analisis struktur statis tak tentu lo 2
DESCRIPTION
Analisis Struktur Statis Tak Tentu LO 2TRANSCRIPT
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