Fundamentals of Structural Analysis Episode 2 Part 1 pdf

... +7.50 DM −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 ... 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-m M F cb = 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m (d) Compute DF at b: 2 m 2 m 4 m a c b...

Ngày tải lên: 05/08/2014, 09:20

... 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 the ... = 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 Method, Part...

Ngày tải lên: 05/08/2014, 09:20

... 0 .22 0.45 0 1 0 .23 0.46 0.31 0 MEM M ab M ba M bc M be M eb M ab M ba M bc M be M eb EAM FEM − 62. 5 62. 5 − 62. 5 62. 5 DM 62. 5 62. 5 COM 31.3 31.3 DM −31.0 20 .6 − 42. 2 21 .6 −43 .2 29 .0 COM 0.0 21 .1 0.0 −14.5 SUM 0.0 ... (2) (3) (4) (5) (6) (7) Problem 1. 2 m 2 m 3 m a c b 3 kN/m 4 kN 2EI E I 2 m 2 m 3 m a cb 8 kN 2EI E I 2 kN-m 50 kN ab cd 50 kN 2 m 2 m 2 m 2...

Ngày tải lên: 05/08/2014, 09:20

... M F − 8 PL 8 PL − ab 2 PL − a 2 bPL − a(1-a)PL a(1-a)PL −(6−8a+3a 2 ) 12 22 wLa (4-3a) 12 23 wLa − 12 2 wL 12 2 wL − 20 2 wL 12 2 wL − 96 5 2 wL 96 5 2 wL − b(2a-b)M a(2b-a)M Note: Positive ... 0 .22 0.45 0 1 0 .23 0.46 0.31 0 MEM M ab M ba M bc M be M eb M ab M ba M bc M be M eb EAM FEM − 62. 5 62. 5 − 62. 5 62. 5 DM 62. 5 62. 5 COM 31.3 31.3 DM −31.0 20 .6 − 4...

Ngày tải lên: 05/08/2014, 09:20

... (6) Problem 2. 2 m 2 m 3 m a c b 3 kN/m 4 kN 2EI E I 2 m 2 m 3 m a cb 8 kN 2EI E I 2 kN-m 50 kN a b c 4 m 4 m 4 m 8 m E I 2EI2EI 50 kN ab c 4 m 4 m 4 m 8 m E I 2EI2EI 2 m 2 m 2 m 4 kN a b c 2 m 2 m 2 m 4 ... kN a b c d 2. 77 kN 3. 36 kN-m 1 . 23 kN 0 .27 kN-m 0 .27 kN-m 1.64 kN-m 0.69 kN 0.69 kN c b 1.64 kN-m 0. 82 kN-m 1 . 23 kN 1 . 23 kN 1 . 23 kN 0.69 kN 0.6...

Ngày tải lên: 05/08/2014, 09:20

... kN a b c d 2. 77 kN 3. 36 kN-m 1 . 23 kN 0 .27 kN-m 0 .27 kN-m 1.64 kN-m 0.69 kN 0.69 kN c b 1.64 kN-m 0. 82 kN-m 1 . 23 kN 1 . 23 kN 1 . 23 kN 0.69 kN 0.69 kN 0.69 kN 0.69 kN 0.69 kN 1 . 23 kN c 1 . 23 kN 1 . 23 kN 0.69 ... (6) Problem 2. 2 m 2 m 3 m a c b 3 kN/m 4 kN 2EI E I 2 m 2 m 3 m a cb 8 kN 2EI E I 2 kN-m 50 kN a b c 4 m 4 m 4 m 8 m E I 2EI2EI 50 kN ab c 4 m 4 m 4...

Ngày tải lên: 05/08/2014, 09:20

... = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − +−−−−− −−+−−−−− −− −−−−−−+− −−−−−+ EK L EK C L EK S-EK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L EK L EA CS L EK S L EA C L EK S L EK L EA CS L EK S L EA C EK L EK C L EK SEK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L...

Ngày tải lên: 05/08/2014, 09:20

... = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − +−−−−− −−+−−−−− −− −−−−−−+− −−−−−+ EK L EK C L EK S-EK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L EK L EA CS L EK S L EA C L EK S L EK L EA CS L EK S L EA C EK L EK C L EK SEK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L...

Ngày tải lên: 05/08/2014, 09:20

... −10[0 .5( 1. 82) (0. 6 25 )+0 .5( 9)(0. 6 25 )-0 .5( 4. 82) (0. 6 25 )(4. 82/ 9)] = −33.0 kN 10 kN/m 6 m 1 . 25 1 . 25 0. 6 25 0.41 F CJ 6 m 9 m 1.18 1. 82 Influence Lines by S. T. Mau 26 8 (F CJ ) max = − [2( 0. 6 25 )+1(0. 6 25 )(7/9)+1(0. 6 25 )(6/9)]= ... maximize F CJ . 1kN 1kN 2 kN 1m 2m 2 kN 10 kN/m 10 kN/m x 6 m 1 . 25 1 . 25 0. 6 25 0.41 F CJ 2 kN 1 . 25 1 . 25 0. 6 2...

Ngày tải lên: 05/08/2014, 09:20

... maximize F CJ . 1kN 1kN 2 kN 1m 2m 2 kN 10 kN/m 10 kN/m x 6 m 1 . 25 1 . 25 0. 6 25 0.41 F CJ 2 kN 1 . 25 1 . 25 0. 6 25 0.41 F CJ 2 kN 1kN 1kN 1m2m 7 m 9 m Influence Lines by S. T. Mau 26 4 S = ( L aL − ) S i ... =10[0 .5( 1.18)(0.41)+0 .5( 6)(0.41)−0 .5( 1.18)(0.41)(1.18/6)] = 14. 65 kN 10 kN/m 6 m 1 . 25 1 . 25 0. 6 25 0.41 F CJ 10 kN/m 6 m 9 m 1.18 1. 82 1 . 25 1 ....

Ngày tải lên: 05/08/2014, 09:20

... 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 0.0 06 0 0.013 0 0.0 32 Total σ (kN/cm 2 ) 0.0 72 0.078 0.088 0.101 0. 125 0.1 56 Error ... depth of the beam, h, is the same for both members. The fixed-end moments are obtained from the above table: P L /2 L /2 w L L /2 L /2 L 0.2L0.2L 0.6L h h b a 0....

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): ... 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

... kN-m Problem 2. (1) M ba = − 525 kN-m, M bc = 525 kN-m, M ab = 26 3 kN-m, M cb = 2, 363 kN-m. (2) M ba = 28 .8 kN-m, M bc = 28 .8 kN-m, M ab = 28 .8 kN-m, M cb = 28 .8 kN-m 12 Matrix Algebra ... 29 6 Gasset plate, 28 4 Gaussian elimination method, 29 3 Geometric non-linearity, 28 8 Global Coordinate, 4, 5, 23 8 Hardening, 28 8 Hinge, 93 Influence lines, 24 9...

Ngày tải lên: 05/08/2014, 09:20

... +15 +15 COM +7.50 +7.50 DM −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 ... 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-m M F cb = 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m (d) Compute DF at b: 2 m 2 m 4 m a c b 3 kN/m 4 kN...

Ngày tải lên: 05/08/2014, 11:20