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...

Ngày tải lên: 05/08/2014, 11:20

... -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...

Ngày tải lên: 05/08/2014, 09:20

... -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...

Ngày tải lên: 05/08/2014, 09:20

... 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 2 4m 3m 3 3m 1 2 3 1 kN 1 kN 1 kN 1 kN 1 kN 1 kN Truss Analysis: Force ... 6 −0.8 3 31 1. 0 2 0.0 5 1. 0 6 −0.0 Contribution of Reaction Forces Equation Number and Value of EntryReaction Number Force Number 2i -1...

Ngày tải lên: 05/08/2014, 09:20

... =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. 25 0.6 25 0. 41 F CJ 6 m 9 m 1. 18 1. 82 Influence Lines by S. T. Mau 256 Solution. ... location of the group load. Placing the group load to maximize F CJ . 1kN 1kN 2 kN 1m 2m 2 kN 10 kN/m 10 kN/m x...

Ngày tải lên: 05/08/2014, 09:20

... 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: Matrix ... 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 x...

Ngày tải lên: 05/08/2014, 09:20

... 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: Matrix ... and Frames 15 3 Table: Beam Deflection Formulas 16 8 Problem 6. 17 3 Beam and Frame Analysis: Displacement Method, Part I 17 5 1. Introducti...

Ngày tải lên: 05/08/2014, 09:20

... equation: ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 66 656 46362 61 56 555 453 52 51 46 454 44342 41 36 353 43332 31 26 252 42322 21 1 6 15 1 413 1 211 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 ... solved by the method o...

Ngày tải lên: 05/08/2014, 09:20

... 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 3 5 4 3 4 5 Truss Analysis: Force Method, Part I by S. T. Mau 51 In this ... 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 2m 9...

Ngày tải lên: 05/08/2014, 09:20

... 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 3 5 4 3 4 5 Truss Analysis: Force Method, Part I by S. T. Mau 49 This particular ... 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...

Ngày tải lên: 05/08/2014, 09:20