Contoh Soal Metode CROSS
18
Soal-soal dan penyelesaian METODE D istribusi M om en { M etode C ross } No.01.Analisislah Portalberikut dengan M etode Cross 2 a.Perhitungan M O M EN PRIM ER (M O ) M o AB = -Pab 2 /L 2 = - 600.3.2 2 /5 2 = -288 kg.m M o BA = Pa 2 b/L 2 = + 600.3 2 .2/5 2 = 432 kg.m M o BC = -qL 2 /12 = - 300.6 2 /12 = -625 kg.m M o CB = qL 2 /12 = + 300.6 2 /12 = 625 kg.m M o CD = -Pa 2 b/L 2 = - 600.3 2 .2/5 2 = -432 kg.m M o DC = Pab 2 /L 2 = + 600.3.2 2 /5 2 = 288 kg.m b.Perhitungan Koefisien Distribusi Titik Sim pulA m AB = 0 Titik Sim pulD m DC = 0 5.00 m A B C D P = 600 kg P = 600 kg q = 300 kg/m ’ EI EI EI 3.00 m 2.00 m P = 600 kg P = 600 kg q = 300 kg/m ’ 288 432 625 625 432 288 B A C D
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Transcript of Contoh Soal Metode CROSS
Soal-soal dan penyelesaian METODE Distribusi Momen { Metode Cross }
No. 01. Analisislah Portal berikut dengan Metode Cross
2
a. Perhitungan MOMEN PRIMER (MO)
MoAB = - Pab2/ L2 = - 600.3.22/ 52 = -288 kg.m
MoBA = Pa2b/ L2 = + 600.32.2/ 52 = 432 kg.m
MoBC = - qL2/ 12 = - 300.62/ 12 = -625 kg.m
MoCB = qL2/ 12 = + 300.62/ 12 = 625 kg.m
MoCD = - Pa2b/ L2 = - 600.32.2/ 52 = -432 kg.m
MoDC = Pab2/ L2 = + 600.3.22/ 52 = 288 kg.m
b. Perhitungan Koefisien DistribusiTitik Simpul AmAB = 0Titik Simpul DmDC = 0
5.00 m
A
B C
D
P = 600 kg P = 600 kg
q = 300 kg/m’
EI EI
EI
3.00 m
2.00 m
P = 600 kg P = 600 kg
q = 300 kg/m’
288
432
625 625
432
288
B
A
C
D
5 mTitik Simpul B
B C2.EI
5 m EI
A4.EIBA 4.EI
LBA 5
4.EIBC 4.2EI
LBC 5maka
Titik Simpul C5 m
B C2.EI
EI 5 m
D4.EICB 4.2EI
LCB 5
4.EICD 4.EI
LCD 5maka
c. Proses Distribusi MomenLihat halaman selanjutnya….
1.00000
0.666672.40
mCD =0.80
= 0.333332.40
mCB =1.60
=
= = = 0.80mCD =kCD
kCDkCB + kCD
= = = 1.60mCB =kCB
kCBkCB + kCD
2.40
0.33333
0.66667
1.00000
1.60
mBA =
mBC =
0.802.40
=
=1.60
= =
kBA = = =mBA =
mBC =
kBA
kBA + kBC
kBC
kBA + kBC
0.80
kBC =
Tabel Distribusi Momen
A DAB BA BC CB CD DC
0.80 0.80 1.60 1.60 0.80 0.80No. DF 0 0.3333333 0.66666667 0.6666667 0.33333333 0
FEM -288.00000 432.00000 -625.00000 625.00000 -432.00000 288.00000Bal 0.00000 64.33333 128.66667 -128.66667 -64.33333 0.00000CO 32.16667 0.00000 -64.33333 64.33333 0.00000 -32.16667Bal 0.00000 21.44444 42.88889 -42.88889 -21.44444 0.00000CO 10.72222 0.00000 -21.44444 21.44444 0.00000 -10.72222Bal 0.00000 7.14815 14.29630 -14.29630 -7.14815 0.00000CO 3.57407 0.00000 -7.14815 7.14815 0.00000 -3.57407Bal 0.00000 2.38272 4.76543 -4.76543 -2.38272 0.00000CO 1.19136 0.00000 -2.38272 2.38272 0.00000 -1.19136Bal 0.00000 0.79424 1.58848 -1.58848 -0.79424 0.00000CO 0.39712 0.00000 -0.79424 0.79424 0.00000 -0.39712Bal 0.00000 0.26475 0.52949 -0.52949 -0.26475 0.00000CO 0.13237 0.00000 -0.26475 0.26475 0.00000 -0.13237Bal 0.00000 0.08825 0.17650 -0.17650 -0.08825 0.00000CO 0.04412 0.00000 -0.08825 0.08825 0.00000 -0.04412Bal 0.00000 0.02942 0.05883 -0.05883 -0.02942 0.00000CO 0.01471 0.00000 -0.02942 0.02942 0.00000 -0.01471Bal 0.00000 0.00981 0.01961 -0.01961 -0.00981 0.00000CO 0.00490 0.00000 -0.00981 0.00981 0.00000 -0.00490Bal 0.00000 0.00327 0.00654 -0.00654 -0.00327 0.00000CO 0.00163 0.00000 -0.00327 0.00327 0.00000 -0.00163Bal 0.00000 0.00109 0.00218 -0.00218 -0.00109 0.00000CO 0.00054 0.00000 -0.00109 0.00109 0.00000 -0.00054Bal 0.00000 0.00036 0.00073 -0.00073 -0.00036 0.00000CO 0.00018 0.00000 -0.00036 0.00036 0.00000 -0.00018Bal 0.00000 0.00012 0.00024 -0.00024 -0.00012 0.00000CO 0.00006 0.00000 -0.00012 0.00012 0.00000 -0.00006Bal 0.00000 0.00004 0.00008 -0.00008 -0.00004 0.00000CO 0.00002 0.00000 -0.00004 0.00004 0.00000 -0.00002Bal 0.00000 0.00001 0.00003 -0.00003 -0.00001 0.00000CO 0.00001 0.00000 -0.00001 0.00001 0.00000 -0.00001Bal 0.00000 0.00000 0.00001 -0.00001 0.00000 0.00000CO 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000Bal 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000CO 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000Bal 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
-239.75000 528.50000 -528.50000 528.50000 -528.50000 239.75000Check Statika 0.00000 0.00000
16
17
18
15
Total
11
12
13
14
7
8
9
10
3
4
5
6
J oint B C
2
BatangK
1
Proses Distribusi Momen
A DAB BA BC CB CD DC
kAB kAB kBC kCB kCD kDCNo. DF (m) 0 mBA mBC mCB mCD 0
FEM MoAB MoBA MoBC MoCB MoCD MoDC
Bal 0 (MoBA+MoBC).(-mBA) (MoBA+MoBC).(-mBC) (MoCB+MoCD).(-mCB) (MoCB+MoCD).(-mCD) 0
CO A = Bal.1 BA / 2 B = Bal.1 AB / 2 C = Bal. 1 CB / 2 D = Bal. 1 BC / 2 E = Bal. 1 DC / 2 F = Bal. 1 CD / 2
Bal 0 (B + C).(-mBA) (B + C).(-mBC) (D + E).(-mCB) (D + E).(-mCD)
CO G = Bal.2 BA / 2 H = Bal.2 AB / 2 I = Bal. 2 CB / 2 J = Bal. 2 BC / 2 K = Bal. K DC / 2 L = Bal. 2 CD / 2
Bal 0 (H + I).(-mBA) (H + I).(-mBC) (J + K).(-mCB) (J + K).(-mCD) 0
CO Proses tersebut diatas diulang hingga hasil bagi ( CO ) sudah mendekati nilai 0 , ( 0,00000 )
Bal
CO 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
Bal 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000MAB MBA MBC MCB MCD MDC
Check Statika
MAB Diperoleh dari hasil penjumlahan kebawah (menurut kolom) dimulai dari Momen Primer MOABsampai pada kondisi balance (bal) sudah bernilai 0,00000.Untuk Momen Total pada Titik simpul yang lain dilakukan dengan cara yang sama.
jumlah satu titik simpul harus = 0 jumlah satu titik simpul harus = 0
5
Total
4
2
3
CBatang
K
1
J oint B
d. Perhitungan Momen DesainKarena kondisi beban dan rangka simetris, maka dianalisis Portal tanpa Pergoyangan.untuk itulah proses distribusi momen (cross) hanya satu kali yaitu akibat beban luar (tanpa pergoyangan).Harga momen desain sama dengan harga cross dari beban luar (lihat jumlah momen dari proses cross).MAB = -239.8 kg.mMBA = 528.5 kg.mMBC = -528.5 kg.mMCB = 528.5 kg.mMCD = -528.5 kg.mMDC = 239.8 kg.m
l. Free Body untuk Perhitungan Reaksi PerletakanBatang AB Batang CDSMB = 0 RCH = - RBH = -182.3 kg gRAH ={(600.2)+(239,75)-(528,5)}/ (5) = 182.25 kg g RDH = - RAH = -417.7 kg gSH = 0RAH + RBH = PRBH = P - RAH = 417.75 kg g
Batang BCSMC = 0RBV ={(300.5.5/ 2)+(528,5)-(528,5)}/ (5) = 750.00 kg gSH = 0RBV + RCV = QRCV = Q - RBV = 750.00 kg g
528,50
528,50
239,75
P = 600 kg P = 600 kg
q = 300 kg/m’
239,75
528,50
528,50B C
DA
528,50
q = 300 kg/m’
528,50
B C
P = 600 kg
528,50
239,75
P = 600 kg
239,75
528,50
B C
A D182,25
417,75
750 750
417,75
182,25
m. Perhitungan Momen, Gaya Lintang dan Gaya NormalPotongan AB interval Potongan BC interval
Mx1 =RAH.X - MAB 0 < X < 2 Mx = RBV.x - 1/ 2.q.x2 - MBC 0 < X < 5Dx1 = RAH Dx = RBV - q.xNx1 = - RAV = RBV Nx = - RBHMx2 =RAH.(3+X) - P.X - MAB 0 < X < 3Dx2 = RAH - PNx2 = - RAV
Pot. AB Pot. BCx Mx Dx Nx x Mx Dx Nx0 -239.8 182.3 -750 0 -528.5 750 -4181 -57.5 182.3 -750 0.5 -191 600 -4182 124.7 182.3 -750 1 71.5 450 -4183 307.0 182.3 -750 1.5 259 300 -4180 307.0 -417.7 -750 2 371.5 150 -4181 -110.7 -417.7 -750 2.5 409.0 0 -4182 -528.5 -417.7 -750 3 371.5 -150 -418
3.5 259 -300 -4184 71.5 -450 -418
4.5 -191 -600 -4185 -528.5 -750 -418
Gbr. Bidang Momen Gbr. Bidang Lintang
Gbr. Bidang Normal
Interval0 < X < 5Interval
0 < X < 3
0 < X < 2
B C
DA
409,0
+
--
+
-
+
--
-
528,5
528,5
528,5
528,5
307 307
239,8 239,8
B C
DA
750
750
417,75
182,25
417,75
182,25
B C
DA750
417,75
750
417,75
No. 02. Analisislah Portal berikut dengan Metode Cross
a
36.87 o
a. Perhitungan MOMEN PRIMER (MO)
MoBE = PL = + 400.1 = 400 kg.m
MoBC = - PL/ 8 = - 1000.4/ 8 = -500 kg.m
MoCB = + PL/ 8 = + 1000.4/ 8 = 500 kg.m
Momen primer akibat pergoyangan
MAB=MBA MXAB = MXBA = (6.EIAB.Dh)/ LAB2
= -0.375 EI.DHmisal EI.DH = X, maka
MXAB = MXBA = -37.5 X
MBC=MCB MXBC = MXCB = (6.EIBC.Dv)/ LBC2 Dv = Dh / Tan a
= 1.000 X
MCD=MDC MXCD = MXDC = (6.EICD.Dr)/ LCD2 Dr = Dh / Sin a
= -0.400 XMAB=MBA = -37.5 .XMBC=MCB = 100 .XMCD=MDC = -40 .X
b. Perhitungan Koefisien DistribusiTitik Simpul AmAB = 0Titik Simpul DmDC = 0
sudut ( a )=2.00 m
P1 = 600 kg
P2 = 400 kg
EI
EI
2 EI
3.00 m
1.00 m
P4 = 400 kg
P3 = 1000 kg
4.00 m2.00 m
A
B C
D
1 m
4 mTitik Simpul B
B C2.EI
4 m EI
A4.EIBA 4.EI
LBA 4
4.EIBC 4.2EI
LBC 4maka
Titik Simpul C4 m
B C2.EI
EI 5 m
D4.EICB 4.2EI
LCB 4
4.EICD 4.EI
LCD 5maka
c. Proses Distribusi MomenLihat halaman selanjutnya….
0.80
=
1.00000
0.714292.80
0.28571
= = =
mCD =0.80
=2.80
mCB =2.00
=
mCD =kCD
kCDkCB + kCD
= = 2.00mCB =kCB
kCBkCB + kCD
3.00
0.33333
0.66667
1.00000
2.00
mBA =
mBC =
1.003.00
=
=2.00
= =
kBA = = =mBA =
mBC =
kBA
kBA + kBC
kBC
kBA + kBC
1.00
kBC =
Tabel Distribusi Momen, akibat Beban Luar
A DAB BA BE BC CB CD DC
1.00 1.00 0.00 2.00 2.00 0.80 0.80No. DF 0 0.33333333 0 0.66666667 0.71428571 0.28571429 0
FEM 0.00000 0.00000 400.00000 -500.00000 500.00000 0.00000 0.00000Bal 0.00000 33.33333 0.00000 66.66667 -357.14286 -142.85714 0.00000CO 16.66667 0.00000 0.00000 -178.57143 33.33333 0.00000 -71.42857Bal 0.00000 59.52381 0.00000 119.04762 -23.80952 -9.52381 0.00000CO 29.76190 0.00000 0.00000 -11.90476 59.52381 0.00000 -4.76190Bal 0.00000 3.96825 0.00000 7.93651 -42.51701 -17.00680 0.00000CO 1.98413 0.00000 0.00000 -21.25850 3.96825 0.00000 -8.50340Bal 0.00000 7.08617 0.00000 14.17234 -2.83447 -1.13379 0.00000CO 3.54308 0.00000 0.00000 -1.41723 7.08617 0.00000 -0.56689Bal 0.00000 0.47241 0.00000 0.94482 -5.06155 -2.02462 0.00000CO 0.23621 0.00000 0.00000 -2.53077 0.47241 0.00000 -1.01231Bal 0.00000 0.84359 0.00000 1.68718 -0.33744 -0.13497 0.00000CO 0.42180 0.00000 0.00000 -0.16872 0.84359 0.00000 -0.06749Bal 0.00000 0.05624 0.00000 0.11248 -0.60257 -0.24103 0.00000CO 0.02812 0.00000 0.00000 -0.30128 0.05624 0.00000 -0.12051Bal 0.00000 0.10043 0.00000 0.20086 -0.04017 -0.01607 0.00000CO 0.05021 0.00000 0.00000 -0.02009 0.10043 0.00000 -0.00803Bal 0.00000 0.00670 0.00000 0.01339 -0.07173 -0.02869 0.00000CO 0.00335 0.00000 0.00000 -0.03587 0.00670 0.00000 -0.01435Bal 0.00000 0.01196 0.00000 0.02391 -0.00478 -0.00191 0.00000CO 0.00598 0.00000 0.00000 -0.00239 0.01196 0.00000 -0.00096Bal 0.00000 0.00080 0.00000 0.00159 -0.00854 -0.00342 0.00000CO 0.00040 0.00000 0.00000 -0.00427 0.00080 0.00000 -0.00171Bal 0.00000 0.00142 0.00000 0.00285 -0.00057 -0.00023 0.00000CO 0.00071 0.00000 0.00000 -0.00028 0.00142 0.00000 -0.00011Bal 0.00000 0.00009 0.00000 0.00019 -0.00102 -0.00041 0.00000CO 0.00005 0.00000 0.00000 -0.00051 0.00009 0.00000 -0.00020Bal 0.00000 0.00017 0.00000 0.00034 -0.00007 -0.00003 0.00000CO 0.00008 0.00000 0.00000 -0.00003 0.00017 0.00000 -0.00001Bal 0.00000 0.00001 0.00000 0.00002 -0.00012 -0.00005 0.00000CO 0.00001 0.00000 0.00000 -0.00006 0.00001 0.00000 -0.00002Bal 0.00000 0.00002 0.00000 0.00004 -0.00001 0.00000 0.00000CO 0.00001 0.00000 0.00000 0.00000 0.00002 0.00000 0.00000Bal 0.00000 0.00000 0.00000 0.00000 -0.00001 -0.00001 0.00000CO 0.00000 0.00000 0.00000 -0.00001 0.00000 0.00000 0.00000Bal 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
52.70270 105.40540 400.00000 -505.40540 172.97297 -172.97297 -86.48649Check Statika 0.00000 0.00000
16
17
18
15
Total
11
12
13
14
7
8
9
10
3
4
5
6
J oint B C
2
BatangK
1
Tabel Distribusi Momen, akibat Pelepasan Pendel 1
A DAB BA BE BC CB CD DC
1.00 1.00 0.00 2.00 2.00 0.80 0.80No. DF 0 0.333333 0 0.66666667 0.71428571 0.285714 0
FEM -37.50000 -37.50000 0.00000 100.00000 100.00000 -40.00000 -40.00000Bal 0.00000 -20.83333 0.00000 -41.66667 -42.85714 -17.14286 0.00000CO -10.41667 0.00000 0.00000 -21.42857 -20.83333 0.00000 -8.57143Bal 0.00000 7.14286 0.00000 14.28571 14.88095 5.95238 0.00000CO 3.57143 0.00000 0.00000 7.44048 7.14286 0.00000 2.97619Bal 0.00000 -2.48016 0.00000 -4.96032 -5.10204 -2.04082 0.00000CO -1.24008 0.00000 0.00000 -2.55102 -2.48016 0.00000 -1.02041Bal 0.00000 0.85034 0.00000 1.70068 1.77154 0.70862 0.00000CO 0.42517 0.00000 0.00000 0.88577 0.85034 0.00000 0.35431Bal 0.00000 -0.29526 0.00000 -0.59051 -0.60739 -0.24295 0.00000CO -0.14763 0.00000 0.00000 -0.30369 -0.29526 0.00000 -0.12148Bal 0.00000 0.10123 0.00000 0.20246 0.21090 0.08436 0.00000CO 0.05062 0.00000 0.00000 0.10545 0.10123 0.00000 0.04218Bal 0.00000 -0.03515 0.00000 -0.07030 -0.07231 -0.02892 0.00000CO -0.01757 0.00000 0.00000 -0.03615 -0.03515 0.00000 -0.01446Bal 0.00000 0.01205 0.00000 0.02410 0.02511 0.01004 0.00000CO 0.00603 0.00000 0.00000 0.01255 0.01205 0.00000 0.00502Bal 0.00000 -0.00418 0.00000 -0.00837 -0.00861 -0.00344 0.00000CO -0.00209 0.00000 0.00000 -0.00430 -0.00418 0.00000 -0.00172Bal 0.00000 0.00143 0.00000 0.00287 0.00299 0.00120 0.00000CO 0.00072 0.00000 0.00000 0.00149 0.00143 0.00000 0.00060Bal 0.00000 -0.00050 0.00000 -0.00100 -0.00102 -0.00041 0.00000CO -0.00025 0.00000 0.00000 -0.00051 -0.00050 0.00000 -0.00020Bal 0.00000 0.00017 0.00000 0.00034 0.00036 0.00014 0.00000CO 0.00009 0.00000 0.00000 0.00018 0.00017 0.00000 0.00007Bal 0.00000 -0.00006 0.00000 -0.00012 -0.00012 -0.00005 0.00000CO -0.00003 0.00000 0.00000 -0.00006 -0.00006 0.00000 -0.00002Bal 0.00000 0.00002 0.00000 0.00004 0.00004 0.00002 0.00000CO 0.00001 0.00000 0.00000 0.00002 0.00002 0.00000 0.00001Bal 0.00000 -0.00001 0.00000 -0.00001 -0.00001 -0.00001 0.00000CO 0.00000 0.00000 0.00000 -0.00001 -0.00001 0.00000 0.00000Bal 0.00000 0.00000 0.00000 0.00000 0.00001 0.00000 0.00000CO 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000Bal 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000CO 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000Bal 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
-45.27027 -53.04054 0.00000 53.04054 52.70270 -52.70270 -46.35135Check Statika
J oint B C
2
BatangK
1
3
4
5
6
7
8
9
10
15
Total
11
12
13
14
0.00000 0.00000
16
17
18
Perhitungan Reaksi Akibat Beban Luar
Tabel Distribusi Momen
A DAB BA BE BC CB CD DC
52.70270 105.40540 400.00000 -505.40540 172.97297 -172.973 -86.48649-45.27027 -53.04054 0.00000 53.04054 52.70270 -52.7027 -46.35135
158.10811 416.89189B SMB = 0 -172.973 C SMD = 0
RAH.h + MAB + MBA = 0 RCH.h + MCD + MDC - RCV.4 = 0RAH = 39.527027 Ton g RCH = 642.34 Ton g
h = 4 SH = 0 r = 5 SH = 0RAH + RBH = 0 RCH + RDH = 0
RBH = -39.52703 Ton g RDH = -642.3 Ton gA D
-86.486
1000 Kg172.9730
B 2 2 C4
SMC = 0RBV.L + MBC + MCB - P.2 = 0
RBV = 583.10811 Kg
SV = 0RBV + RCV = P
RCV = 416.89189 Kg
583.1081 416.8918918 HOBC = 602.815 Ton g642.34234
B C
HOAB =-602.82 Ton g D -642.3423
A
-505.40540
39.52703
-39.52703
105.40540
52.70270
CBatang
M. Beban LuarM. Pendel 1
J oint B
Perhitungan Reaksi Akibat Pendel 1
Tabel Distribusi Momen
A DAB BA BE BC CB CD DC
52.70270 105.40540 400.00000 -505.40540 172.97297 -172.97297 -86.48649-45.27027 -53.04054 0.00000 53.04054 52.70270 -52.70270 -46.35135
26.43581B SMB = 0 -52.7027 C SMD = 0
RAH.h + MAB + MBA = 0 RCH.h + MCD + MDC - RCV.4 = 0RAH = -24.57770271 Ton g RCH = 68.2658 Ton g
h3 = 4 SH = 0 r = 5 SH = 0RAH + RBH = 0 RCH + RDH = 0
RBH = 24.57770271 Ton g RDH = -68.266 Ton gA D
-46.35135
52.70270B 2 2 C
4
SMC = 0RBV.L + MBC + MCB = 0
RBV = -26.43581078 Kg
SV = 0RBV + RCV = 0
RCV = 26.43581078 Kg
-26.4358 26.43581078 HxBC = 92.84347 Ton g68.26576573
B C
HXAD = -92.843 Ton g D -68.26576573
A
CBatang
M. Beban LuarM. Pendel 1
J oint B
53.04054
-24.57770
24.57770
-53.04054
-45.27027
Perhitungan Momen DesainKarena kondisi beban dan rangka simetris, maka dianalisis Portal tanpa Pergoyangan.untuk itulah proses distribusi momen (cross) hanya satu kali yaitu akibat beban luar (tanpa pergoyangan).Harga momen desain sama dengan harga cross dari beban luar (lihat jumlah momen dari proses cross).MAB = 52.7 kg.mMBA = 105.4 kg.mMBC = -505.4 kg.mMCB = 173 kg.mMCD = -173 kg.mMDC = -86.49 kg.m
Perhitungan Harga X
Tabel Distribusi Momen
A DAB BA BE BC CB CD DC
52.70270 105.40540 400.00000 -505.40540 172.97297 -172.97297 -86.48649-45.27027 -53.04054 0.00000 53.04054 52.70270 -52.70270 -46.35135-193.6664 -226.90762 0.00000 226.90762 225.46235 -225.4623 -198.2912-140.9637 -121.50221 400.00000 -278.49779 398.43532 -398.4353 -284.7777
-140.9637 -121.50221 400 -278.4977864 398.43532 -398.4353205 -284.7777
Harga X = 4.2780035
HOBC = -397.1847 kg g
HXBC = 92.843468 X g
Syarat Statika :
HOBC + HXBC = 0-397.1847 + 92,84347.X = 0
Maka Harga XX = 4.2780035
J oint
M.P1 . XMomen Akhir / Desain
Check Statika
BatangM. Beban LuarM. Pendel 1
B C
0.000000.00000
Perhitungan Reaksi Akhir
Tabel Momen Akhir
A DAB BA BE BC CB CD DC
52.70270 105.40540 400.00000 -505.40540 172.97297 -172.97297 -86.48649-45.27027 -53.04054 0.00000 53.04054 52.70270 -52.70270 -46.35135-140.96367 -121.50221 400.00000 -278.49779 398.4353 -398.43532 -284.77773
529.984B SMB = 0 -398.435 C SMD = 0
RAH.h3 + MAB + MBA = 0 RCH.h + MCD + MDC - RCV.4 = 0RAH = -65.61647217 Ton g RCH = 934.3835278 Ton g
h3 = 4 SH = 0 L = 5 SH = 0RAH + RBH = 0 RCH + RDH = 0
RBH = 65.61647217 Ton g RDH = -934.3835278 Ton gA D
-284.7777
1000 Kg398.435
B 2 2 C4
SMB = 0RBV.L + MBC + MCB - P.2 = 0
RBV = 470.0156166 Kg
SV = 0RBV + RCV = P
RCV = 529.9843834 Kg
HBC = 1000.0 Ton g470.0156166 529.9843834
934.383528B C
HAD = -1000.00 Ton g
D -934.3835278A
870.015617 529.984383
M. Akhir
CBatang
M. Beban LuarM. Pendel 1
J oint B
-278.4978
-65.616
65.616
-121.5022
-140.964
Free Body untuk Perhitungan Reaksi PerletakanBatang AB Batang CDSMB = 0 RCH = - RBH = -229.5 kg gRAH ={(600.2)+(239,75)-(528,5)}/ (5) = 229.46 kg g RDH = - RAH = -370.5 kg gSH = 0RAH + RBH = PRBH = P - RAH = 370.54 kg g
Batang BCSMC = 0RBV ={(300.5.5/ 2)+(528,5)-(528,5)}/ (5) = 816.49 kg gSH = 0RBV + RCV = QRCV = Q - RBV = 683.51 kg g
Perhitungan Momen, Gaya Lintang dan Gaya NormalPotongan AB interval Potongan BC interval
Mx1 =RAH.X - MAB 0 < X < 2 Mx = RBV.x - 1/ 2.q.x2 - MBC 0 < X < 5Dx1 = RAH Dx = RBV - q.xNx1 = - RAV = RBV Nx = - RBHMx2 =RAH.(3+X) - P.X - MAB 0 < X < 3Dx2 = RAH - PNx2 = - RAV
Pot. AB Pot. BCx Mx Dx Nx x Mx Dx Nx0 -52.7 229.5 -816.5 0 -505.4 816.5 -3711 176.8 229.5 -816.5 0.5 -134.7 666.5 -3712 406.2 229.5 -816.5 1 161.1 516.5 -3713 635.7 229.5 -816.5 1.5 381.8 366.5 -3710 635.7 -370.5 -816.5 2 527.6 216.5 -3711 265.1 -370.5 -816.5 2.5 598.3 66.49 -3712 -105.4 -370.5 -816.5 3 594.1 -83.51 -371
3.5 514.8 -233.5 -3714 360.5 -383.5 -371
4.5 131.3 -533.5 -3715 -173 -683.5 -371
Interval0 < X < 5Interval
0 < X < 3
0 < X < 2
Coba anda selesaikan Gambar Bidang Momen, Lintang dan Normalnya.
