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

    ???????

    ???

    ??????

    ?

    ????

    ????????????????

    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

    ????[

    (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%

    ?????

    ?????????

    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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    0

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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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    ??

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    | +0|12

    ??????

    ?

    ???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

    ??????

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

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