./ -f
6sr,ot = (sr'?g!ls8'o ?J'E8,0f=-rf-
(8r020,0 ) 20,0tLoz,o = ,/
00zt + 0609
= -Pq ,c
=
u=fif
=n
=U
=tll3 nTET
=o/ tlquv ' c
'sl'0
(98'0) (Sa'96[)-E8-0-
^Jq)=o 3' , J '98'0
- t9r
V9LZ0'0 = (
J + 0609 ,
l60e--- 'L/, q/ uptnluauon .q
364
Jawab :
tangkah perhitungan 3
anggap berat sendiri balok = 0r5 E/n
i$i
tm
tm
'l
I1
= I
(1,2 + 0,61112f = 32,4
tzitlzf = 35
'i\l:it:
rii,:ii,
ri
ii,b.
C.
Mr, =
e. Dlenentukan
iri
il,
fiir$l
r4i
l18l
tffi
fl'
ffil
fri
ili
ffil
lr!,
Mug
d2.b
b d,2 =p=',l?:t;*otambil b = 35 cm ----) d
b = 40 cm ----) d
coba uk. 40/81f. Check berat sendiri balok 3
23171 1 ,6
81 cm
76 Cm ----)
Berat sendiri balok = 0r4 x 0r8t x 2r4 =
yang ditaksirg. Cari lagi Mr, 3
Mn = 1/A (1,2 + 0178)(M; = 1/a (2)( 1212
Mu = 1r4 MD * 1r7 MlMu a 1111096*n= T =--T';96-
h. Cari la*i ukuran balgL :
, -2 123,44.105^A &vE 51r0gg
1212 = 35,64 tm= 35 tm
'' = 111r096tm
= 123r44 tm
= 241575 ,0 1
fl'= 78 ctnr h = 83
MD
Mr,
Mu = 1r4 MD * 1r7 Ml
= 106156 tm
Uenentukan harga /b :
fA =
0,02764 (lihat contoh soal No'' 0,75 (u = 0,02073
ambil f= 0,020l{enentukan Ro dan m :
M = 20,15g (lihat eontoh soal No. 4).R,,= f .f .. l1-l /z . f. m) .ut y
- 0,020( 320a) ( 1 -1 /2(0r020) (20r 159 )
d. I{enentukan }1r, :
rafi;f= IrB,4 tm
dan ukuran balok :
4).
= 51 ,098
[ = 8t cm
0,78 L/m
anbll b = 40 cm ----, cm ----? uk. 40/83.
,ifr,
'"v
61
(------- Zr" lS, ?Z = gzfig ==v rnqele{Tp ue>16uepes
^ Zr"96'0Z ffi=-'sv^o
d^6>1 zrggoLg = =c+--cD= o,
?v.(?r Es,o or) = or,
61 oeL6v = ([s6,8)(st)( 9L,98r)sB,o =
^='o' ?r -sg,; = ^,, tttc IE6rg = (tSr0[) SgrO = F* .[ =
d"
luc tgro[ = d*
800r0 3 gLS[00,0_= d* : (S ^*) gLgloo,o=^a==J
upT{Tutapas[,^ 'qoTaTau uptp up{a1 ue6ue1n1 e66utqas
( sv= ) >tr:e+ 'Tnr spor upltnluouaur up{e plTx . p
gL9[00'0 =90 ['t.0'Z =g d=T=
J3 00zE
di{}
: -QEatec
' luc S = uoleq lnurT TeS. 'ZEn e[eq ' gZZ>t uoleq nlnry . gZfrS _
tTrPl 'Tnl ' 6liz = ue{ol ue6ue1ng, . ( Og/gt ) r6as:ad 6uedureu-ad {nrun ( un=
) splpq uppppa{ rueTep Teurruou uaruou up{n1uefr,(SZ6JJ 1e1ed) ,nrc Zg'Zg = gL .0?. 6910r0 = V
s
)r---Er<-+
:-T TpoE
-, r 69lt0Z vt t
poz-r .-= ezl uwu Pueurp
-r -r.) f =
6910'0 = r -t
z(e8) (or)
s9E
96L'Vt =sol' ll
: uEbUeTn& - T
Karena as y(=0,001576)maka x x--1 sehingga I
y v':-- tulangan tekan sudah meleleh
TPenielasan : Jika A^ A- , maka T
vakibatnYa C" c
akibatnya Is
s=ya
cv
jadi xvv
b. Dtenentukan tulaugan tarik rnaksimumy'
A:s
dalam keadaan seimbang :
6090 ^*b = rrry'Y
= eogSo?o,zoo (ss) = 36;05 cm
**.* = O ,75 *b = 27 ,04 cm
a =, ".X---- = 0185(271041 = 221984 cm-max I maxa.-*"*= 0 ,85 . i; - b. ,*.*
0 ,85 ( 18 6 ,75 ) ( 3 il tz2 ,9841 = 127694,8 k9
: 01003= (27,04 5) = 27r04
s sdh leleho,8s f;).A;
+ C = 145030 kgmax s
2= 45 ,32 cm'145030-mT-
('3200
m=A^max -c
A=s max
Catatan : Jika A=
l{enentukan Mr, :
T =C.+C"78528 = 5555,8
x =a/ 1=
A" yang diketahui1=5fr25 = 24r 54cm2 )
balok diPerbesar -A= maxuk.
C.
A = 5g25 = 24 r54 "^2s)[,2Ar = 2919 = 5170 cm'
S
T- = O". f y= 24 ,54 ( 3200 ) .
= 78528 kg
C" = (f, - 0r85.f;)'o;
c = 0r85.f; .b-a= 0,85 (185,75) (35)(a) = 5555,8 a
kg
a + 1733512 a = 11101 cm'
11191 -- 12,96 cm.0 r 85.
..^ a-- 'uI? lgt|Z = tuc6>1 LgEg€,?, = (t?,t,,g Z/L - EE) O0gg? = uW
xO 61 O09S? 6>1 g6SE? = =J * 5-,=.)*aJ=.[
61 z' szlsl = (gtv, L) v, z2tv = ,D
6{ E,ELVO1 = (Stt[,g'gl _# ) L,S= "Jurc ezzerg = (gt?,Ll1grg = x,[ = P
ruc ge1'L = x qaTo-radlp
(EttL'gsl -L
- '(s..:I)-d6dE tL's + x?'ZZL7c_ sD + -c
(002t)(s8'zls.dsl' v
x
= 009S?
u,
=
dJ
(s8'z)(
( Eeg L'89t[FTIT'66E ) L's =
Z)((sl'e8r) sB'O IT;;'.IbEq ) = =D
x( s--Tm'05') e0 t 'e o'Z =
s_s g' r -l eueuTp
?v'(?r-EB,o-J) = "Dx(s8'0) (st) (EL,98L )E8'0 =
e.q- ?;-sgro = "J
x
-
( C-xftoo?ox :(g-x)t00r0
x' [ =cJ
t
X V,ZZLN =
s
s o
s
/{
?r-g,, tr+l ToIIoS a
ef,u11re
= 6 L frS = TnqPtatTp 6ueI "V ue>16uepas[,
S6r0e = ,"V Tnr{E1a{Tp 9 .oN Teos Treg T-EEIEe Tn{TdTp ledep 6uBd uW uee^t(uplrad . uapT uTpT
1e>1edp {Trel ue6ue1n1 1de1a1 , 9 . oN Tpos pped
sv z*'
?1ep -P1Pp
r{aTaTau qps
szt v t
tucz
'6tfiE? L Tro$ru1 t6rge = urc6>g LeTLEE?6gt
(s-9s) Z'ltEt[ + (([0,11) f sg)([0,[[) B,Esss ,Q
( , P-P ) '3 + (z/e-p, "O
96'ZL . (s-g6'z,L) = 800,0 :
u=H
L9t
- 358
ggal g : Diketahui balok berpenampang persegi ukuran 40 /7 0 - It{utu be
ton K225, baja v32. Memikul momen akibat beban mati= 30tm,
momen akibat beban ntd": :
50^tm. Pertanyaan tul' ba (ra
Jawab :
Er. l{enentukan Mn :
M,.r= 114 MD+l17 MlK225 --- f'
C= 0r83(2251
= 186, 75kg/cm2
= 3200 Yg/cmZ
3200E" 2,03.106
)utk f; , 280k9/cm' ,=0,85
= 1r4(30)M
it1 = ,,U"ng127= -T;36-
b. [lenentukan A
-f=J=
= 141 ,1 tm
jika balok= 0r001575
ditulangi tarik saja :
K.15-,x,=
D
x=maxt-c max
"T
s max
n max
s max
.d =0,003 (55) = 42,61 cm.
0,001576+0r003
0,75 *b = 31 ,96 cm
= 0r85.f; .b.a*a*
= 172490r5 kg
= u?!9_9,5 = s3,e cm3200
= 172490.5 (d-a/21
2A
Iv[
= 172490,5 (55- I ) = 8868944 kgcm =. 88 ,69 t;
Karena momen Mn Mn max (utk tul. tarik tunggal) rmaka
balok harus ditulangi rangkaP.
c. Menentukan tulanqan tekan minimun.
Mn yang diketahui = 141 ,1 tm
Mn max fang dPt diPikul oleh
Msisa 'u 141'1 - 88'69 - $2'41
c" = k.#,= uzi**-*!?- e7350
8s ( 31 e5 )
tuL. tarik tunvlaL=8tif,?
tm
kg
c . 1 check apakah tqlanqa$ tekan qdh lqlel! '
'Zutc/b>l IsE9oZ = gLtgglz'"/6,{^ g0 t00[SL = il
s
- TzsTql-
J
' t.0' z00 [ s I
5tr
g
_-pr,t$
Fp ssarls 6u11rorn,, prpc ue6uap , ue6uprnl p[eq upp ""*.rtj, -ed e[rarlaq 6ueI uBeduBlred . gzfrv = 1e66un1 {Trp1 ue6ueln;,
;:tn_:f ' szzX uolaq n?n!.I 'url oz = r ruetuoru ppoT ocrAros-: )t{ Tn{Tuau og/oe ue'rntn 16as-red 6updureued >torpq Tnqple}tra : 6 T=uE ' :lerueq dn>1nc szi Lt {Trpl -
Tnl qelurn t pua-rp{ .rps.-aqradlp nlred {oreq up-rntn rde1e1 xPtu =v =v undr}tsaw :rrc-TEE5 . o
( atuc av' 6Z = gzf,g teqed ) aruc
z L, gz = ?o(atuc Ll'tB, = szfltt te>1ed)
l,r, Z, Lg = =v Tpp1 .p
Zr" Zg' L 8=sV qalo;ail Tp E. c qey6ueT , T:ep
Z*" Z9'Zg = ZL'TZ + 6rES =s 1e66un1 . Tnl ,xpru s__ xpru s rv + -".- LL-- -V = v
', + 11e66un1 ue6ue1n1 I q SLrO = xPtu
6>1 s, o vg6sz = ogtLg + s, o 6?zLL =
xoxefi s
d
+,J
zwc?'ro = #868f, ="vs xeu
, . 3+t'c
er" zL,gz = ((Et'98!)gg.0-00ztl = &,rr,g 5 = ?; ;:
c1TrJ rIJ
' rrra ./ A.' ^
J' -wc / b>l oozt - o, rpseqes re>1ed Z' -T'p ptpq ue6ue6al e66u1qr=", qalalaur qepns up{a1 ue6ue1n1
(eslooro=)fr t,gzoo,o=+ = s
(s-96,[--)ed!-zT' = r
x.: (,p-x) = c
Ixeux
l$
iIl').
u.
698
tuc 96'Lt =
- 370 -
Tqqahqan izin :
T- = 0,45. f I = 0145(-c ' c
f- =,(ambil utk gradecKeurbali kepersoPlan :
c = + f .x. b = I f. v 2 _C.... - Z _C
T = A_.f_ = (4. 4rg1lSS
T-C19r04 f= = 15 f.
f: ls = 15xf . 19,64
pada diagram regangans d-x
=cx
'185,751 = 84,04 Yg/cm?
260) = 24.000 psi = 1580 kg/cm'
.x..30 = 15.*.f"
.f = 19,54 f_S S
.X
a.
ES
-irrE.c
-E - zr0:.t09fl . =r" c 206351
= 9r83
f .Es = s 's =f .Eccc15 x A A^ 55-xW = 9t65 x
diPeroleh x =
lenganmomen=z=M* = T. z
tr20. 10' = T (48 r02
Teqanqan :
F = _41649'c : j3Eo;gZ'I
f = !16*2s 1 9 ,04
20,94 cm
d-l /3 x=
)-- rC
= I33 Yg/cm?
= 2187 , kg/ cn
55-1 /3 (20,941 = 48,02 cm
artqg kts
T = 41649 kg
/cm2 |( =84 r 04 kg,
( = 1 580 Yg/cm2 |
f c)
fs
Balok tsb tidak kuat memikul M,
- ditamb€rh, atau Pakai tul .
Soal 1 0 : Diketahui balok penampang persegi memikul
K225, ba ja u32. Tulangan tarik tilngga)- =
= 30/50.Pertanyaan M* yang
Jawab :
rangkap.M . Mutu beton
w3925 . Uk. balok
diizinkan dipikul oleh balok ?
630-1
I
-ap sserls-6ue1n1 upp
,, u6T o c* ,-41
6uT {.ro/ulr p.r,cas rXurl{oTEq uerntu ueBluelrad
.(apoc rCV pd 09 appr6 Tq:lpdTp 1dp) Ztn ptpq,SZZ X uolaq nlnu,ur 0l = {oTEq u/7 C = anpfq r"q"q J"p()toT"q TrTpuaE ?eraq tnaputrel unTaq) u/l S.! =r?Eru u.o.o an,iruatu ,1e66un1 ue6ue1n1:aq 16es.rad 6uBdtueuad {ofeq Tnqpl.nrq _ff_S . url lz,g =
a14 qaloiad p?Trt c upp q qp{6upT TrEq .pqaroq )rrprl ------ l rtrclo{ to,le ) -cJ.
?:wc/brt 9166 = (6,0tt)= 1t I;P = tJ Tpe[
6>1 BBB0Z = (0t) (Lg'9t)(vo,vrt = q.x.cJ f = J
l: 1r xo
wc/b>l 6' OLL = Z
,wc/bx Z,'V?L = (?0r?g)
Eg'5 u a6E'9T=s: =-J J
'gt _L-9V L
xx:F
LStrro'c
I
;E7q-?-d?lE--?t-Ei5-=5-t-rqnra-"*-=arua.tF-E6.Tilq-E5EE55r,.qruc Lgrgl = x
(x-E?)(EL'lL) Egr6 = oxgL(x-p) sv.u
= "Zt[.x.ot 0 = Tex?au sr6 pql uoruoru sT1BlS
EB,6 =u t ,wc/6>r90L.eo,z= ""tft.::ri:;r::i:= l;.EEfEr .p
l et
.dl
1
372
Jawab .:
K 225 ----)
u 32 ----)
-{3_A_,Jk/n
S'.
P-r- ,tf c
-T,.=r..E
As
f: = 186,75 Xg/cnzc . J..f,
= 0,45 f; = 84,04 kg/cmzpakai grade 50 \
il = 24000 psi = 1 680 kg/crn?S
n = 9183
iiil
ll
lt
il
,ffi
ffi,fl
Dalam keadaan seimbang ( "ba1ance" ) .
k = 1- =
1+ts= 0 r 32961680
l=R=
gr83(94,04)n.f c
1 -1 /3 k = 0,8901-*. f .i.kZCJ
1 .A ^A,^ . , 2= ; (84,04) (0,890) (0,32961 = 12,3263 kg/cm'
MwL=*,(3)(10)2 37,5 rm ,,
be 'at sendiri balok ditaksir uk. 40/70 = 0 14.O 17 .2 14 = 0 1672 t/nIuI,^, = + (115 + 0,672\(10)2 = 27t15 tm*LuM* = **, + **" = 64,65 tm
Mw)2
= R. b d' ----' b d =64r65.10512,3263 = $24488'3
cbba b= 40 cm ----l h= 114,5 cm .(perbandingan b,/h terlatru kecil)b= 50 cm ----) h= 002 cm
b= 55 cm ----i h = 98 cm, ambil d= l00.cmjadi ukuran balok = 55/105.
Check laqi berat sendiri balok :
berat sendiri = 0,9s l,Os. 2,4 = 1,386 t/m
h, = * (l,s + 1,385)(to)2 = 36,075 tm
%"= 37 ,5 tm
73,575 tmtr
Jl,'?J!:10- = ss6ls4,41 2 ,3263
ryIw
bd
*
- rwc/b>l vo,vg = ]l ='J (--
,, pedecf al paciroluTe.r iraAoil ueepEa{ (--o*(* edullre ,6uequrros uepppetu (u Euaxex
( ,vtc / 6>1 2Zt ' Z L=1 6uequ1
-os uPPPea{ *TPU ( ,r"yfl,
ggg. VL =
=
4d = ueltnT:adtP 6uPd u
Zw)/bi gze,Za =U ! 068,0 =!
wc ZLt,16. = ( S6lg6zt'0 =9x 962€,0 =1 : qaloradlp'(,,acueTeqr ) 6uequTes uerpea{ {n1un qsl lt TpoS ':-fEnpf
, e {oTEq ue6uelnuad: ueezluel.rad . 00 t,/SS : {of eq .
{n TpB[ G- urg 00 turi>11pe[1p 1o1eq 166u11 elueq ,utapT p?Ep - Elrp ,lt Tpos epe6 Tlfft6S
lrw./St1O89t=) "f ).rr7O1 ,,rrtt Ujtr, =":
)to c----(r0,rB =l i) irrc/6a tLt?,s =c! (---- (ss)6o,se , ", | = szL6L
q.*.rrf=,
( TTC
w/t
?ar U, tJ 3W
M
( ,tuc L' 6V
A
p' L' I-s I/{ eueuTp ----T=V
3 tT.rPl urburTnlu/IZSV'L =V'Z'1, L 'Sg,0 ={ofeq s'q (---- OLL/gg = }toTpq uern>tn
tuc Eot = TTquP 'urc ?0t -p f---- ruc sE = q Eqoc
ctc
6v ezLe.= - ((to'98)i-Eot)
--s0 [' (s' Lt. + zol' (zgv' L + g'I ) f I
=gZfiA L te>1ed ) atuc V' LV =s0 [' ( g'Lt + zol' (zgv' L + E' t )f )
=**+or^=rr,l
luc ?0'gg = x(x E0t)(!r6?) Egr6 = x z/l.x.ss
: Terlau sT.rec3 ELEq upp uolaq pped E[ra{aq buBA uebuebal IcarIJ
a
(E0r ) (058'0) (089r )
- ,71 -
Soal 13 :
Gamba$ z
r€r
,ffrr"E-llenentukan lokasi grs netral :
M, = Q (d-1 /3 x)7z,lslto5 = + (84,04).x.55(95-t/l x)
diperoleh x = 3812 cn ----) *b (= 31 1312 cm).Menentukan tulanoan balok :
ft r f.= (95 x) ! x
r = 9I:39,2 (84,04) = 125 kg/cm2tt 38 12 ' -r 'rY/7 v'r'
f = r.r. ft = 9,83 (125) = 1228 xg/crnz< { (=168a*g/cm2lT = C = + (84,04)(38,2)(55) = 88284 k9
As = t?388 = 71 ,s "r2 ----, pakai 15fl25 = 73 t75 "*2Pada soal 1 1 , data-data saDna , hanya ukuran di jadikan 55 /1 20. Pertanyaan penulangan balok secaraoworking stress de-sign" ?
.fawab :dalam keadaan "balanceo
R Yang diPerlukan tE
,! k.d = 0r3296 (115)=37 r904cm.
( *(0r55 .l rz.z,4+1,5).r02 +37,5). r05
( 5s ) ( 1 1s ) 2
- 10,45 xglcm2 ( Rkeadaan
seimbang (=12,3261<9/crn2 |
der reinfolced terca-pai' .
1680 kg/cmz
'-bl{w
b.d4
= 7',f, ,05 . 10:s5. t r52
karena n (Rr"adaan
----) jadt f, !r
Menentukan lokasi qrs nstfal t
t^til=x:(d-x)ct,
{t
a
' lllcZ
/ 6>1 gaetz,J{'[' ].
cJuJ.s'J
962t',0 == ------__ = )tL
t8'6 = u
Tsd ooovz = =Jc_
= ,J SVtO = "J:_!EeEr
e .. ssarls 6u1>1:oaa,, e.recds toTeq ( dpt-6;e:t .Tnl uep,{uelrad .Z€n eEeq,SZZ1 uo+Jq .,+nw ;r; ;;-.1 n uauour fn{Tuaur gg,1g7 lbielslrad 6uedurBued {oTEq Tnqe?o)trq T ,i_TEoS)ro (---- trui16>l rorla) cr
)rruc7o1 6,EL = iff+f#* = 3* =.,( rtuo 6J'lV = E zfrG'Tp)tpd)
aurc g i tf = ##
(9628'0) (68'0) (ro' vll 9 = ,L06gro =
+%€#s+ r
t
Zwc/6>l ?Ot ig = (rrz':::: rlT :,;
f = uL = [
xo c----OA1 Z'eS =(x)l (-----l*c ttrgE
z(tm=
Sv
,I
c
t sv upp 'J uerlnluau;'luc ttrst = x Tppf
rxcS Zt,' O + g€ =x pqoc
6>1 Ltgt L =
tuc gZtrEt =
q-q3e sr r
fuc getrg = [x
16JE,Z r.srgtgz = (t*-e) , l*
'ror.Io
96LeZ- = (x)J uc gt =x,s'8vgz + = (x)J uc st =x
s'LzTev + = (x)J wc 0t =x pqoopuP TPTrl up{n{pITp 'x e6.req qaTo-rodueu {n1un
0 = g'8t098t + xt0,g[9[ - ZxsLL tx etE,,(e/x Err)
Fffi = sor.sotsL(t/x - p) J =MW
x-g t I= ss.x. ,!";+,&,f =
q'x'cJ + = c L
v8z
:1.
ji,:t
ii':tl
.iir
lir
lilx
ilrl
id, '
flsrE
2* Eo'oot,
- 376
(momen yg dpt dipikul oleh tu1. tarik tunflgal).Irlenentukan tt{ 3wl
= Q. b d21
= 12,326 (25)t45)zM* yang bekerr." )**,
l.tenentukan A". 3
t-=A- = /.o.u = ( +-= ).b.d'-s1 c---- z q
-o4 : - . z
t{enentukan tulangran tambahan A- 2
-: --- ---'2M = M- M-- = 15 5124 = 8176 tm--w^ w .*l E2 Mw; (g,-16).105 n,r,.,n^ t--',.1 '-z =rz=;h=k= zleoa ke
T^A"- = += +*S = 13'04
"'n2'2 f=Tulanqan tarik total :---- A =A +A---s -s1 ,2
= gr274 + 13r04 = )2r31 cm2
l.lenentukan tulangan tekan A; i
r_ = T+ .% = W (84,04) = 55r7r ks/em21 Kq
, 22n f- = 2(9,83)(55,71) = 1095'3 kg/cm-
C^"1 U=#karena 2n t., ( q (=1680) maka "1
? 1900
Mw
= 6240'03175 kgcm = 6124 tm
artinya pakai tul. trangkap.
@(ss,71)21 oA7 .*2
= zz ,31 cm2 (5g25 = 24 r55 "*2 )
= 21 ,07 cm' (5g25 = 24 ,55 "*2 )
Diketahui balok z perletakan, bentang = 12 rn. ukuran balok
= 35/60 ditulangi tuI. tarik= 5925
beban mati ( berat sendiri balok )
= qD = 0'504 t/mBeban hidup terpusat ( tengah benta-
ng) = 5t.
Jadi
Soal I 5
AsArc,.
a
5925
( E'0r0tBt ) (
'tuc LeTL, L
r Er90z ) ?8t ' Zrr/
6>1 l Se 902 =sL'ggLi<|Uw/t
00 t 9 [ =
00[S[ =
0E'0 =
?(oozr)(ro'e)= CJ
ub
ecr ' fl ?8€ o
(eee'ea€)(s( -T#33* ) -r)ruc 0t = r{ Z/L
, _wc/b>1 t L l' LZ
sry vr6'lwc/b>1 n/ VB6'L
v" ggg'988 =
z( 6V
'JZ-ss) ( vs'?z)t8,6*€ (6v, LZ) (ie lf =
: TrTpuas lpraq +pqT>te ue+nDua-f'c
?*'f 'oIo'1"8t =
(ooo'oE9) ' oozLogv ) = "t 1 t( Etr69g ) - r-^ ruc64 ett69s --
0t=_r
+
=
:
lc
1-^ Jc
= vrlb-rr'l
I 9=P
xPruW
6= f euEtUTp
xeurr(,#) - "r 9W
ruc 6V' LZ = x(x-ss)(?s'vZ) eg'6 = -x9'LL
(x-p)sv -(r = x ztl-*.gt0 = Terlou s:6 pr{1 uauou sTle+s
ffi r-' ( ,, uoTlcas pa>fcprc
ru1 Z L0' gV =
z(zt)(?0s'0) f + (zll(tr) +:?*, 000.0f9 = t(09)(9S) +
"r-(E(}ff,-r) *u: "1 ,rrr1n1ar"a.r"p . q
. : ne!u11tp n1.reduplnpuaT T?rpraq
uTuq .pdrp TTca{ qrqaT rnqe1a11p 6z( q eua:e>1qt qt uTtu uc SZ = 6I!-2T = i.= -'-q : 1e.rer{s
: ue>16un1rq;edtp n1:ed uelnpueT qe4ede nTnp>[caq3 'E: qEl|eIt
e g00D
IJV ue>1oqed ue6uap, 6uelueq qe6uelrp >lepuade46ue I uelnpuaT uee^due1-red. VZn p[eq 'gZZ>t, uoloq nln6
.1 g = 6ueluaq qe6u-eaTp ()toleq :1eue uelelelred Ts{par rrEp Tesptaq) 1psnd.rol TlEur ueqag
LLE
378
Lendutan akibat beban2
1 PL-6=
terpusat ( beban mati + hidup )
= 51859 cm
48 E .Ice1 (13000)(1200)3
= 48 (2063s1)(387070,3)
Lendutan total = 1,703:1 + 5,859 = 7,5627 cm
""ndot.n .kibat beban hiduP P = 5t :
Syarat ACr tabel 9.5 (b) :
F= * = -:3 = 3'33 cm (akibat beban hidup)
Soal'15 : Pada soal 15, tentukan jangka panjang dan check thd lendu-tan izinJawab : t.i (
1
?
Akiba
f= 'a 17037 + 31606 - 5,3097 cm
Akibat creep dan shrinkage ( susut ) .
f -- f,o"o.r, mati jangka pendek.' lb.-
dimana
^- T -tET--
ot A; 0(= ^ =ffi =o
q = z= ( j"'rk' waktu ) 'jadi t=z (5,30g7 cm) 10,6194 cm
Check thd lendutan izin :
t beban ma
=' '1 t7037 +
5o"o.r,f,"r".p[tot.t?LA=v 480
karena
jangka pendek ) .(8000)(1200)3
48 (2063s1)(387070,3)
= 2,25 cmh l_dup
+ shrinkage = 10 '6194 cm
jangka panjang = 12 '8694 cm
1200480
artinya ukuran ba-Iok ( t inggi balok )
kurang besar.----) tidak boleh. .
61
sa g6'E a
& r e BuIluTp
969Sel - (002t) Be'Ot'16esrad {oTrq 5qe
- fi, '8vde65ur1p ,t
!aE
r,[
{oT18
'c. ( eua13 Ouofo*",
rE.r+au sr61 l6asrad {orBq 6qs dr66uelp torBq s E{prr( qa& s suartr uBselBqrad 6uo10uau lBda1 TETXaU C.r6EltT[ ] 8v)atuc gZ' 6[i = ( S [dg- ] Tnq?xatTp Eupl( "v ?uare{
z*t z' lar s E"StFFf ="v
Cr=,I61 s' ? z,gecc -,
(002) (tt t (st'98r)s8,0 =
"q 'x ''d.?r Egro r )
,Z
Tsd
ooo? ) ?r leo-vtc / 6>15 L, gg tc
rra/bl 00e€ ra
'UlO ZL a 1 E
fiIx
sg'0c
= ,l-r/ (!-- s?,Z>l
-eleqrad Euolouau lpdalt ge$ uBp EuaTJ uEs
1u:15u s1.rn6 re&e "v ur{nxueuan . q'uc 00e a rTce{rel tq ITqu"
ogtlt-- ------- q) tq
002) (---- ( 008 ) ?/ I I1 r/r > ,q
3,?,?,> f---- (e[)gt + 0E>
1 9r + ^q ) tq,t
._:ffif
a ( un- ) In{TdTp rdp 5.t f DuTG)u rElrq uauou uE:(nluiit,
szr8=fE65un1 {rrpl ue6uu1n.tr .z[n rceq ,szzi soqeq n?qf .s g :
- 6ueluaE 'urlexalred z srlBTp 6 Eucdrruad IoTEq trnqrla:11g i-TTiE
-61€-
tEc
tuc
380
Soal. 1 I :
LenganMn
iil
,i1:
ll'
ffi
momen = d,-a/Z = 67 -l /2 ( 3 ,96 ) = 65 ,02 cm
= "125.696 (65,02') = 8172754 kgcm
= 81 ,73 tmTentukan luas tul . tarik dari balok T seperti tergambar, me
mikul momen akibat beban mati = 50 tm dan akibat beban hi-
Mutu beton K225 | ba ja U32. Gunakan cara lcekuatan batas .
Ir .l
+$=TF:ryI o^ | Id=eocm
5cm
Jawab : lgnqfan pernitunqanEr . Menentukan luas tul . tarik maksimum .( eCr CODE ) .
ACr ps.10.3.3 , ((o,zs Fb ,
atau A" < 0,75 A
[:ffii
*b : d = 01003 ; ( 0,003
( e0 )
f+#)S
0,003*b = 59 cm3200( o , oo3 +2'r03. 106
.b = P,t -*b = !0,t5 ) ( 59 ) = 50, 15 cm
karena .b ( =50i rrl ) a (=17 ,51 ----> barok T murniCt = 0,85 fl .br..b = 0,85 (185,75)(35)(50,15) = 278624kg
CZ = 0,85 f; (b" b\^r).t0,85 (185,75)(75-25) (17,5) = 111.116 kg
A=-= u?'?r = W = 87 'o7 t*2!b Y
1 LL1ll 6 = 34 ,7 2 "*23200
A=s.."b
A= A
"b =ro
b =75lr ." ,
-2b = g1r34 a*2.
.[ >{oTeq ueJn>[n ]esaq-redueu nele de>16ue:'Tn1 rr?{euou ledep e1T>t , TuT >tr-re1 -
Tn1 Trco{-redueu {ntrun : uple+Ec(,tuc tt'tB = 6zfrtl re>1ed1 (
(rutc ?e 't6-) xPlu
"v -urc Z0rl?: --09.9!- 'Z t Y '( 'J E'gtTtTiT'uo1 tgtt'LLL ="6>[ ttgl[[[[ = (s'Lt)(0?)(gL'gg[)gg,0
(+)(Aq "q) ?t sB,o
-urc t' 6v = =9919- Z ggLLEL
'up+ sgL'LgL =6>1 gBLLsr = (v'gz)(st)(EL'ggt)sg,0 = e-Aq-.?r gg,0= [J
'ruc V, gZ = e r{aToredlp(z/s'LL-06)
(s' Lt ) (9t-Et) (9 L' gBt )EB'0 +(z/e-06) (e) (st) (E L' gBL )E8,0 = s0t : 1LZ
(z/7 -p)(+)(Aq-aq) ?l EB'0+(z/e-p)(e'*q) ?t sB,0 =uw: uEmln+
zs=:V
=
zn
Is=V
ue{nluaual{ . cT U-rnu ,t
rul gZ'6gL = nlred 1 = p e{T[ "^(f ( e Tpe[
= TnqplaltTp 6^'1,,1
6'0 tu1 ULZ =(OL) Lt +fosl v\fr
Tw l'l+or,,.r v'ltul,gZ,69L = ruc6>1B98LZ69L = ((S',LtlZ/ t-06) ere80z
(z/e-p ) )b>1 t?tBoz =
( Et ) (g' L t ) ( s L'98 t ) ( SB,0 ) =
z*' 6s'oz
acq'P' ,J
= -Eg'q = 9, LL
=u
=W
sg'0tg-E
sv
-t-, -v
-t-aq
ruc A'LL = 1 = p de66ue.f-f6TEE-€-Fq-
'q
- r8e -
382
Soal 19 :
t
Tentukan momen
tion moment ofcara "working
inersia "cracked" ( "cracked transformed sec-inertia"). I., dan mompn kapasitas M, dengan
stress" untuk balok t berikut :
Mutu beton R225
Ba ja U32
Jawab :tr-Egdl cm.w
E. Menqecek apakah_grs netral memotonq badan atqu fIens.te- " b"= ?oo ,
I
c
1
d=6 0,1
=fSS
Statis momen thd perbatasan flens dan web :2
Penampang beton tekan : 200 (12)(6) = 14400 cm'Baja tulangan : 9,83 (50-12)(48,241
= 22161 ,6 "rn3
karena statis momen dari 'ba ja turangan lebih besirdaripada statis momen untuk penampang beton tekan ma
ka qrs netral memotonq badan.data - data z K225 ---:l T. = 0 r45 f l= 0,45 ( 186,75 )= 84,04ce
u 32 ----, ,, = 32oo kg/cnz kg/t*2
)f _ = 24000 psi = 1680 kg/cm'
SLJ
A- = 6g32 - 48,24.*2S
n = E=/*" = 9183
b. Beton di daerah tekan baqian badan diperhitunqkan :
b. 1 qaris netral.Statis momen thd grs netral = 0J--
35.x .1/2 + (200-35l- (12) (x-6)= 9 r83(48,24) (60-x)diperoleh x - 14,85 cm ) t (=12cft)---, T murni
b.2 Menentukan I", :
= i (2oo)(12) 3+(2oo)(
(14,85-12)3 + *. 35
= 216983 ,7 cm4 .
12)(1 4,g5-6 l-2+
(14,85- 12f
1n ii
#cr ( 3s )
'ru1 gL'?v= ruc6{ ?,v 69 L' v t =(go't,) gt'BtE + (9EE'Eg)oo?oB =t,' Z,.ZJ + [r. [C = ^W
ruc E0, Lv'ruc 9St'ES
s6'zl-09?vg'v-09
zr-p = ze
[r-p = [e r uaurolu ue6ua1
t-S8'Z=1-xzl
Z,
'ruc g6'ZL = ZL + E6r0 =uo vvg'v =
(ffiffit(zl)i=Z5
s-=) 'l(-wc/b4 L, c
,wc/b4 S, ggC, =
7 =_-.---. : "D TSPIIOT
TsEtoT{ceqc.[
z)
rrrt I^I uE{n]uauan D. q
SSj = ,-J ueepeo{ te4edlp Tpe[
1.,, Ic ,JI t-tlc- lZ +
, * ,r) . g= Lz
I
LZ' gg = ";XO (---- 6>1 t?0 lB et 6>1' Z, Bt60g = ZC + [
6>1 E?otg = (og9t) vz'gv = "J-=v6>1 gt'gts = (Et) (zt-sg,tt)(61'ot) i
^q-(1-x) t", t
6>1 00?0s (zt (ooz) ( 61,0 t + LZ, gg,) r
--a .lc o z1."q .( "J + "J) t
,'tc/b4 6L'oL = ar=ffi:trl
cJ
tc
J.
=
=
=tc
Ic
LIaToq {EpT 1(
rrnclb>4 089 t
xo c----(r0'?g=l '; )
LLgZ = (E,E9Z)€B'6ggr?!
,vtc/bl VOtUg = 3
(Eg'?[-09) c
sT,u
=-sJc
= J {nlun
=
(x - 0g) : x = t
Zrr/bf 0g9L = =J.
J
c 3J
s*{n1un
r8g
,vtc/b4 LZ,gg =
3 uEbuEbal {caqc €:q
384
C. Beton daerah tefaqc.1 qaris netral.
.=1Tb =200l, a \l
Statis momen thd grs netral = 0
' zoo (12)(x_5) = 9,83 (48,241(60-x)x = 14r9.1 'cm
c.2 I :crI
c.3 Menentukan M-- :w
' o)(r2)3 + 2oo(t2l(14,g'l.r=m(20 ? 1i::1"-"n,(50-,,a4,91)'= 1183429t7 cm .
sl2 + 9,83(48,241
ct = * (r
d imana
+f ).b .tc "1 e
f :f ="1
'e
F="1
= 11,03 t<g/cmz
(x-t) : x
l4r-91-121 (5.6,51)14,91 "
=; (56,21 + 11,03)(zoo)12) = 81048 kg
,1= + (12)( ;*f+l###lr ) - 4,6s3 cm
1
Catatan
= 60-4,563 = 55,347 cm.
= 81048(55,3471 = 4485763 kgcm = 44,858 tm
Kalau l<ita bandingkan hasil M, untuk keadaan b dan
c maka terlihat jika beton daerah tekan diabaikan,hasil M-- tidak begitu berbeda dengan keadaan dimana
wbeton daerah badan diperhitungkan.
boal 20 : Diketahui balqk portal menerus spt tergambar dibawah ini :'
k ??7 )l
a
M1
wa
t
triil
i;i
tii
litffi
&
#
penampang
.rc
_ruc 99'00 [C
'_tuc l'L6ttog v' -(z '8 t -t6 ) 90'zL L +cr-
-(g-z'g[)99'oot + -(z'B[)(s?) P
z'- t'- t'tllc Z'BL = x
xeur'(.(-rrl L I/{
= (?'tt)(
xeu,(-#) =-W
1IE--T I' l
6-t) +. r'
t-tg'6\=?vt r-u)=*
"rq : Pueurp
a : 'I uP>fnluauaw
.rc-rI
rucl69-Z
qes_r +06=P
tucgO'Z I L=(?'L[)
(x-t6)90'zl l.=(9-x)99'00[ + xz/L' x'9? f 8'6=0 = Tellau s:6 pq? ueuou sTlels =V'r=
'lllc S = uolaq lnuTTeS' ( Tbasrad >IoTEq lebeqas
1s6un;Iaq .f, toTtq v uE{P1aTr9d Eppd )'?*' 0LL6L69 :
z( 9-9'9t.1(Zt''SZZ+, ( Z r ) (
+ z(il-n'Sg) ( 06) ( E? )+, ( o6 )
'ruc V' 99 =
- (ilVl(OLl +_!-ZL)1ZZ -- =X ( s? ) ( s?) ( 06 )+( 9+06 ) ( zL)szz
't{eareqrel '1eras Trep x >[e-rp[:req :
I
5ZZ= q
: V ue{E1a1rad {nlunb
f uep f ue>[n+uaual,{ ' I: uebunlrqrad qe>1bue1
3 qeAec
leqT>te ) >lapued e4oueI>t1n rJV epolau uP{euncB[eq ' 7ZZX uoloq nlnl^l
xeu tu1 0L = TT + TC
tuf 09 = TCI
+PqT{E gI{ e6reg
I up>fnluaua]il
szzt +(s?) +
6=: I
( ue{}tnseulTp p6nI qaToq ) ue>tTeqe
Terlau strebe1T>{ ue6ueTn,l
rc
. (TT
unuTxeu uelnpuer ue{n?uauou'ztn
uPueqequao leqT)tP
'snrauau{oTeq >{1n TTS
Eq -rp qalo-rad
-Tp TUT e6.reH
: uElElPJ
1 Et =TT
1 ZL =TC
:d
*rH* ?z = (dnplq
TT + TC (T1r'ruulf 0 t =TCI
: lEqT{E vn e6regw/4 V' 0 =TTu/l Z't =TC
: E=e6--H
ueqoq ) ,, PeoTueqaq ) ,,PeoT
OATT,, =TT'pea'p,, =TCI
v' s,,g=x
s8t
'ef ra:1eq bupA ueueqaquod
- 385
dimana f, = 'l ,g84 4I = 697977A cmgyt = 90 + 12-65,4 = 36,6 cm
M = 27 tl!9-g7-g77ol= 5168081,07 kgcm = 51,68 tm"cr 3616
M max ditumpuan A: akibat DL = 10 tm. akibat LL + DL = 24 tm.
M
akibat DI.' cr - 51'68 - q-- 10 -'17> 1
maxmaka nienutut ACI ps .9.5..2.3 '.
jika ," ) ,n maka re diambil = Ig.
akibat DL + LL, ts = s!168 -2,1s3 > 1 ----> r" = rs"aa"
z4
2. llenentukan Iq dan Icr untuk perletakan B : Momen = -
9=17 , 1cmAt = n.As
-- =-1 68r09 em
'=461 9= 1D.
1 ,4cm
=5g7 g77 0' cm4g
1 ,gg4 2= 27,1 kg/cm-=4
L
Garis netral :
Statis momen thd grs netral = 0
5'x "1
/2 il"::l"lt'"-:7'r:*'u8'oe
(e7-x)
I=cr
M=crM max
51,68 tm: akibat DL =
M_-s0tm:--)#)t---)
maxM
I -I =6979770cmeg
-s cr (1akibat DL*LL=7Otm --, **.*a ,
MT , CY IJr = ( ,:--)- ,, + (t-e max
----) I ditentukan see-M ^bagai
berikut.( # )r).r.,
max
' 70
(1134317,57)) = 2003867,49 cm4.
196 ,7 5
Ai=(n-1 )'A;= 1 00, 662
"^2
+ (4sl(zz)3 + 100,062(zz-512 + 168,0e (s7-zzl21 1 34 3'17 ,57 "*4
.
u1
lu1
z'og = Ll 09 +
LL- = (02+?t) ?ru1 09 = d
z'L z(z[)(r'o)
ZrL = TT leqTltp
= der I^r f '? oz1r+ vt
n =dBTW (----
?=dprw (----
u1 z6'gzurc61 Eg,gZZZ;yZ
v' sg
.Te1o? ue6uedel uauo6
= 09-0t =Bw
= o L-vz =vw lpqT>tp
1 S[ =d leqT{Pu/+ ,'0 =b leqT{e
3 TT leqT{E
=llIC V' E9=X
.rc
1eueurD - [ 's b -r
r-J,*' z9' 6 Lv L6 L
' ( 88 Z(ZL*Ag,E I I +{zLl(szzt + = "r
' ( uepeq qe-raep )
lul
ru19'6f= 0t-g?+9'LZ
ulfot- =(os+or) fur18?=d?=(z)d- (g)dru19'LZ=Z(ZLl (Z, t )f
To leqT>te Tp1ol ue6uedel uauohl
1 u1 o t =vw leqT{p
: TO lpqT{e: unurxeu ue6uedel uauou ue>[nfuaua6
(0116169) ( t'LZIruc I' Sg
'€ [- L6l gEg'gsz +r(zt-8g't [ ) ( s? l \ *
) (s? | (zt-88'et ) + z(g-88'r t) (zLl (szz)
,{
1^
z(+
-tc=W
ue>16un11qrad1p ueta+ ue16eq
Tu-rnu 1. C---- (uroZ[=)(x-t6)8s8'g9z
0
rc uo+aq uBEpeat {ln - -r ue{nluauan1 ( (---- ruc 8B'E[ = x= z/ L' -(zl-x) s?+( g-x l (ztlgzz c
= Te-rleu s-r6 pq1 uauoul sT1plsuepeq 6uolouau dB66ue G----
= T P-r1au s:6 T se{oT ue{n?uaual^l,uc 8s8'992- c sv'u=f
.L {oT eq
r8t
bI uPp I uB{nluauan
-fUP Fl'oz =st\dt +616v=
uauo6 : qe6ua1 6ueluaq )tln€ TPOS
388
tlenentukan I - :e
Untuk DL :
Untuk DL +
cr 28,92Mrvr- =
a\trra = 0r73 ----, ( lM 39,6
maxMM^
= 1-tr;J. ro + (t- ( l^gE )J). r",max r max
= (W)3(e s7s77ot + (r-( W )3us7476s,6t
= 3924225 ,6 "*4 ?
LL : M ^ -:!'rr ' "cr = 28 t92
maxr" = (0;3zzf (6s7g77ol + (1-(0,322)3)(r97476s,61
= 21 4'l 867 ,8 "*4Soal 4 : Harga- harga I^ ditabelkan sbb l-e
Bagianperletakan A
lapangan AB
perletakan B
( kanan )
Ivlenurut ACII
akibat DL :
akibat ot, ( cm4 )
6.979.7702.141 .867,9
2.A03.857 t49
ps .9 .5 .2 .4 harga I" diambil rata-rata.
6.97g.770 + 6.g7g.770 + 3.924.22516]-
akibat DL + LL ( cm= )
I=e
akibat
*( 245.451.997 r8 cm
DL+LL3
I=e *(6979770 + 2.003.867 49 + 2.141 .867 rBl
3 .316 .8 43 ,3
24
cm
So.jrl 5 : teqdEtan akib-e.lE-Pr,, :
Ma 106 kg cm= l0 tm =
= 50 tm= 1 ,2 t/m =. 12 kg/cm
= 12 L
Mg
q
P
f,t.ng.h bentang akibatf- 23 P.L3v - 648 4-4
( rumus ini silahkan
,hat pada IVIEKANIKA
NIK II).
,i,
,'i'
ilr
]ll
1iTEK
'6.979,770
3.924.225,6
5.979.770
fe-sg+lSZfr
61frt,
TUT| 61fr
I{Pr'rPqr-p -requle6ra1t1;ades ue{pfeT_rZ >toTeq Tnr.{e1o>tTq : tZ TG
rucEr,r=33h=ry=gq. E. 6 raqel rJV 1n]nuew
- ruc ggg' z = LZV, | _ Lgz' ? =TT +pqT>{e . L. ruc LgZ' V =
- 9[
- r lnrtzt\ gL000.000.1 + z(0021) 6dfddr=Z * v(oozr)(er)
r(00zLl(0o0LZ) fftL.T
. TCI feqT>te ue6updeT Of LZ= d !r/l'g'L =biruf 0L =BWlruf VZ_O^5
'ruc LZV'L _9t
- -! l^t\r.\t000.000. [ + ?(oozL)(zL\
r(oozLt(ooozL)w)
' ( zlooz t )
,88 rEI
(z(ooe[)
?8e4
sr
'g
9L000'000's ' z + - ( 002 t
TC leqT{e ue6ueruc r 66-[ g?. E :u:19 ? :"" vrv :1
00[st = iy 00[9[ ='s
za '8w
acrggtZA
'VI^I
acr u ?gE
;ET -E_
,wc/b>1 [, tEtgAZ =
ue6uedBr=3
uB6updpr -=3
=ue6ueU"k
t Bw leqr >te
, vw lpqrlte
s 9[
SL,98L
68€
: b leqT{p
-390-
llutu beton K225, baia u32. LL/DL.= 1,5
Tentukan beban mati dan beban hidup maksi4um yg diizinkan menurut ca-
ra kekuatan batas ACI. talu rendanakan penulangan geser .balok ?
Jawab !a. Irlenentukan apakah luas tul . tarik melebihi naks-imum.
.u = eo##5 d = E*r+e$b6 (60) = 3e,33 cm
y', = o,8sx = 0,75max
= 29 r50
x. ( syarat ACI ) .D
CM
5fi25
k_qL= = W (o,oo3) = o,oo24s )'o (: #:ft = o,oolsB)
a = A.* = 0185 (2g 15) = 25 1075 cm
C_=-0,85 f; b (Fr. xmax)"= 0,85 ( 18 6 ,75 ) (35 ) (0,85 ) (2g ,5) = 139312 kg
,C== A; (tO 0,85 f;)= 8,55 (3200 - 0,85(186.75)) = 26003 kg
f=C-+C-= r Es:1s=ks
b . I'lenentukan Mr, : Anggap tulangan tekan sdh le l eh .
letak garis netral :
0,85 f; .b.Fj.* + A; (t, 0,85 f;) = o=.t,
0,85( 18 6,75 ) ( 35 ) (0,85x+8,55(3200-0,85( 186,75) )
= 24,55 (3200)
x = 11'13 cm
Cr t "'=
(11,13-5):
Af=3il19s=8r55
C" = 0185 f; .b.f ,rx
11,13 --, 6'== 0,00165) €V(=0r00158)
---, tul . tekan sdh leleh -
.X
OK
lir
l1
'i1rii:iliii{t'
,$i,
frC-sc tc P=fll/
(11,13) = 52560,8 kg.
1 sgv' g ='nn1e6'
'9[=nn
' 1:asa6 {n1un )
( 0e ) ( ss )
p
E8'0 = fr
uoa tsz'gt = tA
9'e : (6',0-g,g) = gLg'"lz , nA
1 6uedueued JT1{o} Ja -r66ut1 1
urcog -p PuPrurp uoro{ e>{nu Tfepp >1erefjroq qelepe np[uTlTp 6^uo? gLg,tz, = tz)t)LZo'g .f = T.
uA fr )"n : {ceLIJ
uo1 t6'ZLsL'ggL}
IvICq ' ,Jl. _Joeueurp ( "
= 6>1
€g'0tE'0
cA+
Lz
8Z6Z L
' ggr0'E9,0
^) / >cAfr
L9, LZ
6ueluTT L.Z- [.[[.sd fJV Jn-rnueu tb f = tn : .ruoTo>t se pd 6ue1ut1
w/l vzg,L = TCb
w/l ggz, z = TTb Tpe[
L'l'TCb grl + TCb v,l =TTb L'L + Tab vrL = w/+ LZo,g =.b
(-rn+uaf {1n) 06,0 = f, eueurp fr.(Se,f?) = Z(Z,L).bf: .c
ru1 Etrt? =
90'zE t Ett? _
09 ) 09EZE =(z/e-p) 'J = 'w scJ+ J
^s = J V= J00zt) 99'8 =
'p
, ruc6>t
(S-09) g'20092 + ((tt,It .sg,0) z/L(rxc s p) =J +
xo (____ ,t a6>1 ogsgl = (oozt) s9'vz
b>1 g'zoo9z = ((gL,ggL) 9g,o
f'0 f'0
t6t
c ^_.
s s (,J sB'o "J) iv = J
.,J
Tulangan geser dipasang,= 1 ditentukan' sbb :
,1 = (3r6-0r9) -d + ,1 = 21226 m
tuianqan creser :
Pakai sengkang :
fr Vs = Vr, fr.Vc
sepanjq4g d + ,1
= 16,257 12,93 = 3,327 ton = 3327 kgPakaisengkangg 8ffirrlr 2 penampang
fr A .f .d---_- gV
= o,gs=(2. t,/a (o,g)2)(lzoo)(oo)3327
= 4913 cm
( s--* = k: = z' r /q \ !0,9I?(320g) = 26,26 cm|
-max 3,5bw 3,5 ( 35 )
1 atau dl- =2 =JUcm'L max
3327 3327\/ = "-=' = = 3914 kg"s g 0185
1 ,06 ,fl .a* = 1 ,06 1186,?5 ( 35 ) (G0 ) = 3041 9,7 kg
s\2pakai sengkang g8-20 untuk daerah tumpuan dan g8-25 untuklapangan.
gambar :
8-202 ,226m 2 ,226m
Soal 22 : Tentukan harga P dan e dalam keadaan seimbang ("balance" )
untuk penampang lingkaran seperti tergambar. Mutu betonR225, baj a 1J32. Selesaikan secara kekuatan batasr guDakan
ACi code.Jawab :
1. Tentukan lokasi garis ngtral .
x. = 9f9-a,="b 6090 *-fv
i'lli'
,rr{$,
'
3 ( s.6zzoilo -Zv,8Z
d 3< t6r00,0 =
too'o'Z?' 8Zv t-sz) -zv , z'gv6l
(oozs) %#ss,s
Zwc/ 64 Z' t.gv
(oozr) tffirq
g, 6vvz
(ooze) Hifffi
o3 ) oq6ooo,o =
zv' gz coo'o:
^3 ) se zooo,o =
gLgLoo,o.ffi-zv,gz-Ee,t?)
^t) Lozroo,o =
btsroo'o'ffi-zv,Bz-Et,t?)
r9too,o
Z?
? ='?
edu-re s eqeueN
gLgrOo'o
Ct't.o'z =s00zs rI
qqs ue>fTeqefTp
=
= t.f {nfun O?
oaoE
a(,l
l1lr--
rlI rJi(,\A
i
,J}.r,ffi
,",o1J-
..S
z,e , l.f,
urc LE L
tucL gt g =
{ln e[eq ue6ue6e1 uetnluaual^l . p'VZ = (ZVrgZ)Sg,0 _qx.ry'=e" .coZL soc(Se'8t) =
, n=" ,'=, , {c,J E8'
too'o=
ruca t ,tucsg, V
vtc. Lgt
azfr
v o{ _ t a>t ueJarecrucg8'vt=ogt soc (strgt) =
g e{ _( e{ uPleJecruc Sgrg[ =
(6, zl f -2, L-v-sz =g e{ _ t o}t uelaloc
,rl
'q
ruc at, E?= tuc zv' 8z = zril
(6'zr+ -z,t-v-os=p ffi --e*
t6t
H
'[lillrll
prrds
3=Eftpd
394
€. !{enentukan besarnya Tl ,TZrT3ra"nr"rrr""a :
T1 =A. "1
=z
"3C =A-s4 --s4
' C =As- s-55C =A=6 "6
. r, = 6r41 (3200) =
. f - = 2(6,41)(2449,8) =,2
. f- = 2(6 t41 ) (483 ,2) ==3
. fo = 2(6r411(1948r21 ="4
. fo = 2(6,411(3200) =92
5. fo = (6 r411 (3200) =
"6
az = A
Tl =A
20512 k9
31 406 ,4 kg
51.9 4 ,6 kg
24975 ,g kE
41024 kg
20512 kg
f . t{enentukan besarnya Co dan lokasinya.
tuliskan rumus-rumus ( anda dapat nielihatnya pd Analisaseri DELTA ) .
ni kamiilid 2
Dis iII J
I
fh cer
rI
J*h
0
g,Luas
o A=12.2=n
pusat
cos d0
- segmen yang, diarsir t
n' Iqrrr,?o do
, O o( -s in?( cos a( r,
i-- L:, -*do.si, e .
Momen inersia segmen tsb ttkr =t- f1,r^rnto E- ol
Drrr .,
Kembal,i ke persoalan ,kita' :
Perkalian luas segmen yg diarsir danjarak ttk berat segrmen tsb thd pusatlingkaran = Qo
2o(^ h- f =
ir,2 o eoso do=9o = Tro
T
X=
cC
Qo
7t== 0 r 85
= 0r85
10398.77':-1-i-r:;,' = 111068939 ,48f: - Iuas segmen
C( 186 ,75) (g3g ,49)
2/= 25-W = oro3372cosa(--75--=urUJ.
4 = 88 r05o = 1 1537 radian.2
o= t- ( 4 -s inaf cos a( )
+ (so )' (,,537-(sin 88,060)( cos gB , o5o ) ) = g3g ,48 "*2
Qo= # =irt{h (so)3("in BB,06o)3
- 10398r77 "*3.cm
yang diarsir.= 1 49130 ,7 kg .
*,ink
I ingkaran,4 ,4o< sin 4qrr rF 255
l.lenentukan :
rucBev'Lz = (Bz'zt) sB,o -_,9e
ruc gz,zt:- 00t/z+0602 =
qx=Ts?fo o og
:
E,, pac:roJ uToJ lapun,, ueEpee{
o'e nTnp ue{ntrua,l. e.e
r pocJOJUTOJ JOAO,, flE1
qx uep
fPE[:"+ qe{ede nTnp {caqclo . B
-----
' uebun{Tr{i5d qLLrEuEt : @eT
#l
TD
opoD IJV ue4Eung- VZ O e[eq upp SZZ x
uof oq ntrnu p>{T [ 6uedureuad Tn{Tdldp 6ued '.{ edu-resaq ; rFe.ffiE
ruc 0 t = 'd >t1n sel trsTrf uos{o
lur -uo1 9Er9e =urc6>1 zg L' L6ggEgt Ie+o,l,
Tn{Trxour og/ 0t, 6upI ued t6os.rocl 6uedureuocl
tuc gg,OZ = ru Bg0ZrO = |}'*LLL9 s-te
rnqefa{Ta: tZ 1eosq^
=d
t Q" ue>1n{uaual^l
tf' zJ,
,:*9='?='
Jc
JeurEN
6unlTqTp qa
. Lg'gsg'?[Et'gIgt'B I
sg'? [
Lg, g
B9O, I L
(g'b6 I gr ?, g\v Lt+ZLS0u ) -
'6updrueued f e_req {+f pqf,q- ('o
' uo1 tg, t.LL = 6>1
( z t soZ+vZo L v+6, S L6vZ )
.t
9,6Z9LLL =+L'0tL6? L =
scJ+ c=Qd
out3 = o^=o- : .w uP{nluauow'q
zgt'tzl5tvo 'Egf ggv
z's6tgl"tz's6tgt.t
v'g0z060gte t'tLgLVt
r 8s'Bts0E9 t
6>1 g, v6lg6>1 v'go?tt64 zlgoz6>1 z L soz64 vzalv6>1 6'EL6vz61 L'ott6vL
1 ulc64 ) -W( uc ) uoruolu ue6ue1ede6 edu:esaq
E6e
r Qa ue{nluauaw .6
396
t'EE.^EE
HE fr=s
dU0)cat{
.t (,t a.2 ""q".."!::_:!_:C":0,85.f; . /1.*b b
= 0,85( 186,75 ) (0,85 (32,28 ) )30= 1 30663 kg
C = A' (f 0,85 f ')ssyc( 3 ( 4 ,91) (2400-0,85 ( 186,751I
_ 33014 kg. T-A f
= 3(4,91)(2400) = 35352 kg
P._ C + C TDCS= 130663 + 33014 35352
= 128325 kg = '128 ,325 tonMomen thd grs berat penampang :
Pb."b= T (4S-25)+Cc(25-1/2 aO)+c_(25-d,).
S
128325(eo)=35 352(20)+1306 63(25- )r17,43Bl + 33014 (2s-5)
438 cm
I
CLCI
I
P -0n(45-15) + C
b."b = 22114 cm
Karena e (=1Ocm) < "b(= 22,14reinforced rr " Compression
cm) ----) terjadi keadaan
C"_0,8s f; .b. F 1.x
y: (5-y)+ 11068,1 : gTgt,55y= 2 t72
jadi: x=.40+2,72 42,V2 cm
tt over
= 0,85( 186,75 ) (25) (0,85)x=3373,17x
0,85 f;)l(2q00-0,85(186,7511kg
S
= (d-x) : x
= 0,003(45-x)x
2Kg/cm
0r003 (45-x) .2,03. 106 )x
89705 7 (45-x )x
^(1S-1/2 a) + C_(15-d')C S' -
. 30 + 337'3,'l7x (1S-0,425x) + 33014 (15-5)
= A' (fsy= 3(4 r9",_ 33014
=A f
,2,E = 2,03.106
S
T = 3(4,91)(
berikut :
cm ----, f {x) *i .1 1068,1cm ----) f (x) = + 9291,55
Ts
9 zCD
fMomen'0 .=
'0=
rhdT
89 705 ,7 (45-x )
diperoleh : 1433,6x 3 505 97 ,isx2 + 236 r 031x 1z1102695 = 0.
diselesaikan dengan carar
@\oo
x = 40
,(= 45
LnLo
45\ro\c!o\
l*lltlt_..],l
-r
le-
(00t) V'L =
L + oa v'L =.tdu : d uPtnluauaw .e
-
:ue3 qPl[PIl
e apoc lcv. uebuap , qsl uoTolt up{puEcuau , ztn etuolaq nlnW ' TEf,Tds te>1edrp ,, saTJ Tp-ralET !, te6eqeg
'uoloq 6updueuB 'uo? AZL = Teu-rou ,dnpTt{
Tleu uPqaq lPqT{V - sr.rluas6updurpuad uof o{ up{pupcuag
'uo1 VVe = (OZL) L't +Ta L,
Pq, E ZZ>I
s3 -( (sz-a')
-ed spnT t Z/J t unursteu .fn1upqeq lpqT{p ,1 00 t = Tpurou :
uE1a1 Tprurou Tn{Tuaru 6^ -repunq( lEZr' 8z,t=) 9d) uo+o Bo' Z6=6r{6 g0z6=1. 0tt?-? t 0EE+ L, sLtZo I =
6>1 [oEe? =gt'08
. gz rPos
6>1 ,tott6>1 L'sLt.zoL = (gt,oE) LL,tttt
ruc Str0t = x
( sr0t-9v)L,90[66
c
'rrL
s^ .v
c= 3 TpPfqa To-radlp
I
ls? e00r0=0=
31
6'06L0vL xL6rggtt+ z*gL,LL +tx0 = ((sz-ot)+s) ?r0€t
(s=xszl' o)x[t' t,Lte-og .ffi((sz-a) + ,P)
+. e Z/L)'c-(a+oz) ,f, = o
0 = 'd prtr1 uauow!
'- T r (x-EnlL.goL6g - d,
vl0Et "Dx LL,Ertt =o3
- Tp ledep bi 'd ue{n?ua& . ruc 0 g = a( lszt'g zJ=) ea (UOl TZE'ZLL = 6>1 L'}ZT,ZLI = L'LILV
64 L' LgLn =
lle=+t l.t' cz
: qpaaeF
'Tn{TdTPos pped mS
bcccI
I
I
,l
I
I
dI
- ( ,, TOrluoc uoTsualr ) r pacroJUTA-r f,epE uueppea{ redec:a1 (---- (ruc;,L!ZZ=) q"(a eua.rpy
P{T[ 'T,Z'ON
?tOEe + g'[0 Lv?L =.f, =o+"J=
2,L,, ZV=
d
,l
cc
6>1 ? t oEt6>t g'LoLvvl = (zL'rzv,) LL'u.Et
s
c
l-6t
398
. D :P''n fr
ACT
b. lt{enentukan ukuranACI ps.10.3.5 :
dimana Po =
on=+
= #U= 45g,7 ton
ps.9.3r2.2 z fr = Ot75kolomP.- = 0185n max
0,8s f; (on
A (0,95 f , +gc
Po
A=t) + fy.A"t
s ( fy-o,8s f; ) )
Ast=crAJg
A=g', ,24n=
0,95 on (0,95(2 r035 cm'
2035 .*2 ----:
(55)2 - 2375,8 cm 2
c; Dlenentukan penulanqan memaniang.(458,71.1 000 = 0,85(2375,8)(0,g5( 1g6,751
= 0,022Ast = 0,022 (237518) =
pakai 12925 = SB ,92sefimut beton ambil
, d iameter tt core ,, =
52 ,27 "in?2
cm
= 4 cm
D" = 55-8 = 47 cm
d. Tulanqan spiral :A
,o = 0,45 ( f _1)ls c
dimana AgAc
frcr-v
= 2375 ,81'7Tfv
= 0145 (
2CM
147 ')2 =
237 5 ,8'1134,g4
= 010097jarak lilitan s
a.s
'1734,g4 "*2'
_11 186r75''. 3200
ral yang satu( D"-du )
pr-
tL
=3 *rP. max = 0185 o, (0,85 f; + g (fy-0,95 f;))
458 ,7.1000= 186 ,75 ) +0,035 ( 3200-0,95 ( 196 ,75 ) )
h = 5019 cm ambil diameterkolom = 55 em.
+ g (3200-0,95(185,7s ) )
5- A" '/,
ke yang lain=s
p 0[9,0 = o0qt=l=q609 = L+=gs!, =
qxp 060e o-om'
[.p eFPs TPuf,rouup{nluauaw -p'ur1 LVg'ttz =rffin)(92,tl 6z'8v
*u^ ir^
+v^u
W
xu-W
^ )
aL,LSL =
OW
oA ii Pueurp
q) (q
x(
/T
W+
^u
I^l Pua]PX )
'JlE-p-x
Sg'O = / e6req
= , ol'o f, =ffi=W
--r4- =*'^
uPtnluaual^InTnp fTs{q.f,
u1 6z, gl
rua ?l'Lsl =
uol t L' L6L =
r.+(0t)?
oL'o
oL'o(0v) Llxn -W
=il=d d(q' Z'Z' t. ' 6'sd ICV ) 0L ' O=i
. upEu.P
L 6urpueq TpTxp_Tqpql tcaqc nrpT 'Bz(uueouelnuad upp ruoTo{ uprn{n ue{n1uaJ,
.6e n p[eq ,00e X uofaq nln6
^ ru1 0t = --W ! rul
rul ZL = ^^ J rul
1
1
o? = *w 1=qT]= :
oe = *w rroT$, , uauouOV=TT0s = TO leqT{P : TPrurou
fn{ruau r6es:ad bupdueuad rnqela{TC : 9ZT65g, g- O L fr Tpf,Tds re>1ed ,nri
ruc Sr9 = s TTqurp' ruc vL,g = -.--tLAOO'Ilu-A'-vgti_. __c L_Lv)7 .z(L) 11 nn _. .
ruru 0 t = TTqurp 1__ uunu g, 6 =urlurluTlx 1e-ztds fr'uT [ = unururur ! ur e = unurs{pru : uTpT 6Uez{
alt nles 6ue^f, TP-rTd: uelTTTT t{rsreq {e:e[ rsel=qureu ?.0 [ . L. sd rcv
u
400
P. =C +C Tbcs
= 0r85 f: .b.
= 0r85(0r83)(Anggap Pb = Pr, =
Tu"T
9rsberApena
= C_ Karena luas tuI. tarikctul. tekan dianggap sdh
Brz| 1 ^b
diambil = tul. tekan danIeleh.
thd grs berat penampa
D9.tekan= A_ ( tul . tarik )
S
e
300) (b) (0,85) (0,610 d) = 109,74 b d
197 r',a4 ton
bd=wdiperoleh.,
= 1796,4 cm'. selimut beton ambil = 10t . h
b(,'e ht = 17s6'4 ----' :?,;,1;::=;llu ----,,0 = 24,7 = 25
th= 3b= 8l r25coba ukuran 25/8O uiuk keadaan balance ini.
Ukuran kolom kita sesuaikan clengan keadaan "under reinforced" atau ',Over reinforced" yang diderita koiom :
Jika A- yang diperoleh )A^(keadaan balanced) --) tercapai keadaan9- / g'under reinfotced"
Ad yang diperoleh ( A- ( keadaan balanced) --) tercapai keadaan9- \ g"over reinforced".
Catatan :
Utk keadaan " under re inf orced r' :
Pr,= 0,8s f;.b.d( -/*, - * +
utk keadaanb.h.ft c
"over reinf orced tt :
n 3.h.ed2
+ 1r18
At .f+s-v
6ft'+0,5Keteranqan :
e = jarak P--n
Ar=S
f=el=
m=
Iuas tul.A^/(b.h
5
d,-1 /2 h +
f v0,85 f ,
C
(r- flr2+2 (ttm-r)(r- S: )
+'(e'/dll
li@*jr r si Pnetrai, n
n
.! 'j.
,
:t:
llt1l
,'i
..r.liltii
,Li
.a ffi
'
leduaa{ pppd -reqesTp ( -r.rrc A, gL = (.
-uc SL'gL = SL '0t C
x uod/ w
rul 09, Llz,sg'o , t- t
'89'o-['Ete 'zl 6z' gt+v L' LgL
,j* I tfil ^'^ + *o^
= x
tet' Z =
'lUJ 0L = t{
luoTo>t ue6ue1n1 sen';' r.pac.roJuTal lapun.,' ueepea.>[. e__
(pqvL'60[=lqa)(pqEv,L6-)'apAqPqfeI.ITrfa.I,:@SL/ Ot. 9L -q TTqure Lg -p (-___ 0t _q pqoc
z vvvv F I ! r-r av Lo
pq svt L6 =
+ gvL' z-L+ZO, O-)p-q. (00t-egr0)Egr0 ='dru Eor, r - yr# =
' eduBs t s
SZfrg L re>1ed 1
'Eg't = q'q'Esr€= t"+"a) eqoo
=A
oW
qt{
0[['0" 9VL,Z
99
irir
i;l
ffirfll
.i'iil.r;
]Ii
iii
lll
riF
=p,P
=pra
c----tuc
Z
ott + Ee 99-p
%: r6e1 6un11q1p
99 =p +--- 0t = q Eqoc
t0'2002 = p q
pq LV'96: 000[ ' vL'L6LP q LVtg6 =
l(L'Z+(Lr'0-t ) ( r-t?:8t )l tzo,0lz+r(L'z-L) /+ l,z-L + 20,0-xp) (qx00e .f g.0)Egr0 =t
crro, _(ooe) (sg'o)ggro@6€
d
tu ?r sg'o
=-P,P LT,O
---r-J
r{6'0q L'0
SL LOL,Z =' tu gzz' L
VL, L6LGM
udA
xoW
. TSTS
9ZZL + OV-iLp
I
AII
enpa{
08 / st^
m=;5Z9'O =dl
eped 6uesedtp 'zv = Te?.o?.a-Lqo3
: ecueTeq ueeppe>[ {n?un . fffiIE66-e1
L-ev' g [ )J (zo' o)z+.(9V L' Z- L,)
'rpocroJuTor rapunrueBpPe{ urTp uoro}tuPrntn uP{nluauahl z.P
!ril;l!ii:;i
u,rlh,
MI,iilir,
r$l
{!
ao2
f.
Po =.0r85
= 0r85
= 766.1l{enentukan
cl -l- c
7r=tul .
e/h
f,
280 kg/"*2 i0r90disebar pada
= 0r035
ex
ex6-
MNV==Pn
0 ,2450r300 ,035
Check thd bi-axial bendinq eara 'Bressrer reciprocal method".Check apakah Pr,)0,1 f; . on
197.140>0,1 (0,83.300) (30) (75)197 1 40 ) SO OZ5 ----, metode ini berlakuf; ( o, A=t ) + A=t. fy( 0,83.300 ) ( 30.75 79,5 ) + 7g ,6 ( 3900 )
16,8 kg = 766,116 ton ,
P:-ox
P., = 197 r14, tonMr, = 157 r'l 4 tm
xM"nx 157 .1 4a = _- = .- r - = 01797 m-y P_ 197r14'n
+=W=1,06dr diagram interaksi utk
( lihat halaman 350 )
6Pdiperoleh n
Ag
ox
Cara memperoleh gpn/Aq
cari garis e/1= 1 106 (pd grafik. terah ada angka e/h = ljadi diperkirakan saja angka 1 r06)
garis tsb akan memotong /g= 0r035, dr ttk perpotongan tsbtarik grs horizontal, akan memotong sb vertikal pada ang-ka 0,75 (perkiraan)).
Menentukan Po :v
##; o,zls
= 0r817{E-ri.4;
'6 L 2= 0,75 ksi = 750 = 52,5 kg/cm'
( 30 ) ( 75 ) /0 ,7 0= 1 68750 kg = 1 68,750 ton
f --= 42OO kg /cm?v
keempat sisinya.
fn=il.P
diperoleh -# = z:,;l;)",,,ir
lrIil
dari diagram interaksi
xo
. TSTS
S'Lt.L = gZf,gZ te>1ed1
de;eq) .OL/OV TpeIueurV L, L6 [= ) 'e[.ra>[eq 6A
gVIzOO'0 = ry
ffi92fr 8Z
rEqurPD
ledruea>1 eped leqasrp,uc 0?t = OL'OV .ES = .Tn1L
T rce>[-redTp +dp up]n>[nuu
( uol E' vzz = Td
( atuc
' ( {caqcTp
wc/64 ZtTS{ 9'
ztg-t
^og
I
ogL
xOT. d 'd r=t
: >[car{Duol ZLg
>t 000219 = L0'o/ (08) (ov,)zLL^oTo 6
t-t-
6V
% : qaro,edTe
{
uo1 ?0t = 64 000?0eL'O/(08)(0?)s'99
,wc/6>1 Stgg = TSlt E6r0
uo1 gLV'LgzL -_ 6>1 gLVLgzL( 006e ) 09[+( 09t-08'0? ) ( 00t. eB, 0) S8, 0
z" 09 [
E0'0
sZLg'0 = 4gvz' o
o : uP{nlUAUOn
xod
6v6- . qefo-radlP
d.' frS0'0 =
6
- xT o. . o d : d up{n+uaualil08'0V'90'0 = V
EE = rTqur, y
6) =d
=gxa
>leIueq:edrp ' TnJ upp lesaqJadTp . {n , ,a
) Ta eua-rpx
j (ur+ ?l'L6[=) 'a)ro+ E,lg[ = Td
i 98600,0 = gLL'9L:. sL:tLZ + st:g9[ = l,tt
^x "o_ o --od d -dr r+T
T
= + : tcaqc t
n 08'0 f 0'[ = 966'0 = ,!Y,9 = a u T6L'o :
0g/ 0v eqoc: ruoTot uErntn eqoc
^ uo1 sL'tLZ = 6>1 ogttLZ = L'O/(Et)(0e )E'gg = --od
' 9.
404
Check thd bi-axia1 cara "Parme load contour method'.
M(#E)
ox
loq 9 ,.I ML"sf + (#r)Ioq 0r510s F
?s(pn
6P,nA
9y
0,85 f: (on - A=r) +
o,g5 (0r93.300) (40.701099768 kg = '1099,77
O r 05
197 r14 ton
0r7(197,14).1!go40 . -70
interaks i
A . .fsty137,5)+137,5(3900)
ton
gDari diagram
= 0r7 ksi
= 0 r 05
ox
0t847 + 0r4"127
1 ,2597
o, g5 ( 0,93. 3oo ) ( 40,go-157,121+157,12(3900)1256793,55 kg
1256 ,79 ton
(halaman 350) :
ilMdiperoleh il:* = 0,g25 ksi
-'9 "' = g25 psi = 64,75 kg/crn2
= 64,75 (40)(70)(70) /0,7= 18130000 kgcm = 181,3 tm
= 64,75(40) (70) (401/0,7
= 49 ,2g kg/cm2 = 704 ,1 psi = 0r7 ksi.
58. 40.80= 160 "*2
32g25 =- 15'7 , 12 "*2'.
ilp,nAg
(,
.,?-, n^ 1g__Q-.5-gheck : ( ffi) T0;6 T;55
v= 10350000 kgcm = 103,6 tm
I P., ', 97)F-=#=o'18aol-Lq= 6 fu =o,oudi{%no=0,783
cdari diagram konstantay' halaman 34g utk fkita peroleh ,f= 0,55
42OO kq/em?YJ|
loq 0r5 )
ukuran kolom diperbesar,jadikan 40 /.80 .
r t 4gt?gr-' * 10316'
k, 1
dihitung kembali harga ,ro* r ,ro , ( 3
-vI o=.=
I pakai
ffi|]l,1
+
j,ll
i
ili
.t'III
iI
tri
I
II
-J,
3 .rEquEs
- ( e+eJaur ) eduTsTs ?r?duaa{prJ reqosTp szil e[ = u*buerrrl ,og"/ou rucTo{ u,rn{n :*:EET* T6'axr "rI
1;sE'0 6r:r *-6-'0 -ffiT
, t'BLl. ,( '6zr6t t
r ) B?e'o
+
SEE, O 6OIE,O bOT
+ LL9, O
, B'gtz, I TT,ZET I
'' !
fgl'0 =c,J b)
')$
4caqc
6\I
o7dluld)
u!'l
ul^I
frCI
xo
[,,
iffi.
ifi lll
lii
ll,
t E$ 'i3= qo rori.rd r p {
ru?|
u3
F'el* e I
s'$tH
t"sr'o -'##irernfi:1 il000Fftt !E {..,0/(0?)(0s)(0t,} 5L.Vg =
rusfr>i flii{}fiBi}tf E {n(},,/(ilS}t,$S}(CIU} gL,Vg =:
$ ' Iu,?r'bx S[r$'g #i T${ Se6-0 = -5:-^g- z/trda-).rr-'"tp("Llut:%;
i ( 0E I r.{BtrrtsI ur{ } Tfr{p;a}u1 uiB.lbuTp TrEpr{sTo.radlp
E0'o = ui \- --,..(tg)(or)-- - -
tn I 0 0 0, r -T ? i,?6n*Ir"O =
="fr )4 uoa VL, L6L =
ud
E0'0 = 5l/
:ll
ili
ilii
i[l
ffi
fi
ffirffi
E E0u
Ts{ Z9r0 = TSd gLg,;, ,walti4 el,ep =
406
SoaI 27 : Diketahui konsol sePerti
'//
menentukan a : a = 2r5
tergambar .
Gaya horizontal yg bekerja =
1 0 t - Mutu beton k225 , ba jau32
Beban yg berasal dr balok :
-akibat DL = 12 tonLL = 18 ton
Pertanyaan rencanakan konsoltsb dengan ACI Code ?
Jawab : Lanqkah Perhitunqan -:
€1. !,Ienentukan ukuran "bearing p1alg " -dari pelat baja.
Kekuatan bearing plate( gt0,B5 f;.A1 )
Lebar bearing= 35 cm (selebar kolom, arah tegak Iu-rus bidang gambar ) .
V-- = 114 DL + 1,7 LL = 1,4(121 + 117(18) = 47,4 ton.u
vu(fl(0,85 f;)- A1
47,4. 1000 0,7 (0r85.186,75).A1or ) 426,58 .*2
^.r tro
Jadi ukuran bearino plate= 13 x -35 x I ,2 ( tebal am-
bil 12 rnm)
anggap diameter tulangan A==25mm
d' = 1/2 (2,5) + 1 - 2,25 cm
Tul. A- dilas pd baja tul. (ambilS
finya= 25mm) yg tegak lurus pdnya.
Jarak antara ujung tul. A=dengan
baja tul. yg tegak lurus padanya
= 2 cm.
Jarak ujungdepan konsol
tul. A- thd bagianS
ambil = 5 em.
b.c.
+ 1/2 (13)
,75 kg/cm2
56.b*.d b,
=9 cm.
= 35 cm
V, = 47 tq ton= 47400 kg
ftC
Vn
Vn
= 186
=max
$s v,,
konso
ba l-ok
bear ingplate tebal12 mm.
35 x 35
max 1"=2t
lrlenentukan d :
.;j
z*"2t.' z =( sv' gzl (se ) s€zoo' o
tt.Z00'0 = =99?9 sL' ggL
=p'9.tt200r0 = unuTuTru s__IV
?0'0 = * ?0'0 = unur "l* / ,J
3 UInIuTUTIII Sv pq? >tcar{J
(Z*, Zt'Zi = gZfrZ te>1ed;
z*, gt'LL = gz'g+(f'gl r='v +'no 2= "vP{eru Jv <
t ", ,
puare>[ u=p
u-ruJAc -V + 'v uep *v + "v Z e;p1ue -resaq-ra1 qTTldlp =V
z*' 9z' 9 =z*' 9L'9 =
(sv'zrl € = z*" t'8 =
uV
JvJ^
Vg.
Z
'(..luauacroJuTar uoTsual ^f,reru1rd,. ) "V ue{n+ua;, .6
-ruc gz,g - (oozt)s8'0z--fL:2-m=
+# ='vN
' ( {Trpl eu6ue1n1 spnT ) tv uptnluauan.J'gg'0 =
p 610 = ueuou ue6ue1 =ruc gV, gZ =
luc It =
z*' sL' 9 =( 9?' gz' 6, 0) ( 001e )99, 0
:ffi= :,[J fr J--
r=vW
uI1 9669, V =( ErBZt o-[t' o) Ll.+( 60'0 )v, Lv =
c( p-q ) tn + "'*A =*w
frz
p
t{. ( dnpTq upqaq
>{1n peof -ro1)teJ 1e6ur ) c1 LL = (01)L'L =
tN
tu 60'0 = ruc6 = e
'ruoTo:[ uPp Tosuot uenrualr
(t'?' L. ll.sd. fCV leqTT)lrL =(1)?_'l=vtr't =/
,wc/6>10028 =^, -----
Tqepeqra1 uauou uet-n+Earuatf'a
ztn e[eqn
-rasab pr{p6
ruc gl'ZL = Z(?'r)(002t)E8,0
00? Lv -.7
)/'IJ'f, JA=V
A
Tn{rrueu {nlun Tn+ sPnT
JA3 v uP{nluaual{
'luc tt TTqurp ,uclr0e = ,p +
luc g?rgZ(p
t06Z,O = 19 = E 6eurc gzrZ =rp fTqurp
ge 'gE ' sg,o)oot?P=t{ €----(---- p '
i
i
J,
ri,
&
.P
408
A" (=12 t32l > A= minimum (=2,321 ----)
h. tlenentukan "h ( sengkang )
Ar,=o'5(A=-_1. I _ z= 015 (11178-61251 = 5r53 cm-
pakai sengkang 2 penamPang 910 mm
luas 1 sengkang = 2.1/4 tt(tl2
pakai ?# = 3,52 ----? Pakai
Gambar penulanqan konsel.
itr'xl,i{il
SHdil
1lrl
I
'*.1OK
= 1 ,57
4910 (2
2cm
penampang ) .
A, didistribusikan merata sepanjang 2/3 d dari tulangan As'n
a.
TfT#L
i)-l[t o**
12
I
4t,g
25k*t g
2 p:nampang
4g 10
baja tulanqantulangan A= (
tegak lurus"cross bar" ) 925.-
soal.28 : Diketahui balok "spandrel" sepertikut :
tergambar pd denah be-
4
om
om
o4m
l.@
:ii
7 -rr1 9,0? = *6{ 00g0? = Z(Ot)(96gV1 !-I_ Cd,^,
,t/bx g6gv = (002t) L,L + (ovaz) v,L =' t - g'*
TT L'l + TC VrL = n ,,pEoT pe.rolcp,;l,' ur/6>l 002[ = (t) 00t = dnpTq ueqaq
,. t ,..--
i , w/6>{, OOVZ = Tab
vt/6>t ZtV = OOVZ x 9,0 x t,0 = (Og/Otl {oTeq TrTpues 1e-raqw/b>t g09 t = (, ) ZOV = 1e1ad T_rpp Tleur ueqeg
,vt/6>1, 00t = dnpTq upqoq c_w/ b>t ZOV = T leru ueqaqZ_w/ bl, zvw/bx zLw/ 6>t ggz
'(r TP = eTp 6uedueuad r{aT 0 1e>1edtp ;
IU] L ZS, Zruc6>1 Z' Z LOZSZ
I t' --ttoor r lt , ?gt6E sW L'[ )88'o" Eg'0 = fr
z - ( rucz Teqel ) ueltnpe le.reqt =(rucg Teqel 1 1e6a1 feraqZ L' O = 1e1ed T-rrpues 1e-raq
3 TPlueT uBqaguauou uetn+uauon -q
' ( t' g' [ [ rcv )
pe-roaceJ unurxeru,r ) t,l ue{nf ueue6 . E
Z
Z
Z
LZ X
vzx00? z x
,l
_ruc ?gl6E = t rvTvd
(9t'rZL+09'Z0E) = ([Z*) eueuTP
,frZU r'r) fr ',r
IJ {oIeq {nlun ( .r lueuou Teuor s-ro1
JPnra{ uep
: uebunl r Q.a_-d. qe$ue'1T*EenEr'
fJV epolaru ue6uap IJ >toTeq ue1eupcue6-Zto e[eq i SZZx uolaq nlnp
,vt/6q 00t = re+ueT 1e1ed eped dnpTq ueqeg L
0t/0t = (foTrolur) ueTep qEraqos ruoTo>t ue.rn{oOV / Ot = ( roT -ralxa ) -ren T qef oqes uof o>[ ue.rn{O
0.9/ 0E = )tof Eq pnuas uetn{n-re1.au g' e = TelueT t66ur1 ! ruc Z L -1e1ad Teqait
'rs-ro1 Tn{Tuau ue{euEuTp Htr upp rD >{oTpg
{nseu n1u1d 1edue1 >{1n ue>l6ueTTqTp H uep,{ Tp ruoToy
{orEqrauTr
cnI,
4r0
c. lte$entuka{r reaksi di -F dari balok FG vq bekeria pd balok CI dangeSer pada balok CI.
Torsi yg bekerja di C = 2,521 tm , di Itorsi di F pada balok CI = 2 (2 ,521 I
Cross irntuk balok FG
F q
= 2 ,521 tm5,042 tm
C.O = t'carry overtt= induksi
+35,759---)c-o(il-1 -17 '879:/ 2 :58;378
t__+40 r 8
-5 t042
=010
J--40r8
momen :
fr"RF.
Tiniau balok CI :
-5,, A 42 + 58 ,67 g
0,6048VmRF=l911163t
58,679
(4,896)( 1O)2 = 0Rp = 19.[6? Ibht
berat sendiri balok (30/60)= 0r3 x 016 x 2400
= 432 kg /n"faetored l-oad" U=1, 4DL
= 1,4(4,32)=604,}kg/mCatatan :
beban pelat dipikul 01ehbalok as 1 ,2 ,3 , 4 karena"one way slabtt.
atau (d+t5)= 70cm dr as kolom
V = 19'1163 + 0r604g (4-0,7)"u 2
= 11r554 ton.
1
2
Bidan k CS
lbid - qeserrkritiS.ritis utk g
Vr17
*$.4m-D'l li2 t52'.,
2 ,521
er berjangka
I
I
a
d. r+uas tulanqan senqkanq untuk memikur torsi .
T g Tl-;u-c"t U f .o(..x-
4,896
F. tom f
g ,0. d".*1.yl S At= luas 1 penampang sengkang
vtlso'o = 6lsro,o + E fo_ = 1; +
tr
iffi
ffi
iii
h
.ii
=9Av
# ="4A
c+ A)fr =
tA
'-InluaT -Iasa+ TSrOl
t 0E0' 0
{nlun :esa6 -TffiFt.J(sE)(002r)z'vggg '6ue>16ues
6uedurBuedsenT =^V(-- (fCV L-LL ueeurps;ed, n ; 4,A
6>1 Z,1EBB = L'sELv gg:q- = ,,i.Egtt = A
(=n
b>t t' gt Lv
, V99LLZ( ffi'6L20'0 'g,Z + L)/
a
EE'ot w =
n, L a
.,* *J s'Z + ,,
/c=n
uoloq qeTo Tn{TdTp ldp 6^ rasa6 e[eg'.InlUaT .IasaTn{Tluolu ,{r,+,ro . a
tZ,
(z'rulu
tutu
=[I
=[*uoleq
>16ues*:,TI"=,ffifr dB66ue
'luc 9t LV
t +9 ) z-09UC 9, LL
[ +E ) Z-0tE =f T quIP
Zt = 6ue
i+tt'o *':;:; =lxf rr,o + ee,o = ,p
ru? OgVrl = tuc6>1 tL'0009?t = r,tr qofoxadlp
_utc/b4 SL,ggl = ?l ZI
';; ;:::; = l:.=ru1 LZgrZ =
t;,
Glzo'o = #iS: = ft
= ?c pueurp
nL, ,l t 'J r-\
z' n ,A V'O
^r*2. U sLZ, o
1, = uolaq qafo In{TdTp
+tc
ldp 6r{ Ts.rolrul LZg, z =t,l
tlv
412
Coba sengkang g 10 mITr r luas 1 penampang =
jarak senskans = d:#ffi 12,s em.
!f r Check. thd jarak senqkanq maksimum.
1
47D (t)z = 0,785
*r + Y1= '1716 t 4716 = 1613 cm
4D
maK
jadi ambil sengkang 2 penampang g1O 12,5
h. Check senqkanq pada bentanq tenqah.Lintang ditengah bentang :
v= = ry = 9,55815 ton
9558,1 5 = 0r0543s f_. d (3200)(55)v
VAv=
0 ,054 3
- + * = o'03619 + =-Z" - = o'05334s
s = =o*J9?' = 12,4 cm0106334 vr.,
sedangkan jarak sengkang maksimum = 1613 cm
jadi jarak sengkang ambil g12,5 cm.
i. Cheek luas senqkang minimum.
315 br.=
A.At'v
A . = A + 2 A. =ml-n v t fv
s(30)(1s) = O ,4g2 "*23200
sedangkan sengkang yang dipakai 2g1O, luasnya = 1,57 cm2
,. or." *r""ol"Tllli""o-"1,1*,,.,*,*,r*,, .",=r.
Check
maka
At
2A.
+ (*r + yr)
2(0,03169)(17,6 + 47,6) = 4,71g23'5 b*'s 3r5(30)(15) = 0,4g2 cmf=W=U,+YZ
v
) -2 A. ) (t
u
dimana s = 15 0,03169 (15)
0r54285 "*2.
2CM
2<z Ar(=r,oB57 "*2):
,28 xs ,
-\---F- tl-v
T *1 + Y1).S
^-tS
Vu3C.tAcm i= = 0,03169 ; At
E'Z [-0 td bue>t6uas
t@tZLdZ+SZil
T,,
(rtucgZ'Z-ZLfrZ-re>1ed)-urc ggE,[ : T\, !- - rJ .z : vr*,v ---r--t(, - - v T = v
( Z*,
gZ' Z-ZLfrZ re>1ed ) Z*rEgg, I TV + = V
: ue:1a
Z9,gI, = ZLf,Z + ;ZfrL tC>1Ed
-ruc tgrEt V
c = iT\ + vz'vt =
tuc
9Lt Tv + = venv - t6rTSeu -6urseu TsTs ledurea>1eped ele-reu uE>{T snqTrlsTpTp Tv
as 1e66un1 . Tn? r e>1ed el T >t e>[eu
ru1
(
. TSJotr + ;n+ueT leqTlte Tef 01 6uef ueuaur . f n,tr
Tv + * "v -Te1o1v
.
.ruc iZrVt. sEnTTnqele{Tp 6^'r,uU?"xPtu tw euerex
.T
(z/e-p)
z*' vz'vt =
6>1 st'?6g6ot
ruc L,gt =ffi=(Z'e'6'sd ICy) 06,0 --rnfueT {fn fr
rul Lsg, gv =
6'0TTyW =
S LZOLZ 'E8'O
(qLZO' LZ\ (98,0) ( 0t) ( sr'98[ )E8,0 =xeuq.?l ggr0 = xeu ,c
tuc gLO, LZ = qx gL, O -xeux
69, LV = ruc6>1Lt' t LgggLv =gtt v6860 [ =
gt'v6g6o t -xBtu0 \zt xeu
9t'v 6E60 t6>1 st.'v6g6ot =
J + 0609
ss )
u
sV
,I
o-
avp 0609
f,u= nW
W
z(B)f + (8)
ru1 L L0, tv =
(tgt L'61) 9-= ueEuedel
t-(8?09'0)'rn+uarTn{Turaur
tucZ
xPtuW
{nlun {oTeq 6uC%
uPuau upeTn.f, '{Vtr g = "rpseq-raJTv TTquB
Z*" Vt'g = (6r1Q'0)t,
-tgE t r +00 LZ9Z
tt,
, EL'g' Lv+g' LL'
( s8zv9'0)z-00zt . T-_(ETJJogJTZ,, = v
,utc/tr6>1 L0,0- pdlt 6,g=Tsd T fesTu c
'rTe6u
Ed000T=
Tsas?r-ed{000T=
,w/No L' 689 = z
-lu,/N sgg0g'6= c_ u,/Nf _.Z
3 Up?e1PJ
Pd{ T (S)pdw I (?)
ur/qr r (e)
z*/ J6lt (z')BdT (T)
: upqPquP.l
vgv' 0l6>tzzgvv' vNJqIPrfd
l0'0,wc / 1b>16'g( rec
sedoTT>{=PdxTsduffi'
88?000'09T88?, T
mc/6yZ,}w/ l6>t
6' LV
g' vr
Tec sPd= ) ed
ur,/mzal /qrAJ /qTuPqaq
;6 L9 LZ' O
S8TO,9Tt*/
6tI
t*/ 6l66 L9 LZ, O
g8r0'gre */6rt
tt/ ble uT,/qr,
JJ/qrirlgv' at6ts'v
( uo1/{aN- ) NqIleJr.aq
BgTeBZ0'0
,g0tge'grlu
t tuct
89TE8Z0'
v9aL8t'97lu
ttucE
t1lu3u r t': aurnTi6T'
0e 0 6260' 0
9.rE?'9
luZuIc
Z
0t0 6260'0grgv' g
luZuIc
Z
ZAJUCU T
Z': senr8?08'0
vd,z
tu
tuc
8?0e'0vs'z
tu
luc
1ft{cuT6iiETEA
- Tre6d ao rol{rJuPnlesiLe6uad ro?:tPJirehaes.. trleur
olsnJ so.,uBn+Es
ue6ueraleN sxn uPnlesIS uPn?PS
NVO.f,VS Xft.f,Nn ISUflANOX UO.IXVJr Nvurdrilv'r
-2-LATiPIRAN II
LUAS PENAI{PANG BAJA TULANGA POLOS& TULANGAN DEFOT{U( =UI,IR }
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Sifat-sifat mekanis baja tulangan deform:UJI LENGKUNGKelas ba
ja tul.deform
diameterpelengkung
sudutlengkun
Batas u-lyt .*i2
Bj rD30
No. 2No. 3
Bj rD40:No. 2No. 3
LUAS TULANGAI{ & JARAK TUI,AT{GAN
UNTUK PEI,AT LEBAR lOO CM
Ii1
Jarakske
as(cm
7 ,07 ,58ro8r59r09r5I0 ,010 r 511,011, 5
L2 ,0L2 ,513,01.3 ,514 r 0
14,5I5 r 015 r 516r016r5L7 ,0L7 ,5IB, O
19,519,019,52A ,0
L4 ,2913,3312,50LL,7611,I110,5310,009,539,108r708,348,007 ,707,4L7 ,L56 ,946 ,676,466 ,255 ,065,895,725,565 ,4L5 ,275 ,135,00
Untuk tulangan pelat lantai kita biasanya memPergunakan tulangandiameter 6mm; 8mm; 10mm; 12mm; 14mm dan kalauterpaksa dapat digunakan 16mm;19mmi22mmUntuk poer("pile cap")biasanya ditulangi diamter 15mm19mm ,22mm, 2 5mm.
jumlahbatangtiap m
Diameter tulanqan(rnm)
2L ,9920,52L9,2418,1r17,10L6,2415,3914,6613
' 9913,3912,83Lz ,3211' 8411,4011,00Lo ,6219,269,9J'9 ,629r339,058,79'8,35
8 ,328r1'07 ,897 ,69
.22 ,342L ,16
1g ,1518,28L7 ,49L6,7616 ,08L5,4714,8914,3613,8713 r 41L2 ,97L2 ,57L2,L9LL ,8211,49r1,1710 ,8710,5810,3110,05
28 ,36
24 ,6623 ,6322 ,692L ,8221,0120,26r9;5619,9118,30L7,7317 ,1916 ,58L6 ,2Lr5,7515,33L4 ,92L4,5414 ,18
16,1615,08L4,L413,31L2 ,5711,9011,31L4,7710i289,83
9 ,429,056,70I ,388,087 ,807 ,547 ,'307 ,076 ,856 ,656 ,466 ,28
5 ,80
7 ,Lg3,773,533r333 ,142 ,982 ,832 ,692 ,572.'r 45
2 ,362 ,262,L72 ,092 ,021r951,89L ,82L,77L,7LL ,661m5 2L ,571,53L ,49L ,451,41
7 ,186,706 ,285r9I5,595 ,295r034,794 ,574 ,374,L94 ,423 ,873,723,593,473 r 353 ,243,L43r052 ,962 ,872,792,72
2 ,582 ,5L
LL ,22L0 ,479 ,829 ,248 r 73
8 ,277 ,857,487 ,L46 r 83
6,545 ,286 r045 ,825,615,425 ,245 ,074 ,9L4,764 ,624,494 ,364 ,254 rL44r033,93
54,3050,8147 ,5L44,72'42 ,2340 r 0138,0136 ,2434,5533,053L ,6730 r 4129 ,2428,L627 ,L326 ,2L23 ,3424 ,5223,7623,0422 ,382L,722L,L220,5520,01L9 ,4919,01
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2 ko. ObkiA=obh.-t *O-au
I .t -rr.ro-r\/1 )25A = *t (0,36083)(3s-5)(60) #ffi(pakai 7fl22=266,cm2 )
=25, 3 0cm2
At=A=79226 . Cek luas tulangan .
Jumlah luas tul€rrrgdn= 7 +7+5+ 5=24g22=g 3 ,6"m2
t = T#tg .rooB:3,,98E--->oK M
7.Cek kembali apakai anqgapan tadi bahwa v(M\oy
Mx - ??2)oxo
[-3, e88
, q=0,0398 ry -0,4gL7514
J : -15 -^ ?q I6, x =utt) I,M lternyatx 18 ^ --/Ml = 3lft6rg -o ' s6s
ox Yang
Mdoy
aa
be
\/
sar
pan diatas
adalah :
SALAH
Sekarang kita tentukan kembali be M
r {rrr #,=r5 +r8( # )( L3-*; )=2o,G5rm
( hasilnya tak begitu berbeda denqan M yg diperoleh dariIangkah ke 5)
oY rJ ---E--
,M Ilto^=*',=0,726|uu''grafik(haIaman344)dipero1ehL"'f:0,6s ) M
-X =O ,57Mox
a
,,I
l(--r--:l
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f T seq ue6uaP Ppaq-raq n1T6aq
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8TT,0= (99) (St) (9ZZ)(Str6?'gt-98'0) =l*
.,
ooost .* tI
I Lt'6V'0=b I
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