soal no.1SOAL NO. 1MENENTUKAN TITIK BERAT

abcbdDiketahuiABa=20cmeb=7cm53Yc=12cm56XEFMfd=10cmHIe=8cmLgf=5cmg=25cmh=8cmGJKXNR=15cmRP=70kg

DRhC

No.1SOAL NO. 1MENENTUKAN TITIK BERATYabcbdDiketahuiABa=20cmIeb=7cm53YFc=12cm56XEIIIIIMfd=10cmHIe=8cmIVVLgf=5cmg=25cmh=8cmGJKXNR=15cmVIIRP=70kgVI

DRhCX

Titik BeratBangun IBangun IIAi=56x8Ai=26x5=448cm2=130cm2Xi=0.5x56Xi=20+13=28cm=33cmYi=45+4Yi=40+2.5=49cm=42.5cmBangun IIIBangun IVAi=10x5x0.5Ai=7x25=25cm2=175cm2Xi=46+3.3333333333Xi=20+3.5=49.3333333333cm=23.5cmYi=40+3.3333333333Yi=15+12.5=43.3333333333cm=27.5cm

Bangun VBangun VIAi=7x25Ai=0.25x3.14x225=175cm2=176.625cm2Xi=39+3.5Xi=11.3694267516cm=42.5cmYi=6.3694267516cmYi=15+12.5=27.5cm

Bangun VIIAi=36x15=540cm2Xi=18+20=38cmYi=7.5=7.5cm

Tabel titik beratLetak titik beratiAiXiYiAi.XiAi.Yi(cm2)(cm)(cm)(cm3)(cm3)X= Ai . Xi=52145.458I448.00028.00049.00012544.0021952.00 Ai 1669.625II130.00033.00042.5004290.0005525.000=31.232cmIII25.00049.33343.3331233.3331083.333Y= Ai . Yi=43360.333IV175.00023.50027.5004112.5004812.500 Ai 1669.625V175.00042.50027.5007437.5004812.500=25.970cmVI176.62511.3696.3692008.1251125.000VII540.00038.0007.50020520.004050.0001669.62552145.4643360.33

Lokasi titik berat20712710DiketahuiI853Ya=20cm51Xb=7cmIIIII5c=12cmd=10cmIVV25e=8cmf=5cm(31.23,25.97)g=25cmXh=8cmVII15R=15cmVIP=70kg

158

MENENTUKAN MOMEN INERSIABangun IBangun VIxo=1/12x56x512Ixo=1/12x7x15625=2389.3333333333cm4=9114.5833333333cm4Iyo=1/12x175616x8Iyo=1/12x343x25=117077.333333333cm4=714.5833333333cm4Ixoyo=0Ixoyo=0Bangun IIBangun VIIxo=1/12x26x125Ixo=0.0549x50625=270.8333333333cm4=2779.3125cm4Iyo=1/12x17576x5Iyo=0.0549x50625=7323.3333333333cm4=2779.3125cm4Ixoyo=0Ixoyo=0.0164x50625=830.25cm4Bangun IIIIxo=1/36x10x125Bangun VII=34.7222222222cm4Ixo=1/12x36x3375Iyo=1/36x1000x5=10125cm4=138.8888888889cm4Iyo=1/12x46656x15Ixoyo=1/72x100x25=58320cm4=34.7222222222cm4Ixoyo=0Bangun IVIxo=1/12x7x15625=9114.5833333333cm4Iyo=1/12x343x25=714.5833333333cm4Ixoyo=0Tabel rekapitulasii(Xi - X)(Yi - Y)Ai(Xi - X)2Ai(Yi - Y)2Ai(Xi-X)(Yi-Y)IxoIyoIxoyo

I-3.23223.0304679.261237608.516-33344.1502389.31170770II1.76816.530406.43135520.8743799.579270.87323.30III18.10117.3638191.6037537.0447857.51134.7138.934.7222222222IV-7.7321.53010461.732409.602-2070.0609114.6714.60V11.2681.53022220.006409.6023016.8479114.6714.60VI-19.862-19.60169681.28367856.94368763.0632779.32779.3830.25VII6.768-18.47024736.326184218.142-67504.66610125.058320.00140376.642533560.724-19481.87833828.4187068864.9722222222Ixx=33828+533561Iyy=187068+140377Ixy=864.97+-19482=567389=327445=-18617

MENENTUKAN MOMEN INERSIAImin/max=567389.1+327444.67678132567389.1-327444.676781322+-18616.9222=894833.814393330587+3465891752=447417121408

Imin=447417-121408Imax=447417+121408=326009=568825

tg2=2x-18617327445-567389=0.155182=8.8208=4.4104Koordinat titikA=-31.232;27.030F=-11.232;19.030K=7.768;-10.970B=24.768;27.030G=-11.232;-10.970L=14.768;14.030C=24.768;-25.970H=-4.232;14.030M=24.768;19.030D=-26.232;-25.970I=7.768;14.030N=24.768;-10.970E=-31.232;19.030J=-4.232;-10.970X=14.768;-10.970Kedudukan x atau gaya PMxx=P(-10.970)Myy=P(14.768)=-767.9072146939=1033.771305932tegangan yang terjadi pada titik-titik bangun :=P+Myy - Mxx(Ixy/Ixx)X+Mxx - Myy(Ixy/Iyy)YAIyy - (Ixy2/Ixx)Ixx - (Ixy2/Iyy)P=70A1669.625=0.0419255821Myy - Mxx(Ixy/Ixx)=1033.771305932--767.9072146939(-0.0328115325)Iyy - (Ixy2/Ixx)327444.67678132-(610.8492043714)=0.0030858957Mxx - Myy(Ixy/Iyy)=-767.9072146939-1033.771305932(-0.0568551177)Ixx - (Ixy2/Iyy)567389.1-(1058.466360691)=-0.001252152=4.19E-02+3.09E-03X+-1.25E-03Y

dari persamaan diatas diperoleh :A=-8.83E-02F=-1.66E-02K=7.96E-02B=8.45E-02G=2.10E-02L=6.99E-02C=1.51E-01H=1.13E-02M=9.45E-02D=-6.50E-03I=4.83E-02N=1.32E-01E=-7.83E-02J=4.26E-02X=1.01E-01mencari titik dimana tegangan = 0X=-4.2E-02--0.001252152YY=-0.0419255821-0.0030858957X0.0030858957-0.001252152=-13.5861955725cmdimana Y = 0=33.4828216478cmdimana X = 0koordinat garis netral :(-13.5861955725;0)and(0;33.4828216478)

No. 2aSOAL 2 Bagian AP3q2

q1FBG2.0P2

P12.0

DA

2.03.04.05.0

Data :a=2.0P1=3.0Q1=q1.x=2.0.5.0=10.0tonb=3.0P2=0.0Q2=q2.x=3.0.5.0=15.0tonc=4.0P3=5.0d=5.0h1=2.0q1=2.0h2=2.0q2=3.0

Menentukan panjang batang A-FPerbandingan SegitigaF2.0x =2Batang AF =AO2 +OF234 =32 +42Ex =6 =5m2.04Batang AE =x2 +32 =1.5m =1.52 +22AOy =3 -1.5 =2.5mxy =1.5mBatang EF =AF -AE3.0 =5 -2.5 =2.5m1. Keseimbangan Gaya LuarTinjau Balok E-C-H:P2 =0.0ECHEHVEVC5.55.0

SME = 0SMC = 0P2(10.50) -VC(5.50) =0P2(5.00) +VE(5.50) =00.0(10.50) -VC(5.50) =00.0(5.00) +VE(5.50) =00.00 -VC(5.50) =00.00 +VE(5.50) =05.50VC =0.005.50VE =0.0VC =0.00tVE =0.00t

Check/Kontrol :SV = 0SH = 0VE +VC -P2 =0HE =00.00 +0.00 -0.00 =00.00 =0(..Okay)

Tinjau Balok UtamaP3 sin aP3 =5.0tq2 =3.0t/m

q1 =2.0t/mFBG2.0P3 cos aHBVBP1 =3.0tE2.0VE

DAVA2.03.04.05.0

Q1 =q1*5.0 =2.0*5.0 =10.0tSin a =4 /5 =0.8sin a =0.8Q2 =q2*5.0 =3.0*5.0 =15.0tCos a =3 /5 =0.6459.040.86sin0.51cos4.29SMA = 02.57 - P1(2.00) -VE(1.50) +P3 sin a(3.00) -VB(7.00) +Q1(0.5*3)Q2(0.5*5.0 +7.0) -HB(4) -P3 cos a(4) =0-3.0(2.00) -0.0(1.50) +4.00(3.00) -VB(7.00) +10.00(0.5*3)15.00(0.5*5.0 +7.0) +HB(4.00) -3.00(4.00) =0-6.00 -0 +12 -7.00VB +15 +142.5 +4.00HB -12 =0151.50 +4.00HB -7.00VB =0.Pers. 1SV = 0VA +VE +VB -Q1 -Q2 -P1 -P3 sin a =0VA +0.00 +VB -10.0 -15.0 -3.0 -4.00 =0-32.0 +VA +VB =0.Pers. 2

SH = 0HB -P3 cos a =0HB -3.00 =0HB =3.00t

Subtitusi Nilai HB pada Persamaan (1)

151.50 +4.00HB -7.00VB =0151.50 +12.00 -7.00VB =0163.50 -7.00VB =07.00VB =163.50VB =23.36tSubtitusi Nilai VB pada Persamaan (2)

-32.0 +VA +VB =0-32.0 +VA +23.36 =0-8.6 +VA =0VA =8.64tCheck/Kontrol :SMB = 0VA(7.00) +VE(5.50) -P1(9.00) -Q1(0.5*3.0 +4.0) -P3 sin a(4.00) +Q2(0.5*5.0) =08.64(7.00) +0.00(5.50) -3.00(9.00) -10.0(5.50) -4(4.00) +15.0(2.50) =060.50 +0.00 -27.00 -55 -16 +37.5 =00.00 =0(..Okay)

2. Keseimbangan Gaya Dalam

Tinjau Bentang DA (0 x 2 ) / Struktur dari Kiri - KananP1 =3.0tMx = - P1xx012 =-3.0xMx0.00-3.00-6.00Dx-3.00-3.00-3.00DDx =Mx /dxNx0.000.000.00x =-3.0t

Nx =0

Tinjau Bentang AE (0 x 2,5 )q1 =2.0t/mMx = -MA -DA cosax +VA cosax -1/2q1 cosax2 =-6 -1.80x +5.19x -0.6x2 =-6 +3.39x -0.6x2xDx =Mx /dxMAQcosa =3.39 -1.2xQx022.5DA cosaNx = - VA sina +DA sina +q1 sinaxMx-6.00-1.63-1.29VA sinaVA cosa =-6.9142857143 +2.4 +1.6xDx3.390.990.39VA =-4.5142857143 +1.6xNx-4.5142857143-1.3142857143-0.51

Tinjau Bentang EF (0 x 2,5 ) q1 =2.0t/mMx = -ME +DEx +VE cosax -1/2q1 cosax2 =-1.29 +0.39x +0.00x -0.6x2 =-1.29 +0.39x -0.6x2xDx =Mx /dxMEQcosa =0.39 -1.2xQx012.5NEDENx = - VE sina +NE +q1 sinaxMx-1.29-1.50-4.07VE sinaVE cosa =0.00 +-0.51 +1.6xDx0.39-0.81-2.613.41VE =-0.51 +1.6xNx-0.511.093.492.79

Tinjau Bentang DA (0 x 5 ) / Struktur dari Kanan ke Kiriq2=3.0t/mMx = -1/2q2x2x01245 =-1.5x2Mx0.0-1.5-6.0-24.0-37.5Dx0.003.006.0012.0015.00GDx = -Mx /dxNx0.000.000.000.000.00x =3.0x

Nx =0

Tinjau Bentang BF (0 x 4 )Mx = - MB1 -VB1x +VBxxMB1 =-37.5 -15.00x +23.36x =-37.5 +8.36xBHBx0124Dx = -Mx /dxMx-37.5-29.1-20.8-4.070.000(..Okay) =-8.4tDx-8.4-8.4-8.4-8.4-4.9VBVB1Nx3.003.003.003.00Nx =HB =3.00t

Tinjau Bentang HC (0 x 5 ) P2 =0.0tMx = - P2xx01245 =0.0xMx0.00.00.00.00.0Dx0.000.000.000.000.00HDx = -Mx /dxNx0.000.000.000.000.00x =0.0t

Nx =0

Tinjau Bentang BF (0 x 5.2 )Mx = - MC1 -VC1x +VCxxMC1 =0.0 -0.00x +0.00x =0.0 +0.00xCx01245.5Dx = -Mx /dxMx0.00.00.00.00.000.000(..Okay) =0.0tDx0.000.000.000.000.000.0VCVC1Nx0.000.000.000.000.00Nx =0

P3 =5.0q2 =3.0

q1 =2.0FBG2.0P2 =0.0

P1 =3.02.0

DA

2.03.04.05.0

-37.5

-4.07-4.07

FB0.0G

-6.00-6.00ECH

AD15.00

F-8.4G

-2.61B0.00

E3.39

DA

-3.003.003.49

FBG-0.51E

-0.51

DA-4.51

ERROR:#DIV/0!

No. 2b

SOAL 2b:

SCDEFGH3.0

HAA2.0HBVABVB2.02.02.03.03.02.02.03.0

Data :a=2.0g=2.0q=2t/mb=2.0h=3.0a=60oc=2.0h1=3.0Sina=0.8660254038d=3.0h2=2.0Cosa=0.50e=3.0P1=2.0Q1= .q.x= .2.0.6.0=6.0tonf=2.0P2=3.0

1. Keseimbangan Gaya LuarReaksi PerletakanSMA = 0 - Q1 (2.00) +P1 sina(8.00) -VB(10.00) -P1 cosa(3.00) -P2(3.00) -HB(2.00) =0-6.00 (2.00) +1.73(8.00) -VB(10.00) -1.00(3.00) -3.0(3.00) -2.00HB =0-12.0 +13.8564064606 -10.00VB -3.00 -9.00 -2.00HB =0-10.14 -10.00VB -2.00HB =0.Pers. 1

SMB = 0 - Q1(2/3*6.00 +8.00) +VA(10.00) -P1 sina(2.00) -P1 cosa(5.00) -P2(5.00) +HA(2.00) =0-6.00 (12.00) +VA(10.00) -1.7320508076(2.00) -1.00(5.00) -3.0(5.00) +2.00HA =0-72.0 +10.00VA -3.4641016151 -5.00 -15.00 +2.00HA =0-95.5 +10.00VA +2.00HA =0.Pers. 2

SMSKanan = 0 - VB(5.00) -HB(5.00) +P1 sina(3.00) =0-5.00VB -5.00HB +1.73(3.00) =0-5.00VB -5.00HB +5.20 =0.Pers. 3

SMSKiri = 0VA(5.00) -HA(3.00) -Q1(2/3*6.00 +3.0) =05.00VA -3.00HA -6.00(7.00) =05.00VA -3.00HA -42.00 =0.Pers. 4

FPenyelesaian Persamaan dengan Metode Matriks ;

-10.14 -10.00VB -2.00HB =0.Pers. 1-95.5 +10.00VA +2.00HA =0.Pers. 2-5.00VB -5.00HB +5.20 =0.Pers. 35.00VA -3.00HA -42.00 =0.Pers. 4

VAVBHAHB0.00-10.00.00-2.00VA =10.140.00-10.000.00-2.00VA =10.1410.000.002.000.00VB =95.510.000.002.000.00VB =95.460.00-5.000.00-5.00HA =-5.200.00-5.000.00-5.00HA =-5.205.000.00-3.000.00HB =42.05.000.00-3.000.00HB =42.00

[ A ]*[ B ] =[ C ][ B ] = [ A ]Inverse*[ C ]0.000.080.000.05MINVERSE9.2598076211-0.130.000.050.00-1.5277568136VA =0.000.080.000.05x10.140.000.130.00-0.251.4330127019VB =-0.130.000.050.0095.460.130.00-0.250.002.5669872981HA =0.000.130.00-0.25-5.20HB =0.130.00-0.250.0042.0

VA =9.2598076211tVB =-1.5277568136tHA =1.4330127019tHB =2.5669872981t

SVKanan = 0VB +VS -P1 sina =0-1.5277568136 +VS -1.73 =0VS =3.26t

Check/Kontrol :SV = 0VA +VB -Q -P1 sina =09.2598076211 -1.53 -6.0 -1.73 =00.00 =0(..Okay)

Tinjau Simpul A

S1S2Perhitungan sudut (a)AS1 = AS2 =3.02 +2.02 =9.00 +4.00 =3.61mSin a =3.0 /3.61 =0.8320502943aaCos a =2.0 /3.61 =0.5547001962

HAASH = 0HA - S1 cosa + S2 cosa =0VA1.4330127019 - S1 0.55 + S2 0.5547001962 =0.Pers. (1)

SV = 0VA + S1 sina + S2 sina =09.2598076211 + S1 0.83 + S2 0.8320502943 =0.Pers. (2)

Penyelesaian Persamaan dengan Metode Matriks1.4330127019 - S1 0.55 + S2 0.5547001962 =0.Pers. (1)9.2598076211 + S1 0.83 + S2 0.8320502943 =0.Pers. (2)S1S2-0.550.5547001962S1 =-1.4330127019-0.550.5547001962-0.9010.601-1.4330.830.8320502943S2 =-9.25980762110.830.83205029430.9010.601-9.260[A][B][C]-4.2727516694-6.8561520569[B] =[A] Inverse*[C]0.000S1 =-4.2727516694S2 =-6.8561520569

Diperoleh Nilai ;S1 =-4.2727516694t(tekan)S2 =-6.8561520569t(tekan)

Tinjau Simpul B

S3S4Perhitungan sudut (b)AS3 = AS4 =5.02 +2.02 =25.00 +4.00 =5.39mSin b =5.0 /5.39 =0.9284766909bbCos b =2.0 /5.39 =0.3713906764

HBBSH = 0HB + S4 cosb - S3 cosb =0VB2.5669872981 + S4 0.37 - S3 0.3713906764 =0.Pers. (1)

SV = 0VB + S3 sinb + S4 sinb =0-1.5277568136 + S3 0.93 + S4 0.9284766909 =0.Pers. (2)

Penyelesaian Persamaan dengan Metode Matriks2.5669872981 + S40.37 - S30.3713906764 =0.Pers. (1)-1.5277568136 + S40.93 + S30.9284766909 =0.Pers. (2)S3S4-0.370.37S3 =-2.5669872981-0.370.3713906764-1.3460.539-2.5670.930.93S4 =1.52775681360.930.92847669091.3460.5391.528[A][B][C]4.2786346372-2.6331901919[B] =[A] Inverse*[C]0.000S3 =4.2786346372S4 =-2.6331901919

Diperoleh Nilai ;S3 =4.2786346372t(tarik)S4 =-2.6331901919t(tekan)

2. Keseimbangan Gaya Dalam

Bentang SF (0 x 3 ) / Struktur dari Kiri-Kanan

Mx =VSxx0123x =3.3xMx0.003.266.529.78Dx3.263.263.263.26SDx =Mx /dxNx0.000.000.000.00 =3.3tVSNx =0

Bentang FG (0 x 4 )

MFxMx =MF +DFx -P1 sinax -S3 sinbx =9.8 +3.26x -1.7320508076x -3.9726125294x =9.8 -2.44xS3 cosbx0124DFS3Dx =Mx /dxMx9.787.334.890.00S3 sinb =-2.44tDx-2.44-2.44-2.44-2.44Nx-0.59-0.59-0.59-0.59Nx =P1 cosa -S3 cosb =1.00 -1.5890450118 =-0.59t

Bentang HG (0 x 3 ) Mx =0x0123P2=3.0tMx0.000.000.000.00GDx =0Dx0.000.000.000.00Nx-3.00-3.00-3.00-3.00xNx = - P2 =-3.0t

Bentang SE (0 x 3 ) Mx = -VSxx0123VS =-3.26xMx0.00-3.26-6.52-9.78Dx3.263.263.263.26SDx = -Mx /dxNx0.000.000.000.00x =3.26t

Nx =0

Bentang ED (0 x 4 ) DE4.00 =q1ME6.002.006.0q1 =8.00ECD4.00Eq1 =1.33t/mx6.00S2

x =qxQx = 1/2qx .x4.001.33 = 1/20.33x24.0qx =1.33x =0.1666666667x2xqx =0.33x4.00

Mx = -ME -DE.x +S2 sina . x - 1/3Qx.x =-9.78 -3.26x +5.70x - 1/30.1666666667x3 =-9.78 +2.44x -0.0555555556x3

Dx = -Mx /dx =-2.44 +0.1666666667x2x01234Mx-9.78-7.39-5.33-3.94-3.56Nx = S2 cosaDx-2.44-2.28-1.78-0.940.22 =3.80tNx3.803.803.803.803.80

Bentang CD (0 x 2 ) q2 =q - q1 =2.00 -1.33 =0.67t/m

CxDqx =x2.000.672.002.0qx =0.67xqx =0.33x

Q1 =qxMx = - 1/2Q1x + 1/3Q2x =2.00x = -1/22.00x2 + 1/30.17x3 =-1.00x2 +0.06x3Q2 = 1/2qxx =0.17x2Dx =Mx / dx =-2.00x +0.1666666667x2

Nx =0

x012Mx0.00-0.94-3.56Dx0.00-1.83-3.33Nx0.000.000.00

Bentang CD (0 x 2 ) xq2 =q - q1 =2.00 -1.33 =0.67t/m

CCxDqx =x2.02.002.002.0qx =2.00xqx =1.00x

Q1 =q2xMx = -1/2Q1x + 1/3Q2x - 1/2Q3x =0.67x = -1/20.67x2 +1/30.50x3 -1/21.33x2 =-0.3333333333x2 +0.17x3 -0.67x2Q2 = 1/2qxx =-1.00x2 +0.17x3 =0.50x2

Dx =Mx / dxQ3 =q1x =-2.00x +0.50x2 =1.33x

Nx =0

x012Mx0.00-0.83-2.67Dx0.00-1.50-2.00Nx0.000.000.00

MxNxDxS2cosaS2si naq1qq2qxQ1Q3Q2Q1Q2Q3q1qq2qxQ1Q2

No. 4SOAL 4 BbcdefgC4E8G12Iha135791113153.5 mAababB2D6F10H14abcdefgh4.00 m3.50 m4.00 m3.50 m

Data :L1=4.00 mh=3.50 mL2=3.50 mL3=4.00 mL4=3.50 m

1).Perhitungan Suduta =arc tan(3.5 /2.0 ) =60.2551187031b =arc tan(3.5 /1.8 ) =63.4349488229sin a =0.8682431421sin b =0.894427191cosa =0.4961389384cos b =0.4472135955

2).Reaksi PerletakanSMB = 0SMA = 0VA x15.00 -1(15.00 -x) =0 -VB x15.00 +x =0VA =(15.00 -x)VB =x15.0015.00

3).Gaya - Gaya Batanga. Potongan a - aBeban P=1 ton Bentang (4 x 15 ) NBatang -1 (S1)SMD = 0VAx4.0 +s1sin a x4.0 =0Untuk ;x =4s1 =-0.845s1 = -VA /sin ax =15s1 =0s1 = -(15.00 -x)x115.000.8682431421

NBatang-2 (S2)SMC = 0VAx2.0 +s2x3.50 =0Untuk ;x =4s2 =0.419047619s2 =VAx2.0 /3.50x =15s2 =0s2 =(15.00 -x)x0.571428571415.00

b. Potongan b - bBeban P=1 ton Bentang (4 x 15 ) NBatang -3 (S3)SV = 0VA -s3sin a =0Untuk ;x =4s3 =0.8446174784s3 =VA /sin ax =15s3 =0s3 =(15.00 -x)x115.000.8682431421

NBatang -4 (S4)SMD = 0VAx4.0 +s4x3.50 =0Untuk ;x =4s4 =-0.838s4 =-4.0VA /3.50x =15s4 =0s4 =(15.00 -x)x-1.1415.00

c. Potongan c - cBeban P=1 ton Bentang (0 x 4 ) NBatang -5 (S5)SV = 0VB -s5sin b =0Untuk ;x =0s5 =0.000s5 =VB /sin bx =4s5 =0.298s5 =xx115.000.894427191

NBatang -4 (S4)SMD = 0 -VBx11.00 -s4x3.50 =0Untuk ;x =0s4 =0.000s4 = -(11.00 /3.50)xVBx =4s4 =-0.838s4 =xx-3.142857142915.00

NBatang -6 (S6)SME = 0 -VBx9.25 +s6x3.50 =0Untuk ;x =0s6 =0.000s6 = -(9.25 /3.50)xVBx =4s6 =-0.705s6 =xx-2.642857142915.00

Beban P=1 ton Bentang (7.5 x 15 ) NBatang -4 (S4)SMD = 0VAx4.00 +s4x3.50 =0Untuk ;x =7.5s4 =-0.571s4 = -(4.00 /3.50)xVAx =15s4 =0.000s4 =(15.00 -x)x-1.1415.00

NBatang -5 (S5)SV = 0VA +s5sin b =0Untuk ;x =7.5s5 =-0.559s5 = -VA /sin bx =15s5 =0.000s5 = -(15.00 -x)x1.0015.000.894

NBatang -6 (S6)SME = 0VAx5.75 -s6x3.50 =0Untuk ;x =7.5s6 =0.821s6 =VAx(5.75 /3.50)x =15s6 =0.000s6 =(15.00 -x)x1.6415.00

d. Potongan d - dBeban P=1 ton Bentang (0 x 4 ) NBatang -6 (S6)SME = 0 -VBx9.25 +s6x3.50 =0Untuk ;x =0s6 =0.000s6 =VBx(9.25 /3.50)x =4s6 =0.705s6 =xx2.642857142915.00

NBatang -7 (S7)SV = 0VB +s7sin b =0Untuk ;x =0s7 =0.000s7 = -(VB /sin b)x =4s7 =-0.298s7 = -xx115.000.894427191

NBatang -8 (S8)SMF = 0 -VBx7.5 -s8x3.50 =0Untuk ;x =0s8 =0.000s8 = -VBx(7.5 /3.50)x =4s8 =-0.571s8 = -xx2.142857142915.00

Beban P=1 ton Bentang (7.5 x 15 ) NBatang -6 (S6)SME = 0VAx5.75 -s6x3.50 =0Untuk ;x =7.5s6 =0.821s6 =VAx(5.75 /3.50)x =15s6 =0.000s6 =(15.00 -x)x1.642857142915.00

NBatang -7 (S7)SV = 0VA -s7sin b =0Untuk ;x =7.5s7 =0.559s7 =(VA /sin b)x =15s7 =0.000s7 =(15.00 -x)x115.000.894427191

NBatang -8 (S8)SMF = 0VAx7.5 +s8x3.50 =0Untuk ;x =7.5s8 =-1.071s8 = -VAx(7.5 /3.50)x =15s8 =0.000s8 = -(15.00 -x)x2.142857142915.00

e. Potongan e - eBeban P=1 ton Bentang (0 x 7.5 ) NBatang -8 (S8)SMF = 0 -VBx7.5 -s8x3.50 =0Untuk ;x =0s8 =0.000s8 = -VBx(7.5 /3.50)x =7.5s8 =-1.071s8 = -xx2.142857142915.00

NBatang -9 (S9)SV = 0VB -s9sin a =0Untuk ;x =0s9 =0.000s9 =(VB /sin a)x =7.5s9 =0.576s9 =xx115.000.8682431421

NBatang -10 (S10)SMG = 0 -VBx5.5 +s10x3.50 =0Untuk ;x =0s10 =0.000s10 =VBx(5.5 /3.50)x =7.5s10 =0.786s10 =xx1.571428571415.00

Beban P=1 ton Bentang (11.5 x 15 ) NBatang -8 (S8)SMF = 0VAx7.5 +s8x3.50 =0Untuk ;x =11.5s8 =-0.500s8 = -VAx(7.5 /3.50)x =15s8 =0.000s8 = -(15.00 -x)x2.142857142915.00

NBatang -9 (S9)SV = 0VA +s9sin a =0Untuk ;x =11.5s9 =-0.269s9 = -(VA /sin a)x =15s9 =0.000s9 = -(15.00 -x)x115.000.8682431421

NBatang -10 (10)SMG = 0VAx9.5 -s10x3.50 =0Untuk ;x =11.5s10 =0.633s10 =VAx(9.5 /3.50)x =15s10 =0.000s10 =(15.00 -x)x2.714285714315.00

f. Potongan f - fBeban P=1 ton Bentang (0 x 7.5 ) NBatang -12 (S12)SMH = 0 -VBx3.5 -s12x3.50 =0Untuk ;x =0s12 =0.000s12 = -VBx(3.5 /3.50)x =7.5s12 =-0.500s12 = -xx115.00

NBatang -11 (S11)SV = 0VB +s11sin a =0Untuk ;x =0.0s11 =0.000s11 = -(VB /sin a)x =7.5s11 =-0.576s11 = -xx115.000.8682431421

NBatang -10 (S10)SMG = 0 -VBx5.5 +s10x3.50 =0Untuk ;x =0.0s10 =0.000s10 =VBx(5.5 /3.50)x =7.5s10 =0.786s10 =xx1.571428571415.00

Beban P=1 ton Bentang (11.5 x 15 ) NBatang -10 (S10)SMG = 0VAx9.5 -s10x3.50 =0Untuk ;x =11.5s10 =0.633s10 =VAx(9.5 /3.50)x =15s10 =0.000s10 =(15.00 -x)x2.714285714315.00

NBatang -11 (S11)SV = 0VA -s11sin a =0Untuk ;x =11.5s11 =0.269s11 =(VA /sin a)x =15s11 =0.000s11 =(15.00 -x)x115.000.8682431421

NBatang -12 (S12)SMH = 0VAx11.5 +s12x3.50 =0Untuk ;x =11.5s12 =-0.767s12 = -VAx(11.5 /3.50)x =15s12 =0.000s12 = -(15.00 -x)x3.285714285715.00

g. Potongan g - gBeban P=1 ton Bentang (0 x 11.5 ) NBatang -12 (S12)SMH = 0 -VBx3.5 -s12x3.50 =0Untuk ;x =0s12 =0.000s12 = -VBx(3.5 /3.50)x =11.5s12 =-0.767s12 = -xx115.00

NBatang -13 (S13)SV = 0VB -s13sin b =0Untuk ;x =0.0s13 =0.000s13 =(VB /sin b)x =11.5s13 =0.857s13 =xx115.000.894427191

NBatang -14 (S14)SMI = 0 -VBx1.75 +s14x3.50 =0Untuk ;x =0.0s14 =0.000s14 =VBx(1.75 /3.50)x =11.5s14 =0.383s14 =xx0.515.00h. Potongan h - hBeban P=1 ton Bentang (0 x 11.5 ) NBatang -14 (S14)SMI = 0 -VBx1.75 +s14x3.50 =0Untuk ;x =0s14 =0.000s14 =VBx(1.75 /3.50)x =11.5s14 =0.383s14 =xx0.515.00

NBatang -15 (S15)SV = 0VB +s15sin b =0Untuk ;x =0.0s15 =0.000s15 = -(VB /sin b)x =11.5s15 =-0.857s15 = -xx115.000.894427191

3.5 m

4.00 m3.50 m4.00 m3.50 m

-0.845 t0.419 t

0.845 t

-0.838 t0.298 t

0.705 t-0.559 t

0.559 t

-0.298 t

-1.071 t0.576 t

0.786 t-0.269 t

0.269 t

-0.576 t

-0.767 t0.857 t

0.383 t

-0.857 t

GP-S1GP-S2GP-S3GP-S4GP-S5GP-S6GP-S7GP-S8GP-S9GP-S10GP-S11GP-S12GP-S13GP-S14GP-S15S1S2S3S4S5S6S7S8S9S10S11S12S13S14S15AB

no.4 popoNO. 4

P31SIN0.857493COS0.514496P21IP4253.8873012632HJ3.33333333331.6666666667K22G2.6034165586SIN0.640184COS0.7682212796CDXEFABP1P2P3P4P41222222P1 =5tonP2 =4tonP3 =3tonP4 =2tona =2mb =5mModulus Elastisitas :E =2x106kg/cm2KESEIMBANGAN GAYA-GAYA LUARMA=0 -VB .12 +2P1 +4(P2 + P21) +6(P3 + P31) +8P4 +10(P4 + P42) =0 -VB .12 +2x5 +4(4 +4) +6(3 +3) +8x2 +10(2 +2) =0 -VB .12 +134 =0-12VB =-134VB =11.1666666667TonMB=0VA .12 -10P1 -8(P2 + P21) -6(P3 + P31) -4P4 -2(P4 + P42) =0VA .12 -10x5 -8(4 +4) -6(3 +3) -4x2 -2(2 +2) =0VA .12 -166 =012VA =166VA =13.8333333333TonKontrol :V=0VA +VB -P1 -P2 -P21 -P3 -P31 -P4 -P41 -P42 =011.1666666667 +13.8333333333 -5 -4 -4 -3 -3 -2 -2 -2 =025 -25 =00 =0(OKAY)

Menghitung Sudut pada Struktur

5y =(52 +62) 1/2 =7.8102496759sina =0.6401843997cosa =0.76822127966

3.33y =(3.33333333332 +22) 1/2 =3.8873012632sinb =0.8574929257cosb =0.51449575542MENCARI GAYA BATANG DENGAN METHOD KESEIMBANGAN TITIKTITIK AS1V=0Va +S1 Sin a =013.8333333333 +0.6401843997S1 =0S1 =-21.6083574367t(Tekan)S2H=0VaS1 Cos a +S2 =0-16.6 +S2 =0S2 =16.6t(Tarik)TITIK CV=0S3S3 -P1 =0S3 =5t(Tarik)S2S4H=0S4 -S2 =0P1S4 =16.6t(Tarik)TITIK GV=0S6S6 Sin a -S5 Sin a -S1 Sin a -S3 =0 --13.8333333333 -5 -0.6401843997S5 +0.6401843997S6 =0G0.6401843997S6 -0.6401843997S5 =-8.8333333333(Pers. 1)H=0S1S3S5S6 Cos a +S5 Cos a -S1 Cos a =00.7682212796S5 +0.7682212796S6 --16.6 =00.7682212796S5 +0.7682212796S6 =-16.6(Pers. 2)

Eliminasi Pers. 1 & 20.64S6 -0.64S5 =-8.8333333333x0.770.49S6 -0.49S5 =-6.78595463640.77S6 +0.77S5 =-16.6x0.640.49S6 +0.49S5 =-10.6270610344 --0.98S5 =3.841106398S5 =-3.905124838t(Tekan)Subtitut Nilai S5 =-3.905124838pada pers. 10.64S6 -0.64S5 =-8.83333333330.64S6 --2.50 =-8.83333333330.64S6 =-11.33S6 =-17.70t(Tekan)TITIK DV=0S5 Sin a +S7 -P2 =0S5S7-2.5 +S7 -4 =0S7 =6.5t(Tarik)S4S8DH=0 -S5 Cos a -S4 +S8 =0P2 --3 -16.6 +S8 =0S8 =13.6t(Tarik)TITIK HV=0P21S10S10 Sin a -S6 Sin a -S9 Sin b -S7 -P21 =00.6401843997S10 --11.3333333333 -0.8574929257S9 -6.5 -4 =0H0.64S10 -0.8574929257S9 =-0.83(Pers.3)

H=0S6S7S9S10 Cos a +S9 Cos b -S6 Cos a =00.7682212796S10 +0.5144957554S9 --13.6 =00.7682212796S10 +0.5144957554S9 =-13.6(Pers. 4)Eliminasi Pers.3 & 40.64S10 -0.86S9 =-0.83x0.770.49S10 -0.66S9 =-0.64018439970.77S10 +0.51S9 =-13.6x0.640.49S10 +0.33S9 =-8.7065078354 --0.99S9 =8.0663234358S9 =-8.1633326528t(Tekan)Sutitut Nilai S9 =-8.1633326528pada pers. 30.64S10 -0.86S9 =-0.830.64S10 --7.00 =-0.830.64S10 =-7.83S10 =-12.2360578256t(Tarik)TITIK IH=0P31S14 Cos a -S10 Cos a =00.7682212796S14 --9.4 =0GS14 =-12.2360578256t(Tarik)

V=0S10S11S14 -S10 Sin a -S14 Sin a -S11 -P31 =0 -0.6401843997S10 -0.6401843997S14 -S11 -3 =0 --7.83 --7.8333333333 -3 -S11 =0S11 =12.67t(Tekan)

TITIK XV=0S11S9 Sin b +S13 Sin b +S11 -P3 =0S9S130.86S13 +-7 +12.67 -3 =00.86S13 +2.67 =0S8S12S13 =-3.1098410106t(Tarik)XH=0S13 Cos b -S9 Cos b +S12 -S8 =0P3-1.6 --4.2 +S12 -13.6 =0S12 =11t(Tarik)TITIK JH=0S18 Cos a -S13 Cos b -S14 Cos a =0S140.7682212796S18 --1.6 --9.4 =0S18 =-14.3187910725t(Tarik)JV=0S13S14 Sin a -S13 Sin b -S18 Sin a -S15 =0S15S18-7.83 --2.67 --9.1666666667 -S15 =0S15 =4t(Tekan)

TITIK EV=0S15S17 Sin a +S15 -P4 =0S170.64S17 +4.00 -2 =0S17 =-3.1240998704t(Tarik)S12S16EH=0S17 Cos a +S16 -S12 =0P4-2.4 +S16 -11 =0S16 =13.4t(Tekan)TITIK FV=0S19S19 -P41 =0S19 -2 =0S19 =2t(Tarik)S16S20FH=0S20 -S16 =0P41S20 -13.4 =0S20 =13.4t(Tekan)TITIK KH=0P42S21 Cos a -S18 Cos a -S17 Cos a =0S180.7682212796S21 --11 --2.4 =0S21 =-17.4428909429t(Tarik)KV=0S17S18 Sin a -S17 Sin a -S21 Sin a -S19 -P42 =0S19S21-9.17 --2.00 --11.1666666667 -2 -2 =013.17 -13.17 =00.00 =0OKMenghitung Gaya Batang Akibat Beban Vertikal 1 ton di Titik X

SIN0.857493COS0.514496I5HJK3.33333333331.6666666667G222.6034165586SIN0.640184CDXEFCOS0.7682212796ABP =1t222222a =2mb =5mModulus Elastisitas :E =2x106kg/cm2KESEIMBANGAN GAYA-GAYA LUARMA=0 -VB .12 +6P =0 -VB .12 +6x1 =0 -VB .12 +6 =0-12VB =-6VB =0.5TonMB=0VA .12 -6P =0VA .12 -6x1 =0VA .12 -6 =012VA =6VA =0.5TonKontrol :V=0VA +VB -P =00.5 +0.5 -1 =01 -1 =00 =0(OKAY)

Menghitung Sudut pada Struktur

5y =(52 +62) 1/2 =7.8102496759sina =0.6401843997cosa =0.76822127966

3.33y =(3.33333333332 +22) 1/2 =3.8873012632sinb =0.8574929257cosb =0.51449575542MENCARI GAYA BATANG DENGAN METHOD KESEIMBANGAN TITIKTITIK AS1V=0Va +S1 Sin a =00.5 +0.6401843997S1 =0S1 =-0.7810249676t(Tekan)S2H=0VaS1 Cos a +S2 =0-0.6 +S2 =0S2 =0.6t(Tarik)TITIK CV=0S3S3 =0S3 =0tS2S4H=0S4 -S2 =0S4 =0.6t(Tarik)TITIK GV=0S6S6 Sin a -S5 Sin a -S1 Sin a -S3 =0 --0.5 -0 -0.6401843997S5 +0.6401843997S6 =0G0.6401843997S6 -0.6401843997S5 =-0.5(Pers. 1)H=0S1S3S5S6 Cos a +S5 Cos a -S1 Cos a =00.7682212796S5 +0.7682212796S6 --0.6 =00.7682212796S5 +0.7682212796S6 =-0.6(Pers. 2)

Eliminasi Pers. 1 & 20.64S6 -0.64S5 =-0.5x0.770.49S6 -0.49S5 =-0.38411063980.77S6 +0.77S5 =-0.6x0.640.49S6 +0.49S5 =-0.3841106398 --0.98S5 =0S5 =0tSutitut Nilai S5 =0pada pers. 10.64S6 -0.64S5 =-0.50.64S6 -0.00 =-0.50.64S6 =-0.50S6 =-0.78t(Tekan)TITIK DV=0S5 Sin a +S7 =0S5S70 +S7 =0S7 =0tS4S8DH=0 -S5 Cos a -S4 +S8 =0 -0 -0.6 +S8 =0S8 =0.6t(Tarik)TITIK HV=0S10S10 Sin a -S6 Sin a -S9 Sin b -S7 =00.6401843997S10 --0.5 -0.8574929257S9 -0 =0H0.64S10 -0.8574929257S9 =-0.50(Pers.3)

H=0S6S7S9S10 Cos a +S9 Cos b -S6 Cos a =00.7682212796S10 +0.5144957554S9 --0.6 =00.7682212796S10 +0.5144957554S9 =-0.6(Pers. 4)Eliminasi Pers.3 & 40.64S10 -0.86S9 =-0.50x0.770.49S10 -0.66S9 =-0.38411063980.77S10 +0.51S9 =-0.6x0.640.49S10 +0.33S9 =-0.3841106398 --0.99S9 =0S9 =0tSutitut Nilai S9 =0pada pers. 30.64S10 -0.86S9 =-0.500.64S10 -0.00 =-0.500.64S10 =-0.50S10 =-0.7810249676t(Tekan)TITIK IH=0S14 Cos a -S10 Cos a =00.7682212796S14 --0.6 =0IS14 =-0.7810249676t(Tarik)

V=0S10S11S14 -S10 Sin a -S14 Sin a -S11 =0 -0.6401843997S10 -0.6401843997S14 -S11 =0 --0.50 --0.5 -S11 =0S11 =1.00t(Tekan)

TITIK XV=0S11S9 Sin b +S13 Sin b +S11 -P =0S9S130.86S13 +0 +1.00 -1 =00.86S13 +0.00 =0S8S12S13 =0tXH=0S13 Cos b -S9 Cos b +S12 -S8 =0P0 -0 +S12 -0.6 =0S12 =0.6t(Tarik)TITIK JH=0S18 Cos a -S13 Cos b -S14 Cos a =0S140.7682212796S18 -0 --0.6 =0S18 =-0.7810249676t(Tarik)JV=0S13S14 Sin a -S13 Sin b -S18 Sin a -S15 =0S15S18-0.50 -0.00 --0.5 -S15 =0S15 =0t

TITIK EV=0S15S17 Sin a +S15 =0S170.64S17 +0.00 =0S17 =0tS12S16EH=0S17 Cos a +S16 -S12 =00 +S16 -0.6 =0S16 =0.6t(Tarik)TITIK FV=0S19S19 =0S19 =0t

S16S20FH=0S20 -S16 =0S20 -0.6 =0S20 =0.6t(Tarik)TITIK KH=0S21 Cos a -S18 Cos a -S17 Cos a =0S180.7682212796S21 --0.6 -0 =0S21 =-0.7810249676t(Tekan)KV=0S17S18 Sin a -S17 Sin a -S21 Sin a -S19 =0S19S21-0.50 -0.00 --0.5 -0 =00.50 -0.50 =00.00 =0OKMenghitung Gaya Batang Akibat Beban Horisontal 1 ton di Titik X

SIN0.857493COS0.514496I5HJK3.33333333331.6666666667G222.6034165586SIN0.640184CDXEFCOS0.7682212796AP =1tB222222a =2mb =5mModulus Elastisitas :E =2x106kg/cm2KESEIMBANGAN GAYA-GAYA LUARH=0HA +P=0HA=1.00t()

Menghitung Sudut pada Struktur

5y =(52 +62) 1/2 =7.8102496759sina =0.6401843997cosa =0.76822127966

3.33y =(3.33333333332 +22) 1/2 =3.8873012632sinb =0.8574929257cosb =0.51449575542MENCARI GAYA BATANG DENGAN METHOD KESEIMBANGAN TITIKTITIK AS1V=0Va +S1 Sin a =00 +0.6401843997S1 =0HaS1 =0tS2H=0VaS1 Cos a +S2 -Ha =00 +S2 -1 =0S2 =1t(Tarik)TITIK CV=0S3S3 =0S3 =0tS2S4H=0S4 -S2 =0S4 =1t(Tarik)TITIK GV=0S6S6 Sin a -S5 Sin a -S1 Sin a -S3 =0 -0 -0 -0.6401843997S5 +0.6401843997S6 =0G0.6401843997S6 -0.6401843997S5 =0(Pers. 1)H=0S1S3S5S6 Cos a +S5 Cos a -S1 Cos a =00.7682212796S5 +0.7682212796S6 -0 =00.7682212796S5 +0.7682212796S6 =0(Pers. 2)

Eliminasi Pers. 1 & 20.64S6 -0.64S5 =0x0.770.49S6 -0.49S5 =00.77S6 +0.77S5 =0x0.640.49S6 +0.49S5 =0 --0.98S5 =0S5 =0tSutitut Nilai S5 =0pada pers. 10.64S6 -0.64S5 =00.64S6 -0.00 =00.64S6 =0.00S6 =0.00t(Tekan)TITIK DV=0S5 Sin a +S7 =0S5S70 +S7 =0S7 =0tS4S8DH=0 -S5 Cos a -S4 +S8 =0 -0 -1 +S8 =0S8 =1t(Tarik)TITIK HV=0S10S10 Sin a -S6 Sin a -S9 Sin b -S7 =00.6401843997S10 -0 -0.8574929257S9 -0 =0H0.64S10 -0.8574929257S9 =0.00(Pers.3)

H=0S6S7S9S10 Cos a +S9 Cos b -S6 Cos a =00.7682212796S10 +0.5144957554S9 -0 =00.7682212796S10 +0.5144957554S9 =0(Pers. 4)Eliminasi Pers.3 & 40.64S10 -0.86S9 =0.00x0.770.49S10 -0.66S9 =00.77S10 +0.51S9 =0x0.640.49S10 +0.33S9 =0 --0.99S9 =0S9 =0tSutitut Nilai S9 =0pada pers. 30.64S10 -0.86S9 =0.000.64S10 -0.00 =0.000.64S10 =0.00S10 =0tTITIK IH=0S14 Cos a -S10 Cos a =00.7682212796S14 -0 =0IS14 =0t

V=0S10S11S14 -S10 Sin a -S14 Sin a -S11 =0 -0.6401843997S10 -0.6401843997S14 -S11 =0 -0.00 -0 -S11 =0S11 =0.00t

TITIK XV=0S11S9 Sin b +S13 Sin b +S11 =0S9S130.86S13 +0 +0.00 =00.86S13 +0.00 =0S8S12S13 =0tXH=0PS13 Cos b -S9 Cos b +S12 -S8 +P =00 -0 +S12 -1 +1 =0S12 =0tTITIK JH=0S18 Cos a -S13 Cos b -S14 Cos a =0S140.7682212796S18 -0 -0 =0S18 =0tJV=0S13S14 Sin a -S13 Sin b -S18 Sin a -S15 =0S15S180.00 -0.00 -0 -S15 =0S15 =0t

TITIK EV=0S15S17 Sin a +S15 =0S170.64S17 +0.00 =0S17 =0tS12S16EH=0S17 Cos a +S16 -S12 =00 +S16 -0 =0S16 =0tTITIK FV=0S19S19 =0S19 =0t

S16S20FH=0S20 -S16 =0S20 -0 =0S20 =0t(Tarik)TITIK KH=0S21 Cos a -S18 Cos a -S17 Cos a =0S180.7682212796S21 -0 -0 =0S21 =0tKV=0S17S18 Sin a -S17 Sin a -S21 Sin a -S19 =0S19S210.00 -0.00 -0.00 -0.00 =00.00 -0.00 =00.00 =0OK

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REKAP GB POPO

Tabel Rekapitulasi Gaya Batang

BatangGaya Batang ( ton )Panjang S.Uv.LS.Uh.LS.LBeban P (S)Beban UvBeban Uh(L)1-21.608-0.7810.0002.603416558643.9370-56.256216.6000.6001.000219.92033.20033.20035.0000.0000.0001.66666666670.00008.333416.6000.6001.000219.92033.233.2005-3.9050.0000.0002.60341655860.0000-10.1676-17.703-0.7810.0002.603416558635.9970.000-46.08976.5000.0000.0003.33333333330.000021.667813.6000.6001.000216.32027.227.2009-8.1630.0000.0003.88730126320.0000-31.73310-12.236-0.7810.0002.603416558624.8800.000-31.8561112.6671.0000.000563.333063.3331211.0000.6000.000213.200022.00013-3.1100.0000.0003.88730126320.0000-12.08914-12.236-0.7810.0002.603416558624.8800-31.856154.0000.0000.0003.33333333330.000013.3331613.4000.6000.000216.080026.80017-3.1240.0000.0002.60341655860.0000-8.13318-14.319-0.7810.0002.603416558629.1150-37.278192.0000.0000.0001.66666666670.00003.3332013.4000.6000.000216.080026.80021-17.443-0.7810.0002.603416558635.4670-45.411359.12994-31.667

untukE =2 x 106 kg/cm2=2 x 105 t/m2DDx=S.Uv.LDDx=S.Uh.LE.AE.A=359.1289549939=942.00E+052.00E+05=1.80E-03m=4.68E-04m

Sheet1