LATIHAN SOAL
PORTAL BERTINGKAT DENGAN METODE CROSS
Disusun Untuk Memenuhi Tugas Mata Kuliah Mekanika Teknik IV
Dosen Pengampu :
Drs. Waluyo, M.Pd.
Oleh :
EKO EAHYU RULLY PRABOWO
K1515023
KELAS A
PROGRAM STUDI PENDIDIKAN TEKNIK BANGUNAN
FAKULTAS KEGURUAN DAN ILMU PENDIDIKAN
UNIVERSITAS SEBELAS MARET
SURAKARTA
2017
PERHITUNGAN MOMEN JEPIT / MOMEN PRIMER
P3P3P1P1Skala 1 : 200
P3
P3
P1
P1
JJKKLLq1q1q1q1
J
J
K
K
L
L
q1
q1
q1
q1
6,92 m6,92 m6,92 m6,92 m3,46 m3,46 m
6,92 m
6,92 m
6,92 m
6,92 m
3,46 m
3,46 m
q2q2q2q2HHP2P2P2P2P2P2P2P2P3P3P3P3Dik:
q2
q2
q2
q2
H
H
P2
P2
P2
P2
P2
P2
P2
P2
P3
P3
P3
P3
aaaaaaaaaaaaaaaaIIGGP1 = 1 kN a = 1,73 m
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
I
I
G
G
3,46 m3,46 mP2P2P2P2P2P2P2P2P3P3P3P3P2 = 2 kN b = 0,86 m
3,46 m
3,46 m
P2
P2
P2
P2
P2
P2
P2
P2
P3
P3
P3
P3
q2q2q2q2DDFFEEP3 = 3 kN c = 2,60 m
q2
q2
q2
q2
D
D
F
F
E
E
aaccbbaaaabbccaa
a
a
c
c
b
b
a
a
a
a
b
b
c
c
a
a
AA3,46 m3,46 mq1 = 1,23 kN
A
A
3,46 m
3,46 m
BBCCq2 = 2,23 kN
B
B
C
C
q2q2P3P3P2P2P2P2Batang FE
q2
q2
P3
P3
P2
P2
P2
P2
E E F F aabbccaa
E
E
F
F
a
a
b
b
c
c
a
a
MFE = Pab2l2+q12l2 ab6l2x2-8lx3+3x4
= 2 . 1,73 . 5,22 +3 . 2,6 . 4,332 +2 . 5,2 . 1,7326,922+q12 . 6,922 6 . 6,922 .5,22-8 . 6,92 . 5,23+3 . 5,24 -0" }
= 93,55 +146,24+31,1247,88+2,23574,63 {7.769,08-1.496,93+2.193,48 -"0"}
= 5,65 + 32,85
= 38,5 kNm ( )
MEF = Pba2l2+q12l2 ab4lx3-3x4
= 2 . 5,2 . 1,732 +3 . 4,33 . 2,62 +2 . 1,73 . 5,226,922+q12 . 6,922 4 . 6,92 .5,2³-3. 5,24 - 0" }
= 31,12 + 87,81+93,5547,88+2,23574,63 {3.892,02-2.193,48-0}
= 4,43+ 6,59
= 11,02 kNm ( ) - 11,02 kNm
q2q2P3P3P2P2P2P2Batang ED
q2
q2
P3
P3
P2
P2
P2
P2
D D E E aabbccaa
D
D
E
E
a
a
b
b
c
c
a
a
MED = Pab2l2+q12l2 ab6l2x2-8lx3+3x4
= 2 . 1,73 . 5,22 +3 . 2,6 . 4,332 +2 . 5,2 . 1,7326,922+q12 . 92 6 . 6,922 .6,922-8 . 6,92 . 6,923+3 . 6,924-6 . 6,922 .1,732-8 . 6,92 . 1.733+3 . 1,734" }
= 93,55+146,24+31,1247,88+2,23574,63 {13.758,64-18.344,85+6.879,32-"859,91-281,69+26,87"}
= 4,43 + 6,55
= 10,98 kNm ( )
MDE = Pba2l2+q12l2 ab4lx3-3x4
= 2 . 5,2 . 1,732 +3 . 4,33 . 2,62 +2 . 1,73 . 5,226,922+q12 . 6,922 4 . 6,92 . 6,923-3. 6,924- 4 . 6,92 . 1,733-3. 1,734" }
= 31,12 + 87,81+93,5547,88+2,23574,63 { 9.172,42-6879,32-143,31-26,87}
= 4,43 + 8,44
= 12,87 kNm ( ) - 12,87 kNm
q2q2P3P3P2P2P2P2Batang GH - HI
q2
q2
P3
P3
P2
P2
P2
P2
H H G G aaaa
H
H
G
G
a
a
a
a
aaaa
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a
a
a
MGH = Pab2l2+q12l2 ab6l2x2-8lx3+3x4
= 2 . 1,73 . 5,22 +3 . 3,46 . 3,462 +2 . 5,2 . 1,7326,922+q12 . 6,922 6 . 6,922 .5,22-8 . 6,92 . 5,23+3 . 5,24-6 . 6,922 .1,732-8 . 6,92 . 1.733+3 . 1,734" }
= 93,55 +124,26 +31,1247,88+2,23574,63 {7.769,08-7.784,05+2.193,48-"859,91-286,63+26,87"}
= 5,19 + 6,12
= 11,31 kNm ( ) = MHG = - 11,31 kNm ( )
MHI = MGH = 11,31 kNm ( ) = MIH = MHG = - 11,31 kNm ( )
q1q1P1P1Batang LK
q1
q1
P1
P1
K K L L
K
K
L
L
6,92 m6,92 m
6,92 m
6,92 m
MLK = 18Pl+112ql2
= 181 .6,92+1121,23 .6,922
= 0,86 + 4,90 = 5,76 kNm ( )
MKL = - 5,76 kNm ( )
q1q1P3P3Batang KJ
q1
q1
P3
P3
J J K K
J
J
K
K
6,92 m6,92 m
6,92 m
6,92 m
MKJ = 18Pl+112ql2
= 183 .6,92+1121,23 .6,922
= 2,59 + 4,90 = 7,49 kNm ( )
MJK = - 7,49 kNm ( )
PERHITUNGAN FAKTOR DISTRIBUSI
Perhitungan Dimensi Kolom dan Balok
Balok
ht=110 6,92=69,2 cm ht=115 6,92=46,1 cm
diambil ukuran 60 cm (lt. 1), 50 cm (lt. 2), dan 45 cm (lt. 3)
l. lt.1=3460=45 cm lt. 1 = 45/60
l. lt.2=3450=37,5=40 cm lt. 2 = 40/50
l. lt.3=3445=33,75=35 cm lt. 3 = 35/45
Kolom
Lt. 1 = llt.1 + 2x5 cm = 55 cm 55/55
Lt. 2 = llt.2 + 2x5 cm = 50 cm 50/50
Lt. 3 = llt.3 + 2x5 cm = 45 cm 45/45
Skala 1 : 200
35/4535/4535/4535/45KKJJLL
35/45
35/45
35/45
35/45
K
K
J
J
L
L
45/4545/4545/4545/4545/4545/453,46 m3,46 m
45/45
45/45
45/45
45/45
45/45
45/45
3,46 m
3,46 m
40/5040/5040/5040/50HH
40/50
40/50
40/50
40/50
H
H
IIGG
I
I
G
G
50/5050/5050/5050/5050/5050/503,46 m3,46 m
50/50
50/50
50/50
50/50
50/50
50/50
3,46 m
3,46 m
45/6045/6045/6045/60DDFFEE
45/60
45/60
45/60
45/60
D
D
F
F
E
E
3,46m3,46m55/5555/55
3,46m
3,46m
55/55
55/55
55/5555/5555/5555/55AA
55/55
55/55
55/55
55/55
A
A
BBCC
B
B
C
C
6,92 m6,92 m6,92 m6,92 m
6,92 m
6,92 m
6,92 m
6,92 m
Perhitungan Momen Inersia
IAF = IBE = ICD = 112 55 . 553 = 7,62 x 105 cm4 = 7,62 x 109 mm4
IFG = IEH = IDI = 112 50 . 503 = 5,21 x 105 cm4 = 5,21 x 109 mm4
IGL = IHK = IIJ = 112 45 . 453 = 3,42 x 105 cm4 = 3.42 x 109 mm4
IFE = IED = 112 45 . 603 = 8,10 x 105 cm4 = 8,10 x 109 mm4
IGH = IHI = 112 40 . 503 = 4,17 x 105 cm4 = 4,17 x 109 mm4
ILK = IKJ = 112 35 . 453 = 2,66 x 105 cm4 = 2,66 x 109 mm4
Perhitungan Faktor Kekakuan
K=EIL (Terjepit di 2 tempat) E beton = 4700 MPa
K=0,75 EIL (Terjepit di 1 tempat)
KAF = KBE = 4700 . 7,62 x 1093460 = 10,35 x 109 mm5
KCD = 0,75 4700 . 7,62 x 1093460 = 7,76 x 109 mm5
KFG = KEH = KDI = 4700 . 5,21 x 1093460 = 7,07 x 109 mm5
KGL = KHK = KIJ = 4700 . 3,42 x 1093460 = 4,64 x 109 mm5
KFE = KED = 4700 . 8,10 x 1096920 = 5,50 x 109 mm5
KGH = KHI = 4700 . 4,17 x 1096920 = 2,83 x 109 mm5
KLK = KKJ = 4700 . 2,66 x 1096920 = 1,80 x 109 mm5
Perhitungan Faktor Distribusi
ρ=K K
Titik F ( KF = KAF + KFE + KFG = 10,35 + 5,50 + 7,07 = 22,92)
ρFA = 10,35 x 10922,92 x 109 = 0,45 ρFE = 5,50x 10922,92 x 109 = 0,25
ρFG = 7,07x 10922,92 x 109 = 0,30
Titik E ( KE = KBE + KFE + KEH + KED = 10,35 + 5,50 + 7,07 + 5,50 = 28,42)
ρEB = 10,35 x 10928,42 x 109 = 0,36 ρEH = 7,07 x 10928,42 x 109 = 0,24
ρEF = 5,50x 10928,42 x 109 = 0,2 ρED = 5,50 x 10928, x 109 = 0,2
Titik D ( KD = KCD + KED + KDI = 7,76 + 5,50 + 7,07 = 20,33)
ρDC = 7,76 x 10920,33 x 109 = 0,38 ρDE = 5,50 x 10920,33 x 109 = 0,27
ρDI = 7,07x 10920,33 x 109 = 0,35
Titik G = Titik I
( KG = KI = KFG + KGH + KGL = 7,07 + 2,83 + 4,64 = 14,54)
ρGF = 7,07 x 10914,54 x 109 = 0,48 = ρID ρGH = 2,83 x 10914,54 x 109 = 0,2 = ρIH
ρGL = 4,64 x 10914,54 x 109 = 0,32 = ρIJ
Titik H ( KH = KEH + KGH + KHI + KHK = 7,07 + 2,83 + 2,83 + 4,64 = 17,37)
ρHE = 7,07 x 109 17,37 x 109 = 0,41 ρHG = 2,83 x 109 17,37 x 109 = 0,16
ρHI = 2,83 x 109 17,37 x 109 = 0,16 ρHK = 4,64 x 109 17,37 x 109 = 0,27
Titik K ( KK = KHK + KLK + KKJ = 4,64+ 1,80 + 1,80 = 8,24 )
ρKH = 4,64 x 109 8,24 x 109 = 0,56 ρKL = 1,80 x 109 8,24 x 109 = 0,22
ρKJ = 1,80 x 109 8,24 x 109 = 0,22
Titik L = Titik J
( KL = KJ = KGL + KLK = 4,64 + 1,80 = 6,44)
ρLG = 4,64 x 1096,44 x 109 = 0,72 = ρJI ρLK = 1,80 x 1096,44 x 109 = 0,28 = ρJK
TABEL CROSS
PERHITUNGAN SFD, BMD, DAN NFD
Perhitungan SFD
Batang LK – KJ
R = (1 2 ql+12 p)
DL = RLK + MLK - MKL L = 4,755 + ( - 0,39) = 4,36 kN
DP1.1 = DL – q1 . ½ l = 4,36 – 4,255 = 0,107 kN
DP1.2 = DP1.1 – P1 = 0,107 – 1 = - 0,892 kN
DK.1 = DP1.2 – q1 . ½ l = - 0,892 – 4,255 = - 5,148 kN
DK.2 = DK1 + RKL + MKL - MLK L + RKJ + MKJ - MJK L
=]/- 5,148 + 4,755 + 0,39 + 5,755 + 0,31 = 6,06 kN
DP3.1 = DK2 – q1 . ½ l = 6,06 – 4,755 = 1,808 kN
DP3.2 = DP3.1 – P3 = 1,808 – 3 = - 1,191 kN
DJ.1 = DP3.2 – q1 . ½ l = - 1,191 – 4,476 = - 5,447 kN
DJ.2 = DJ.1 + RJK + MJK - MKJ L
= - 5,447 + 5,755 + ( - 0,31) = 0
Batang GH – HI
R = P - Q 2
DG = RGH + MGH - MHG L = 7,357 + ( - 0,17) = 7,18 kN
DP2.1 = DG – P2.1 = 7,18 – 2 = 5,18 kN
DP3.1 = DP2 – q2 . 1,73 = 5,18 – 3,85 = 1,32 kN
DP3.2 = DP3.1 – P3 = 1,32 – 3 = - 1,73 kN
DP2.2.1 = DP3.2 – q2 . 1,73 = - 1,73 – 3,85 = - 5,52 kN
DP2.2.2 =2P2.2.1 – P2.2 = - 5,52 – 2 = - 7,52 kN
DH = DP2.2.2 + RHG + MHG - MGH L + RHI + MHI - MIH L +
= - 7,52 + 7,35 + 0,17 + 7,35 + 0,29 = 7,65 kN
DP2.1 = DH – P2.1 = 7,65 – 2 = 5,65 kN
DP3.1 = DP2.1 – q2 . 1,73 = 5,65 – 3,85 = 1,79 kN
DP3.2 = DP3.1 – P3 = 1,79 – 3 = - 1,20 kN
DP2.2.1 = DP3.2 – q2 . 1,73 = - 1,20 – 3,85 = - 5,06 kN
DP2.2.2 = DP2.2.1 – P2.2 = - 5,06 – 2 = - 7,06 kN
DI = DP2.2.2 + RIH + MIH - MHI L
= - 7,06 + 7,35 + ( - 0,29) = 0
Batang FE – ED
RFE = P2 . 5,2 + P3 . 4,33 +P2 .1,73+q2 . 5,2 . 4,32 L REF = P+Q- RFE
RED = P2 . 5,2 + P3 . 4,32 +P2 .1,73+q2 . 5,2 . 2,6 L RDE = P+Q- RED
DF = RFE + MFE - MEF L = 11,03 + 2.05 = 13,09 kN
DP2.1.1 = DPF – q2 . 1,73 = 13,09 – 3,85 = 9,23 kN
DP2.1.2 = DP2,1,1 – P2.1 = 9,23 – 2 = 7,23 kN
DP3.1 = DP2.1.2 – q2 . 0,86 = 7,23 – 1,92 = 5,31 kN
DP3.2 = DP3.1 – P3 = 5,31 – 3 = 2,31 kN
DP2.2.1 = DP3.2 – q2 . 2,6 = - 2,31 – 5,80 = - 3.49 kN
DP2.2.2 = DP2,2,1 – P2.2 = - 3,49 – 2 = - 5,49 kN
DE = DP2.2.2 + REF + MEF - MFE L + RED + MED - MDE L +
= - 5,49 + 7,49 + (-2,06) + 8,24 + 0,57 = 8,75 kN
DP2.1 = DE – P2.1 = 8,75 – 2 = 6,75 kN
DP3.1 = DP2.1 – q2 . 0,86 = 6,75 – 1,92 = 4,83 kN
DP3.2 = DP3.1 – P3 = 4,83 – 3 = 1,83 kN
DP2.2.1 = DP3.2 – q2 . 2,6 = 1,83 – 5,80 = - 3,96 kN
DP2.2.2 = DP2.2.1 – P2.2 = - 3,96 – 2 = - 5,96 kN
DD.1 = DP2.2.2 – q2 . 1,73 = - 5,96 – 3,85 = - 9,81 kN
DD.2 = DD.1 + RDE + MDE - MED L
= - 9,81 + 10,39 + ( - 0,700) = 0
BATANG AF – FG – GL
DAF = MAF - MFA L = - 1,23 kN
DFG = MFG - MGF L = - 0,69 kN
DGL = MGL - MLG L = - 0,17 kN
BATANG BE – EH – HK
DBE = MBE - MEB L = - 0,065 kN
DEH = MEH - MHE L = - 0,005 kN
DHK = MHK - MKH L = - 0,157 kN
BATANG CD – DI – IJ
DCD = MCD - MDC L = - 0,743 kN
DDI = MDI - MID L = 0,101 kN
DIJ = MIJ - MJI L = - 0,289 kN
GAMBAR DIAGRAM SFD
Skala Jarak ( 1 : 75 cm) Skala Gaya ( 1 : 5 kN)
*Keterangan besar gaya dapat dilihat pada perhitungan
Perhitungan Momen Maksimum
q1q1P1P1Batang LK
q1
q1
P1
P1
K K L L
K
K
L
L
6,92 m6,92 m
6,92 m
6,92 m
Mmax = 14Pl+18ql2 = 141 . 6,92+18 1,23 .6,922 = 9,09 kNm
q1q1P3P3Batang KJ
q1
q1
P3
P3
J J K K
J
J
K
K
6,92 m6,92 m
6,92 m
6,92 m
Mmax = 14Pl+18ql2 = 143 . 6,92+18 1,23 .6,922 = 12,55 kNm
q2q2P3P3P2P2P2P2Batang GH - HI
q2
q2
P3
P3
P2
P2
P2
P2
H H G G aaaa
H
H
G
G
a
a
a
a
aaaa
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a
MP2.1 = MP2.2 = RGH . 1,73 = 7,35 . 1,73 = 12,72 kNm
MP3 = RGH . 3,46 – P2.1 . 1,73 – q2 . 1,73 . 0,86
= 7,35 . 3,46 – 2 . 1,73 – 2,23 . 1,73 . 0,86 = 18,65 kNm
q2q2P3P3P2P2P2P2Batang FE
q2
q2
P3
P3
P2
P2
P2
P2
E E F F aabbccaa
E
E
F
F
a
a
b
b
c
c
a
a
MP2.1 = RFE . 1,73 – q2 . 1,73 . 0,86
= 11,03 . 1,73 – 2,23 . 1,73 . 0,86 = 15,76 kNm
MP3 = RFE . 2,6 – P2.1 . 0,86 – q2 . 2,6 . 1,3
= 11,03 . 2,6 – 2 . 0,86 – 2,23 . 2,6 . 1,3 = 19,42 kNm
MP2.2 = - {- REF . 1,73}
= - {- 7,49 . 1,73} = 12,95 kNm
q2q2P3P3P2P2P2P2Batang ED
q2
q2
P3
P3
P2
P2
P2
P2
D D E E aabbccaa
D
D
E
E
a
a
b
b
c
c
a
a
MP2.1 = RED . 1,73
= 8,24 . 1,73 = 14,25 kNm
MP3 = RED . 2,6 – P2.1 . 0,86 – q2 . 0,86 . 0,43
= 8,24 . 2,6 – 2 . 0,86 – 2,23 . 0,86 . 0,43 = 18,87 kNm
MP2.2 = - {- REF . 1,73 + q2 . 1,73 . 0,86 }
= - {- 7,49 . 1,73 + 2,23 . 1,73 . 0,86 } = 16,27 kNm
GAMBAR DIAGRAM BMD
Skala Jarak ( 1 : 75 ) Skala Gaya ( 1 : 8 )
*Keterangan besar gaya dapat dilihat pada perhitungan dan tabel cross
Perhitungan NFD
Batang AF – FG – GL
NLG = - {RL} = - {RLK + MLK - MKL L-} = - 4,343 kN
NGF = NLG – RG = NLG – {RGH + MGH - MHG L-} = - 4,343 – 6,719 = - 11,062 kN
NFA = NGF – RF = NGF – {RFE + MFE - MEF L-} = - 11,062 – 10,353 = - 21,415 kN
Batang BE – EH – HK
NKH = - {RK} = - {RKL + MKL - MLK L- + RKJ + MKJ - MJK L-}
= - {5,237 + 7,747} = - 12,984 kN
NHE = NLG – RH = NKH – {RHG + MHG - MGH L- + RHI + MHI - MIH L-}
= - 12,984 – {7,871 + 7,869} = - 28,724 kN
NEB = NGF – RE = NHE – {REF + MEF - MFE L- + RED + MED - MDE L-}
= - 28,724 – {8,032 + 9,031} = - 45,787 kN
Batang CD – DI – IJ
NJI = - {RJ} = - {RJK + MJK - MKJ L-} = - 1,833 kN
NID = NJI – RI = NJI – {RIH + MIH - MHI L-} = - 1,833 – 6,721 = - 8,554 kN
NDC = NID – RD = NID – {RDE + MDE - MED L-} = - 8,554 – 9,354 = - 17,908 kN
Batang LK – KJ
NLK = - RLG = - 0,025 kN
NKJ = - RJI = - 0,031 kN
Batang GH – HI
NGH = - {RGL + RGF}
= - {- 0,025 + -0,031} = -0,056 kN
NHI = - {RHK + RHE}
= - {0,003 + (-0,016)} = 0,013 kN
Batang FE – ED
NFE = - {RFG + RFA}
= - {- 0,031 + 0,314} = - 0,283 kN
NED = - {RDI + RDC}
= - {0,002 + 0,393} = - 0,395 kN
GAMBAR DIAGRAM NFD
Skala Jarak ( 1 : 75 cm) Skala Gaya ( 1 : 10 kN)
0,0130,013-0,056-0,056-0,395-0,395-0,283-0,283-0,031-0,031-0,025-0,025-8,554-8,554-1,833-1,833-21,15-21,15-11,062-11,062-4,343-4,343-12,984-12,984-28,724-28,724-17,908-17,908-45,787-45,787
0,013
0,013
-0,056
-0,056
-0,395
-0,395
-0,283
-0,283
-0,031
-0,031
-0,025
-0,025
-8,554
-8,554
-1,833
-1,833
-21,15
-21,15
-11,062
-11,062
-4,343
-4,343
-12,984
-12,984
-28,724
-28,724
-17,908
-17,908
-45,787
-45,787
