NAMA : MEILISA ASMARANI
NPM :12811fi150
1. Perhiturqan Momen Primer
= -th2qL2 = -Llt2(2X6F
=tlL2qt =tltlt2x6f= -LUt92qLz = -trlL92(2X8f
=5lL92qL2 =sltszt2x8f
Moo.
Mo.o
Mor.
Moo
Mo*
Moo.
Moo,
Moro
= -1l8 PL
=v8PL
= -6,0(X) tm
- 600O tm
= -7,333 tm
= 3,333 tm
= -3,000 tm
= 3,(XX) tm
= 0,000 tm
= 0,0(X) tm
= -V8 {aX6)
= VB (aX6)
=2(L+ 1,21
=2(1,2+Q9)
=2(O9+46)
2. Momen Primer pada titlk kumpul
dititik B :
TB = Mom+ Moac
= 6 tm + (-7,333 tm) '
dititik c :
Tc =Moca+Moco
= 3,333 tm + (-3 tm)
dititik D :
TD =Mooc+Mooe
=3tm+0
3. Jumlah Koeflsien Xekakuan pada titlk kumpuldititik B :
Ps =2(klg+kBc)dititik c :
9c =2(kac+kco)
dititik D :
Po =2(ko+koe)
=-kac/Ps
=-ksc/Pc
=-ko/Qc
='W/ 9s
=-t,214,4
= - L,2l 4,2
= -0,9 / 4,2
=-O,913
= -1,333 tm
0,333 tm
3,000 tm
4r40o
4,2O0
3,00o
= 4,273
= -0,286
= 4,214
= {,300
4. Koefisien Momen Takabeya pada ujung batang
Vsc
Vce
Vco
Yoc
Perhitungan Momen Rotasi {anggapan belum teriadi perputaran sudut}M(ot,
M{o).
M(olo
=-Ts/ Pe
=-Tcl Pc
=-To/Po
Perataan Momen Rotasi
M(1),
M(il.
M(1lo
M(r),
M(r).
MPlo
M(r),
M(t)c
M(3)o
M(o),
M(o).
M{olo
M(rl,
M(t).
M(t)o
= M{o}s+ Vr. (M{o).}
= M(olc+ Vo (M(t)r) * Y.o (M(olo)
= M0)o+ Y* (M(tl.)
= M(ols+ Yr. (M(t).)
= M(olc+ Ycg (M{t)r) * Y. (M(t}o}
= M@lo+ V* (M{t'.}
= M(ole+ YBc (M(tl.)
= M{o}c+ V., (Mtt}r) + V.o (M(ao)
= M{o}o+ Vo (M(').)
= M(o)g+ Vr. (M(tl.)
= M(olc+ V., (M{n}r) * V.o (M(t)o)
= M(olo+Y*(M(o).) .
= MPle+ Vo (MFld
= M(olc+ Vo (Mttlr) + V* {MF)o)
= M{o)o+V* (M(t).}
Momen Final
Mee = kna (2 na(tlo * M(s)r1+ Moo,
= 1 (2 (0) + Q288) + (-5)
MBA = ksr (2 "(sl,
+ M(slo1 + Mo*
= 1 {2 (0,288) + 0} + 6
MBc = kac (2 "(s).
+ M{s).; + Mor.
= L,212 {0,288} + 0,055} + (-7,333}
Mca = ke (2 "(s).
+ M(s!r} + Mo.,
= t,? (2 (0,056) + O288) + 3,333
M.o = kcD (2 "(s).
+ M(ulo) * Mo.o
= 0,9 (2 (0,056) + (-0,017)) + (-3)
MDc = koc (2 M($o + M{s}.1 + Moo.
= 0,9 (2 (-1,017)+ 0,056) + 3
Moe = kor (2 M(u)o * M{s)r1+ Moo,
= 0,6 (2 (-1,017) + 0) + 0
Meo = kro (2 M(t), * M(s)o; + Mo.o
= 0,6 (2 {0} + {-0,017}} + 0
= - l-t,33?114,4 = Q303 tm
= -a333/4,2
= -313
Q325 tm
Q042 tm
-1,013 tm
a,292
0,054
-1,016
0,288
0,056
-!,ot7
0,288 tm
0,056 tm
-1,017 tm
0,288 tm
0,056 tm
-1,017 tm
-5,712 tm
6,575 tm
= -5,575 tm
3,814 tm
= -3,814 tm
1,220 tm
= -1,220 tm
= -0,610 tm
= -0,079 tm
= -1,000 tm
tm
tm
tm
tm
tm
tm
FREE BODY DIAGRAM
f5,8561
batang AB :
q:2t/m
Afibatq l:"'lr:6,
ru"'ruakibat Mo, l=0,952 t
1-6,375 1l- t6
akibat M* t ,.,0r, *
A
l:"u'rr:6,rt''ul"to,gsz,l=t't"/"l= 1,096 t
tt6,rlgt lo,3+s ,,ruJ Tr,o*
l'5671 lno* L,*,
u,"XW;u'u
fs,assRo.l *J't*
batans BC:
batans CD :
3,914
akibat q
akibat M6p
akibat Ms6
3,814
fibatq:
akibat Mo
akibat M.t
Fu*%: u,
?"'t"1,I=0,822 tl=t'tt%Y=O,4771
l:z<tt1r: ztl=''t'u l"*=0,822 t
t''"*,,'=O,477 t
+6,345ln,.
t1,6ssRcBl
Batang DE:
L,zn 0,610
P:4t
E
akibat Mo.
akibat M.o
l=t''*lnf=0,305
l=ntto/Lo,rsg
lo,+sg
R*+
f t,sezt-
nJlz,+tt
R.ol
*J'ott
a
Diarram Lintann
Lxr =0 Lxz =0Lxr =5,856-2x1 =6 51, =6,345 -2x2 =g
-2x1 =-5,855 -2x2=-6,345
\ =2,928m I M1 xz =3,!73m * M2
Mr =Mm+ {Rls.Xr}-(q .Xr."lrl/.r,l
= l-5,7L21+ (5,855 .2,9281- 12 . 2,928 . t,4ill.
= (-5,7t21+ t7,146 - 8,573
=3,401 tm t ,
Mz = Msc + (Rec .xz)- {q .xr.tlrXrl
= {-6,575} + (6,345 .3,1731- 12 .3,L73. 1,587}
= (-6,5751 + 2Q133 - ilO,O7L
=3,487 tm
Ms =Mco+ (R6o.3)
= (-3,814)+ (2,433 . 3)
= (-3,8141 +7,299
=3,485 tm
Dlarram Momen