Soal-soal dan penyelesaian METODE Distribusi Momen { Metode Cross

Soal-soal dan penyelesaian METODE Distribusi Momen {

Metode Cross

}

No. 01. Analisislah Portal berikut dengan Metode Cross

2

a. Perhitungan MOMEN PRIMER (M O) MoAB o

=

- Pab2/L2 2

M BA

=

Pa b/L

MoBC

=

- qL2/12

MoCB

=

qL2/12

M CD

=

2

- Pa b/L

MoDC

=

Pab2/L2

o

= - 600.3.22/52

2

2

= + 600.3 .2/5

2

0

mDC

0

=

=

-288 kg.m

=

432 kg.m

= - 300.62/12

=

-625 kg.m

= + 300.62/12

=

625 kg.m

= - 600.32.2/52

=

-432 kg.m

= + 600.3.2 2/52

=

288 kg.m

b. Perhitungan Koefisien Distribusi Titik Simpul A = mAB Titik Simpul D

2

5m Titik Simpul B B

C 2.EI

5m

EI A

mBA

=

mBC

=

kBA kBA + kBC kBC kBA + kBC

kBA

=

kBC

=

4.EIBA LBA 4.EIBC LBC

= =

4.EI 5 4.2EI 5

=

0.80

=

1.60

=

1.60

=

0.80

maka

mBA

=

mBC

=

0.80 2.40 1.60 2.40

=

0.33333

=

0.66667 1.00000

Titik Simpul C 5m B

C 2.EI EI

5m D

mCB

=

mCD

=

kCB kCB + kCD kCD kCB + kCD

kCB

=

kCD

=

maka

mCB

=

mCD

=

1.60 2.40 0.80 2.40

=

0.66667

=

0.33333 1.00000

c. Proses Distribusi Momen Lihat halaman selanjutnya….

4.EICB LCB 4.EICD LCD

= =

4.2EI 5 4.EI 5

Tabel Distribusi Momen Joint Batang K No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

DF FEM Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal

Total Check Statika

A AB 0.80 0

B C BA BC CB CD 0.80 1.60 1.60 0.80 0.3333333 0.66666667 0.6666667 0.33333333

D DC 0.80 0

-288.00000 0.00000 32.16667 0.00000 10.72222 0.00000 3.57407 0.00000 1.19136 0.00000 0.39712 0.00000 0.13237 0.00000 0.04412 0.00000 0.01471 0.00000 0.00490 0.00000 0.00163 0.00000 0.00054 0.00000 0.00018 0.00000 0.00006 0.00000 0.00002 0.00000 0.00001 0.00000 0.00000 0.00000 0.00000 0.00000 -239.75000

432.00000 -625.00000 625.00000 -432.00000 64.33333 128.66667 -128.66667 -64.33333 0.00000 -64.33333 64.33333 0.00000 21.44444 42.88889 -42.88889 -21.44444 0.00000 -21.44444 21.44444 0.00000 7.14815 14.29630 -14.29630 -7.14815 0.00000 -7.14815 7.14815 0.00000 2.38272 4.76543 -4.76543 -2.38272 0.00000 -2.38272 2.38272 0.00000 0.79424 1.58848 -1.58848 -0.79424 0.00000 -0.79424 0.79424 0.00000 0.26475 0.52949 -0.52949 -0.26475 0.00000 -0.26475 0.26475 0.00000 0.08825 0.17650 -0.17650 -0.08825 0.00000 -0.08825 0.08825 0.00000 0.02942 0.05883 -0.05883 -0.02942 0.00000 -0.02942 0.02942 0.00000 0.00981 0.01961 -0.01961 -0.00981 0.00000 -0.00981 0.00981 0.00000 0.00327 0.00654 -0.00654 -0.00327 0.00000 -0.00327 0.00327 0.00000 0.00109 0.00218 -0.00218 -0.00109 0.00000 -0.00109 0.00109 0.00000 0.00036 0.00073 -0.00073 -0.00036 0.00000 -0.00036 0.00036 0.00000 0.00012 0.00024 -0.00024 -0.00012 0.00000 -0.00012 0.00012 0.00000 0.00004 0.00008 -0.00008 -0.00004 0.00000 -0.00004 0.00004 0.00000 0.00001 0.00003 -0.00003 -0.00001 0.00000 -0.00001 0.00001 0.00000 0.00000 0.00001 -0.00001 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 528.50000 -528.50000 528.50000 -528.50000 0.00000 0.00000

288.00000 0.00000 -32.16667 0.00000 -10.72222 0.00000 -3.57407 0.00000 -1.19136 0.00000 -0.39712 0.00000 -0.13237 0.00000 -0.04412 0.00000 -0.01471 0.00000 -0.00490 0.00000 -0.00163 0.00000 -0.00054 0.00000 -0.00018 0.00000 -0.00006 0.00000 -0.00002 0.00000 -0.00001 0.00000 0.00000 0.00000 0.00000 0.00000 239.75000

Proses Distribusi Momen Joint A Batang AB K kAB 0 No. DF (m)

BA kAB mBA

BC kBC mBC

CB kCB mCB

CD kCD mCD

D DC kDC 0

FEM

MoAB

MoBA

MoBC

MoCB

MoCD

MoDC

Bal

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

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

bagi ( CO ) sudah

mendekati nilai 0 ,

( 0,00000 )

0.00000

0.00000

0.00000

1 2 3 4 5

CO

B

C

(MoBA+MoBC).(-mBA) (MoBA+MoBC).(-mBC) (MoCB+MoCD).(-mCB) (MoCB+MoCD).(-mCD)

Proses tersebut diatas diulang hingga hasil

0 F = Bal. 1 CD / 2

Bal CO

0.00000

Bal Total Check Statika

0.00000 MAB

0.00000

0.00000

0.00000 0.00000 MBA MBC jumlah satu titik simpul harus = 0

0.00000 0.00000 MCB MCD jumlah satu titik simpul harus = 0 O

MAB Diperoleh dari hasil penjumlahan kebawah (menurut kolom) dimulai dari Momen Primer M AB sampai pada kondisi balance (bal) sudah bernilai 0,00000. Untuk Momen Total pada Titik simpul yang lain dilakukan dengan cara yang sama.

0.00000 MDC

d. Perhitungan Momen Desain Karena 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.m MBA = 528.5 kg.m MBC = -528.5 kg.m MCB = 528.5 kg.m MCD = -528.5 kg.m MDC = 239.8 kg.m

l. Free Body untuk Perhitungan Reaksi Perletakan Batang AB SMB = 0 RAH = {(600.2)+(239,75)-(528,5)}/(5) SH = 0 RAH + RBH = P RBH = P - RAH

=

182.25 kg

f

=

417.75 kg

f

750.00 kg

h

750.00 kg

h

Batang BC SMC = 0 RBV = {(300.5.5/2)+(528,5)-(528,5)}/(5) = SH = 0 RBV + RCV = Q RCV = Q - RBV =

Batang CD RCH = - RBH = RDH = - RAH =

g -417.7 kg g -182.3 kg

m. Perhitungan Momen, Gaya Lintang dan Gaya Normal Potongan AB interval Mx1 = Dx1 = Nx1 = Mx2 = Dx2 = Nx2 =

RAH.X - MAB RAH - RAV = RBV RAH.(3+X) - P.X - MAB RAH - P - RAV

Pot. AB Interval 0
0
x 0 1 2 3 0 1 2

Mx

0
Potongan BC 2 Mx = RBV.x - 1/2.q.x - MBC Dx = RBV - q.x Nx = - RBH

interval 0
0
Dx

-239.8 182.3 -57.5 182.3 124.7 182.3 307.0 182.3 307.0 -417.7 -110.7 -417.7 -528.5 -417.7

Nx -750 -750 -750 -750 -750 -750 -750

Pot. BC Interval 0
Gbr. Bidang Momen

x Mx Dx 0 -528.5 750 0.5 -191 600 71.5 450 1 1.5 259 300 2 371.5 150 2.5 409.0 0 3 371.5 -150 3.5 259 -300 71.5 -450 4 4.5 -191 -600 5 -528.5 -750

Gbr. Bidang Lintang

Gbr. Bidang Normal

Nx -418 -418 -418 -418 -418 -418 -418 -418 -418 -418 -418

No. 02. Analisislah Portal berikut dengan Metode Cross

a

sudut ( a )= 36.87

a. Perhitungan MOMEN PRIMER (M O) MoBE

=

PL

= + 400.1

=

400 kg.m

o

=

- PL/8

= - 1000.4/8

=

-500 kg.m

o

=

+ PL/8

= + 1000.4/8

=

500 kg.m

M BC M CB

Momen primer akibat pergoyangan MXAB = MXBA = MAB=MBA (6.EIAB.Dh)/LAB2 = -0.375 EI.DH misal EI.DH = X, maka MXAB = MXBA = MBC=MCB

MXBC = MXCB =

MCD=MDC

MXCD = MXDC =

MAB=MBA = MBC=MCB = MCD=MDC =

-37.5 .X 100 .X -40 .X

b. Perhitungan Koefisien Distribusi Titik Simpul A

mAB

= Titik Simpul D

0

mDC

0

=

-37.5 X Dv = Dh / Tan a

=

(6.EIBC.Dv)/LBC2 1.000 X

Dr = Dh / Sin a

=

(6.EICD.Dr)/LCD2 -0.400 X

o

4m Titik Simpul B B

C 2.EI

4m

EI A

mBA

=

mBC

=

kBA kBA + kBC kBC kBA + kBC

kBA

=

kBC

=

4.EIBA LBA 4.EIBC LBC

= =

4.EI 4 4.2EI 4

=

1.00

=

2.00

=

2.00

=

0.80

maka

mBA

=

mBC

=

1.00 3.00 2.00 3.00

=

0.33333

=

0.66667 1.00000

Titik Simpul C 4m B

C 2.EI EI

5m D

mCB

=

mCD

=

kCB kCB + kCD kCD kCB + kCD

kCB

=

kCD

=

maka

mCB

=

mCD

=

2.00 2.80 0.80 2.80

=

0.71429

=

0.28571 1.00000

c. Proses Distribusi Momen Lihat halaman selanjutnya….

4.EICB LCB 4.EICD LCD

= =

4.2EI 4 4.EI 5

Tabel Distribusi Momen, akibat Beban Luar Joint Batang K No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

DF FEM Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal

Total Check Statika

A AB 1.00 0

BA 1.00 0.33333333

B BE 0.00 0

C BC CB CD 2.00 2.00 0.80 0.66666667 0.71428571 0.28571429

0.00000 0.00000 400.00000 -500.00000 500.00000 0.00000 0.00000 33.33333 0.00000 66.66667 -357.14286 -142.85714 16.66667 0.00000 0.00000 -178.57143 33.33333 0.00000 0.00000 59.52381 0.00000 119.04762 -23.80952 -9.52381 29.76190 0.00000 0.00000 -11.90476 59.52381 0.00000 0.00000 3.96825 0.00000 7.93651 -42.51701 -17.00680 1.98413 0.00000 0.00000 -21.25850 3.96825 0.00000 0.00000 7.08617 0.00000 14.17234 -2.83447 -1.13379 3.54308 0.00000 0.00000 -1.41723 7.08617 0.00000 0.00000 0.47241 0.00000 0.94482 -5.06155 -2.02462 0.23621 0.00000 0.00000 -2.53077 0.47241 0.00000 0.00000 0.84359 0.00000 1.68718 -0.33744 -0.13497 0.42180 0.00000 0.00000 -0.16872 0.84359 0.00000 0.00000 0.05624 0.00000 0.11248 -0.60257 -0.24103 0.02812 0.00000 0.00000 -0.30128 0.05624 0.00000 0.00000 0.10043 0.00000 0.20086 -0.04017 -0.01607 0.05021 0.00000 0.00000 -0.02009 0.10043 0.00000 0.00000 0.00670 0.00000 0.01339 -0.07173 -0.02869 0.00335 0.00000 0.00000 -0.03587 0.00670 0.00000 0.00000 0.01196 0.00000 0.02391 -0.00478 -0.00191 0.00598 0.00000 0.00000 -0.00239 0.01196 0.00000 0.00000 0.00080 0.00000 0.00159 -0.00854 -0.00342 0.00040 0.00000 0.00000 -0.00427 0.00080 0.00000 0.00000 0.00142 0.00000 0.00285 -0.00057 -0.00023 0.00071 0.00000 0.00000 -0.00028 0.00142 0.00000 0.00000 0.00009 0.00000 0.00019 -0.00102 -0.00041 0.00005 0.00000 0.00000 -0.00051 0.00009 0.00000 0.00000 0.00017 0.00000 0.00034 -0.00007 -0.00003 0.00008 0.00000 0.00000 -0.00003 0.00017 0.00000 0.00000 0.00001 0.00000 0.00002 -0.00012 -0.00005 0.00001 0.00000 0.00000 -0.00006 0.00001 0.00000 0.00000 0.00002 0.00000 0.00004 -0.00001 0.00000 0.00001 0.00000 0.00000 0.00000 0.00002 0.00000 0.00000 0.00000 0.00000 0.00000 -0.00001 -0.00001 0.00000 0.00000 0.00000 -0.00001 0.00000 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 0.00000 0.00000

D DC 0.80 0 0.00000 0.00000 -71.42857 0.00000 -4.76190 0.00000 -8.50340 0.00000 -0.56689 0.00000 -1.01231 0.00000 -0.06749 0.00000 -0.12051 0.00000 -0.00803 0.00000 -0.01435 0.00000 -0.00096 0.00000 -0.00171 0.00000 -0.00011 0.00000 -0.00020 0.00000 -0.00001 0.00000 -0.00002 0.00000 0.00000 0.00000 0.00000 0.00000 -86.48649

Tabel Distribusi Momen, akibat Pelepasan Pendel 1 Joint Batang K No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

DF FEM Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal CO Bal

Total Check Statika

A AB 1.00 0

BA 1.00 0.333333

-37.50000 -37.50000 0.00000 -20.83333 -10.41667 0.00000 0.00000 7.14286 3.57143 0.00000 0.00000 -2.48016 -1.24008 0.00000 0.00000 0.85034 0.42517 0.00000 0.00000 -0.29526 -0.14763 0.00000 0.00000 0.10123 0.05062 0.00000 0.00000 -0.03515 -0.01757 0.00000 0.00000 0.01205 0.00603 0.00000 0.00000 -0.00418 -0.00209 0.00000 0.00000 0.00143 0.00072 0.00000 0.00000 -0.00050 -0.00025 0.00000 0.00000 0.00017 0.00009 0.00000 0.00000 -0.00006 -0.00003 0.00000 0.00000 0.00002 0.00001 0.00000 0.00000 -0.00001 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 -45.27027 -53.04054

B BE 0.00 0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000

C BC CB CD 2.00 2.00 0.80 0.66666667 0.71428571 0.285714 100.00000 -41.66667 -21.42857 14.28571 7.44048 -4.96032 -2.55102 1.70068 0.88577 -0.59051 -0.30369 0.20246 0.10545 -0.07030 -0.03615 0.02410 0.01255 -0.00837 -0.00430 0.00287 0.00149 -0.00100 -0.00051 0.00034 0.00018 -0.00012 -0.00006 0.00004 0.00002 -0.00001 -0.00001 0.00000 0.00000 0.00000 0.00000 0.00000 53.04054

D DC 0.80 0

100.00000 -40.00000 -40.00000 -42.85714 -17.14286 0.00000 -20.83333 0.00000 -8.57143 14.88095 5.95238 0.00000 7.14286 0.00000 2.97619 -5.10204 -2.04082 0.00000 -2.48016 0.00000 -1.02041 1.77154 0.70862 0.00000 0.85034 0.00000 0.35431 -0.60739 -0.24295 0.00000 -0.29526 0.00000 -0.12148 0.21090 0.08436 0.00000 0.10123 0.00000 0.04218 -0.07231 -0.02892 0.00000 -0.03515 0.00000 -0.01446 0.02511 0.01004 0.00000 0.01205 0.00000 0.00502 -0.00861 -0.00344 0.00000 -0.00418 0.00000 -0.00172 0.00299 0.00120 0.00000 0.00143 0.00000 0.00060 -0.00102 -0.00041 0.00000 -0.00050 0.00000 -0.00020 0.00036 0.00014 0.00000 0.00017 0.00000 0.00007 -0.00012 -0.00005 0.00000 -0.00006 0.00000 -0.00002 0.00004 0.00002 0.00000 0.00002 0.00000 0.00001 -0.00001 -0.00001 0.00000 -0.00001 0.00000 0.00000 0.00001 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 52.70270 -52.70270 -46.35135 0.00000

Perhitungan Reaksi Akibat Beban Luar Tabel Distribusi Momen Joint A B C Batang AB BA BE BC CB CD M. Beban Luar 52.70270 105.40540 400.00000 -505.40540 172.97297 -172.973 M. Pendel 1 -45.27027 -53.04054 0.00000 53.04054 52.70270 -52.7027

105.40540 158.10811 SMB = 0 B RAH.h + MAB + MBA = 0 RAH = 39.527027 Ton SH = 0 h= 4 RAH + RBH = 0 RBH = -39.52703 Ton A 52.70270

416.89189 -172.973 C

SMD = 0 RCH.h + MCD + MDC - RCV.4 = 0 RCH = 642.34 Ton g SH = 0 RCH + RDH = 0 RDH = -642.3 Ton f

g r=

D DC -86.48649 -46.35135

5

f D -86.486

1000 Kg -505.40540 B

172.9730 2

2

C

4 SMC = 0 RBV.L + MBC + MCB - P.2 = 0 RBV = 583.10811 Kg SV = 0 RBV + RCV = P RCV = 416.89189 Kg

583.1081

HOBC =

416.8918918

-39.52703

602.815 Ton

g

642.34234 B

C

HOAB = -602.82 Ton f D 39.52703 A

-642.3423

Perhitungan Reaksi Akibat Pendel 1 Tabel Distribusi Momen A Joint Batang AB BA M. Beban Luar 52.70270 105.40540 M. Pendel 1 -45.27027 -53.04054

B C D BE BC CB CD DC 400.00000 -505.40540 172.97297 -172.97297 -86.48649 0.00000 53.04054 52.70270 -52.70270 -46.35135

-53.04054 SMB = 0 B RAH.h + MAB + MBA = 0 RAH = -24.57770271 Ton SH = 0 h3 = 4 RAH + RBH = 0 RBH = 24.57770271 Ton A -45.27027

53.04054 B

26.43581 SMD = 0 RCH.h + MCD + MDC - RCV.4 = 0 RCH = 68.2658 Ton g SH = 0 RCH + RDH = 0 RDH = -68.266 Ton f

-52.7027 C

f r=

5

g D -46.35135

52.70270 2

2

C

4 SMC = 0 RBV.L + MBC + MCB = 0 RBV = -26.43581078 Kg SV = 0 RBV + RCV = 0 RCV = 26.43581078 Kg

-26.4358

x

H BC = 92.84347 Ton

26.43581078

24.57770

g

68.26576573 B

C

X

H AD = -92.843 Ton f D -24.57770 A

-68.26576573

Perhitungan Momen Desain Karena 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.m MBA = 105.4 kg.m MBC = -505.4 kg.m MCB = 173 kg.m MCD = -173 kg.m MDC = -86.49 kg.m

Perhitungan Harga X Tabel Distribusi Momen Joint Batang M. Beban Luar M. Pendel 1 M.P1 . X Momen Akhir / Desain Check Statika

A AB 52.70270 -45.27027 -193.6664 -140.9637

BA 105.40540 -53.04054 -226.90762 -121.50221

-140.9637 -121.50221 Harga X = 4.2780035

O

-397.1847 kg

X

92.843468 X

H BC = H BC =

f g

Syarat Statika : O X H BC + H BC = 0 -397.1847 + 92,84347.X = 0 Maka Harga X X=

4.2780035

B BE 400.00000 0.00000 0.00000 400.00000 0.00000 400

C

D DC -86.48649 -46.35135 -198.2912 -284.7777

BC CB CD -505.40540 172.97297 -172.97297 53.04054 52.70270 -52.70270 226.90762 225.46235 -225.4623 -278.49779 398.43532 -398.4353 0.00000 -278.4977864 398.43532 -398.4353205 -284.7777

Perhitungan Reaksi Akhir Tabel Momen Akhir A B C D Joint Batang AB BA BE BC CB CD DC M. Beban Luar 52.70270 105.40540 400.00000 -505.40540 172.97297 -172.97297 -86.48649 M. Pendel 1 -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 M. Akhir

-121.5022 SMB = 0 B RAH.h3 + MAB + MBA = 0 RAH = -65.61647217 Ton SH = 0 h3 = 4 RAH + RBH = 0 RBH = 65.61647217 Ton A -140.964

529.984 -398.435

SMD = 0 RCH.h + MCD + MDC - RCV.4 = 0 RCH = 934.3835278 Ton g SH = 0 RCH + RDH = 0 RDH = -934.3835278 Ton f D -284.7777

C

f L=5

g

1000 Kg -278.4978 B

398.435 2

2

C

4 SMB = 0 RBV.L + MBC + MCB - P.2 = 0 RBV = 470.0156166 Kg SV = 0 RBV + RCV = P RCV = 529.9843834 Kg

HBC = 470.0156166

1000.0 Ton g

529.9843834 934.383528

65.616 B

C

HAD = D

-934.3835278

-65.616 A 870.015617

-1000.00 Ton f

529.984383

Free Body untuk Perhitungan Reaksi Perletakan Batang AB SMB = 0 RAH = {(600.2)+(239,75)-(528,5)}/(5) SH = 0 RAH + RBH = P RBH = P - RAH

=

229.46 kg

f

=

370.54 kg

f

816.49 kg

h

683.51 kg

h

Batang BC SMC = 0 RBV = {(300.5.5/2)+(528,5)-(528,5)}/(5) = SH = 0 RBV + RCV = Q RCV = Q - RBV =

Perhitungan Momen, Gaya Lintang dan Gaya Normal Potongan AB interval Mx1 = Dx1 = Nx1 = Mx2 = Dx2 = Nx2 =

RAH.X - MAB RAH - RAV = RBV RAH.(3+X) - P.X - MAB RAH - P - RAV

Pot. AB Interval 0
0
x 0 1 2 3 0 1 2

Mx

0
Batang CD RCH = - RBH = RDH = - RAH =

g -370.5 kg g -229.5 kg

Potongan BC 2 Mx = RBV.x - 1/2.q.x - MBC Dx = RBV - q.x Nx = - RBH

interval 0
0
Dx

-52.7 229.5 176.8 229.5 406.2 229.5 635.7 229.5 635.7 -370.5 265.1 -370.5 -105.4 -370.5

Nx -816.5 -816.5 -816.5 -816.5 -816.5 -816.5 -816.5

Pot. BC Interval 0
x 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Mx

Dx

-505.4 816.5 -134.7 666.5 161.1 516.5 381.8 366.5 527.6 216.5 598.3 66.49 594.1 -83.51 514.8 -233.5 360.5 -383.5 131.3 -533.5 -173 -683.5

Coba anda selesaikan Gambar Bidang Momen, Lintang dan Normalnya.

Nx -371 -371 -371 -371 -371 -371 -371 -371 -371 -371 -371