Fundamentals Of Structural Analysis Episode 2 Part 5 pot
Fundamentals Of Structural Analysis Episode 2 Part 5 pot
... 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 2 Problem 1. (1) Construct ... Mau 27 0 Placing finite length uniform load for maximum compression in member CJ. (F CJ ) max = −10[0 .5( 1. 82) (0. 6 25 )+0 .5( 9)(0. 6 25 )-0 .5( 4. 82) (0...
Ngày tải lên: 05/08/2014, 11:20
Fundamentals of Structural Analysis Episode 2 Part 5 pptx
... −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
Fundamentals of Structural Analysis Episode 2 Part 5 pdf
... 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
Fundamentals of Structural Analysis Episode 2 Part 3 potx
... (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 ... nodal a b c 4 kN a b 2 kN 2 kN 2 kN-m 2 kN 2 kN 2 kN-m 2 kN-m + a b c 2 kN 2 kN-m 1 2 3 2 kN 2 kN-m x y 2 kN-m Beam and Frame Analysis:...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 7 potx
... kN 5 kN 1.8 kN 3 .2 kN 2. 4 kN −4 kN−3 kN 4 kN 1. 12 kN 2. 88 kN −4.8 kN −1.4 kN 5 kN 0.84 kN3.84 2. 4 4kN 5kN 4. 32 kN 4.68 kN 5. 4 kN −7.8 kN 6 .24 3 .24 3 .24 5 5 2 kN 1 kN 0 .5 kN 0 .5 kN 1 kN 1 kN 0 .5 ... Solutions Member Force (MN) Elongation (m) Member Force (MN) Elongation (m) 1 0.3 75 0.011 8 0 0 2 0.3 75 0.011 9 0. 6 25 0.031 3 0.3 75 0.011 10 0 0 4 0.3 75...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals Of Structural Analysis Episode 2 Part 6 potx
... 17.63 0 42. 98 σ due to axial force (kN/cm 2 ) 0.0 72 0.0 72 0.088 0.088 0. 1 25 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. 1 25 0. 156 Error ... table: P L /2 L /2 w L L /2 L /2 L 0.2L0.2L 0.6L h h b a 0.2L0.8L h h b a 0.2L0.2L 0.3L c 0.2L0.8L h h b a h h 0.3L 100 kN 10 kN/m w P Other Topics by S. T. Mau 28...
Ngày tải lên: 05/08/2014, 11:20
Fundamentals of Structural Analysis Episode 1 Part 5 pdf
... -0.0 32 0. 026 2 0 33,333 0 -0.6 0 -0.018 0.011 3 0 25 ,000 0 -0.8 0 -0.0 32 0. 026 4 0 .50 33,333 0.0 15 -0.6 -0.009 -0.018 0.011 5 -0.83 20 ,000 -0.0 42 1.0 -0.0 42 0. 050 0. 050 6 0 20 ,000 0 1.0 0 0. 050 ... (kN-mm) 1 80.00 25 ,000 3 .20 0.00 -0.33 0.00 -1.06 2 80.00 25 ,000 3 .20 -0.71 -0.33 -2. 26 -1.06 3 40.00 25 ,000 1.60 0.00 0.33 0.00 0 .53 4 -113.13 17,680 -6.40...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 5 pps
... is α =5( 10 -6 )/ o C. Problem 5- 3. 1 2 4m 3m 3 3m 1 2 3 1.0 kN 0 .5 kN 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 4m 3@4m=12m 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 4m 3@4m=12m Truss Analysis: Force Method, Part II ... -0.0 32 0. 026 2 0 33,333 0 -0.6 0 -0.018 0.011 3 0 25 ,000 0 -0.8 0 -0.0 32 0. 026 4 0 .50 33,333 0.0 15 -0.6 -0.009 -0.018 0.011 5 -0.83 20 ,000 -0.0 42 1.0...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 1 ppsx
... + 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 2 m...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 1 pdf
... Distribution for a Two-DOF Problem Node ab cd Member ab bc cd DF 0 0. 6 25 0.3 75 0 .5 0 .5 0 MEM M ab M ba M bc M cb M cd M dc EAM 30 DM + 15 + 15 COM +7 .50 +7 .50 DM −4.69 2. 81 COM 2. 35 −1.41 DM +0.71 +0.70 COM ... below. FBDs of the two members. 1 .5 kN-m 2 m 2 m 4 kN 3 kN-m 2. 38 kN 1. 62 kN 3 kN-m 4 m 3 kN/m 4 .5 kN-m 6.38 kN 5. 62 kN Beam and Frame Analysis: Displ...
Ngày tải lên: 05/08/2014, 09:20