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 ... ∆ 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...
Ngày tải lên: 05/08/2014, 09:20
... 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 6 Example 21. ... ∆ 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 EI...
Ngày tải lên: 05/08/2014, 11:20
... 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
Fundamentals Of Structural Analysis Episode 1 Part 4 ppsx
... 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 ... equation. Contribution of Member Forces Equation Number and Value of Entry Member Number Force Number 2i -1 Coeff. 2i Coeff. 2j -1 Coeff. 2j...
Ngày tải lên: 05/08/2014, 11:20
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: ... below. L L w c 5wL/4 3wL /8 3wL /8 3wL /8 −3wL /8 5wL /8 -5wL /8 9wL 2 /12 8 ∆ M V L L w L 3/8L 3/4L wL 2 /8 Beam and Frame Analysis: Force Method...
Ngày tải lên: 05/08/2014, 11:20
Fundamentals of Structural Analysis Episode 1 Part 1 docx
... is: ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− −−− −−− −− −−− 8 .12 6. 98 .12 6.900 6 .19 9.236.92.706 .16 8 .12 6.96.25 08 .12 6.9 6.92.704 .14 6.92.7 0 08 .12 6. 98 .12 6.9 06 .16 6.92.76.99.23 ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎧ 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 where ... as ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 1 pptx
... is: ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− −−− −−− −− −−− 8 .12 6. 98 .12 6.900 6 .19 9.236.92.706 .16 8 .12 6.96.25 08 .12 6.9 6.92.704 .14 6.92.7 0 08 .12 6. 98 .12 6.9 06 .16 6.92.76.99.23 ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎧ 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 where ... as ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 2 pot
... beginning of the last section now becomes ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− −−− −−− −− −−− 8 .12 6. 98 .12 6.900 6 .19 9.236.92.706 .16 8 .12 6.96.25 08 .12 6.9 6.92.704 .14 6.92.7 0 08 .12 6. 98 .12 6.9 06 .16 6.92.76.99.23 ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎧ 0 0 0 3 2 2 u v u = ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎧ − 3 1 1 0 0 .1 5.0 y y x P P P Note ....
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 3 pps
... 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 4 Truss Analysis: Force Method, Part I by S. T. Mau 43 Problem 1. Use the method of joint ... m 5kN 1. 2 m 1m 1m 1 kN 2 m2 m 1m 1m 2 kN 2 m2 m 1m 1m 2 kN 1 kN Truss Analysis: Force Method, Part I by S. T. Mau 49 This particular c...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 3 pps
... forces. ΣM E =0, (12 ) 8 – (3) 16 – (8) F GJ =0 F GJ = 6 kN. ΣM G =0, (12 ) 8 – (3) 16 + (8) F EH =0 F EH = −6 kN. B D 8 kN C A F EH E H G F GJ J 8m 6m 16 kN 8 kN J H E G F EH F GJ 8m 3m 9m Truss Analysis: ... 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 4 Truss Analysis...
Ngày tải lên: 05/08/2014, 09:20