Bài Giảng KC Nhà Dân Dụng

8/10/2019 Bi Ging KC Nh Dn Dng

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Bi ging: Kt cu btng cng trnh dn dng

Chng 1. Tnh ton sn btng ct thp ton khi Trang 1

Chng 1 TNH TON SN BTNG CT THP TON KHI

1.1. KHI NIM(Concept)

Sn l kt cu chu trc tip ti trng sdng, hsn c bi hdm, dmtruyn ti ln ct v ct truyn xung mng.

Sn BTCT (Reinforced concrete floor) c sdng rt phbin v nhng uim ca n nh: chu lc ln, chng chy tt, n nh ln, nhng sn BTCT vn

c nhng khuytim nh: cch m cha tht tt (cn phi hp vi cc vt liu cchm), thi cng phc tp, trng lng bn thn ln. Sn BTCT c phn thnh nhngloi sau:

1.1.1. Theo phng php thi cng:

Theo PP thi cng ta c thchia sn BTCT thnh cc loi sau:

Sn BTCT ton khi: sn, dm c lin khi cng lc, y l dng thngdng v n nh cao v tui thln, nhng thi cng phc tp v ko di.

Sn BTCT lp ghp (Precast concrete floor): hdm c BT trc, sau lp ghp cc panel sn (c ch to ti xng), sn lp ghp c thi gian thi cngnhanh, ph hp vi qui m xy dng ln, thi cng hng lot, nhng n nh khng

cao.

Phn tip sau ta chnghin cu dng sn BTCT ton khi.

1.1.2. Phn loi theo s kt cu:

Theo s kt cu ta phn thnh cc loi sn nhsau: Sn loi bn - dm: (sau ny ta gi l sn 1 phng) l dng sn chu un theo 1

phng hoc 2 phng nhng phng cn li chu un rt nh. Lin kt c thl k lntng hoc lin khi vi dm, nhng ch 2 cnh i din.

Sn loi bn k bn cnh(sau ny ta gi l sn 2 phng): l dng sn chu untheo 2 phng, lin kt c thl k ln tng (gi) hoc lin khi vi dm (ngm),cc lin kt vi dm c 2 cnh k.

Hay ta c bng so snh nhsau phn bit r hn vsn 1 phng v 2 phng:

Sn 1phng(ng mt trong 2 sau)

Sn 2 phng(ng c2 sau)

Tlcnh di trn cnh ngn > 2.

Lin kt c 2 cnh i din nhau

Tlcnh di trn cnh ngn 2.

Lin kt c 2 cnh knhau.

Tisao c yu cu thnht, ta stm hiusau y:


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Ta tin hnh tnh ton kho st i vi 1 bn k n 4 cnh, c kch thc cnhngn l L1, cnh di l L2, nhhnh 1.1.

Ti trng tc dng ln bn lq(kN/m2), gi s ct 1 dy bn rng 1m(hoc 1 n v chiu di) theo 2 phng

kho st, ta c:o Ti tc dng ln dy bn theo phng

ngn (L1) l q*1m=q (kN/m), theophng L2cng vy.

o Ta xem cc dy bn lm vic nhcc

dm n gi 2 u v c moment theotng phng l M1, M2; vng theo

tng phng lf1,f2.

o Theo SBVL ta c vng ca dm kn c tnh nhsau:

JE

LM

JE

Lqf

.

.

48

5

.

.

384

5 24

Vy ta c:JE

LMf

.

.

48

52

111

JE

LMf

.

.

48

52

222

o V thc cht 2 dy bn lm vic ng thi vi nhau, tc l ti gia bn ta c

f1=f2, hay:

JE

LM

JE

LM

.

.

48

5

.

.

48

52

222

11

M1L12 = M2L2

2M1= M22

1

2

L

L t =

1

2

L

L

M1 = 2 M2 ( 1. 1)

o Tcng thc (1.1) ta thy:

Nu L1=L2th =1, tc l M1=M2.

Nu =2, th M1= 4.M2.Nu =3, th M1= 9.M2.

Tc l nu cng ln th Moment theo phng ngn cng chnh lch ln so vi

moment theo phng di. Qui phm xy dng cho php ly 2 th xem nhbn chlm vic theo phng ngn, cn phng di moment l rt nhnn khng cn tnh ton.Trong vic btr thp cng c qui nh thp cu totheo phng di khng c nh

hn 1/4 lng thp theo phng ngn.

Hnh 1.1


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1.2. TNH TON SN DNG BN DM

Xem thm ti liu.1.3. TNH TON SN DNG K BNCNH

1.3.1. S h thng sn 2 phng:

Hthng sn 2 phng gm

cc sn c lin kt vi dm(ngm) hoc k ln tng (ta n)hoc tdo, nhng vn m bo cc

sn lm vic 2 phng, hnh 1.2.

H thng sn 2 phng rtthng dng, thng p dng cho

nhng cng trnh c ti trng va phi( 1000kG/m2) v nhp 6m.

Thng thng chu vi cacng trnh cng l hthng dm - ctchkhng phi tng nhhnh 1.2, Hnh 1.2.h thng sn 2 phng

hnh 1.2 cho v dthy c tnh tng qut ca cc sn.

1.3.2. Tnh ton sn:a). c im cu to:

Chiu dy sn chn trong khong 1501

40

1L

v ph thuc vo ti trng,

chiu dy sn nn chn chn n cm, chng hn nh6, 7, 8, 9, 10cm; thng thng sn2 phng nn chn chiu dy nhsau:

o hs= 1501

L cho sn cc tng c ti trng va,

o hs= 1401

L cho sn cc tng c ti trng ln,

o hs= 6 8cm cho sn mi.

Kch thc dm (cdm ngang v dm dc) chn trong khong

h= L

121

81 ; b = h

41

21 .

Thp sn b tr di dng li khong cch u nhau trong khong@=1020cm, dng thp CI hoc AI, ng knh thp t6 12m.m; lp bo v atrong khong 1,5 2cm.


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Hnh 1.3.Moment sn ngm 4 cnh

b). Tnh ton ni lc sn:

Tu theo lin kt 4 cnh bn m ta chia thnh 11 loi bn nhsau:

Tu theo loi bn m ta c cng thc tnhmoment khc nhau, di y ta xt bn ngm 4cnh, c cc moment nhhnh 1.3:

Cc gi trmoment c tnh bng cng thc sau

M1= m91.P

M2= m92.P Vi cc hsm91, m92, k91, k92

MI= k91.P tra bng phlc 1;

MII= k92.P P = (p+g)L1.L2= q.L1.L2.

p: l hot ti ca sn (daN/m2 hoc kG/m2), ly theo TCVN 2737-1995.g: l tnh ti sn, tnh t cc lp cu to sn (daN/m2hoc kG/m2), cng ly theoTCVN 2737-1995hoc trang 38 - quyn [4].

Tng qut ta c nhsau:

M1= mi1.P i: l loi ssn (1 11)

M2= mi2.P Cc hsmi1, mi2, ki1, ki2

MI= ki1.P tra bng1-19, trang 32 quyn [4]; (1.2)

MII= ki2.P P = (p+g)L1.L2= q.L1.L2(daN hoc kG)

Cng tc tnh ton ta c thlp thnh bng tnh nhsau:


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c). Tnh v btr thp:

Tnh ton trn 1m brng sn theo phng ngn v theo phng di, tnh nhcu kin chu un tit din chnht, vi b = 1m = 100cm, h = hs.

Cng tc tnh ton ta c thlp thnh bng tnh nhsau:

Vic btr thp cn ch vtr gi chung gia 2 sn, nu chnh lch t th sdng thp ln btr chung, thp c btr ra n 1/4 chiu di nhp.

Thp chu moment dng cng c thdng thp ln ko qua nu cc sn clng thp chnh lch t dthi cng, xem hnh 1.4.

Hnh 1.4Btr thp sn nhhnh (a) c

ththay thbng cch btrnhhnh (b) (a)

(b)


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1.3.3. Tnh ton dm:a). Skt cu:

H thng chu lc ca sn l dm ngang v dm dc, cc h thng dm ny c

tnh nh dm lin lc nhiu nhp hay tnh chung vi khung cn ph thuc vo kchthc cng trnh (phn ny sc ni r hn trong chng 2 - Khung BTCT). Thng

thng nu tnh khung phng th hthng dm ngang c tnh chung vi ct to thnhhthng khung, cn hthng dm dc c tnh nhdm lin tc nhiu nhp gi ln

ct, c nhim vlin kt cc khung ngang vi nhau v tm sn.

b). Ti tc dng:

Ti tc dng ln dm bao gm:

o Tnh ti: do bn thn dm, do

tm sn truyn vo v dotng xy trn dm

o Hot ti: do sn truyn vo.

Ti ca sn truyn vo c dnghnh thang, tam gic hay hnh ch nhttu thuc vo kch thc sn, nichung dng truyn ti da vo gc

truyn lc ca tm sn vo dm, gcny c xc nh tng phn gicca cc gc tm sn (hnh 1.5), c ththy nu gc tm sn vung th ng

phn ti l gc 45o so vi dm v tacng c nhn nh l :

L1L2

1 2

B

B

L2L1

43 5

A

B

B

C

B1

D

ES1

S2 S2S1

S3 S4 S5 S5 S3

S6 S6

Hnh 1.5. truyn ti sn vo dm

Hnh 1.4c.Btr thp sn thc t


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o Ti truyn theo phng cnh ngn l hnh tam gic,

o Theophng cnh di lhnh thang,

o Sn 1 phng ( >2) th ti truyn ch yu theo phng di v c dng hnhchnht (ng phn ti chia i tm sn) nh sn S6 trong hnh 1.5

Trong tnh ton c thginguyn ti tam gic v hnh thang gii ni lc chodm, nhng sgp rc ri nu trn cng on dm c nhiu dng ti tc dng - nhon dm 2-3, 3-4 ca dm trc D trong hnh 1.5. Ta c thqui cc ti tam gic v hnh

thang thnh hnh chnht tng ng theo cc cng thc chuyn i sau y, xemhnh 1.6:

o Ti hnh thang truyn t1 pha dm: qt= kqL1/2 (1.3)

o Ti tam gic truyn t1 pha dm: qt=8

5qL1/2 (1.4)

vi : q l ti tc dng ln sn (c thl hot ti hoc tnh ti) (kG/m2)

k l hsqui i, c thtra bng I.1 bn di hoc tnh theo cng

thc sau: k = (1- 22 +3), vi=2

1

*2 L

L

L1l kch thc cnh ngn ca sn.

Nu ti truyn t2 pha dm ging nhau (cng tam gic hoc hnh thang)th nhn 2.

* Ch : trnh nhm lnkch thc L1, L2l qui c cnh ngn v cnh di ca snvi cc kch thc L1, L2, L3,l cc kch thc khc nhau ca cc nhp. Chng hn

nhtrn hnh I.5: sn S1c kch thc cnh ngn l L1=L2; cnh di L2=B1.

Bng 1.1. Tra h s k trong cng thc 1.3

L2/L1 1,00 1,02 1,04 1,06 1,08 1,10 1,12 1,14 1,16 1,18

k 0,625 0,637 0,649 0,660 0,671 0,681 0,690 0,700 0,709 0,717

L2/L1 1,20 1,22 1,24 1,26 1,28 1,30 1,32 1,34 1,36 1,38

k 0,725 0,733 0,740 0,748 0,754 0,761 0,767 0,773 0,779 0,785

Hnh 1.6.Qui ti tam gic v hnh thang thnh ti tng ng


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Hnh 1.7.Cc trng hp hot ti nguy him

L2/L1 1,40 1,42 1,44 1,46 1,48 1,50 1,52 1,54 1,56 1,58

k 0,790 0,796 0,801 0,806 0,810 0,815 0,819 0,823 0,827 0,831

L2/L1 1,60 1,62 1,64 1,66 1,68 1,70 1,72 1,74 1,76 1,78

k 0,835 0,839 0,842 0,846 0,849 0,852 0,856 0,859 0,862 0,864

L2/L1 1,80 1,82 1,84 1,86 1,88 1,90 1,92 1,94 1,96 1,98 2,00

k 0,867 0,870 0,872 0,875 0,877 0,880 0,882 0,884 0,886 0,889 0,891

c). Tnh ton ni lc:

Gii ni lc cho dm ta c thgii bng phng php tnh tay (tra bng) hoctnh bng cc phn mm my tnh nhSAP2000, STAAD, ETAB,

lng trc nhng trng hp ti trng nguy him c thxy ra cho dm tacn phi thp ti trng (xem hnh 1.7), cc bc thc hin nhsau:

o Trc ht gii ring trng hp tnh ti (TT).

o Tch hot ti thnh cc trng hp nguy him, ta c cc trng hp sau:

Hot ti cht y (HT1): cho phn lc gi ln nht.

Hot ti cch nhp l (HT2): cho moment dng(moment nhp) lnnhtti nhp l.

Hot ti cch nhp chn (HT3): cho moment dng(moment nhp) lnnhtti nhp chn.

Hot ti 2 nhp lin tc 1 (HT4): cho moment m (moment gi) lnnhtti gi k2 nhp t ti.

Hot ti 2 nhp lin tc 2 (HT5): cho moment m (moment gi) lnnhtti gi k2 nhp t ti

o Thp ti trng: ly tnh ti cng ln lt cho cc hotti, ta c cc thp thnh phn, nhtrn ta c:

THp 1 = TT + HT1, . THn= TT + HTn.


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o Biu Baoni lc: chng tt c cc trng hp thp thnh phn tac biu bao ni lc : Bao = max/min {TH1THn}

o V dsau y vdm 3 nhp cho thy r hn vbn cht vn thp titrng:

d). Tnh ton v btr thp:

Tnh ct dc: tnh theo bi ton cu kin chu un, trn mi ondm ta lygi trmoment max (nhp) v moment min (gi) tnh thp cho nhp v gi. Nh

hnh trn th t biu BAO moment ta c moment max nhp 1 l 25513 kG.m vmoment min gi B l -28941 kG.m.

Nu l dm T, I hoc dm lm vic chung vi sn th ta c th tnh theo titdin chT vi nhng vtr c cnh nm trong min nn, gisnhdm hnh trn msn nm trn dm th ta tnh vi tit din chT cho moment dng (nhp), tnh vitit din chnht cho moment m (gi).

Tnh ct ai: ly lc ct max trn mi on dm tnh ct ai cho tng ondm hoc c thly lc ct max trn ton dm tnh v btr ct ai cho ton dm.

Vn ct ct dc theo tnh tonsgp nhiu kh khn do kh xc nh chnhxc vtr ct l thuyt, nn thng ta c thct thp theo cu to:

o Thp nhp ct cch gi 1 on = L/5.o Thp gi ct cch gi 1 on = L/4.o Khi ct thp cn ch dng biu .

TNH TI

HOT TI 1

HOT TI 2

HOT TI 3

TH1 = TT + HT1

TH2 = TT + HT2

TH3 = TT + HT3

BAO = CHNG 3 BIU TH1TH3

A B C D

(1) (2) (3)


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Chng 2. Tnh ton khung btng ct thp ton khi Trang 10

Chng 2 TNH TON KHUNG BTCT TON KHI

2.1. KHI NIM:

Khung l hthng gm ct v x (dm), c thlp ghp hoc ton khi.

Trn thc tt gp khung lm vic ring lm thng kt hp vi hthng dm

dc to thnh hton khi chu lc chnh cho cng trnh.

Cng trnh dng khung chu lc (tng xy chen) thng p dng cho cc cng

trnh nhv va, tng ti chn ct khong 500T, i vi cng trnh ln ngi

ta thng p dng dng chu lc vch cng (concrete diaphragm), li cng

(rigidity core) hoc kt hp.

Tu theo dng mt bng cng trnh m ta c th tch khung ring l (khung

phng chc ct v dm ngang) cho dtnh vi chnh xc c thchp nhn

c hoc tnh khung khng gian (c ct, dm ngang v dm dc)

L

BDMN

GANG

DM DOC

Vi nhng cng chtrgii kt cu nhhin nay ta nn gii khung khng gian

s cho chnh xc cao hn (mc d kt qu ni lc c hi nhhn). Ch nn giikhung phng trong nhng trng hp sau:

o Khi chiu di cng trnh L 2,5 ln chiu rng B, lc ny ct chyu

chu lc theo phng ngn.

o Khi khu 1 phng ln hn phng kia gp 2,5 ln.

Khung cc cng trnh c khu ln nh: nh ht, hi trng ngi ta c th

lm x ngang gy khc hoc cong.

nhip khung 15 - 18m nhip khung 18 - 25m


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2.2. CHN SB KCH THC TIT DIN

2.2.1. i vi dm:

C thchn sbchiu cao dm h=m

1L, vimcho trong bng 2.1:

Bng 2.1:Hsm chn kch thc dm

Hnh dng dm Mt nhp nhiu nhp

1. Thng

2. Gy khc

Khng thanh cng.

C thanh cng.

3. Cong

Khng thanh cng

C thanh cng

10 12

12 16

16 20

18 24

30 35

12 16

12 18

16 24

18 30

30 40 Nu bit trc ni lc (thng cha bit do cha gii kt cu) ta c thchn nh

sau: ho= 2bR

M

b

(ly trn s); h = ho+ a.

2.2.2. i vi ct:

Chn sbtit din ct theo cng thc sau: F = k*b

R

N

o K: l h siu chnh = 1,2 1,5 (cho ct chunn lch tm).

o N: tng lc dc tc dng ln ct, do cha gii kt

cu nn ta cha bit chnh xc lc dc ny m chc thc lng bng

cch tnh sbti tc dng ln sn, dm ri truyn vo ct theo nguyn

tc chia i.

Ch :

o Ta tnh dn ti n chn ct tng trt ca cc ct in h nh (ct bin, ct

gc, ct giacng trnh) ri tnh tit din theo cng thc trn,

o Sau ta s gim tit din ct theo tng tng (hoc 2 tng), mi ln gim

5cm hoc 10cm tuc im cng trnh,

o Bi v y ch l bc tnh sb c tit din nhp vo chng trnh

gii kt cu nn tnh chnh xc cha cao, sau khi gii kt cu xong ta c

c lc dc chnh xc cc ct, tin hnh tnh thp iu chnh tit

S


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din cho hp l (v mt hm lng nh hc trong phn btng c

s), sau c thiu chnh li tit din nhp li chng trnh gii kt

cu tnh li thp n khi tho.Thng thng vi mc chnh lch

tit din t, ngi ta khng cn gii li kt cu, bi v sai skhng ln.

o Tit din ct b,h nn chn theo tlh=(1- 2)b. Cng trnh dng mt bng

vung, lch tm t nn chn tit din vung h~b, nu dng mt bng

chnht chnh lch nhp theo 2 phng nhiu (lch tm nhiu) nn chn

tit din chnht nhng h/b khng nn vt qu 3 ln slm cho ct c

mnh ngang ln.

V d: ta c mt cng trnh cao 6 tng, ti trng

chn ct bin sb tnh c l 100tn, ct gia l 150tn,

ct chn btng B20 Rb = 110kG/cm2. Ta tnh c tit

din ct nhsau:

Ct bin F= k*b

R

N( chn k=1,4 - lch tm nhiu)

F = 1,4*110

000.100= 1272 cm2.

o Ta chn tit din l 30*40(1200cm2) cho 2 on ct tng 1, 2.

o Tng 3,4 sl 25*35cm.

o Tng 5,6 sl 20*30cm (c thct tng 6 gim cn 20*20cm).

Ct gia : F = 1,4*110

000.150= 1909 cm2.

o Ta chn tit din l 30*60(1800cm2) cho 2 on ct tng 1, 2.

o Tng 3,4 sl 25*50cm.

o Tng 5,6 sl 20*40cm (c thct tng 6 gim cn 20*35cm).

2.3. XC NH STNH

im quan trng u tin l phi xc nh vtr lin kt cng chn ct, vtr

ny c th l nh mng hoc mt trn ging mng (ground sill), c mt s

quan im nhsau:

o Nu mng t khng su ( 1,5m) thng chc 1 h king (kt hp

ging mng, c thnm hi cao hn nh mng) th ta ly lin kt cng l

nh mng.

30*40

30*40

25*35

25*35

20*30

20*30

20*30

20*30

25*35

25*35

30*40

30*40

30*60

30*60

25*50

25*50

20*40

20*35

20*35

20*40

25*50

25*50

30*60

30*60


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o Nu mng t kh su (> 1,5m)

cn c 2 h king v ging

mng ring bit, ta c th ly lin

kt cng l mt trn ging mng,

bi v v tr ging mng lc ny

nm ngay gn trn nh mng.o Vic xc nh su chn mng

phthuc vo a cht thu vn,

tng ti trng cng trnh gii

php mng, chng hn: mng c

trm cn t su hn mc nc

ngm, cn mng cc btng c th

t cn hn.

o Trong chai trng hp trn, ging mng khng nn tnh vo khung,cn king th c thtnh hoc khng tnh vo hkhung.

Chuyn vca mng xem nhkhng c, biv trong tnh ton mng ta khng

ch ln lch ca cc mng trong khong cho php (t gy ph hoi kt cu bn

trn).

Gc xoay (do mng ln nghing) cng xem nhkhng c, v y gc xoay kh

nhv c hging mng khng ch.

Lin kt ca dm ct c xem l lin kt cng (ngm) khi cng ca ct(Ejct)> 6lnEjdmv ngc li nuEjdm> 4lnEjctth c xem l lin

kt khp, nhng trn thc tt c tiu ch ny rt kh, thng thng ta

quan nin nhsau:

o Nu tnh khung phng: lin kt ct v dm ngang (chu lc chnh) l

ngm (to thnh khung cng), cn ct v dm dc lin kt khp vi nhau

(xem nhdm dc lin kt cc khung vi nhau).

o Nu tnh khung khng gian th lin kt ca cdm ngang, dc v ct l

lin kt nt cng.

o Vi nhng qui c trn ta thy: vi khung phng scho ta moment trn

ct ln hn vi khung khng gian (do schng ko ca cc thanh dc),

nhng ta khng bit c moment theo phng vung gc vi phng ta

tnh v khng thlng c slm vic theo phng ny.

TANG TRET

MONG

A GIA NG MONG

AKI ENG

MAT NEN

VTR

LI EN KET CNG

MONG

VTR

LIEN KET CNG

A GIA NG MONG

TANG TRET

MAT NEN


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Vic lp stnh cn phi n gin ho n mc t nh hng n kt qugii

ni lc, vic n gin ho nhm mc ch to ra skhung i xng, dnhp

sliu vo my tnh,.c thlm nhng php n gin ho sau y:

o Nu chiu di nhp sai khc khng qu 10% th xem nh c nhp bng

nhau v ly kch thc trung bnh.

o Nu dc ca dm


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Hot ti ngang gm c: p lc t vo tng tng hm, p lc nc v gi. p

lc gi c tnh nhsau:

o p lc gi c tnh tc dng vo ct, dng lc phn b(hoc tp trung

ti nt khung), thay i theo chiu cao cng trnh, cng thc tng qut:

q=Wo*k*C*n*B (daN/m).

- Wo: p lc gi tiu chun, theo phn vng p lc gi,phlc 6.

- K: l hsthay ip lc gi theo chiu cao, phlc 7.

- C : l h s kh ng, ph thuc vo mt hng gi, bng2-8,trang 49 - quyn [4].

- n: l hsvt ti = 1,2.

- B: din hng gi, bng khong cch bc khung mi bn.

o p lc gi c tnh trn cao mi tng hoc mi 2 (3) tng, ta c th

lp bng tnh nhsau:

Q= Wo*K*C*n*BTng Cao

H s

K C=0,8 (gi y) C=-0,6 (gi ht)

Tng 1 -2 - - q1 q4

Tng 3 4 - - q2 q5

Tng 5 6

.

- - q3 q6

Ghi ch: HsC bng trn ng vi mt hng gi thng ng.

o p lc gi ta c thnhp vo khung di 2 dng : dng ti phn blnct hoc dng lc tp trung ti nt.

q1

q2

q3 q6

q5

q4


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2.5. XC NH NI LC (Forces)

C thtnh khung bng tay, bng nhiu phng php (nu l khung n gin).

Hoc tnh bng phn mm (SAP, STAAD, ETAB) vi chnh xc cao v

nhanh chng.

Ta cn thpti trng (combination load) nhm tm ra c gi trni lc thcsnguy him cho kt cu trong cc tnh hung ti c thxy ra. Nguyn tc t

hp nhsau:

o Trc ht tch ring trng hp tnhti, gii ni lc ring,

o Chia hot ti thnh nhiu trng hp c thxy ra trn thc t, chnn

cht ti nhng trng hp thc snguy him, bqua nhng trng hp

khng nguy him gim strng hp ti phi gii, theo phn tch (xem

cc cu kin l n hi) ta c cc trng hp sau l nguy him:

(HT1) Hot ti cht y: cho ta gi trlc dc ct ln nht.

(HT2) Hot ti t cch nhp (cch tng): cho ta gi tr momen

nhp (M+max) ln nht trn nhp t ti.

(HT3) tso le vi trng hp HT2.

(HT4)t2 nhp lin tc, sole tng: cho ta gi trmomen gi ln

nht (M- min) k2 nhp t ti.

(HT5)tsole vi trng hp 4.

(HT6)tsole vi trng hp 4, 5.

(GT) Go tc dng tpha tri cng trnh.

(GP) Go tc dng tpha phi cng trnh.

HT1 HT2 HT3

HT5

HT4

HT6

q4q2

q1q3

GT GP

q2

q1


17/40



o Trn thc t vn cht ti rt phc tp, ta khng th lng ht cc

trng hp thc snguy him, nhng nu cht qu nhiu trng hp s

lm phc tp bi ton c thdn n sai st. Theo cc tiu chun nc

ngoi (Chu u v M) vic cht ti rt n gin, chcht 2 trng hp

cch nhp, cch tngsau ngi ta nhn kt quni lc cho hs1,2

1,4 (cho cdm v ct).

o Bc tip theo l ta cng ln lt cc trng hp hot ti cho tnh ti

theo nguyn tc sau (theo tiu chun VIT NAM TCVN 2737 : 1995):

T hp chnh: gm tnh ti v mt hot ti bt k

TH1 = TT + GT

TH2 = TT + GP

TH3 = TT + HT1

..

TH8 = TT + HT6

Thp ph: gm tnh ti v 2 hot ti

TH9 = TT+(HT1+GT)*0,9.

TH10= TT+(HT2+GT)*0,9.

TH14 = TT+(HT6+GT)*0,9.

TH15 = TT+(HT1+GP)*0,9.

TH16 = TT+(HT2+GP)*0,9.

.

TH20 = TT+(HT6+GP)*0,9.

Hs0,9 khi trong thp c 2 hot ti, v him khi 2 hot ti ny

xy ra cng lc.

Trng hpBAO ni lcthnh lp bng cch bng cch vchng

tt ccc trng hp thp trn vo cng 1 biu, ng vin

bn ngoi l biuBAO ni lc. Vmt tnh ton, ta tnh nhsau: Trng hp BAO=Max/Min (TH1, TH2, , THn).

2.6. TNH V BTR THP2.6.1. Tnh thpa). Tnh thp dm:

Dm c tnh nhcu kin chu un, ni lc dng tnh ton dm gm:


18/40



Momen max, min (M+/- max/min) nhp v gi tnh ct dc dm, ch

momen m ti gi thng c 2 gi trbn tri v bn phikhc nhau, cn tnh

cho c2 nu tit din dm bn tri v phikhc nhau.

Lc ct max tnh ct ai, nu lc ct trn ton bdm t thay i th cho php

ly lc ct ln nht trong tt ccc nhp tnh ct ai v btr chung cho ton b

dm, nu khc nhau nhiu th nn tnh v btr ct ai khc nhau (tit kim).

b). Tnh thp ct:

Ta tin hnh tnh thp nhsau:

Thp ct c tnh nh cu kin

chu nn lch tm, tnh ct dc ta

cn tnh vi 2 cp lc l Nmax Mtv Mmax Nt , trng hp Nt v

Mt cngha l lc dc v momen

tng ng vi trng hp ti gy ra

Mmax v Nmax. Vi 2 cp ni lc ny

ta tnh c 2 gi trAsv ta sly

gi tr ln hn b tr thp cho

ct.

Mi phn tct ta chcn tnh vi1 mt ct ti chn ct hoc u ct,

bi v momen u ct hoc chn

ct l ln nht, cn lc dc th ln

nht chn mi on ct.

Trong trng hp ny lc ct nhp gia tng i nh, nn tnh ring.

< M >

< Q >

< N >


19/40



I

II

III

IV

Ctai ct thng thng khng c tnh ton m chb tr theo cu to (

hc trong mn Btng cs) bi v lc ct trn ct thng rt nh so vi tit

din ct, nu ta c tnh ton cng chcho kt qul gi trcu to.

2.6.2. Btr thp

i vi dm vic ct thpkhng cn tnh ton v rt

kh xc nh c vtr ct

l thuyt, y ta nn ct

theo cu to (xem hnh) v

cng cn ch n dng

biu khi ta ct thp v c

nhng trng hp ta khng

thct thp. Btr thp ai cng cn ch n dng biu .

i vithp ct th trn mi on ta chtnh trn 1 mt ct (chn hoc u ct)

v thp c btr u cho cct tchn n u.

Trong vic btr thp khung ta cn ch nhiu nht nhng vtr I, II, III, IV nh

hnh di.

V TR II

L/4

L/3

L/2

L/5


20/40



VTR IV

VTR III

Btr thp ti nt ct vi dm ngang v dm dc


21/40


Chng 4. Tnh ton cu thang btng ct thp ton khi Trang 21

Chng 3 TNH TON CU THANG BTCT TON KHI3.1. Cc dng thang c bn chu lc:

3.1.1. Thang 2 v gp khc song song:

a). Phn tch kt cu:(nhtrong hnh)

Dng thang ny khng c

dm limon hai bn vthang, bn

chu lc theo phng di, bn

thang k ln 1 u l dm chiu

i (chiu n), 1 u l dm

chiu ngh; lin kt ny c

xem l lin kt ngm khi

hd/hs>3, xem l khp khi

hd/hs3. S tnh ca bn

thang (kcchiu ngh) ta phn tch trn 1 dy brng 1m (nhtrong hnh).

Dm chiu nghta xem nhdm n gin2 u ngm (lin kt vi ct)

Dm chiu n cng l dm n gin nhng lin kt c thxem l khp v lin

kt vi dm sn (cng nh).

b). Xc nh cc ti trng:

- Ti trng tc dng ln v thang l q1: gm c tnh ti v hot ti; tnh tigm c cc lp cu to ca v thang (bc thang, sn btng ct thp, va

trt), bc thang c thtnh trng lng tng bc (kg), nhn sbc, chia u

cho cv thang (m2) ta c ti trng tc dng trn 1m2v thang (kg/m2),


22/40



hoc xem bc c bdy trung bnh l 10cm, nhn cho trng lng ring ca

vt liu xy bc.

- Ti trng tc dng ln chiu nghl q2: cng gmc tnh ti v hot ti, hot

ti ging nh tc dng trn v thang, tnh ti gm c cu to cc lp sn

chiu ngh(khng c bc).

- Ti trng tc dng ln dm chiu nghl q3: gm c ti trng ca phn thnthang (bao gm cchiu ngh) truyn vo, tng xy trn dm v trng lng

bn thn dm, phn ti trng ca thn thang truyn vo ta c thly phn lc

gi ca kt cu thn thang gii trn (n v lc ny l kg/1m b rng

thang).

- Ti trng tc dng ln dm chiu n l q4: cng gm c ti trng ca phnthn thang truyn vo (ly nh trn), ca sn chiu n v trng lng bn

thn dm.- Ta c thxc nh ni lc bng cch tnh tay (tra bng) i vi nhng dm

n gin hoc gii bng SAP2000.

c). Tnh v b tr thp:

- Biu momen nhhnh v, ta tnh thp vi

cc momen Mmaxv Mmintng ng cho nhp

v gi.

- Btr thp nhhnh vbn.

Cc thanh s 1, 2, 3 l thp chu lc; cc thanh

s 4, 5 l thp cu to


23/40



3.1.2. Thang 3 v (three-flight staircase):

a). Phn tch kt cu:(nhtrong hnh)

Thang 3 vdng bn chu lc lm vic cng ging nhthang 2 vtrn; v1 v v

3 l 2 vchu lc chyu, v2 xem nh ta ln 2 v1 v 3. Ta cng tnh ton theo

cch ct 1 dy bn rng 1m xem nhdm, stnh ca 2 dy bn ny cng ging nh

trn, nhng ch thm phn chiu nghngoi nhng ti trng nhtrn cn c ti trng

ca v2 truyn vo, tnh bng cch ly ti tc dng ln v2 nhn vi 1/2 chiu di v

ny. Cn dm chiu ngh c hnh dng nh

hnh bn, ti trng q3 xc nh nh trn, cn

ti trng q5l do sn v2 truyn vo (ton b

v thang), ti trng tng v trng lng bn

thn dm.


24/40



b). Tnh v b tr thp :(nhtrong hnh)

V2 khng cn tnh, ta chbtr theo cu to tnhng thanh thp ca 2 vkia

(bi v vny nh). Trong vic btr thp cn ch n dng biu , ta xem cc biu

momen sau:

3.1.3. Thang xon (helical stair, spiral stair):

Thang xon c th dng bn cng c th dng dm, cch tnh cng gn ging

nhau, c tnh thm mcao thng ta thit kdng bn chu lc nu tng chiu di

thang khng qu 4,5m.

Vmt kt cu ta phi tnh di dng khng gian (kt cu trxon trong khnggian tr), cch tnh ti trng tng tnhtrn trnh by, nhng thng ni lc ca

kt cu xon c tnh ra tng i nhso vi tnh phng, nn vic tnh v btr thp

c tng ln khong 50%.

Tham kho cch tnh 1 thang xon dng bn chu lc di y:

2

1

4

3

1

4

5

3

4

1

7

1

2 2

6COT

M A T CA T 1 - 1

5

DCN

DAM SAN

4

2

4

1

3

M AT C A T 2 - 2

1

4

1

5

2

7


25/40



Ta tnh ra ni lc ca

bn thang xon (xem nh

dm c tit din b rng

bng 100cm v cao bng b

dy bn - hb), ly gi tr

momen un ln nht

(momen 3-3 - thng 2

u cu kin) tnh thp chu

un, tng thm 50% v btr

cho bn (xem bn v). Thp

Momen 3-3 Momen 2-2 xon


26/40



theo phng ngang chu tip lc xon (momen 2-2) nn c hn cht vo ct thp dc

(hoc buc ktng thanh). Ch vtr chu lc nguy him nht l chlin kt vi dm

trn v di (chiu i v chiu n), cn c neo khoc hn vo ct thp dm.

Trong hnh bn: thp s1,2 l thp chu lc, thp s3 l thp cu to nhng b

tr khng t hn 6a150.

3.2. Cc dng thang c dm limon chu lc:3.2.1. Thang 2 v gp khc song song:

a). Phn tch kt cu:

Vmt chu lc thang ny c dm limon 2 bn

bn thang, p dng cho trng hp thang c kch thc

ln. Ta phn tch tnh ton cc cu kin nhsau:

Bn thang:tnh ton nhbn k 4 cnh, ty theo

t s 2 cnh m bn lm vic 1 phng hoc 2

phng.

Bn chiu ngh: cng tnh tng tbn thang.

Dm LM3: tnh nh dm n, 2 u gi ln DC v DCN1. Chu ti trng

gm: trng lng bn thn dm, lan can, bn thang truyn vo.

Dm DCN1: cng tnh nhdm n, 2 u gi

ln dm LM1 v LM2. Chu ti trng gm:

trng lng bn thn dm, bn chiu nghtruyn vo, bn thang(c thkhng c, nu bn

thang l bn 1 phng) tng lc l q1 ; lc

tp trung P ca 2 dm LM3 (ly phn lc gi

hoc lc ct 2 u dm LM3).

q3

DAM KHUNG

COT

P1

DAM LM 1

q2


27/40



Dm DCN2: Cng l dm n gin, nhng 2 u c thxem l ngm v lin kt

vi ct c cng ln. Chu ti trng gm: trng lng bn thn dm, bn

chiu nghtruyn vo v trng lng tng.

Dm LM1 v LM2: l dng dm gy khc, u lin kt vi dm c thxem l

khp, u kia c thxem l ngm (lin kt vi ct), nhhnh v. Chu ti trng

gm :

o Trng lng bn thn dm(go) + bn thang

truyn vo (tc dng vo on thn thang) +

trng lng tng = q2;

o (go) + bn chiu ngh truyn vo (tc dng

vo on chiu ngh - c th khng c nu bn

chiu nghl bn 1 phng) + tng = q3;

o Lc tp trung do dm DCN1 truyn vo (ly phn lc gi nhtrn)=P1.b). Tnh v b tr thp:

i vi sn bn thang v sn chiu ngh:tnh v btr thp nhsn 2 phng,

ch thp theo phng ngn lun nm di.

i vi dm chiu ngh:cn ch n on neo v gia cng ct ai ti vtr

c dm limon LM3 k ln.

i vi dm limon:LM1 v LM2 cn ch on gy khc, c thb tr ct

thp lin tc nu cc on un cong khng qu phc tp, hoc c thct ri ti

cc vtr un cong, nhng phi ch on neo thp phi 30

q3

COT

DA M KHUNGP1

DAM LM 2

q2


28/40



3.2.2. Thang 3 v :

phn tch kt cu gn

ging nhthang 2 v:

Cc v thang 1 v 3

tnh nh bn 1 phng

(hoc 2 phng).

V thang 2 cng vi

chiu ngh lm vic 1

phng (theo phng

ngang).

Cc dm limon LM1,

LM1, LM3 tnh nh

i vi thang 2 v.

Dm limon LM2 v LM2 c snhhnh di.

Btr thp tham kho dng thang 2 v.

3.3. Thang xng c:

L dng kt cu dm limon chu lc, nhng bc thang l dng bn consol

(cantilevel) chu lc; kt cu c phn tch nhsau:

Bc thang dng bn consol chu ti trng l trng lng bn thn bc v ngi

i, xem nhmi bc c 1 hoc 2 ngi (ty theo bc rng hay hp), trng lngtrung bnh 1 ngi ly l 75kg - an tan c thly trung bnh 1 bc chu ti

trng trung bnh l 200kg (v1 bn consol).

Dm limon chu ti trng ca tan b cc bc thang, sn chiu ngh v trng

lng bn thn n.

Dm chiu nghtrong tng phn tch nhtrn, chu ti trng ca 2 dm limon

gi vo, trng lng tng v bn thn dm.

B tr thp ch bc thang l bc consol nn btr thp lp trn, sn chiunghcng vy, nhng an ton btr thp 2 lp cho sn chiu ngh.

Bc thang c ththi cng ton khi hoc lp ghp, ch n thp neo bc thang

vo dm phi m bo khong neo l 30.


29/40




30/40


Chng 4. Tnh ton Mng btng ct thp ton khi Trang 30

Chng 4 TNH TON MNG BTCT TON KHI

4.1. KHI NIM:

Mng (foundation base) l cu kin tip nhn tan b ti trng cng trnh, truyn

ti ny xung nn.

Tnh tan mng v nnphi da trn trng thi gii hn 1 v 2 (TTGH1, 2)

m bo chu lcv bin dng.

Ti trng dng tnh ton cho mng thng l cp ni lc Nmax Mtv Qt

cn cp lc Mmax Ntv Qt kim tra; Qmax kim tra su chn mng.

Nmax+ N king= Ntt

Gi tr tiu chun Ntc= Ntt/1,15

H thng king l khng th thiu khi tnh ton mng, c th c 1 lp

king v 1 lp ging nu mng chn su (xem thm chng Khung).

Phn loi mng: Mng n, mng bng, mng b, mng cc.

4.2. MNG N (single footing, pad foundation):

4.2.1. Cu to:

Mng c th c hnh git bc (mng ln) hoc hnh thp (mng nh), nhtrong

hnh.

Mng git bc: h =200

BTA4x6 M AC >=100

CAT VANG AM CHAT

1

0

0

100

12

h

o1

C

HNH 1


31/40



h=1050 chia h1=300; h2=300; h3=450.

h=1200 chia h1=300; h2=450; h3=450.

h=1500 chia h1=450; h2=450; h3=600.

Thp theo phng di, thp theo phng ngn.

thp 10; khong cch a 200.

Lp bo v a 3,5cm nu c lp lt mng; 7cm nu khng c lp lt.

mng phi t trong lp t chu lc 100.

Mng c th c hnh vung (ng tm) hoc hnh ch nht (lch tm),

c a/b= 1,2-2.

Thp ct neo vo mng lneo(hoc 35), phi c t nht 2 ct ai nm trong

mng.

Thp trong mng nn thp trong ct, ni khng qu 50% lng thp cng v

tr.

4.2.2. Mng n chu lc ng tm:

a).Tnh din tch mng:

Din tch mng c tnh theo cng thc: Fm=Ntc

Rtc-tb*H (4.1)

Trong Rcl cng ca t nn t nhin c tnh bng cng thc :

Rtc= m[(Ab + Bh)+ Dc] (4.2)

Vi:

- A, B, D: tra bng theo ( hc trong

mn nn mng)

- l trng lng ring ca t di y

mng.

- b, h: l b rng v chiu su chn mng .Ban u ta cha bit c b rng b ca

mng, ta c th tnh theo 2 cch sau:

o Cch 1: Gi thit b=1m tnh Rtc

ri tnh Fmsau xc nh c b

hh

o

45o

a

a

h c

h

b

c

c

ho

b

o

o

c

h h hc oo

N

h

h

45

o

HNH 2


32/40



thay vo (4.2) tnh li Rtcn khi tho.

o Cch 2:kim tra iu kin

Ptc=Ntc

Fm+tb*H R

tc. (4.3)

Kch thc mng:

o Mng vung a = Fm;

o Mng chnht th b =Fm , vi = a/b = 1,2 2.

o Mng m rng u theo ct:

b =Fm*hcbc

(4.4)

b). Xc nh chiu cao mng (h)

Mng vung:

Chiu cao mng c xc nh theo iu kin chng m thng (ct

m thng mng):

P 0,75Rbtbtbho. (4.5)

Trong :

- btb: l chu vi trung bnh ca thp m thng.

btb= (bt+ bd)/2

btl chu vi nh thp m thng = 2(bc+hc)bdl chu vi y thp m thng = 2(bc+hc+4ho)

btb= 2(bc+hc+2ho)

- P: lc m thng

P = N Ft*P. (4.6)

N: l gi tr lc dc tnh ton chn ct.

P: l p lc di mng = N/Fm.

Ft: l din tch y thp m thng = (hc+2ho)*(bc+2ho)

C th chn trc hosau kim tra li theo cng thc (4.5) hoc thay P

bng cng thc (4.6) ri tm ho, ta c :

ho

)(5,0

75,02

1cc

bt

bhPR

N (4.7)


33/40



i vi mng bc th chiu cao bc di cng ca mng c xc nh

theo iu kin btng chu ct: Q 0,8Rbtbho1. (4.8)

Vi b=1m b rng mng.

Ho1l chiu cao tnh ton ca bc di cng (xem hnh 1).

Q= P.L1.b vi b=1m Q=P.L1.L1= 0,5(a hc) ho. (xem hnh 1 vi a l b rng mng)

Cng cn tnh ton hotheo iu kin chu un nhsau:

ho= L1btr

d

Rb

bP

4,0

. (4.9)

vi b l b rng mng (nu mng vung th l a), btr= bc.

Mng ch nht:

Cng kim tra iu kin (4.5) nhng ch tnh 1 bn mng (xem hnh 3),

tc l P=P.F1

Vi: - F1c th tnh gn ng nhsau: F1= a1.b; a1= 0,5(a hc) ho

- btb l gi tr trung bnh ca cnh trn (bc)

v cnh di thp m thng bd= (bc+2ho);

nn btb= bc+ ho.

c).Tnh thp:

Xem nhcnh mng ngm ti chn ct,

tnh trn 1m b rng mng, lc tc dng l

p lc t P = N/Fm. Theo 2 phng ta s

tnh nhsau:

Phng ngn: chiu di on consol l Ln= 0,5(b bc); M = P.2nL /2.

Phng di: chiu di

on consol l Ld= 0,5(a

hc); M = P. 2dL /2.

Tnh thp theo cng thc gn

ng sau: As=os

hR

M

..9,0 (4.10)

B tr thp: thp phng di

nm di, phng ngn nm trn

h

h

a

h

h

b

h

c

o

o

45

o

o c

c

bb

h o

c F1

a1

HNH 3

L

a0,1L0,1L


34/40



(ch lp bo v); nu mng c cnh di 3m c th ct bt mi u 0,1L

(xem hnh 4)

4.2.3. Mng n chu lc lch tm:

Mng chu lc lch tm l nhng mng chu cc ni lc M, N, Q hoc nhng

mng ch chu lc dc N nhng t lch tm

so vi tm mng (mng chn vt); cc ni

lc dng tnh ton l Nmax, Mt, Qt; cn

cc lc Mmax, Nt, Qt kim tra.

Cn phn bit ti trng tiu chun

nh mngv y mng(xem hnh 5).

tcdmN = Ntc+ tb.H.Fm

tcdmM = M

tc Qtc.h Ntc.d

(d: l lch ca lc dc Ntcso

vi tm mng)

tcdmQ = Q

tc.

a). Xc nh din tch mng:

Tnh din tch mng theo cng thc (4.1).

Kim tra din tch y mng theo cc iu kin sau:

tcPmax

1,2 Rtc.

tctbP R

tc vi tctbP = 2

minmaxtctc PP

Trong : tcP minmax/ = )6

1(L

e

F

No

m

tc

dm (4.11)

eo= tcdm

tc

dm

N

M

Tu thuc vo ln ca eom Pminc th m, dng hoc bng 0:

Nu eo L/6 th Pminm, p lc y mng c min ko (hnh bn)

ptc

M T N

M

min ptcmax

N

Q

tctc

tc

Mtc

Q

Ntc tc

dmdm

dm

HNH 5

h

H

ptcmax

ptcmin


35/40



Thng thng ta nn khng ch Pmin0, tc l eo L/6.

b). Xc nh chiu cao mng (h):

Tnh nhmng ch nht chu lc ng tm, nhng thay Pbng Pmax. Pmax

tnh tng t nhcng thc (4.11) nhng gi tr N v M l gi tr tnh ton ti

nh mng. Chn h theo iu kin cu to sau kim tra bng iu kin m thng

c). Tnh v b tr thp:

n gin v an ton ta ly p lc di mng l Pmax tnh momen

cnh mng.

Tnh ton tng t nhtrn theo c 2 phng b v L.

4.3. MNG BNG(strip foundation):

4.3.1. Khi nim Mng bng c th di tng (nh c tng chu lc hoc mng tng k)

hoc di ct (nh dng khung chu lc).

C th thit k mng bng 1 phng (ngang hoc dc cng trnh) hoc hai

phng (c ngang v dc, cn gi

l mng bng giao nhau).

Mng bng thng c

hai b phn: sn mng (dm

mng) v cnh mng; i vi

mng di tng c th khng

cn sn mng.(xem hnh 6)

4.3.2. Tnh ton mng bng 1 phng di ct:

a). Xc nh kch thc mng:

C th xem nhmng chu lc ng tm (nu momen nh), lc ny kch

thc mng xc nh nhsau: b =Ntc

L(Rtc-tb*H) (4.12)

Trong :

Ntctng lc dc tiu chun tc dng trn nh mng

b l b rng cnh mng


36/40



L l chiu di mng bng c

xc nh theo bc ct (xem hnh); vi Lo

xc nh sao cho Mo~ M1v 1,5m.

Cc bc tnh tng t nh tnh

mng n chu lc ng tm.

Tnh nh mng chu lc lch

tm:

Tnh din tch mng theo

cng thc (4.12).

Kim tra din tch y mng theo cc iu kin sau:

- tcPmax 1,2 Rtc.

- tc

tbP Rtc

vitc

tbP = 2 minmaxtctc PP

- tcPmin 0

Trong : tcP minmax/ = )6

1(L

e

F

No

m

tc

dm ; eo= tcdm

tc

dm

N

M

T hnh 7 ta c: tcdmN = tc

iN + tb.H.Fm

tc

dmM = tc

iM tc

iN *Yitc

iQ *h

(vi Yil khong cch t im t lc dc th i n trng tm mng O)

tc

dmQ = tc

iQ

*Ch : cc ni lc Ni, Mi, Qiphi cng trng hp t hp (tc l xy ra ng thi)


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b). Xc nh chiu cao mng (h):

Chiu cao cnh mng tnh nhcng thc

(4.8), (4.9), vi b=1m b rng cnh v phi

tho mn nhng iu kin cu to, nu cnh

mng ln m ta khng th tng chiu cao h

th c th b tr thm ct xin trong cnh chu lc ct (xem hnh 8).

Chiu cao sn mng c th c tnh s

b theo ti trng (p lc t) v khong cch

cc ct (nh i vi dm), sau tu theo ta tnh theo quan nim mng cng

hay mng mm (mng na cng) m iu chnh cho hp l.

c). Tnh v b tr thp:

i). Cnh mng:

- Cnh mng c tnh nhbn consol ngm vo sn mng v tnh trn 1m

b rng cnh mng (xem hnh 6). Chiu di on consol l bc= 0,5(b bs).

- Ti tc dng ta ly Pmaxcho an ton, Pmaxc tnh vi gi tr tnh tonca

M, N, Q.

- B trthp nhhnh 6, vi thp theo phng ngn l thp chu lc c tnh

ra nh trn, khong cch b tr t 10 20cm, dng thp 10, v c th ct

50% lng thp ti v tr cch sn mng 1 on l bc+ 20. Cn thp dc l

thp cu to 6 v a 300.

ii).Sn mng:

- C th tnh theo quan nin mng cng, na cng hay mng mm, da

theo iu kin sau: L1 4.4 dkb

EJ (4.13)

Trong : EJ: cng ca tit din ngang ca mng.

B: b rng mng.

L1: khong cch 2 ct

k: h s nn (c th ly nh hng khong 300 400 T/m3

i vi t khng qu yu)

- Tnh mng cng: xem mng l mt dm lt ngc (dm o), vi cc ct

l gi ta, chu ti trng l p lc t Pmaxv Pmin, gii tm ni lc v tnh

thp nhtnh cu kin chu un tit din ch T (hoc ch nht), b tr thp

h

bc bs

>=50

13-1

2

Cnh

>=150

50

200

45

o

HNH 8


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ch l dm mng chu lc ngc li so vi dm sn biu moment

ngc b tr thp ngc.

- Tnh mng mm (cng hu hn): xem dm mng l kt cu dm t trn

nn n hi v s b bin dng theo nn (khi nn ln), mt trong nhng thng

s quan trng nht khi tnh mng mm l tm ra c trng n hi S ca dm,

S ph thuc h s nn k, cng ca dm v c tnh nhsau:

S = 4.

4

dkb

EJ (m) (4.14)

k: c th ly nhtrn hoc tnh theo 2 cch:

o Th nghim nn ti hin trng k=S

P(P l ti trng nn, S l ln).

o T kt qu tnh ton mng k=

S

gl2 hay k=

b

Etb2(b l b rng mng,

Etbl modul bin dng trung bnh ca nn t).

o Gii kt cu c th gii tay (xem thm sch) theo quan nim dm di v

hn hoc gii bng SAP vi cng l xo l k(T/m3); ti trng tc dng l

cc gi tr ni lc chn ct tc dng xung mng (ch cn M v N).

4.3.3. Tnh ton mng bng 2 phng di ct:

Vic tnh mng bng hai phng (mng bng giao nhau) l rt phc tp v

s lin kt gia 2 dy mng theo hai phng v mng chu tc dng ca ngthi 2 moment Mxv Mycng vi N.

kim tra p lc di mng ta xem nh2 dy mng tch ri nhau v

b qua nh hngca moment xon, tc l tnh nh mng bng 1 phng.

Khi tnh dm mng theo quan nim mng mm ta cng tnh nh mng bng

1 phng vi ngoi lc l Mx, My v N ti mi chn ct, lin kt di y

mng l lin kt l xo c cng l k.

Khi tnh theo quan nim mng cng (dm o, gi ta l ct) th ti tc

dng l p lc di y mng v ly gi tr trung bnh ca tng dy.

4.3.3. Tnh ton mng bng di tng:

Mng bng di tng ta gp trong trng hp nhsau :

Mng ca cng trnh c tng chu lc.

Mng ca tng k.

Mng di cc tr cng


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Nu phn tng xem nhtuyt i cng th mng bng di tng ch yu

cn tnh ton vi phn cnh mng v c tnh nhtrn.

Trong mt s trng hp tng c nhiu l ca lm gim yu cng v

mng c th bin dng theo phng dc th ta cn tng cng thm thp cho

phng dc v tnh ton nh mng bng di ct .

4.4. MNG B (continuous foundation, mat foundation):

Mng b c cu to gn ging nh sn lt ngc, chu ti trng l p lc

t. Mng b c nhiudng (xem hnh 9): c sn (sn trn, di hoc hp)

hoc khng sn (xem thm trong sch Nguyn l cu to kin trc).

Ch nn thit k mng b trong trng hp dng mt bng cn xng nhm

mc ch cho tng ti trng xung mng khi qui v trng tm mng t sinh ra

moment. Nu mng c hnh dng phc tp ta tnh trng tm mng theo cng

thc sc bn : X =m

y

F

S; Y =

m

x

F

S.

Chiu dy sn v kch thc sn c th ly sb nhsau: i vi mng b khng sn : hs=1/6 1/8 L.

i vi loi c sn : hs= 1/8 1/10 L.

Chiu cao sn = 1/6 1/8 L.

V mt tnh ton mng b c th tnh theo phng php mng cng hoc

mm (quan nim nh mng bng), nu tnh mng cng ta xem bn mng nh

BA N M ON G BE

COT

S N M ON G Pd

Lo

COT

BA N M ON G BE

S N M ON G

M ON G BEK IE U S N TRE N

M ON G BEK IE U S N D I

BA N M ON G BELo

S N M ON G COT

Lo

BA N M ON G BE

M ON G BEK IE U H OP

S N M ON G COT

BA N SAN TA N G H AM

H N H 9


40/40


sn lt ngc, chu ti l p lc t, nu lch tm tng i nh c th xem

p lc t phn b u v bngmF

N .

Kim tra ng sut y mng cng ging nh i vi mng bng.

Tnh thp bn mng v sn mng nhtnh vi sn, dm nhng b tr thpngc li.

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Bài Giảng KC Nhà Dân Dụng