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 Truss ... 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...
Ngày tải lên: 05/08/2014, 11:20
... 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 Truss ... 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...
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 ... 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...
Ngày tải lên: 05/08/2014, 09:20
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 ... diagrams. (1) (2) (3) (4) (5) (6) (7) (8) (9) (10 ) Problem 2. Beam Problems. 3 m5 m 10 kN 3 m 5 m 10 kN 3 m5 m 3 kN/m 3 m 5 m 3 kN/m 3 m5 m 10 kN-m 3 m 5 m 10 kN-m 6 m...
Ngày tải lên: 05/08/2014, 09:20
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 computing ... table. . 1 ( ∆ c ) = ∫ m(x) EI dxxM )( =( EI 1 )[( 3 1 ) ( 2 1 ) ( 4 L )( 2 L ) +( 2 1 ) ( 2 1 ) ( 4 L ) ( 2 L )+( 6 1 ) ( 2 1 ) (...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 4 ppsx
... Mau 243 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 9998979695 94 8988878685 84 7978777675 74 6968676665 646 362 61 5958575655 545 352 51 49 4 847 4 645 444 342 41 3635 343 332 31 2625 242 322 21 1 615 1 41 3 1 211 000 000 000 000 000 000 KKKKKK KKKKKK KKKKKK KKKKKKKKK KKKKKKKKK KKKKKKKKK KKKKKK KKKKKK KKKKKK ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals Of Structural Analysis Episode 1 Part 9 ppsx
... redundant force. P a b 5P /16 11 P /16 3PL /16 V M 5PL/32 ∆ 11 P /16 −5P /16 −3PL /16 I nflection point L /2 L /2 L /2 L /2 P a b Beam and Frame Analysis: Force Method, Part III by S. T. Mau 17 4 Beam and Frame Analysis: ... good approximation to the correct solutions. P 1 2 3 θ 2 ’ M 1 M 1 θ 11 θ 1 ’ θ 3 ’ M 1 θ 21 M 1 θ 31 M 1 M 2 θ 21 M 2 θ 22 M 2 θ 3...
Ngày tải lên: 05/08/2014, 11:20
Fundamentals of Structural Analysis Episode 1 Part 1 docx
... as ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ 2 2 1 1 y x y x F F F F = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 44 4 342 41 343 332 31 242 322 21 1 41 3 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...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 1 pptx
... as ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ 2 2 1 1 y x y x F F F F = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 44 4 342 41 343 332 31 242 322 21 1 41 3 1 211 kkkk kkkk kkkk kkkk ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ 2 2 1 1 v u v u (12 ) 2 1 1 2 x x Truss Analysis: Matrix ... is: ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− −−− −−− −− −−− 8 .12 6.98 .12 6.900 6 .19 9.236.92.706 .16 8 .12 6.96.2508 .12 6.9 6.92.7 04...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 2 pot
... equation: ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 6665 646 362 61 5655 545 352 51 46 4 544 4 342 41 3635 343 332 31 2625 242 322 21 1 615 1 41 3 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 ... 2 44 4 342 41 343 3...
Ngày tải lên: 05/08/2014, 09:20