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