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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

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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

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

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.

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

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