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PERBAIKAN LO 2 ANALISIS STRUKTUR STATIS TAK TENTU (KELAS D) Dosen Pengampu: Miftahul Iman S.T., M.Eng. Disusun Oleh: Fildzah Adhania Johanes Paransa 13 511 178 PROGRAM STUDI TEKNIK SIPIL FAKULTAS TEKNIK SIPIL DAN PERENCANAAN UNIVERSITAS ISLAM INDONESIA FILDZAH ADHANIA JOHANES PARANSA / 13 511 178 PERBAIKAN NILAI LO 2 AS. STATIS TAK TENTU, KELAS D

Analisis Struktur Statis Tak Tentu LO 2

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Analisis Struktur Statis Tak Tentu LO 2

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PERBAIKAN LO 2ANALISIS STRUKTUR STATIS TAK TENTU (KELAS D)

Dosen Pengampu:Miftahul Iman S.T., M.Eng.

Disusun Oleh:Fildzah Adhania Johanes Paransa13 511 178

PROGRAM STUDI TEKNIK SIPILFAKULTAS TEKNIK SIPIL DAN PERENCANAANUNIVERSITAS ISLAM INDONESIA2014/2015UTS NO. 3Determine the internal moment at each support of the beam. The moment of inertia of each span is indicated.

Structure Deformation Caused by the Loads

Primary MomentsFEM AB= -2000 N x 2 m= -4000 NmFEM BA= 2000 N x 2 m= 4000 NmFEM BC= ( 1500 N x 42 ) / 12= -2000 NmFEM CB= 2000 NmFEM CD= 0 Nm

Beam StiffnessKAB = KBA=KBC = KCB= ( 4EI ) / L= ( 4E x 300 x 106 ) / 4= 300 x 106 EKCD = KDC= ( 4EI ) / L= ( 4E x 240 x 106 ) / 3= 320 x 106 E

Distribution NumberDFAB= DFBA = 0DFBC= 1DFCB= 0,484DFCD= 0,516DFDC= 0

Distribution TableJointBCD

MemberBCCBCDDC

DF010.4840.5160

FEM4000.000-2000.0002000.000

Dist-2000.000-968.000-1032.000

CO-484.000-1000.000-516.000

Dist484.000484.000516.000

CO242.000242.000258.000

Dist-242.000-117.128-124.872

CO-58.564-121.000-62.436

Dist58.56458.56462.436

CO29.28229.28231.218

Dist-29.282-14.172-15.110

CO-7.086-14.641-7.555

Dist7.0867.0867.555

CO3.5433.5433.777

Dist-3.543-1.715-1.828

CO-0.857-1.772-0.914

Dist0.8570.8570.914

CO0.4290.4290.457

Dist-0.429-0.207-0.221

CO-0.104-0.214-0.111

Dist0.1040.1040.111

M4000.000-4000.000587.015-587.015-293.563

Joint Reactions

Shearing Force Diagram (SFD)

Bending Moment Diagram (BMD)

TUGAS HARDY CROSS1. Analyze the frame below by moment distribution method.

Structure Deformation Caused by the Load

Pimary MomentsFEM AB= ( wL2 ) / 12= ( -3.6 x 202 ) / 12= -120 kftFEM BA= 120 kftFEM BC= (-PL ) / 8= ( -32 x 20 ) / 8= -80 kftFEM CB= 80 kft

Beam StiffnessKAB= ( 4 / 3 ) x ( EI / L )= ( 4 / 3 ) x ( 2EI / 20 )= 2/15 KBA= ( 4 / 3 ) x ( EI / L )= ( 4 / 3 ) x ( 2EI / 20 )= 2/15 KBF= 4EI / L= 4EI / 10= 2/5 KFB= 4EI / L= 4EI / 10= 2/5 KBC= 4EI / L= 4EI / 20= 1/5 KCB= 4EI / L= 4EI / 20= 1/5 KCD= ( 4 / 3 ) x ( EI / L )= ( 4 / 3 ) x ( EI / 9 )= 4/27KDC= ( 4 / 3 ) x ( EI / L )= ( 4 / 3 ) x ( EI / 9 )= 4/27KCE= 4EI / L= 4EI / 10= 2/5KEC= 4EI / L= 4EI / 10= 2/5

Distribution NumberKB= KBA + KBF + KBC= (2/15) + (2/5) +(1/5)= 11/15KBA= KBA / KB= 2/11Total = 1

KBF= KBF / KB= 6/11KBC= KBC / KB= 3/11

KC= KCB + KCD + KCE= (1/5) + (4/27) + (2/5)= 101/135KCB= KCB / KC= 27/101Total = 1

KCD= KCD / KC= 20/101KCE= KCE / KC= 54/101

Distribution TableFBCE

FBBFBABCCBCDCEEC

K0.0000.5450.1820.2730.2670.1980.5350.000

FEM0.0000.000120.000-80.00080.0000.0000.0000.000

-10.693-21.386-15.842-42.772-21.386

7.99315.986-5.329-7.993-3.996

0.5341.0680.7912.1371.068

-0.146-0.291-0.097-0.146-0.073

0.0100.0190.0140.0390.019

15.694114.574-98.28855.632-15.036-40.597

7.84731.9810.000-20.298

ORFBCEC

FBBFBABCCBCDCEEC

K0.0000.5450.1820.2730.2670.1980.5350.000

FEM0.0000.000120.000-80.00080.0000.0000.0000.000

-32.727-65.45514.54521.818-21.386-15.842-42.772-21.386

2.9165.8331.944-10.69310.909-2.160-5.833-2.916

2.916-2.916

0.3980.7950.265-1.4581.458-0.289-0.780-0.390

0.398-0.390

0.0530.1060.035-0.1950.199-0.039-0.106-0.053

0.053-0.053

--58.720136.790-67.16167.821-18.330-49.491-

-29.36010.9090.000-24.745

OR

Joints Reactions

Normal Force Diagram (NFD)

Shearing Force Diagram (SFD)

Bending Moment Diagram (BMD)

2. Analyze the frame below by moment distribution method.

Structure Deformation Caused by the Load

Primary MomentsFEM AB = FEM BA= 0 kftFEM BC = FEM CB= ( wL2 ) / 12 = ( 4 x 402 ) / 12= 1600/3 kftFEM CD = FEM DC= 0 kft

Beam StiffnessKAB = KCD= ( 3 / 4 ) x ( EI / L )= ( 3 / 4 ) x ( 360 / 18 )= 15KBC = ( 1 / 2 ) x ( EI / L )= ( 1 / 2 ) x ( 600 / 40 )= 7,5

Distribution NumberDFBA = DFCD= KAB / Ks= 15 / (15 + 7,5)= 2/3DFBC = DFCB= KBC / Ks= 7,5 / (15 + 7,5)= 1/3

Distribution TableBC

BABCCBCD

K0.6670.3330.3330.667

FEM0.000-533.333533.3330.000

355.556177.778-177.778-355.556

59.259-88.88988.889-59.259

29.630-29.630

9.877-14.81514.815-9.877

4.938-4.938

1.646-2.4692.469-1.646

0.823-0.823

-0.549-0.4120.412-0.274

0.137-0.137

-0.091-0.0690.0690.091

0.023-0.023

-0.015-0.0110.0110.015

0.006-0.006

-0.004-0.0020.0020.004

0.001-0.001

-0.0010.0000.0000.001

0.0000.000

0.0000.0000.0000.000

425.678-426.664426.664-426.501

-0.9870.164

Joints Reactions

Normal Force Diagram (NFD)

Shearing Force Diagram (SFD)

Bending Moment Diagram (BMD)

3. If girder AB of the rigid frame in figure below is fabricated 1.92 in too long. What moments are created in the frame when it is erected? Given: E = 29,000 kips/in2.

Add 1.92 in to the end of girder AB, and erect the frame with a clamp at joint B to prevent rotation (see figure above). Compute the fixed-end moments in the clamped structure using the slope-deflection equation.Column BC:B = C = 0BC= (1,92) / (12x12)= +0,0133 radFEM BC = FEM CB= 0MBC = MCB= = = -5785,5 kip.in = 482,13 kip.ftNo moments develop in member AB because AB = A = B = 0.

Distribution FactorsKAB = I / L= 450 / 30= 15KBC= 360 / 12= 30K= 15 + 30= 45DFBA= KAB / K= 15 / 45= 1/3DFBC= KBC / K= 30 / 45= 2/3

Moment Distribution Analysis

Distribution TableB

BABC

K0.3333330.666667

FEM482.13-482.13

-160.71321.42

160.71-80.355

-53.5753.57

0

4. Determine the reactions and the member end moments produced in the frame shown in figure below by a load of 5 kips at joint B. Also determine the horizontal displacement of girder BC. Given: E = 30,000 kips/in2. Units of I are in in4.

Structure Deformation Caused by the Load

MAB = MBA= -6EI / L2= -6(30000)(100) / (20 x 12)2= -312 kip.in= -26 kip.ftMCD = MDC= -6EI / L2= -6(30000)(200) / (40 x 12)2= -166 kip.in= -13 kip.ft

Joint B:KAB= ( 3 / 4 ) x ( I / L )= ( 3 / 4 ) x ( 100 / 20 )= 15/4KBC= I / L= 200 / 40= 20/4K= 35 / 4DFAB= KAB / K= 3/7DFBC= KBC / K= 4/7

Joint C:KCB= I / L= 200 / 40= 5KCD= I / L= 200 / 40= 5K= 10DFCB= KCB / K= 5/10= DFCD= KCD / K= 5/10=

Distribution TableBC

BABCCBCD

K0.4285710.5714290.50.5

FEM-26-13

11.1428614.857146.56.5

3.257.428571

-1.39286-1.85714-3.71429-3.71429

-1.85714-0.92857

0.7959181.0612240.4642860.464286

-15.454115.454089.75-9.75

00

Joints Reactions

Normal Force Diagram (NFD)

Shearing Force Diagram (SFD)

Bending Moment Diagram (BMD)

FILDZAH ADHANIA JOHANES PARANSA / 13 511 178PERBAIKAN NILAI LO 2AS. STATIS TAK TENTU, KELAS D