Fundamentals Of Structural Analysis Episode 2 Part 7 doc

... Solutions Member Force (MN) Elongation (m) Member Force (MN) Elongation (m) 1 0. 375 0.011 8 0 0 2 0. 375 0.011 9 0. 625 0.031 3 0. 375 0.011 10 0 0 4 0. 375 0.011 11 -0. 625 -0.031 5 -0. 625 -0.031 12 -0 .75 0 -0. 023 6 0 0 13 -0 .75 0 -0. 023 7 0. 625 -0.031 Case (a): ... kN −1. 12 kN −1. 12 0 kN 2 kN 2. 24 −1 kN 0 kN 2. 24 0 kN 1 kN 1. 12 2 .24 −1. 12 −1 kN 1 kN 1 kN 1 4 kN...

Ngày tải lên: 05/08/2014, 11:20

... Solutions Member Force (MN) Elongation (m) Member Force (MN) Elongation (m) 1 0. 375 0.011 8 0 0 2 0. 375 0.011 9 0. 625 0.031 3 0. 375 0.011 10 0 0 4 0. 375 0.011 11 -0. 625 -0.031 5 -0. 625 -0.031 12 -0 .75 0 -0. 023 6 0 0 13 -0 .75 0 -0. 023 7 0. 625 -0.031 Case (a): ... kN 0.5 kN 1.5 kN −1. 12 kN −1. 12 0 kN 2 kN 2. 24 −1 kN 0 kN 2. 24 0 kN 1 kN 1. 12 2 .24 −1. 12 −1 kN 1 k...

Ngày tải lên: 05/08/2014, 09:20

... Solutions Member Force (MN) Elongation (m) Member Force (MN) Elongation (m) 1 0. 375 0.011 8 0 0 2 0. 375 0.011 9 0. 625 0.031 3 0. 375 0.011 10 0 0 4 0. 375 0.011 11 -0. 625 -0.031 5 -0. 625 -0.031 12 -0 .75 0 -0. 023 6 0 0 13 -0 .75 0 -0. 023 7 0. 625 -0.031 Case (a): ... 0.00 2 0.011 -0. 073 6 0.045 -0. 073 3 0. 023 -0. 129 7 0. 023 -0. 129 4 0.034 -0. 073 8 0.000 -0. 073 The...

Ngày tải lên: 05/08/2014, 09:20

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Ngày tải lên: 05/08/2014, 09:20

... 8.33 0 17. 63 0 42. 98 σ due to axial force (kN/cm 2 ) 0.0 72 0.0 72 0.088 0.088 0. 125 0. 124 σ due to moment (kN/cm 2 ) 0 0.006 0 0.013 0 0.0 32 Total σ (kN/cm 2 ) 0.0 72 0. 078 0.088 0.101 0. 125 0.156 Error ... = 11. 52 28 .80= − 17 .28 kN-m M bc = 17 .28 kN-m M ab = 5 .76 + 28 .80=34.56 kN-m M cb = 8.64 kN-m. The moment and deflection diagrams are shown below. a b c...

Ngày tải lên: 05/08/2014, 09:20

... Mau 326 Warren, 2 Truss influence lines, 26 3 - 27 1 Unit load, 71 -73 , 87, 88, 133 Unit load method, 71 , 77 , 133, 136, 1 52 Virtual displacement, 71 , 25 3 principle of, 71 Virtual force, 69, 71 , 72 principle ... 2. 02 kN-m, M cb = −0.95 kN-m, M cd = 0.95 kN-m, M dc = 0.00 kN-m Moment and deflection diagrams (Problem 7) -8.3 73 .9 - 52. 5 -22 .5 30 -15 2 m 2 m...

Ngày tải lên: 05/08/2014, 09:20

... (Problem 7) -8.3 73 .9 - 52. 5 -22 .5 30 -15 2 m 2 m 2. 02 2.98 0.95 I nflection point 2 m 2 m 2. 36 −1 .73 2. 18 0. 32 I nflection point -58.3 53.9 -33.3 16 .7 -8.3 -16 .7 Index by S. T. Mau 325 Nodal ... 29 3 Curvature, 123 radius of, 123 D’Alembert force, 29 4 Deflection, 67, 82, 121 , 26 0 transverse, 121 truss, 67 Deflection curve, 144 sketch of, 144 Deforma...

Ngày tải lên: 05/08/2014, 09:20

... M o ( 2 2 x −Lx)+C 2 Condition 2: v = 0 at x = LC 2 = M o 2 2 L EIv = M o ( 2 22 Lx + −Lx) Rotation: θ = v’ = EI M o (x−L) at x=0, v’= − EI M o L Deflection: v = EI M o ( 2 22 Lx + −Lx) ... ) −1 −1/ 12 5/ 12 1/6 M oment(Deflection) Diagram ( Unit: Pa 3 /EI ) a/6 27 01 /77 76=−0.35 −1/3=−0.33 P /2 P /2 Beam and Frame Analysis: Force Method, Part I by S. T. M...

Ngày tải lên: 05/08/2014, 09:20

... −4.69 2. 81 COM 2. 35 −1.41 DM +0 .71 +0 .70 COM +0.36 0.35 DM −0 .22 −0.14 COM −0.11 −0. 07 DM +0.04 +0.03 COM +0. 02 +0. 02 DM −0.01 −0.01 COM 0.00 0.00 SUM 2. 46 −4. 92 +4. 92 +14 . 27 +15 .73 +7. 87 In ... loads 2. 46 4. 92 -14 . 27 15 .73 -7. 87 I nflection point Beam and Frame Analysis: Displacement Method, Part I by S. T. Mau 179 M bc = 2 1 ( M ba + M bc ) = 2...

Ngày tải lên: 05/08/2014, 09:20

... +0 .71 +0 .70 COM +0.36 0.35 DM −0 .22 −0.14 COM −0.11 −0. 07 DM +0.04 +0.03 COM +0. 02 +0. 02 DM −0.01 −0.01 COM 0.00 0.00 SUM 2. 46 −4. 92 +4. 92 +14 . 27 +15 .73 +7. 87 In the above table, the encircled ... reversed: M F ab = 2 kN-m M F bc = 2 kN-m 2 m 2 m 4 m a d b 4 kN c 2 m2 m 4 kN 1.33 1.33 1.33 2. 67 2. 67 I nflection point Beam and Frame Analysis: Displacement Me...

Ngày tải lên: 05/08/2014, 09:20