Mekanika Teknik

NO. 1a qP3

ADEIEF

H2H2P2P11.5EI2EI1/2H1H1B

C

L1L2L3q=2.0L1=2.0H1=3.0P1 =4.0L2=3.0H2=4.0P2 =3.0L3=2.0P3 =2.0Konstruksi yang akan dianalisaq=2.0P3=2.0Misalnya:

AEIDEIEEIFL1 =2EI1 =14.04.0L2 =5.5EI2=1.5bbL3 =3EI3 =1P2=3.0P1=4.0L4 =7EI4 =21.5EI2EI1.5aa3.0B

C

2.03.02.0

Struktur dasar yang dikekang

Momen primerMAD=1ql2=12.02.02=0.667 tm0.6671212-0.667MDA=-MBA=-0.667t/m-0.893MDB=-Pa2..b=-31.52(4)=-0.893 tm2.380L25.521.500MBD=Pa.b2=31.54.02=2.380 tm-1.500L25.522.939MDE=1ql2=12.03.02=1.5-3.91812124.000MED=-MDE=-1.5 tmMEC=Pa2..b=43.024.0=2.939 tmL27.02MCE=-Pa.b2=-43.04.02=-3.918 tmL27.02MEF=P.L=2.02.0=4.0

0.6670.6671.51.540.8932.9392.3803.918Derajat ketidak tentuan kinematis : 2D1D2

Gaya eqivalen Q yang koresponding dengan lendutanQ1=-0.059Q2=5.439

Q1=1.500-0.667-0.893=-0.059Q2=4.000-1.500+2.939=5.439

d2D1d5d3Di berikan D1 = 1 satuanD2d6d7Di berikan D2 = 1 satuanDiagram H-D

H1d2H2H5d6H6d1H3d5H7d3d7d4d8H4H800d10010d210A=10d31000d40010d51001d60101d70100d800D1D2

4EI1/L12EI1/L10000002EI1/L14EI1/L1000000004EI2/L22EI2/L20000[S] =002EI2/L24EI2/L2000000004EI3/L32EI3/L30000002EI3/L34EI3/L3000000004EI4/L42EI3/L30000002EI4/L44EI4/L4

2100000012000000001.0910.5450000[S] =000.5451.091000000001.3330.6670000000.6671.333000000001.1430.5710000000.5711.143

[K] = [A]T [S] [A]

[A]T[S][A][K]=011010002100000000011010002.0001.000000000000001101200000010000001101.0002.000000000001.0910.545000010001.0910.5450000000.5451.091000000000.5451.091000000001.3330.667001000001.3330.6670000000.6671.333000100000.6671.333000000001.1430.571010000001.140.570000000.5711.143000000000.571.141.00021.09090909090.54545454551.33333333330.666666666700004.42424242420.6666666667[K]=EI4.4240.66700000.66666666671.33333333331.14285714290.5714285714100.66666666672.47619047620.6672.4761000[K]-1=0.236-0.0631/EI100.2355848435-0.0634266886-0.0630.42101-0.06342668860.4209225701[D]= [K]-1 [Q]00

[D]=0.236-0.063-0.059-0.06-0.0630.4215.4395.44

D1=-0.359-0.3589168931D2=2.2932.2930600426[H] = [S] [A] [D][S][A][D]

2100000000-0.3591012000000102.29320001.0910.5450000101.09090909090[H]=000.5451.0910000000.5454545455000001.3330.66700101.33333333330.666666666700000.6671.33300010.66666666671.33333333330000001.1430.5710101.14285714290000000.5711.1430000.5714285714H1=-0.359-0.3589168931H2=-0.718-0.7178337862H3=-0.392-0.3915457016H4=-0.196-0.1957728508H5=1.0501.0501508376H6=2.8182.8181354615H7=2.6212.6206400487H8=1.3101.3103200244Momen akhir = H - Momen primer

MAD =-0.359-0.667=-1.026 tmMDA =-0.718--0.667=-0.051 tmMDB =-0.392--0.893=0.501 tmMBD =-0.196-2.380=-2.576 tmMDE =1.050-1.500=-0.450 tmMED =2.818--1.500=4.318 tmMEC =2.621-2.939=-0.318 tmMCE =1.310--3.918=5.229 tmMEF =0-4=-4.0Syarat Momen dititik Hubung = 0

TITIK DTITIK EMDA + MDB + MDE = 0MED + MEC + MEF = 0-0.051+0.501+-0.450=04.318+-0.318+-4=00.000=0OK!!!0.000=0OK!!!

q=2q=2P32.0MAD=1.026MAD=0.051MDE=0.450MED=4.318MEF=4HADHDAHDEHEDADDEEVAD2VDAVDE3VEDVEF2

HDBHEC

DMDB=0.501VECEMEC=0.318P2=3VDB4P1=44

B2C3VBDMBD=2.576VCEMCE=5.229

HBDHCE

VDA=-0.051+4-1.026VED=4.318+9-0.450VEF=2.0()23=1.4616246603()=4.2894287664()

VAD=4-1.4616246603VDE=6-4.2894287664=2.5383753397()=1.7105712336()

HDB=3.172195894()HEC=6.289()HBD=3.172195894()HCE=6.289()

VDB=0.501+4.5-2.576VCE=-0.318+16+5.2295.57=0.441()=2.987()VBD=3-0.4409232985VEC=4-2.9872217014=2.559()=1.013()HED=1.013()HDA=0.572()HDE=1.013()HAD=0.572()RVA=2.538()RHA=0.572()MA=1.026RVB=3.172()RHB=2.559()MB=2.576RVC=6.289()RHC=2.987()MC=5.229

Dimana,Q=q(L1+L2)=10.0KontrolV = 0 H = 0 RVA+RVB+RVC-Q-P3=0RHA+RHB+RHC+P2-P1=02.538+3.172+6.289-10-2=00.572-2.559+2.9872217014+3-4=00.000=0OK!!!0.000=0OK!!!MF = 0(Tinjau keseluruhan struktur)+

RVA (L1+L2+L3)+RVB (L2+L3)+RC (L3)+RHB (H2+(1/2H1))-RHC(H2+H1)-P2(H2)+P1(H2)-Q ((1/2 L1+L2)+L3)-MA-MB+MC=02.5387.0+3.1725.0+6.2892.0+2.5595.5-2.9877.0-34.0+44.0-104.5-1.026-2.576+5.229=017.769+15.861+12.579+14.075-20.911-12+16-45-1.026-2.576+5.229=00.000=0(Oke!!!)

Persamaan gaya - gaya dalam

Tinjau Bentang B - D, interval0x 1.5dan x 5.5Bidang MomenBidang LintangBidang NormalB'Mx=- MBD + VBD xDx=VBDNx=-HBDx=-2.576+2.559 x=2.559 t=-3.172 t

Bu/x=0Mx=-2.576 tmVBDu/x=1.5Mx=1.263 tmMBD=2.576

HBD

DBidang MomenBidang Lintang1.5( x -Mx=- MBD + VBD x-P ( x -1.5Dx=VBD-P=1.924-0.441 x=2.559-3P=3=-0.441 t1.5u/x=1.5Mx=1.263 tmu/x=5.5Mx=-0.501 tmBidang NormalBVBDNx=-HBDMBD=2.576=-3.172 t

HBD

Tinjau Bentang C - E, interval0x 3dan x 7

C'Bidang MomenBidang LintangMx=-MCE + VCE xDx=-VCEx=-5.229+2.987 x=-2.987 t

Cu/x=0Mx=-5.229 tmBidang NormalVCEu/x=3Mx=3.733 tmMCE=5.229Nx=-HCE=-6.289 tHCE

EBidang MomenBidang Lintang3.0( x -Mx=-MCE + VCE x-P ( x -3.0Dx=-VCE+P=6.7713126287-1.013 x=-2.9872217014+4P=4 t=1.013 t3u/x=3Mx=3.733 tmu/x=7Mx=-0.318 tmBidang NormalCVCENx=-HCEMCE=5.229=-6.289 t

HCETinjau Bentang A - D, interval0x 2

MAD=1.026q=2HADADVAD2Bidang MomenBidang Lintang

Mx=-MAD + VAD x-1/2 qx2Dx=VDA-qx=-1.026+2.538 x-1x2=2.538-2.00 x

u/x=0Mx=-1.026 tmu/x=0Mx=2.538 tu/x=2Mx=0.051 tmu/x=2.00Mx=-1.462 t

Momen MaxBidang Normal

dM=2.538-2 x=0Nx=-HADdx=-0.572 tx=1.269 mMomen max terjadi pada jarak x =1.269 mu/x=1.269Mx=0.585 tmTinjau Bentang D - E, interval0x 3

MDE =0.450q=2Bidang LintangHDEDEDx=VDE-qxVDE3=1.711-2.00 xu/x=0Mx=1.711 tBidang Momenu/x=3.00Mx=-4.289 t

Mx=-MDE + VDE x-1/2 qx2=-0.450+1.711 x-1x2Bidang Normal

u/x=0Mx=-0.450 tmNx=-HDEu/x=3.00Mx=-4.318 tm=-1.013 t

Momen MaxdM=1.711-2 x=0Momen max terjadi pada jarak x =0.855 mdxu/x=0.855Mx=0.282 tmx=0.855 mTinjau Bentang A - D, interval0x 2Bidang LintangP3 =2 tBidang MomenDx=PMx=- Px=2.0x=-2 xu/x=0Mx=0.0tmu/x=2.00Mx=-4.0tm

Gambar Bidang Momen, Lintang dan Normal4.0

1.5

1.5

2.03.02.04.3184

1.0260.4500.4500.051D0.0000.50100.5010.0510.3180.5850.282

4.0001.2633.7334.318E0.0000.318

2.5765.229Diagram bidang momen

2.53821.7111.4620.4411.013

4.289

2.559

2.987

Diagram bidang Lintang

0.5721.0133.1726.289

Diagram bidang Normal

No 1b qcPde

aba=6.0c=2.0P=2.0b=4.0d=3.0q=1.0e=1.0

Struktur yang akan dianalisa

q =1.0

bD2.0P =2.0C3.0

aB1.0A6.04.0

Akibat translasiAkibat rotasiT =2 joint - ( 2 jepit + sendi + rol + batang )R =joint - jepit =2x4-(2x2+0+0+3)=4-2 =1translasi=2rotasi

Analisa sudut dan panjang batang

tan a =5=1.25CD=2=2=6.325 m4Sin bsin18.43a=51.3469.775

tan b =2=0.3333333333BD=5.0=5=6.403 m6Sin asin51.34b=18.43

Menghitung Momen Primerq =1.0DDMCD=1ql2=11.06.02=3.000 tmC1212MDC=-MBA=-3.000 tm6.0c.Gaya luar ekivalen di titik diskrit yang koresponding dengan lendutanD3D1D2Selanjutnya Q1 dicari dengan cara sbb :-bentang DEq =1.00VDC=VCD=1/2 (q.l)=3

DDC

6.00 m

Distribusi gaya di titik DDik:a=51.34VDC=3PDBb=18.43g=71.57VDCHDaPDB=RDC cos gPDC=RDC sin gb=2.846=0.949gPDCHD=2.098

Sehingga diperoleh matriks {Q} :jadi;Q1 = P- HD =-0.098ton-0.098Q13.000Q2-3.000Q3

Diberikan D1 =1satuand4d5

d3d2d6d1Tinjau segitiga C - C' - C''

C''C'C'' =D( sin( 18.43 ))b=0.316CC'D=1

Tinjau segitiga D - D' - D''g=90-(b + a)DD''=DD'/( COS( 20.22 ))D''=20.22=1.011gD'DD'=D( COS( 18.43 ))D'D''=DD'( Tan( 20.22 ))b=0.949=0.350DD =1

0.3500.316=0.350x6.324555320336758- x)0.3162.00-0.316 x=0.350 xx=2.00=3.004x6.324555320336758- x)0.666

ab

d1 =1=0.25d5 =1.011=0.160.25d1 ( e + d )6.403 m0.25d2d2 =d1 =0.25d6=d11=0.16-0.11d3d3=-C'C''=-0.11-0.11d4a0.16d5d3=-D'D''=-0.110.16d6b

D1

Diberikan D2 =1satuan0d1 d31d21d30d40d5d20d6

D2Diberikan D3 =1satuan0d1 d50d20d3d41d41d50d6

D2Diagram H-Dd4H5H4H3d3d5H2d6d2H6d1H1

Matriks akibat Deformasi [A]

0.2500d10.250.000.000.2510d20.251.000.00A=-0.1110d3-0.11.000.00-0.1101d4-0.10.001.000.1601d50.160.001.000.1600d60.160.000.00Matriks kekakuan bahan [S]

4EI1/L12EI1/L100002EI1/L14EI1/L10000S=004EI2/L22EI2/L200002EI2/L24EI2/L20000004EI3/L32EI3/L300002EI3/L34EI3/L310.500000.250.25-0.1052631579-0.10526315790.15789473680.157894736810.500000.5100000110000.510000=000.6324555320.31622776600000110000.6324555320.31622776600000.3162277660.63245553200000.3162277660.6324555320000000.62469504760.312347523800000.62469504760.312347523800000.31234752380.624695047600000.31234752380.6246950476Matriks kekakuan struktur [K] = [A]T[S][A]

[A]T[S][A]

0.250.25-0.1052631579-0.10526315790.15789473680.157894736810.500000.25000.3750.375-0.0998613998-0.09986139980.14795409020.1479540902[K] =0110000.5100000.25100.510.6324555320.31622776600000110000.6324555320.31622776600-0.1110000.3162277660.6324555320.62469504760.3123475238000.3162277660.63245553200-0.10526315790100000.62469504760.31234752380.15789473680100000.31234752380.62469504760.157894736800

0.25524579690.27513860020.0480926904[K] =0.27513860021.6324555320.3162277660.25524579690.27513860020.04809269040.04809269040.3162277661.25715057960.27513860021.6324555320.3162277660.04809269040.3162277661.2571505796Matriks Lendutan [D] = [K]-1 {Q}4.7881136402-0.81103910410.0208406356-0.098-2.974.7881136402-0.81103910410.0208406356-0.10-2.9666598058[D] =-0.81103910410.7813307037-0.16551192343.000=2.92-0.81103910410.7813307037-0.16551192343.0002.92031214350.0208406356-0.16551192340.8362858043-3.000-3.010.0208406356-0.16551192340.8362858043-3.000-3.0074433365Matriks gaya dalam [H][H] = [S] [A] [D][S][A][D]10.500000.2500-2.970.3480.3750.500.34765864460.5100000.25102.921.8080.375101.8078147164[H] =000.6324555320.31622776600-0.105263157910-3.01=1.192-0.09986139980.6324555320.3162277661.1921852836000.3162277660.63245553200-0.101-0.682-0.09986139980.3162277660.632455532-0.682335589300000.62469504760.31234752380.157894736801-2.3180.147954090200.6246950476-2.317664410700000.31234752380.62469504760.157894736800-1.3780.147954090200.3123475238-1.3782969316Momen akhir = H - Momen primer

MAC=0.348-0=0.348MCA =1.808-0=1.808MCD=1.192-3.000=-1.808MDC=-0.682--3.000=2.318MDB =-2.318-0=-2.318MBD=-1.378-0=-1.378Syarat Momen dititik Hubung = 0

TITIK CTITIK DMCA + MCD = 0MDC + MDB = 01.808+-1.808=02.318+-2.318=00.000=0OK.0.000=0OK!!!

Menghitung Reaksi perletakan

Tinjau bentang CDDimana ;VCAMCA=1.808t/mMAC=0.348t/mMCACHCAHCA=MCA+MAC4.04.0=0.539 t[]

HACAHAC=MCA+MACMAC4=0.539 t[]VAC

Tinjau bentang CDq =1Dimana ;MDCP-HCA=2.539 tMCDQ=6.00tDMCD=1.808t/mCvDCMDC=2.318t/mP-HCA6.00VCD=3.761tOK!!!VCDVDC=2.239t

VCD=--MCD+MDC-Q3.0-(P-HCA)2VDC=Q-VCD6=2.239[ ]=3.761 t[ ]

Tinjau bentang BD

VDBVDB = VDC = 2.239 t[ ]VBD = VDB = 2.239 t[ ]HDBDMDBMBD = 1.378MDB = 2.3185.0

HBD=MDB+MBD+VDB4.0MBDBHBD5=2.530 t[]4.0VBDHDB=MDB+MBD+VBD4.05=2.530 t[]

Menghitung Gaya-gaya dalam

Tinjau Bentang A - C, interval0x4.0

Bidang MomenBidang LintangBidang NormalCMx=MAC - HAC xDx=-HACNx=-VACx=0.348-0.539 x=-0.539 t=-3.761 t

AHACu/x=0Mx=0.348tmMACu/x=4.0Mx=-1.808tm

VACTinjau Bentang F - E, interval0x 4.00

Dimana,a=51.340Sin a=0.78086880941Dcos a=0.6246950476Tan a=1.25Bidang MomenxMx =VBD x-HBD tan a x+MBF=2.239 x-3.163 x+1.378aBHBD=-0.924 x-1.378MBDu/x=0Mx=1.378tmVBDu/x=4.00Mx=-2.318tm

Bidang LintangBidang NormalDx =VBD cos a -HBD sin aNx =- HBF cos a-VBF sin a=-0.577=-3.329

Tinjau Bentang C - D, interval0x 6.00q =1Dimana,a=18.435Sin a=0.316227766cos a=0.9486832981MCDDHCDb

VCDx

Bidang Momen

Mx=-MCD + VCD x - HCD tan b x-1/2 q x2=-1.808+3.761 x-0.846 x-0.500 x2=-1.808+2.915 x-0.500 x2

u/x=0Mx=-1.808tmu/x=6.00Mx=-2.318tm

Momen Max

dM=2.915-1.000 x=0dxx=2.915Momen max terjadi pada jarak x =2.915u/x=2.915Mx=4.908tm

Bidang Lintang

Dx=VCD cos b-qx cos b+HCD sin b =3.568-0.95 x+0.803

u/x=0Mx=4.371u/x=6.00Mx=-1.321

Bidang Normal

Dx=- VCD sin b+qx sin b-HCD cos b =-1.189+0.32 x-2.409

u/x=0Mx=-3.598u/x=6.00Mx=-1.701

Gambar Diagram Momen, Lintang dan Normal

q=1.0

2.0

P=2.0

3.0

1.0

6.04.0

2.3182.318

1.808

1.808

4.908

1.3780.348

Diagram bidang momen

4.371

1.321

0.577

0.539


1.701

3.5983.3293.761


Gambar Diagram Momen, Lintang dan Normal

q=1.0

2.0

P=2.0

3.0

1.0

6.04.0

2.3182.318

1.808

1.808

4.908

1.378

0.348

Diagram bidang momen

4.371

1.321

0.577

0.539


1.701

3.5983.3293.761


No 2a.

Konstruksi yang akan dianalisaq=2.00t/m

ADEIEF4.004.00bbP2=3P1=4.001.5EI2EI1.50aa3.00B

C

232

2 a.Kontol dengan Menggunakan metode Fleksibilitas

Dengan cara yang sama diperoleh gaya luar (Q) yang koresponding dengan lendutanQ1=-0.059tmQ2=5.439tm

Mengubah struktur statis menjadi statis tertentu dengan memberikan gaya rendundat

Diberikan R1 = 1 satuan5

2

5

R1

5

Diberikan R2 = 1 satuan

R2

5


R31

1


3

3R4


5.505.5

5.505.5

R51.50


11

11

1R61

Gaya luar yang koresponding dengan lendutan

Diberikan D1 = 1 satuan11

D11

1

Diberikan D2 = 1 satuanD2

1

1

Matriks statis akibat gaya redundat [P']Matriks statis tertentu akibat gaya Q [P]Matriks [P']Matriks [P]

0010000000100000-20-100000-20-10000000005.50-10000005.5-100P'=000001P=00000001002010-5.501102010-5.5110-50-1-35.5-1-10-50-1-35.5-1-105013-5.51115013-5.5111-5-5-1-3-1.50-1-1-1-5-5-1-3-1.5-1-1-1

Matriks bahan [M]

0.67-0.33000000-0.330.67000000L-L001.22-0.6100003EI6EI=00-0.611.22000000001.00-0.5000-LL0000-0.501.00006EI3EI0000001.17-0.58000000-0.581.17

0-2002-55-50.00-2.000.000.002-55.00-5.00[P']T=0000000-5000.000.000.000.000.00-5.001-1001-11-11-10.000.001.00-1.001.00-1.0000000-33-3000.000.000.00-3.003.00-3.00005.50-5.55.5-5.5-1.5005.500.00-5.505.50-5.50-1.5000-111-11-100-1.001.001.00-1.001.00-1.00

Matriks fleksibilitas akibatgaya luar [Fo] = [P']T[M][Po]0.67-0.330.000.000.000.000.000.001-1005-69-8.752817.5[F]=2817.5-0.330.670.000.000.000.000.000.000000003-5.83333333338.758.758.758.750.000.001.22-0.610.000.000.000.001-1002-22-1.756.53.56.53.50.000.00-0.611.220.000.000.000.0000002-35-5.251510.51510.50.000.000.000.001.00-0.500.000.00007-3-88-61.4583333333-23.5-7-23.5-70.000.000.000.00-0.501.000.000.0000-222-22-1.756.53.56.53.50.000.000.000.000.000.001.17-0.580.000.000.000.000.000.00-0.581.17Matriks fleksibilitas akibat bekerjanya gaya redundat [F'] = [P']T[M][P']0.67-0.330.000.000.000.000.000.001-1005-69-8.75129.166666666743.753070.5-92.75280.9699341021-0.1145724128-1.1466227348-1.1420922570.10513703760.1927512356[F']=129.166666666743.753070.5-92.7528[F']-1=0.9699341021-0.1145724128-1.1466227348-1.1420922570.10513703760.1927512356-0.330.670.000.000.000.000.000.000000003-5.833333333343.7529.16666666678.7526.25-7.29166666678.75-0.11457241280.27695687980.0763816085-0.0149468324-0.1287399572-0.374866230143.7529.16666666678.7526.25-7.29166666678.75-0.11457241280.27695687980.0763816085-0.0149468324-0.1287399572-0.37486623010.000.001.22-0.610.000.000.000.001-1002-22-1.75308.758.515-23.56.5-1.14662273480.07638160851.76441515651.261394838-0.0700913584-0.1285008237308.758.515-23.56.5-1.14662273480.07638160851.76441515651.261394838-0.0700913584-0.12850082370.000.00-0.611.220.000.000.000.0000002-35-5.2570.526.251540.5-45.7515-1.142092257-0.01494683241.2613948381.5232930624-0.0709899656-0.130148270270.526.251540.5-45.7515-1.142092257-0.01494683241.2613948381.5232930624-0.0709899656-0.13014827020.000.000.000.001.00-0.500.000.00007-3-88-61.4583333333-92.75-7.2916666667-23.5-45.75156.0138888889-33.58333333330.1051370376-0.1287399572-0.0700913584-0.07098996560.0873357040.2592890111-92.75-7-23.5-45.75156.0138888889-33.58333333330.1051370376-0.1287399572-0.0700913584-0.07098996560.0873357040.25928901110.000.000.000.00-0.501.000.000.0000-222-22-1.75288.756.515-33.583333333310.16666666670.1927512356-0.3748662301-0.1285008237-0.13014827020.25928901111.0208177325288.756.515-33.583333333310.16666666670.1927512356-0.3748662301-0.1285008237-0.13014827020.25928901111.02081773250.000.000.000.000.000.001.17-0.580.000.000.000.000.000.00-0.581.17Matriks statis struktur [P] =[P] -[P'][F']-1[F]

[P][P'][F']-1[F][P]

00-0.23558484350.06342668860.24-0.060.000.001.000.000.000.00-1021-0-0-0.23558484350.06342668860.24-0.0600-0.4711696870.12685337730.47-0.13-2.000.00-1.000.000.000.00-1011-0-0-0.4711696870.12685337730.47-0.1300-0.25700164740.06919275120.26-0.070.000.000.000.005.50-1.000-0-0-000-0.25700164740.06919275120.26-0.07[P]=00--0.12850082370.0345963756=0.13-0.030.000.000.000.000.001.000-0-0-001-0.12850082370.03459637560.13-0.03100.7281713344-0.19604612850.270.202.000.001.000.00-5.501.0000-0-1-0-00.7281713344-0.19604612850.270.20-10-1.0724876442-0.51894563430.070.52-5.000.00-1.00-3.005.50-1.0000-0-0-0-0-1.0724876442-0.51894563430.070.52111.07248764420.5189456343-0.070.485.000.001.003.00-5.501.00-0-000001.07248764420.5189456343-0.070.48-1-1-0.9637561779-1.2405271829-0.040.24-5.00-5.00-1.00-3.00-1.50-1.00-0-00000-0.9637561779-1.2405271829-0.040.24

[P]-1=0.240.470.260.130.270.07-0.07-0.040.240.470.260.130.270.07-0.07-0.04-0.06-0.13-0.07-0.030.200.520.480.24-0.06-0.13-0.07-0.030.200.520.480.24

Matriks fleksibilitas [F] = [P]T[M][P]Lendutan di titik distrik [D] = [F][Q]-00.23558484350.2355848435-00.2355848435-0.0634266886-0.063426688600.236-0.063-0-0.0634266886-0.06342668860-0.06342668860.420922570.420922570-0.0630.421[F]=0.236-0.063[D]=-0.359-0.0630.4212.293-0.0592286501-0.3589168931Gaya pada Struktur [H] = [P][Q]Momen akhir = H - Momen primer5.43877551022.2930600426

-0.359MAD =-0.359-0.667=-1.026 tmok!!!!0.24-0.06-0.3589168931-1.0255835598-0.718MDA =-0.718--0.667=-0.051 tmok!!!!0.47-0.13-0.7178337862-0.0511671195[F]=-0.392MDB =-0.392--0.893=0.501 tmok!!!!0.26-0.07-0.39154570160.5010162819-0.196MBD =-0.196-2.380=-2.576 tmok!!!!0.13-0.03-0.1957728508-2.575938141.050MDE =1.050-1.500=-0.450 tmok!!!!0.270.201.0501508376-0.44984916242.818MED =2.818--1.500=4.318 tmok!!!!0.070.522.81813546154.31813546152.621MEC =2.621-2.939=-0.318 tmok!!!!-0.070.482.6206400487-0.31813546151.310MCE =1.310--3.918=5.229 tmok!!!!-0.040.241.31032002445.2286873713MEF =0.000-4.000=-4.0ok!!!!-4Karena nilai momen akhir yang didapatkan dari metode Fleksibilatas sama dengan nilai momen akhir yang didapatkan dari metode kekakuan sama maka=terkontrol

nO. 2b.Struktur yang akan dianalisa

q =1.0

bDCD=6.32455532032.0DB=6.4031242374P =2.0C a3.0B1.0A6.04.0

2 b.Kontrol dengan menggunakan metode Fleksibel

c.Dengan cara yang sama diperoleh gaya luar (Q) yang koresponding dengan lendutanQ1=-0.098tmQ2=3.000tmQ3=-3.000tm

Mengubah struktur statis menjadi statis tertentu dengan memberikan gaya rendundat


R1

Diberikan R2 = 1 satuan64641R2

Diberikan R3 = 1 satuan111111R3

Diberikan D1 = 1 satuan22

D1

3Diberikan D2 = 1 satuan111D21Diberikan D3 = 1 satuan1

D2

1

Matriks statis akibat gaya redundat [P']Matriks statis tertentu akibat gaya Q [P]Matriks [P']Matriks [P]

00100000100004-100004-1000P'=0-410100-41010-66-1P=-2-10-66-1-2-106-612116-61211-101-13-1-1-101-13-1-1

Matriks bahan [M]

1.33-0.670000L-L-0.671.3300003EI6EI002.11-1.0500=00-1.052.1100-LL00002.13-1.076EI3EI0000-1.072.13

000-66-100.000.000.00-6.006-10[P']T=04-46-6104-4.006.00-6.001.001-11-11-11-11.00-1.001.00-1.00

Matriks fleksibilitas akibat gaya luar [Fo] = [P']T[M][Po]1.33-0.670.000.000.000.00006-1323-28-10.986149397470.198659860551.2249938995[F]=-1170.198659860551-0.671.330.000.000.000.00-35-1517-149-35.8653364543-54.0337114327-22.410934831-35.9-54.0337114327-220.000.002.11-1.050.000.002-23-33-33.122993201612.72767955786.40312423743.112.72767955786.40312423740.000.00-1.052.110.000.000.000.000.000.002.13-1.070.000.000.000.00-1.072.13Matriks fleksibilitas akibat bekerjanya gaya redundat [F'] = [P']T[M][P']494.2-269.8170.20.010.00-0.02[F']=494.2-27070.2[F']-1=0.010.00-0.02-269.81273.3-62.00.000.020.08-270273.3-62.00.000.020.0870.2-62.016.73-0.020.080.4470.2-62.016.7-0.020.080.44

Matriks statis struktur [P] =[P] -[P'][F']-1[F]

[P][P'][F']-1[F][P]

000-1.39-0.090.071.390.09-0.07-0.020.080.44-1.39-0.090.071.390.09-0.071.390.09000-0.98-0.480.160.980.48-0.160.020.01-0.11-0.98-0.480.160.980.48-0.160.980.48[P]=010-0.980.48-0.16=-0.980.520.16-0.02-0.010.110.980.48-0.16-0.980.520.16-0.980.52-2-10-1.28-1.22-0.47-0.720.220.47-0.010.050.16-1.28-1.22-0.47-0.720.220.47-0.720.222111.281.220.470.72-0.220.530.01-0.05-0.161.281.220.470.72-0.220.530.72-0.223-1-12.29-0.83-1.260.71-0.170.26-0.03-0.07-0.182.29-0.83-1.260.71-0.170.260.71-0.17

[P]-1=1.390.98-0.98-0.720.720.711.390.98-0.98-0.720.720.710.090.480.520.22-0.22-0.170.090.480.520.22-0.22-0.17-0.07-0.160.160.470.530.26-0.07-0.160.160.470.530.26

Matriks fleksibilitas [F] = [P]T[M][P]Lendutan di titik distrik [D] = [F][Q]1.19702841010.3859893059-1.3150510663-0.48317132660.77685857880.75601794324.79-0.810.02-0.10-2.9666598058-0.2027597760.57857092770.866703241-0.0801393861-0.2935707293-0.1280588059-0.810.78-0.173.002.92[F]=4.788-0.8110.021[D]=-2.9670.0052101589-0.1603017645-0.16770567450.83409205320.83957643090.00329062670.02-0.170.84-3.00-3.01-0.8110.781-0.1662.9200.021-0.1660.836-3.007

Gaya pada Struktur [H] = [P][Q]Momen akhir = H - Momen primer

0.348MAC=0.348-0.000=0.348ok!!!!0.3480.34765864461.808MCA =1.808-0.000=1.808ok!!!!1.8081.8078147164[F]=1.192MCD=1.192-3.000=-1.808ok!!!!-1.8081.1921852836-0.682MDC=-0.682--3.000=2.318ok!!!!2.318-0.6823355893-2.318MDB =-2.318-0.000=-2.318ok!!!!-2.318-2.3176644107-1.378MBD=-1.378-0.000=-1.378ok!!!!-1.378-1.3782969316

Karena nilai momen akhir yang didapatkan dari metode Fleksibilatas sama dengan nilai momen akhir yang didapatkan dari metode kekakuan sama maka=terkontrol

NO. 3P2P3P1

t

abba

abc

Data - dataa =3.0P1 =4.0b=6.0P2 =3.0c=3.0P3 =2.0t=4.0

Struktur yang akan dianalisaP2 =3.0P3 =2.0

P1 =4.0

4.0

AabbaB

3.06.03.0

Analisa Suduttan a =4.00a =53.13Sin a=0.83.00Cos a=0.6tan b =4.00b =33.69Sin b=0.5556.00Cos b=0.832

Mencari gaya-gaya dalam

VA=-P14.0-P29.0-P33.0VB=P14.0+P23.0+P39.01212VA=-44.0-39.0-23.0VB=44.0-33.0-29.01212=1.417 t()=3.583 t()

Kontrol SVHA=4.0()VA+VB-P2-P3=01.417+3.583-3-2=00.000=0OK!!!Drajat Ketidaktentuan Kinematis = 9D1D5D2D6EFABD9CDD4D8D3D7

b.Vektor gaya luar dititik diskrit yang koresponding dengan lendutan

Q1=3.0Q5=2.0

Q2=4.0

c.Diberikan D1 = 1 satuanD1

d1 =sin ad3 =sin 90d4=sin 0d5 =sin bd1 =0.8d3 =1d4 =0d5 =0.555

d.Diberikan D2 = 1 satuanD2

d1 =cos ad3 =cos90d4=-cos0d5 =-cos bd1 =0.6d3 =0d4 =-1d5 =-0.832

e.Diberikan D3 = 1 satuan

D3

d6 =-sin bd2=sin 0d3=-sin 90d10=sin 0d6=-0.555d2=0d3=-1d10=0

f.Diberikan D4 = 1 satuan

D4

d6 =-cos bd2=cos0d3=cos 90d10=-cos0d6=-0.832d2=1d3=0d10=-1

g.Diberikan D5 = 1 satuanD5

d4=sin0d6 =Sin bd7=sin90d9=sin ad4 =0d6 =0.555d7=1d9=0.8

h.Diberikan D6 = 1 satuanD6

d4=cos0d6 =cos bd7=cos90d9=-cos ad4 =1d6 =0.832d7=0d9=-0.6

i.Diberikan D7 = 1 satuan

D7

d5 =-Sin bd7=-sin90d8=sin 0d10=sin 0d5 =-0.555d7=-1d8=0d10=0

j.Diberikan D8 = 1 satuan

D8

d5 =cos bd7=cos90d8=-cos0d10=cos0d5 =0.832d7=0d8=-1d10=1

k.Diberikan D9 = 1 satuan

D9d8=cos0d9=cos ad8=1d9=0.6

Perhitungan matriks[A]

0.8000.6000000000d10.8000.6000.0000.0000.0000.0000.0000.0000.000000100000d20.0000.0000.0001.0000.0000.0000.0000.0000.00010-1000000d31.0000.000-1.0000.0000.0000.0000.0000.0000.0000-10001000d40.000-1.0000.0000.0000.0001.0000.0000.0000.0000.555-0.8320000-0.5550.8320d50.555-0.8320.0000.0000.0000.000-0.5550.8320.00000-0.555-0.8320.5550.832000d60.0000.000-0.555-0.8320.5550.8320.0000.0000.000000010-100d70.0000.0000.0000.0001.0000.000-1.0000.0000.0000000000-11d80.0000.0000.0000.0000.0000.0000.000-1.0001.00000000.800-0.600000.600d90.0000.0000.0000.0000.800-0.6000.0000.0000.600000-100010d100.0000.0000.000-1.0000.0000.0000.0001.0000.000D1D2D3D4D5D6D7D8D9

Dari gambar struktur dapat diperoleh matriks kekakuan axial [S] :

0.20000000000.200000000000.3330000000000.333333333300000000000.250000000000.2500000000000.20000000000.166666666700000000000.1390000000000.138675049100000000000.1390000000000.138675049100000000000.250000000000.2500000000000.3330000000000.333333333300000000000.20000000000.200000000000.1670000000000.1666666667

Perhitungan matriks kekakuan

[K] = [A]T.[S].[A]

Matriks [A]T

0.80100.555000000.80100.5547001962000000.600-1-0.832000000.600-1-0.83205029430000000-100-0.555000000-100-0.5547001962000001000-0.832000-101000-0.8320502943000-1000000.555100.80000000.5547001962100.80000100.83200-0.60000100.832050294300-0.600000-0.5550-10000000-0.55470019620-100000000.83200-10100000.832050294300-101000000010.60000000010.60

matriks[A]T.[S]

0.1600.2500.077000000.1600.2500.0769230769000000.1200-0.167-0.115000000.1200-0.1666666667-0.11538461540000000-0.2500-0.077000000-0.2500-0.0769230769000000.333000-0.115000-0.16700.3333333333000-0.1153846154000-0.1666666667000000.0770.25000.1600000000.07692307690.2500.1600000.16700.11500-0.12000000.166666666700.115384615400-0.1200000-0.0770-0.2500000000-0.07692307690-0.2500000000.11500-0.33300.16700000.115384615400-0.333333333300.166666666700000000.3330.120000000000.33333333330.120

matriks[K]

0.4210.032-0.25000-0.0430.06400.42066924590.0319961312-0.25000-0.04266924590.064003868800.0320.335-000-0.1670.064-0.09600.03199613120.3346724699-000-0.16666666670.0640038688-0.09600580320-0.25-00.2930.064-0.043-0.064000-0.25-00.29266924590.0640038688-0.0426692459-0.0640038688000000.0640.596-0.064-0.0960-0.1670000.06400386880.5960058032-0.0640038688-0.09600580320-0.1666666667000-0.043-0.0640.421-0.032-0.25000.09600-0.0426692459-0.06400386880.4206692459-0.0319961312-0.2500.0960-0.167-0.064-0.096-0.0320.335-00-0.0720-0.1666666667-0.0640038688-0.0960058032-0.03199613120.3346724699-00-0.072-0.0430.06400-0.250-00.293-0.0640-0.04266924590.064003868800-0.25-00.2926692459-0.064003868800.064-0.0960-0.16700-0.0640.596-0.3330.0640038688-0.09600580320-0.166666666700-0.06400386880.5960058032-0.333333333300000.096-0.0720-0.3330.40500000.096-0.0720-0.33333333330.4053333333

Perhitungan lendutan[D] = [K]-1.{Q}

[K]-1[Q]8.4823776028-3.49733680387.9823776028-1.68754.3926223972-1.24733680384.8926223972-3.9375-4.5-3.49733680388.1353379606-2.83067013712.25-3.25266319625.1353379606-3.91932986295.2567.982-2.83111.287-1.6884.893-0.1415.197-3.498-4.060-37.9823776028-2.830670137111.2869509924-1.68754.8926223972-0.14096026375.1971957868-3.4977901266-4.0602901266-1.6882.25-1.6883-0.5632.25-0.563334-1.68752.25-1.68753-0.56252.25-0.5625334.393-3.2534.893-0.5638.482-1.0037.982-2.813-4.504.3926223972-3.25266319624.8926223972-0.56258.4823776028-1.00266319627.9823776028-2.8125-4.5[D] = -1.2475.135-0.1412.25-1.0037.146-1.2304.2615.0110-1.24733680385.1353379606-0.14096026372.25-1.00266319627.1459907455-1.22961998954.26065278495.01065278494.893-3.9195.197-0.5637.982-1.23011.287-2.37-4.060-24.8926223972-3.91932986295.1971957868-0.56257.9823776028-1.229619989511.2869509924-2.3727901266-4.0602901266-3.9385.250-3.4983-2.8134.261-2.3738.0118.0110-3.93755.25-3.49779012663-2.81254.2606527849-2.37279012668.01065278498.0106527849-4.56-4.0603-4.55.011-4.0608.01111.0110-4.56-4.06029012663-4.55.0106527849-4.06029012668.010652784911.010652784900000000000000000000

-48.222D1-48.221724817949.539D249.5386886461[D] = -45.055D3-45.055058151215.188D415.1875-43.153D5-43.153275182126.289D626.2886886461-46.32D7-46.319941848838.438D838.437546.5D946.5

Perhitungan gaya Batang[H] = [S].[A].{D}

Matriks[S].[A]

0.1600.12000000000.160.1200000000000.333000000000.3333333333000000.2500-0.2500000000.250-0.250000000-0.20000.20000-0.16666666670000.16666666670000.077-0.11538461540000-0.0770.11500.0769230769-0.11538461540000-0.07692307690.1153846154000-0.077-0.1150.0770.11500000-0.0769230769-0.11538461540.07692307690.115384615400000000.2500-0.2500000000.250-0.25000000000-0.3330.3330000000-0.33333333330.333333333300000.160-0.120000.12000000.16-0.12000.12000-0.16666666670000.20000-0.16666666670000.16666666670

-1.771H1-1.77083333335.063H25.0625-0.792H3-0.7916666667[H] = -3.875H4-3.875-1.427H5-1.42719737991.427H61.42719737990.792H70.79166666672.688H82.6875-4.479H9-4.47916666673.875H103.875

Kontrol gaya-gaya batang

Titik ASV =0H1VA+H1 Sin a=01.417+-1.417=00.000=0OK!!!HAaH2SH =0H1 cos a+H2-HA=0-1.063+5.063-4=0VA0.000=0(Oke!!!)

Titik BSV =0VB+H9 Sin a=0H93.583+-3.583=00.000=0(Oke!!!)H8aSH =0-H9 cos a-H8=0-2.688-2.688=00.000=0(Oke!!!)VBDiagram perpindahan batang

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