jimmy

NO 1 SLOPE nP1q=2.5+10P2

EB40CF60

40404m4040

AD

2m6m2m

Jika material yang digunakan struktur tersebut adalah beton bertulang, hitung dan gambar Bidang Momen, Bidang Gaya Lintang dan Bidang Gaya Normal konstruksi diatas (Selesaikan dengan Metode Slope Deflection dan Metode Cross (MINIMAL 4 SIKLUS)) FREE BODY DAN BIDANG M, D DAN N SATU SAJAAMETODE SLOPE DEFLECTIONSDIMENSI PENAMPANGBalokBC;b=40cm=0.4mh=60cm=0.6m

kolomAB;b=40cm=0.4mh=40cm=0.4m

kolom CD;b=40cm=0.4mh=40cm=0.4mINERSIA PENAMPANGBalokBC;Ix=1*b*h312=0.0833333333*0.4*0.63=0.0072m4kolomAB;Ix=1*b*h312=0.0833333333*0.4*0.43=0.00213m4kolomCD;Ix=1*b*h312=0.0833333333*0.4*0.43=0.0021333333m4PERBANDINGAN INERSIAEbeton=2000000t/m2BalokBC;CB=0.00720m4=14400=3.375EIkolomAB;BA=0.0021333333m4=4266.6666666667=1EIkolomCD;DC=0.0021333333m4=4266.6666666667=1EIPEMBEBANAN DAN JARAKn=9JARAKLBE;LEB=2mBEBANLBC;LCB=6mq=3.4t/m'LBA;LAB=4mP1=2tLCD;LDC=4mP2=2tLCF;LFC=2mMOMEN PRIMERq=3.4t/m'2t2tMFDCMFCBMFCF

EB3.375EICFMFBEMFCDMFBA4m1EI1EI

MFABMFDCAD

2m6m2m

JOINT AMFAB=0( Tidak ada beban )JOINT BMFBA=0( Tidak ada beban )MFBE=+P1*LBE=+2*2=+4tmMFBC=-1*qCD*LBC212=-0.08*3.4*62=-10.20tmJOINTCMFCD=0( Tidak ada beban )MFCF=-P2*LCF=-2*2=-4tmMFCB=10.2tm

jepit d

MFDC=0( Tidak ada beban )

MOMEN AKHIRq=3.4022MbCMCBMCF

EB3.38EICFMBEMCDMBA4m1EI1EI

MABMDCAD

2m6m2m

MOMEN AKHIR DALAM DEFORMASI STRUKTUR

A=0(PELETAKAN JEPIT)D=0(PELETAKAN JEPIT)JEPIT A

MAB=MFAB+EI(B)LAB

=0+1(B)4

=0.25B

JOINT B

MBE=4Tm

MBA=MFBA+EI(2B)LAB

=0+1(2B)4

=0.5B

MBC=MFBC+EI(2B+C)LBC

=-10.20+3.375(2B+C)6

=-10.2+0.5625(2B+C)

=-10.2+1.125B+0.5625CSYARAT BATAS

MB=0MB=MBE+MBA+MBC=(4) + (0.5B) + (-10.2+1.125B+0.5625C)=-6.20+1.625B+0.5625C=0=1.625B+0.5625C=6.20..PERSAMAAN 1JOINT C

MCF=-4Tm

MCD=MFCD+EI(2C)LCD

=0+1(2C)4

=0.5C

MCB=MFCB+EI(2C+B)LCB

=10.2+3.375(2C+B)6

=10.2+0.5625(2C+B)

=10.2+1.125C+0.5625B

SYARAT BATAS

MC=0MC=MCF+MCD+MCB

=(-4) + (0.5C) + (10.2+1.125C+0.5625B)

=6.20+1.625C+0.5625B=0

=1.625C+0.5625B=-6.20..PERSAMAAN 2

JEPIT D

MDC=MFDC+EI(C)LDC

=0+1(C)4

=0.25C

MENYELESAIKAN PERSAMAAN

1.63B+0.56C=6.20(1)0.56B+1.63C=-6.20.(2)

1.625B+0.5625C=6.20x 10.5625B+1.625C=-6.20x 2.88888888891.625B+0.5625C=6.21.625B+4.6944444444C=-17.91-B0B+-4.13C=24.11

-4.13C=24.11C=-5.84mencari B1.625B+0.5625C=6.201.625B+0.5625*-5.84=6.201.625B+-3.28=6.201.625B=6.20+3.28235294121.625B=9.48B=5.8352941176NILAI MOMEN AKHIR

JEPIT A

MAB=0.25B=0.25*5.8352941176=1.4588235294Tm

JOINT B

MBE=4Tm

MBA=0.5B=0.5*5.8352941176=2.9176470588Tm

MBC=-10.2+1.125B+0.5625C

=-10.2+(1.125*5.8352941176) + (0.5625*-5.84)

=-10.2+6.56+-3.28

=-6.92Tm

MB=0.00

MBE+MBA+MBC=0.00

4+2.92+-6.92=0.000.00=0.00

OK

JOINT C

MCF=-4Tm

MCD=0.5C=0.5*-5.84=-2.92Tm

MCB=10.2+1.125C+0.5625B

=10.2+(1.125*-5.84) + (0.5625*5.84)

=10.2+-6.56+3.28

=6.92Tm

MB=0.00

MBE+MBA+MBC=0.00

-4+-2.92+6.92=0.000.00=0.00

OK

JEPIT D

MDC=0.25C=0.25*-5.84=-1.46Tm

BENDA BEBASq=3.40t/m'2t2t6.926.924

EB3.38EICF4

BC2.922.92

4m1EI1EI

1.461.46AD

2m6m2m

MENGHITUNG REAKSI PELETAKANq=3.4BENTANG BC6.926.92

BC

6m*Qbc=20.4t/m'LQbc=3mRB=MC = 06RB-MBC+MCB-QbcxLQbc=06RB-6.92+6.9220.40x3=06RB-6.92+6.92-61.20=06RB-61.20=06RB=61.20RB=10.20RB=MC = 0

-6RC-MBC+MCB+QbcxLQbc=0-6RC-6.92+6.92+20.4x3=0-6RC-6.92+6.92+61.2=0-6RC+61.20=0-6RC=-61.20

RC=10.2TON

kontrolRB+RC-Qbc=010.2+10.2-20.4=020.4-20.4=0OKBENTANG AB

B

2.924m1EI

1.46A

RA=

4RA+MAB+MBA=04RA+1.46+2.92=04RA+4.38=04RA=-4.38RA=-1.09

RA=1.09

KONTROLRB=RB-RA=0-4RB+MAB+MBA=01.09-1.09=0-4RB+1.46+2.92=00.00=0-4RB+4.38=0-4RB=-4.38

RB=1.09

BENTANG CDC

2.92

4m1EI

1.46DRD=

4RD-MCD-MDC=04RD-2.92-1.46=04RD-4.38=04RD=4.38

RD=1.09

KONTROLRC=RD-RC=0-4RC-MCD-MDC=01.09-1.09=0-4RC-2.92-1.46=00.00=0-4RC-4.38=0-4RC=4.38RC=-1.09

RC=1.09BENDA BEBAS DENGAN REAKSI PELETAKANq=3.40t/m'2t2t6.926.924

EB3.38EICF4

12.212.2

1.091.09BC2.922.924m

1EI1EI

1.461.46AD1.091.09

no 1 crossnP1q=2.5+10P2

EB40CF60

40404m4040

AD

2m6m2m

PEMBEBANAN DAN JARAKJARAKLBE;LEB=2mBEBANLBC;LCB=6mq=3.4t/m'LBA;LAB=4mP1=2tLCD;LDC=4mP2=2tLCF;LFC=2mPERBANDINGAN INERSIAEbeton=2000000t/m2BalokBC;CB=0.00720m4=14400=3.375EIkolomAB;BA=0.0021333333m4=4266.6666666667=1EIkolomCD;DC=0.0021333333m4=4266.6666666667=1EIFAKTOR DISTRIBUSI DAN KEKAKUAN1).JEPIT AAB=02).joint BKba=4EI=4*1=1LBA4Kbc=4EI=4*3.375=2.25LBC6+kb=3.25BA=Kba=1=0.308kb3.251.00BC=Kbc=2.25=0.692kb3.253). joint CKcb=4EI=4*3.375=2.25LCB6Kcd=4EI=4*1=1LBC4+kc=3.25CB=Kcb=2.25=0.692kc3.251.00CD=Kcd=1=0.308kc3.254).JEPIT dDC=0M. PRMER SAMA DENGAN SLOPE ( TERMASUK GAMABARNAY)catat dari metode slopeMOMEN AKHIR

q=000MDCMCBMCF

EB3.38EICFMBEMCDMBA4m1EI1EI

MABMDCAD

2m6m2m

TABEL PERHITUNGAN

SIKLUSTITIK BUUL /JOINTABCDBATANGABBABCBECBCDCFDCFAKTOR DISTRIBUSIABBABCBECBCDCFDC0.0000.3080.6920.0000.6920.3080.0000.0001Momen Primer0.0000.000-10.2004.00010.2000.000-4.0000.000M ikat0.000-6.2006.2000.000M.. Distribusi0.0001.9084.292-4.292-1.9080.000

2M. induksi 0.9540.000-2.1462.1460.000-0.954M.Ikat0.954-2.1462.146-0.954M distribusi0.0000.6601.486-1.486-0.6600.000


4M. induksi 0.1140.000-0.2570.2570.000-0.114M.Ikat0.114-0.2570.257-0.114M distribusi0.0000.0790.178-0.178-0.0790.000batas catat lanjut catat tabel yang paling terakhr5M. induksi 0.0400.000-0.0890.0890.000-0.040M.Ikat0.040-0.0890.089-0.040M distribusi0.0000.0270.062-0.062-0.0270.000





m akhir1.462.92-6.924.0006.92-2.92-4.000-1.46

kontrol momen1.460.0000.000-1.46

m akhir siklus 41.402.88-6.884.006.88-2.88-4.00-1.40

m1.400.0000.000-1.40

MENGHITUNG MOMEN AKHIR ( BATAS SIKLUS 4M AKHIR A= M AKHIR SIKLUS -(ik*M)=1.40-(0.000x)=1.40-0=1.40M AKHIR B=

MBA =M AKHIR SIKLUS -(ik*M)=2.88-(0.308x0.00)=2.88-0=2.88MBC =M AKHIR SIKLUS -(ik*M)=-6.88-(0.692x0.00)=-6.88-0=-6.88MBD =M AKHIR SIKLUS -(ik*M)=4.00-(0.000x0.00)=4.00-0=4.00M AKHIR C=

MCB =M AKHIR SIKLUS -(ik*M)=6.88-(0.692x0.00)=6.88-0=6.88MCD =M AKHIR SIKLUS -(ik*M)=-2.88-(0.308x0.00)=-2.88-0=-2.88MCF =M AKHIR SIKLUS -(ik*M)=-4.00-(0.000x0.00)=-4.00-0=-4.00M AKHIR A= M AKHIR SIKLUS -(ik*M)=-1.40-(0.000x-1.40)=-1.40-0=-1.40

bidng M D N

GAMBAR BIDANG M6.924444

10.9210.92

15.304.38

2.922.92

1.461.46GAMBAR BIDANG D12.202212.201.091.090.54705882351.091.09GAMBAR BIDANG N1.091.0912.2012.2012.2012.20

no 2 slopeq=(2+n/4) T/M1.50m

P1=(4+n/4) T

A2.00EIBEIC1.50EID

6.00m4.00m5.00m

Soal :Hit. dan gambar bid. M D N, dengan metode slope deflection dan metode cross.

Dik :N=9.00P1=(4+n/4)=4.00+(9.00/4.00)=6.25tq=(4+n/4)=2.00+(9.00/4.00)=4.25t/m

Penye. :Momen Primer :

MFABq=4.25t/m1.50mMFDCMFBAMFBCMFCBMFCDP1=6.25t

A2.00EIBEIC1.50EID

6.00m4.00m5.00m

Momen Primer / Momen Jepit Ujung :

NB :Untuk cantilever dihitung langsung dari beban ventilever.

MFAB=-1/12q(LAB)2=-0.084.2536.00=-12.75tmMFBA=1/12q(LAB)2=0.084.2536.00=12.75tmMFBC=-1/12q(LBC)2=-0.084.2516.00=-5.67tmMFCB=1/12q(LBC)2=0.084.2516.00=5.67tmMFCD=-Px [ (ab2) /LBC ]=-6.25x [ (1.5012.25) /5.00]=-22.97tmMFDC=Px [ (ba2) /LBC ]=6.25x [ (3.502.25) /5.00]=9.84tm

MABq=4.25t/m1.50mMDCMBAMBCMCBMCDP1=6.25t

A2.00EIBEIC1.50EID

6.00m4.00m5.00m

Sesuai dengan formula umum :

Momen Akhir =MF+I(2.00a+b)L

MAB=MFAB+2.00(0.00a+b)6.00

=-12.75+0.33(b)

=-12.75+0.33b

MBA=MFBA+2.00(2.00b+0.00a)6.00

=12.75+0.33(2.00b)

=12.75+0.67bMBC=MFBC+1.00(2.00b+c)4.00

=-5.67+0.25(2.00b+c)

=-5.67+0.50b+0.25c

MCB=MFCB+1.00(2.00c+b)4.00

=5.67+0.25(2.00c+b)

=5.67+0.50c+0.25b

MCD=MFCD+1.50(2.00c+d)5.00

=-22.97+0.30(2.00c+0.00)

=-22.97+0.60c

MDC=MFDC+1.50(0.00d+c)5.00

=9.84+0.30(0.00+c)

=9.84+0.30cSyarat Batas :

MB =0.00

MBA+MBC=0.00

(12.75+0.67b) + (-5.67+0.50b+0.25c )=0.00

7.08+1.17b+0.25c=0.00

7.08+1.17b+0.25c=0.00

1.17b+0.25c=-7.08.. ( 1 )

MC =0.00

MCB+MCD=0.00

(5.67+0.50c+0.25b) + (-22.97+0.60c)=0.00

-17.30+0.25b+1.10c=0.00

-17.30+0.25b+1.10c=0.00

0.25b+1.10c=17.30.. ( 2 )

Mensubtitusi ke - 2 persamaan diatas :

1.17b+0.25c=-7.08.. ( 1 )

0.25b+1.10c=17.30.. ( 2 )

Mensubtitusi persamaan 1 dan 2 :

1.17b+0.25c=-7.08x1.00

0.25b+1.10c=17.30x4.67

1.17b+0.25c=-7.08

1.17b+5.13c=80.74-

0.00b+-4.88c=-87.83

-4.88c=-87.83c=17.98Memasukan nilai c ke dalam persamaan 1 :

MB =1.17b+0.25c=-7.08

1.17b+0.2517.98=-7.08

1.17b+4.50=-7.08

1.17b=-7.08-4.50

1.17b=-11.58

b=-9.93Momen Akhir :MAB=-12.75+0.33b

=-12.75+0.33-9.93

=-12.75+-3.31

=-16.06tmMBA=12.75+0.67b

=12.75+0.67-9.93

=12.75+-6.62

=6.13tmMBC=-5.67+0.50b+0.25c

=-5.67+0.50-9.93+0.2517.98

=-5.67+-4.96+4.50

=-6.13tm

MCB=5.67+0.50c+0.25b

=5.67+0.5017.98+0.25-9.93

=5.67+8.99+-2.48

=12.18tm

MCD=-22.97+0.60c

=-22.97+0.6017.98

=-22.97+10.79

=-12.18tm

MDC=9.84+0.30c

=9.84+0.3017.98

=9.84+5.40

=15.24tmSyarat Batas :

MB =0.00

MBA+MBC=0.00

6.13+-6.13=0.00

0.00=0.00OK!!

MC =0.00

MCB+MCD=0.00

12.18+-12.18=0.00

0.00=0.00OK!!

FREEBODY

q=4.25t/m1.50m16.066.13P1=6.25t

12.18A2EIBEIC1.50EID6.1312.1815.24RakaRBkiRBkaRCkiRCkaRDki6.00m4.00m5.00m

RAka=1/2Lq+MAB-MBALL

=0.506.004.25+16.06-6.136.006.00

=12.75+2.68-1.02

=9.05tkontrol=RAka+Rbki-QAB=0.00

9.05+16.45-(4.25*6.00)=0.00RBki=1/2Lq-MAB+MBA25.50-25.50=0.00LL

=0.506.004.25-16.06+6.136.006.00

=12.75-2.68+1.02

=16.45t

RBka=1/2Lq+MBC-MCBLL

=0.504.004.25+6.13-12.184.004.00

=8.50+1.53-3.04kontrol=RAka+Rbki-QAB=0.00=6.99t6.99+10.01-(4.25*4.00)=0.00RCki=1/2Lq-MBC+MCB17.00-17.00=0.00LL

=0.504.004.25-6.13+12.184.004.00

=8.50-1.53+3.04

=10.01t

5.00RCka-p1*3.50-MCD+MDC=0.005.00RCka-6.25*3.50-12.18+15.24=0.005.00RCka-21.88-12.18+15.24=0.005.00RCka-18.81=0.005.00RCka=18.81

RCka=3.76

kontrol=Rdki+RCka-p1=0.00-5.00+p1*1.50-MCD+MDC=0.002.49+3.76-6.25=0.00-5.00Rdki6.25*1.50-12.18+15.24=0.006.25-6.25=0.00-5.00Rdki9.38-12.18+15.24=0.00-5.00Rdki12.44=0.00-5.00Rdki=-12.44RdkiRdki=2.49

no crosMOMEN PRIME ( CATAT DARI SLOPE )MFABq=4.25t/m1.50mMFDCMFBAMFBCMFCBMFCDP1=6.25t

A2.00EIB1.00EIC1.50EID

6.00m4.00m5.00m

MOMEN aKHIR( CATAT DARI SLOPE )

MABq=4.25t/m1.50mMDCMBAMBCMCBMCDP1=6.25t

A2.00EIB1.00EIC1.50EID

6.00m4.00m5.00m

FAKTOR DISTRIBUSI DAN KEKAKUAN

1).JEPIT AAB=0

2).joint B

Kba=4EI=4*2.00=1.3333333333LBA6.00

Kbc=4EI=4*1.00=1LBC4.00+kb=2.3333333333

BA=Kba=1.3333333333=0.571kb2.331.00BC=Kbc=1=0.429kb2.33

3). joint CKcb=4EI=4*1.00=1LCB4.00

Kcd=4EI=4*1.50=1.2Lcd5.00+kc=2.2

CB=Kcb=1=0.455kc2.201.00CD=Kcd=1.2=0.545kc2.20

4).JEPIT dDC=0

TABEL PERHITUNGAN

SIKLUSTITIK BUUL /JOINTABCDBATANGABBABCCBCDDCFAKTOR DISTRIBUSIABBABCCBCDDC0.0000.5710.4290.4550.5450.0001Momen Primer-12.75012.750-5.6675.667-22.9699.844M ikat-12.7507.083-17.3029.844M.. Distribusi0.000-4.048-3.0367.8659.4380.000

2M. induksi -2.0240.0003.932-1.5180.0004.719M.Ikat-2.0243.932-1.5184.719M distribusi0.000-2.247-1.6850.6900.8280.000


4M. induksi -0.0990.0000.192-0.0740.0000.230M.Ikat-0.0990.192-0.0740.230M distribusi0.000-0.109-0.0820.0340.0400.000batas catat lanjut catat tabel yang paling terakhr5M. induksi -0.0550.0000.017-0.0410.0000.020M.Ikat-0.0550.017-0.0410.020M distribusi0.000-0.010-0.0070.0190.0220.000




m akhir-16.066.13-6.1312.18-12.1815.24

kontrol momen-16.060.000.0015.24

m akhir siklus 4-16.006.15-6.1512.20-12.2015.21

kontrol momen-16.000.000.0015.21

MENGHITUNG MOMEN AKHIR ( BATAS SIKLUS 4

M AKHIR A= M AKHIR SIKLUS -(ik*M)=-16.00-(0.000x)=-16.00-0=-16.00

M AKHIR B=

MBA =M AKHIR SIKLUS -(ik*M)=6.15-(0.571x0.00)=6.15-0=6.15

MBC =M AKHIR SIKLUS -(ik*M)=0.00-(0.429x0.00)=0.00-0=0.00

M AKHIR C=

MCB =M AKHIR SIKLUS -(ik*M)=12.20-(0.455x)=12.20-0=12.20

MCD =M AKHIR SIKLUS -(ik*M)=-12.20-(0.545x)=-12.20-0=-12.20

M AKHIR D= M AKHIR SIKLUS -(ik*M)=15.21-(0.000x)=15.21-0=15.21

bidang m d

q=4.25t/m1.50m16.066.1312.18P1=6.25t

A2EIBEIC1.50EID6.1312.1815.24RakaRbkiRBkaRCkiRCkaRDki9.0516.456.9910.01

6.00m4.00m5.00m

GAMBAR BIDANG M16.06

6.5625

6.1312.188.5016.0615.24

19.13

GAMBAR BIDANG D

9.05

6.99

3.76

2.49

10.01

16.45

jimmy

Documents