Moment Distribution
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2.25 m 2.25 m A B C 3.5 m 80 kN 65.5 kN/m I 2I CONTOH 1 Lukiskan GDR dan GML bagi struktur di baa!" #. $akt%r Agi!an &$.A' F.A AB = 1 &!u(ung s%k%ng ada)a! *in' JOIN T ANGGOT A K ΣK F.A B BA +,I/L - +,I/+.5 - 0. 0,I &+,I/+.5' &8,I/3.5' - 3.#8,I 0.9EI/3.18EI = 0.28 BC &2'+,I/L - 8,I/3.5 - 2.2 ,I 2.29EI/3.18EI = 0.72 F.A CB = 0 &!u(ung s%k%ng ada)a! diikat t gar' 2. M%m n 1u(ung rikat &M 1 ' − M AB HT - M BA HT - ± Pa b 2 L 2 - ± ( 80 ) ( 2.25 )( 2.25 ) 2 4.5 2 - +5 kNm − M BC HT - M CB HT - − W L 2 12 - 3.5 ¿ ¿ ¿ 2 − ( 65.5 ) ¿ ¿ - 66.86 kNm 3. Agi!an M%m n
description
MOMENT
Transcript of Moment Distribution
CONTOH 1Lukiskan GDR dan GML bagi struktur di bawah:
2.25 m2.25 mABC3.5 m80 kN65.5 kN/mI2I
1. Faktor Agihan (F.A)F.AAB=1 (hujung sokong adalah pin)JOINTANGGOTAKKF.A
BBA4EI/L = 4EI/4.5 = 0.90EI(4EI/4.5) + (8EI/3.5) = 3.18EI0.9EI/3.18EI = 0.28
BC(2)4EI/L = 8EI/3.5 = 2.29EI 2.29EI/3.18EI = 0.72
F.ACB=0 (hujung sokong adalah diikat tegar)
2. Momen Hujung Terikat (MHT)
====45 kNm == ==66.86 kNm
3. Agihan Momen
Anggota ABBABCCB
FBS0.50.50.50
FA10.280.720
MHTAgih -4545456.12-66.8615.7466.860
BSAgih3.06-3.0622.5-6.300-16.27.870
BSAgih -3.153.15-1.530.4301.10-8.100
BSAgih0.22-0.221.56-0.440-1.120.550
BSAgih-0.220.22-0.110.0300.08-0.560
BSAgih0.02-0.020.11-0.030-0.080.040
Momen Akhir067.34-67.3466.66
(45 66.86) x 0.28 = -6.12(45 66.86) x 0.72 = -15.74(22.5 0) x 0.28 = 6.30(22.5 0) x 0.72 = 16.20
4. Tindak Balas
MB = 0;RA (4.5) + 67.34 (80 x 2.25) = 0 RA = 25.04 kNRY = 0;25.04 + RB1 -80 = 0 RB1 = 54.96 kN67.34 kN/m80 kNRentang AB
BA
I
2.25 m2.25 mRB1RA
66.66 kN/m66.34 kN/mMC = 0;RB2 (3.5) + 66.66 - 66.34 - (65.5 x 3.5 x 3.5/2) = 0 RB2 = 114.53 kNRY = 0;114.53 + RC (65.5 x 3.5) = 0 RC = 114.72 kNRentang BC
CB65.5 kN/mRCRB2
3.5 m2I
5. 114.53 kNGambarajah Daya Ricih & Momen Lentur
1.75 m25.04 kN
-54.96 kN
1.75 m-114.72 kN
66.66 kN/m-57.17 kN/m-33.72 kN/m66.49 kN/m
3 m3 mABC1 m32 kN12 kN/mEI2EI CONTOH 2
1. Faktor Agihan (F.A)F.AAB=1 (hujung sokong adalah pin)JOINTANGGOTAKKF.A
BBA4EI/L = 4EI/3 = 1.33EI(1.33EI) + (2EI) = 3.33EI1.33EI/2.33EI = 0.40
BC(2)4EI/L = 8EI/4 = 2EI 2EI/3.33EI = 0.60
F.ACB=1 (hujung sokong adalah rola)
2. Momen Hujung Terikat (MHT)
====9 kNm= - = =-22 kNm= + = + =34 kNm
3. Agihan Momen
Anggota ABBABCCB
FBS0.50.50.50.5
FA10.400.601
MHTAgih -9995.2-227.834-34
BSAgih2.6-2.64.55-177.53.9-3.9
BSAgih 2.5-2.5-1.31.2-1.51.633.75-3.75
BSAgih0.6-0.6-1.251.25-1.881.880.82-0.82
Momen Akhir023.6-23.60
4. Tindak Balas
12 kN/mMB = 0;RA (3) + 23.6 (12 x 3 x 1.5) = 0 RA = 10.13 kNRY = 0;10.13 + RB1 (12 x 3) = 0 RB1 = 25.87 kN23.6 kN/mRentang AB
AB
I
3 mRB1RA
MC = 0;RB2 (4) - 23.6 - (12 x 4 x 2) (32 x 1) = 0 RB2 = 37.9 kNRY = 0;37.9 + RC (12 x 4) - 32 = 0 RC = 42.1 kNRentang BC
32 kN23.6 kN/m
12 kN/mCBRCRB2
1 m3 m2I
5. 37.9 kNGambarajah Daya Ricih & Momen Lentur
10.13 kN
2.16 m1.9 kN
0.84 m
-25.87 kN
-30.1 kN
-23.67 kN/m-42.1kN
66.66 kN/m
4.26 kN/m
36.02 kN/m
CONTOH 3 (KEKUKUHAN UBAHSUAI)
2.25 m2.25 mABC3.5 m80 kN65.5 kN/mI2I
1. Faktor Agihan (F.A)F.AAB=1 (hujung sokong adalah pin)JOINTANGGOTAKKF.A
BBA(3/4)(4EI/L) = 12EI/(4x4.5) = 0.67EI(0.67EI) + (2.29EI) = 2.96EI0.67EI/2.96EI = 0.23
BC(2)4EI/L = 8EI/3.5 = 2.29EI 2.29EI/2.96EI = 0.77
F.ACB=0 (hujung sokong adalah diikat tegar)
2. Momen Hujung Terikat (MHT)
====45 kNm== ==66.86 kNm
3. Agihan Momen
Anggota ABBABCCB
FBS0.500.50
FA10.230.770
MHTAgih -4545455.03-66.8616.8366.860
BSAgih0022.5-5.180-17.338.420
BSAgih000000-8.60
Momen Akhir067.35-67.3566.68
3 m3 mABC1 m32 kN12 kN/mEI2EICONTOH 4 (UBAHSUAI KEKUKUHAN)
1. Faktor Agihan (F.A)F.AAB=1 (hujung sokong adalah pin)JOINTANGGOTAKKF.A
BBA(3/4)4EI/L = (3/4)4EI/3 = EI(EI) + (1.5EI) = 2.5EIEI/2.5EI = 0.40
BC(3/4)(2)4EI/L = (3/4)8EI/4 = 1.5EI 1.5EI/2.5EI = 0.60
F.ACB=1 (hujung sokong adalah rola)
2. Momen Hujung Terikat (MHT)
====9 kNm= - = =-22 kNm= + = + =34 kNm
3. Agihan Momen
Anggota ABBABCCB
FBS0.5000.5
FA10.400.601
MHTAgih -9995.2-227.834-34
BSAgih004.55-177.500
BSAgih 00000000
Momen Akhir023.7-23.70
CONTOH 5 Tentukan tindakbalas pada setiap sokong dan lukiskan GDR & GML jika diketahui pada sokong B turun sebanyak 4 mm. EI malar sepanjang rasuk dengan nilai EI = 3000 kNm2
25 kN45 kN
25 kN/m
BCDA
4 m1 m5 m4 m
Mas
Penurunan pada sokong B akan memberikan kesan kepada nilai momen hujung terikat. Rumus yang digunakan akibat penurunan sokong ini adalah dengan adalah anjakan sokong.
1. Faktor Agihan (F.A)F.ACD=0 (tiada penyokong pada titik D)F.AAB =F.ACB= 1
JOINTANGGOTAKKF.A
BBA(3/4)4EI/L = (3)(4)EI/(4)(5) = 0.6EI 0.6EI + 0.38EI = 0.98EI0.6EI/0.98EI = 0.61
BC(3/4)4EI/L = (3)(4)EI/(4)(8) = 0.38EI 0.38EI/0.98EI = 0.39
2. Momen Hujung Terikat (MHT)
=- - =- =-54.96 kNm= - = =49.2 kNm=- + =- + =-43.87 kNm= += + =46.13 kNm=-25(1)=-25kNm
3. Agihan Momen
Anggota ABBABCCBCD
FBS0.5000.50
FA10.610.3910
MHTAgih -54.9654.9649.2-3.25-43.87-2.0846.13-21.13-250
BSAgih0027.48-10.32-10.57-6.600000
BSAgih0000000000
Momen Akhir063.11-63.1125-25
4. Tindak Balas
Rentang AB
MB = 0;RA (5) + 63.11 (25 x 5 x 2.5) = 0 RA = 49.9 kNRY = 0;49.9 + RB1 (25 x 5) = 0 RB1 = 75.1 kN63.11 kN/m25 kN/m
AB
RB1RA5 m
Rentang BC
MB = 0;-RC1 (8) + (45 x 4) 63.11 + 25 = 0 Rc1 = 17.74 kNRY = 0;17.74 + RB2 - 45 = 0 RB1 = 27.26 kN25 kN/m63.11 kN/m45 kN
CB
4 m4 mRC1RB2
Rentang CD
RY = 0;RC2 - 25 = 0 RC2 = 25 kN25 kN/m25 kN
DC
RC2
5. Gambarajah Daya Ricih & Momen Lentur
49.9 kN
(+)25 kN27.26 kN
(-)
2 m17.74 kN
75.1 kN
63.11 kN/m
(-)(+)25 kN/m
45.93 kN/m
49.9 kN/m
30 kNCONTOH 6 (KERANGKA TEGAR TANPA HUYUNG)
16 kN/m10 kN/m
CA
4EI/3EIB
3 m
EI
D
2 m2 m2 m
1. Faktor Agihan (F.A)F.AAB=0 JOINTANGGOTAKKF.A
BBA4EI/L = 4EI/4 = 1EI(1EI) + (2EI) + (1EI) = 4EI1EI/4EI = 0.25
BC(3/4)(4/3)4EI/L = (3/4)(4/3)4EI/2 = 2EI 2EI/4EI = 0.5
BD(3/4)4EI/L = (3/4)4EI/3 = 1EI1EI/4EI = 0.25
F.ACB=F.ADB =1
2. Momen Hujung Terikat (MHT)
=- - = =- 28.33 kNm= + = + = 28.33 kNm= ==-5.33 kNm= ==5.33 kNm===0 kNm
3. Agihan Momen
Anggota ABBABCBDCBDB
FBS00.5000.50.5
FA10.250.50.2511
MHTAgih -28.3328.3328.33-5.75-5.33-11.50-5.755.33-5.3300
BSAgih-2.88000.67-2.671.3400.670000
BSAgih 0.3400000000000
Momen Akhir-30.8723.25-18.16-5.0800
4. Tindak Balas
30 kNRentang AB
30.87 kN/m10 kN/mMB = 0;-30.87 + 23.25 (30 x 2) + RA (4) (10 x 4 x 2) = 0 RA = 36.91 kNRY = 0;36.91 + RBA (10 x 4) 30 = 0 RBA = 33.09 kN23.25 kN/m
AB
2 m2 mRBARAB
Rentang BC
16 kN/m18.16 kN/mMC = 0;-18.16 + RBC (2) (16 x 2 x 1) = 0 RBC = 25.08 kNRY = 0;25.08 + RCB (16 x 2) = 0 RCB = 6.92 kN
BC
2 mRCBRBC
BRentang BD
HBD5.08 kNm kN/m
MB = 0;-5.08 + HDB (3) = 0 HDB = 1.69 kNFX = 0;HDB = 1.69 kN
HDB
D
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