Fundamentals of Structural Analysis Episode 2 Part 3 pps

Fundamentals of Structural Analysis Episode 2 Part 3 pps

... 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...

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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 ... 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...

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Fundamentals Of Structural Analysis Episode 2 Part 3 ppt

... 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 Analysis: Di...

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Fundamentals of Structural Analysis Episode 1 Part 3 pps

... trusses shown. (1-a)(1-b)(1-c) (2- a) (2- b) (2- c) (3- a) (3- b) (3- c) (4-a)(4-b)(4-c) Problem 1. 3m 4m 3 kN 3m 4m 3m 8 kN 3m 4m 3m 8 kN 3 kN 3m 4m 4 kN 3m 4m 4 kN 3m 2m 6 kN 2m 1.5m 1 .2 m 1.6 m 0.9 m 2 m2 m 1 .2 m 0.9 m 0.7 ... example. F 2 3 F 3 0. 83 kN 1 0.5 kN 0.17 kN F 1 0. 62 kN x y 1 2 3m 2m 3 2m 1 2 3 6 kN 1.5 m 4 5 6 3 4 5 3 4 5 4 5 Truss Analysis:...

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Fundamentals of Structural Analysis Episode 1 Part 3 pps

... in this example. F 2 3 F 3 0. 83 kN 1 0.5 kN 0.17 kN F 1 0. 62 kN x y 1 2 3m 2m 3 2m 1 2 3 6 kN 1.5 m 4 5 6 3 4 5 3 4 5 4 5 Truss Analysis: Force Method, Part I by S. T. Mau 42 We can identify: (1) ... kN 3m 4m 4 kN 3m 4m 4 kN 3m 2m 6 kN 2m 1.5m 1 .2 m 1.6 m 0.9 m 2 m2 m 1 .2 m 0.9 m 0.7 m 5 kN 4 kN 0.9 m 0.9 m 0.7 m 4kN 0.9 m 5kN 1 .2 m 1m 1m 1 kN 2 m2 m 1m 1m...

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Fundamentals of Structural Analysis Episode 2 Part 1 ppsx

... )( 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 2 kN 2 m 2 m 4 m a c b 3 kN/m 4 kN 4 kN-m Beam ... 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 Anal...

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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...

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Fundamentals of Structural Analysis Episode 2 Part 7 pps

... resulting in: ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 010 001 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 33 32 3 1 23 222 1 13 121 1 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − −− 1 13 13/ 43/ 10 33 /1 03/ 31 Or, X= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 33 32 3 1 23 222 1 13 121 1 xxx xxx xxx = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − −− 1 13 13/ 43/ 10 33 /1 03/ 31 Note ... in ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 100 010 001 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 33 32 3 1 23 222 1 13...

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Fundamentals Of Structural Analysis Episode 2 Part 1 pps

... )( 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 2 kN 2 m 2 m 4 m a c b 3 kN/m 4 kN 4 kN-m Beam ... 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 Anal...

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Fundamentals of Structural Analysis Episode 2 Part 1 pdf

... 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 Method, Part I by S. T. Mau 190 Solution. (1) ... )( 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...

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