Fundamentals of Structural Analysis Episode 1 Part 4 pptx

Fundamentals of Structural Analysis Episode 1 Part 1 docx

Fundamentals of Structural Analysis Episode 1 Part 1 docx

... as ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ 2 2 1 1 y x y x F F F F = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 444342 41 343332 31 242322 21 1 413 1 211 kkkk kkkk kkkk kkkk ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ 2 2 1 1 v u v u (12 ) 2 1 1 2 x x Truss Analysis: ... form: 1 2 3 1 2 3 1 2 1 2 3 2 1 3 3 P y2 P x2 (F y2 ) 1 +(F y2 ) 2 (F x2 ) 1 + (F x2 ) 2 P y3 P x3 (F y3 ) 2 +(F y3 ) 3 (F x3 ) 2 + (F x3 ) 3...
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Fundamentals of Structural Analysis Episode 1 Part 1 pptx

Fundamentals of Structural Analysis Episode 1 Part 1 pptx

... as ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ 2 2 1 1 y x y x F F F F = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 444342 41 343332 31 242322 21 1 413 1 211 kkkk kkkk kkkk kkkk ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ 2 2 1 1 v u v u (12 ) 2 1 1 2 x x Truss Analysis: ... is: ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− −−− −−− −− −−− 8 .12 6.98 .12 6.900 6 .19 9.236.92.706 .16 8 .12 6.96.2508 .12 6.9 6.92.704 .14 6.92.7 00...
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Fundamentals of Structural Analysis Episode 1 Part 2 pot

Fundamentals of Structural Analysis Episode 1 Part 2 pot

... equation: ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 66656463 62 61 56555453 52 51 46454443 42 41 36353433 32 31 26 2 524 2 322 21 1 615 1 413 1 21 1 KKKKKK KKKKKK KKKKKK KKKKKK KKKKKK KKKKKK ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎧ 3 3 2 2 1 1 v u v u v u = ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎧ 3 3 2 2 1 1 y x y x y x P P P P P P (15 ) where ... representation: ⎪ ⎪...
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Fundamentals of Structural Analysis Episode 1 Part 3 pps

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 ... joint. 4 F 6 F 3 F 2 3 5 4 3 4 5 1 F 3 F 1 R y1 R x1 3 4 5 F 4 F 2 R y5 5 R x5 3 5 4 10 kN 3@ 2m=6m 1 3 2 4 5 6 7 1 2 3 5 6 7 8 10 11...
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Fundamentals of Structural Analysis Episode 1 Part 3 pps

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 ... = 0, F 5 (3/ 5) – F 4 (3/ 5) –F 1 = 0, F 1 = –6 kN. 2 1 2 3 6 kN 4 5 6 R y1 R x1 1 3 4 5 R x5 R y5 F 5 3 4 5 6 kN 3 F 6 3 4 5 2 F 5 F 4 F 1...
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Fundamentals of Structural Analysis Episode 1 Part 4 pps

Fundamentals of Structural Analysis Episode 1 Part 4 pps

... shown. (1- a) (1- b) (2) (3) (4- a) (4- b) Problem 2. 4m 4@ 3m =12 m 1 kN a b 4m 4@ 3m =12 m 1 kN a b 12 kN 4 @ 4m =16 m 2m 3m a b c 6@3m =18 m 4m 4m 12 kN a b c 3@3m=9m 4m 4m 15 kN a AB b c 3@3m=9m 4m 4m 15 kN a AB b ... form: x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1...
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Fundamentals of Structural Analysis Episode 1 Part 4 pptx

Fundamentals of Structural Analysis Episode 1 Part 4 pptx

... shown. (1- a) (1- b) (2) (3) (4- a) (4- b) Problem 2. 4m 4@ 3m =12 m 1 kN a b 4m 4@ 3m =12 m 1 kN a b 12 kN 4 @ 4m =16 m 2m 3m a b c 6@3m =18 m 4m 4m 12 kN a b c 3@3m=9m 4m 4m 15 kN a AB b c 3@3m=9m 4m 4m 15 kN a AB b ... form: x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1...
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Fundamentals of Structural Analysis Episode 1 Part 5 pdf

Fundamentals of Structural Analysis Episode 1 Part 5 pdf

... -0. 71 -1. 00 -3.40 -4.80 6 -56 .56 17 ,680 -3.20 1. 00 0.94 -3.20 -3.00 7 40.00 25, 000 1. 60 -0. 71 0.33 -1. 14 0 .53 8 -56 .56 17 ,680 -3.20 0.00 -0.47 0.00 1. 50 9 -40.00 25, 000 -1. 60 -0. 71 -0.33 1. 14 ... (mm/kN) 1 -0.33 25, 000 -0. 013 -0.8 0. 010 -0.032 0.026 2 0 33,333 0 -0.6 0 -0. 018 0. 011 3 0 25, 000 0 -0.8 0 -0.032 0.026 4 0 .50 33,333 0. 0 15 -0.6 -0.009 -0...
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Fundamentals of Structural Analysis Episode 1 Part 5 pps

Fundamentals of Structural Analysis Episode 1 Part 5 pps

... -0. 71 -1. 00 -3.40 -4.80 6 -56 .56 17 ,680 -3.20 1. 00 0.94 -3.20 -3.00 7 40.00 25, 000 1. 60 -0. 71 0.33 -1. 14 0 .53 8 -56 .56 17 ,680 -3.20 0.00 -0.47 0.00 1. 50 9 -40.00 25, 000 -1. 60 -0. 71 -0.33 1. 14 ... 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 =12 m 1 2 3 4 5 6 1 2 3 4 5...
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Fundamentals of Structural Analysis Episode 1 Part 6 doc

Fundamentals of Structural Analysis Episode 1 Part 6 doc

... kN/m 2m 6m 6 kN 2m 6 kN 3 kN/m 2m 3m 3m 6 kN 2m 6 kN 15 kN 15 kN 3 kN/m 2m 3m 3m 6 kN 2m 6 kN 15 kN 15 kN 6 kN -6 kN 9 kN -9 kN V V Beam and Frame Analysis: Force Method, Part I by S. T. Mau 11 1 Shear ... 3m 3m 6 kN 2m 6 kN V V 5 kN 5 kN 6 kN 1 kN 6 kN 2m 3m 3m 2m 30 kN-m M 6 kN/m 6 kN/m M 1 kN/m 1 kN/m 15 kN-m 12 kN-m -12 kN-m -15 kN-m Beam an...
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Fundamentals of Structural Analysis Episode 1 Part 6 pdf

Fundamentals of Structural Analysis Episode 1 Part 6 pdf

... right. 3 kN/m 2m 6m 6 kN 2m 6 kN 3 kN/m 2m 3m 3m 6 kN 2m 6 kN 15 kN 15 kN 3 kN/m 2m 3m 3m 6 kN 2m 6 kN 15 kN 15 kN 6 kN -6 kN 9 kN -9 kN V V Beam and Frame Analysis: Force Method, Part I by S. T. ... right. -6 kN-m V P arabolic L inear -6 kN M V -6 kN 4 kN F lat V -6 kN -2 kN F lat 1m 3 kN 2m 10 kN 4 kN 6 kN -6 kN-m L inear 6 kN-m M -6 kN-m 6 kN-m M...
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Fundamentals of Structural Analysis Episode 1 Part 7 ppsx

Fundamentals of Structural Analysis Episode 1 Part 7 ppsx

... Frame Analysis: Force Method, Part I by S. T. Mau 11 9 (11 ) (12 ) (13 ) (14 ) (15 ) (16 ) Problem 2. Frame problems. 5 m 5 m 10 kN 5 m 5 m 10 kN-m 5 m 5 m 10 kN 5 m 5 m 10 kN-m 5 m 5 m 10 kN 5 m 5 m 10 ... Beam P a/2EI P a/EI 2aaa R eactions P a/2EI P a/EI 11 Pa 2 /12 EI 5Pa 2 /12 EI Shear(Rotation)Diagram ( Unit: Pa 2 /EI ) 1 1/ 12 5 /12 1/ 6 M oment(Deflection) Diagram...
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Fundamentals of Structural Analysis Episode 1 Part 8 ppsx

Fundamentals of Structural Analysis Episode 1 Part 8 ppsx

... displaced configuration of the frame as shown below. Displaced configuration. P 2P 2P P 1 2 2 1 1 1/ L1/L 1 1/L 1/ L 1 1 2PL 2L 1 1 2L 2L P Beam and Frame Analysis: Force Method, Part II by S. T. Mau 14 0 The ... ∆ d Load Diagram Moment Diagram (M)(m)(m)(m)(m) a~b EI 1 ( 3 1 ) (2PL)(2L)(2L) = EI PL 3 8 3 00 EI 1 ( 3 1 ) (2PL)(2L)(2L) = EI PL 3 8 3 b~c EI2 1 (...
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Fundamentals Of Structural Analysis Episode 1 Part 4 ppsx

Fundamentals Of Structural Analysis Episode 1 Part 4 ppsx

... shown. (1- a) (1- b) (2) (3) (4- a) (4- b) Problem 2. 4m 4@ 3m =12 m 1 kN a b 4m 4@ 3m =12 m 1 kN a b 12 kN 4 @ 4m =16 m 2m 3m a b c 6@3m =18 m 4m 4m 12 kN a b c 3@3m=9m 4m 4m 15 kN a AB b c 3@3m=9m 4m 4m 15 kN a AB b ... form: x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1...
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Fundamentals Of Structural Analysis Episode 1 Part 5 pptx

Fundamentals Of Structural Analysis Episode 1 Part 5 pptx

... -0. 71 -1. 00 -3.40 -4.80 6 -56 .56 17 ,680 -3.20 1. 00 0.94 -3.20 -3.00 7 40.00 25, 000 1. 60 -0. 71 0.33 -1. 14 0 .53 8 -56 .56 17 ,680 -3.20 0.00 -0.47 0.00 1. 50 9 -40.00 25, 000 -1. 60 -0. 71 -0.33 1. 14 ... (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 -11 3 .13 17 ,6...
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