Fundamentals of proximity analysis

... have a direct bearing on the design of the content of a course in structural analysis: the reduction of credit hours to three required hours in structural analysis in most civil engineering curricula ... four DOFs, since each node has two DOFs and there are two nodes for each member. We may also use the way each of the DOF is sequenced to refer to a particular DOF. For example,...

Ngày tải lên: 06/07/2014, 11:20

... are changed by the addition of one more diagonal member. (9) Concluding remarks. If the number of nodes is N and the number of constrained DOF is C, then (a) the number of simultaneous equations ... The resulting method of analysis is simple and straightforward and is very easy to be implemented into a computer program. As a matter of fact, virtually all structural analysis...

Ngày tải lên: 06/07/2014, 11:20

... 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: ... 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...

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: ... 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...

Ngày tải lên: 05/08/2014, 09:20

... 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: ⎪ ⎪...

Ngày tải lên: 05/08/2014, 09:20

... 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...

Ngày tải lên: 05/08/2014, 09:20

... 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...

Ngày tải lên: 05/08/2014, 09:20

... 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...

Ngày tải lên: 05/08/2014, 09:20

... 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...

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 ... (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

... 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...

Ngày tải lên: 05/08/2014, 09:20

... 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...

Ngày tải lên: 05/08/2014, 09:20

... 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...

Ngày tải lên: 05/08/2014, 09:20

Ngày tải lên: 21/10/2014, 10:39