Fundamentals of Structural Analysis Episode 2 Part 1 ppsx
Fundamentals of Structural Analysis Episode 2 Part 1 ppsx
... +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
Fundamentals of Structural Analysis Episode 2 Part 4 ppsx
... = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − +−−−−− −−+−−−−− −− −−−−−−+− −−−−−+ 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
Fundamentals of Structural Analysis Episode 2 Part 2 pptx
... 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 ... kN a b c d 2. 36 kN-m 1 .27 kN-m 2. 27 kN 1. 73 kN b 1 .27 kN-m 0.37 kN-m 0. 81 kN 0. 81 kN 1. 73 kN 1. 73 kN 0.37 kN-m 0 .17 kN-m 0 .28 kN 0 .28 kN 0. 8...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 3 potx
... 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 ... 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 Ana...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 5 pptx
... =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. 625 0. 41 F CJ 10 kN/m 6 m 9 m 1. 18 1. 82 1 .25 1 .25 0. 625 0. 41 F CJ 6 m 9 m 1. 18 1. 82 Influence Lines by S. T. Mau 25 6 Solution. ... 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. 625 0. 41 F CJ 2 kN 1 .25...
Ngày tải lên: 05/08/2014, 09:20