# Perhitungan momen ujung batang dari struktur dengan metode slope deflection equation [tugas analisis

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• l4:NTA : HITUN MMNM F

t(MRANlMDT o6

I STFF'` JH INl pFNliN

EttiN

LN H FHITNN

MEttHITON MOME PF'MERFEM.= -4C23 -60)2

:2 12

, -3210 = -27O Fh/.FT

FEM oA=

L2

q|)2=3`f3

2

UJN TNpFFLE)N

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12

13750 =

= 4 EI.0+ 2 FI.3-270

`M =

MO =

M 00ElL

: :30

"(2 A) FFM DIA

D 2_l +270 =4 FI.OB+ 2 EI+27030 30 30

(20t FFM

+ 2=Ett ` 185=_LEI .l 1875

3D a

3 FT

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(2a4FEMB

+2 Fl o -20

12

324012

FFM D= _Pab2=12

= _16g750=900?a'b =

12

=2 rN F

-50C152c30,2

-18514.FT

5C:,(30)

:5 NFT

0!

FEM 43 =

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• l4cs -- 2 Et ( z Oe + go) + fEt4 ecL

= 4 e, Oc + 2 El Oo + tAZ,5 = 4 Et 0c + 2 Et OO t tO7,5

, neNap'rrul@ KtrsEt|4bANaAN+ t4ae -- o

Yl>a* Yloc = O-7 YlcO = O

OlrtT 4

4 " NT 24 OrNT c

M43=04 E1 0

+ 2 EI B270 `0

"

: :OAtt M=04 EI B+ 2

3 EI 3+ 2 E 4 2 EI = i37,5-270 0

~

a~

1

M=0 F: c+ 2 l 3 +l50

0El Oa+27o+ 4 trl 0O+ 2 el 0c-187,q=o

b%

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5 lD LM ttFHtt MTel5 ME| :0

%4/3

270

-8

:4 El + 2 : 3_:3z|, 1

_______________i

NH:T N h2nN r,.

lZ,

0

FI

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2/3 0

2/30

40

2/3

0

270

-.

la7r

= 4oX

730 4

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4/

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| +0|12

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???27 2

= 23125

: 2gl, z5 x

= 23125(3/4+3~a3)=23337/E = 22 (Jl4 1 IBr5/EIt = 23i25(1/2+03-58) rl875/I

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Maa =q e7.0a+ z trl.Oo -27o = I ey.23u,379

"M`

M

-- Vtl, zs .r ( - 4t, zs) - zto = o

=1EL.)a+ 2 ET..0a+77O=4 EL"bbtu

--(-a2,5) + i55,625 + z7o : 34?. tzb

+2 EI.-6187_27030

'075) 1.2427a

30

(-rce;,e7e)+ 2 el ('dalo) + t87'e?o

Mc; q I .0o+z Etgc-18715=I EL-G68,7E)+zEIG'togO,9ts)-Bt,"

0

= _325-12gl = 412

M0= 4 EI.+ 2 El 3+'37_ 4 EI

0

.1462-412 107rr0

• E 0

nFT

Vao = 4zrg6z5Yga + Voc6S437+:47fOl37 Fv2o= 13

' r

ZPla = oVas (T) - 3,6 Czo)Ct5)+ 743, t25 = oVao (b) - 16zo + flZ t2, = ovae L*) - PZ6 t Bzg = oVeO , 1276,A75

= l7.lVQzg lO) = oVga= tgo3,r25 = 65, 4>79 f" Cf )

"

z Mo=0

"(1543_tt)

7%125_Vtt `=

406375-V4( 0c= 4C37 = i362, F (,

"

M=073C C="(20c C32)l9)25,0V= :0902 = %47C

FT

4,43J F

h

Vo=V

3.6 Ew irf

42%2

• '

|

:

1/A=42/2 N

I-1

• zV=0Lyc =Lx=yc

Il :,

L =L,=Lla=L =

'%25 v

L625 F:362i32'

ZHx = OMx + 50 Cx) - Vc (xtta)=ol"1x=Vc (,x t,e)- nLx)

!a= l715QzgEw

l4o= l),=oz5 (o+15)-b (O)r Zog.1?79 Y+,I.F'TY1=-- 146629L5+re)- 5o (s)= 21,2e kN. FTF1 o = t?,%25 [to+ ts)- 5O (to) = - t@, W79 tsN'FTYv= i2t,t*2e(l*tb)- 9o ftZ)" - 4?,, t2> k,/.FT7V= OLx+Vc-EO=oLx- 50

-V6

L= 5o :25L= 50- l25Ll=50- 1%25L`=5 l2

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76t4)z> FYv6.4?zs rN76. 4b7> ktvZb.4gr= Yv