Fundamentals of Structural Analysis Episode 1 Part 9 pps
Fundamentals of Structural Analysis Episode 1 Part 9 pps
... 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 32 M 3 M 3 θ 31 M 3 θ 23 M 3 θ 33 θ 2 =0 θ 1 =0 θ 3 =0 P Beam and Frame Analysis: Force Method, Part ... self-evident. L L w P a b 5 P /16 11 P /16 3PL /16 L /2 L /2 Beam and Frame Analysis: Force Method, Part III by S. T. Mau 16 3 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 212 12 11...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 3 pps
... m2 m 1. 2 m 0 .9 m 0.7 m 5 kN 4 kN 0 .9 m 0 .9 m 0.7 m 4kN 0 .9 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 cut ... 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: For...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 4 pps
... 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: ... 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...
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 ... kN/m 9. 6 kN/m 12 kN 9 kN 39 kN 48 kN 60 kN 15 kN 15 kN 60 kN T V -12 kN -48 kN -60 kN 75 kN-m 9 kN - 39 kN 15 kN M -75 kN-m 75 kN-m 48 kN/5m =9. 6 kN/m 36 kN/5m...
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 ... ∆ 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...
Ngày tải lên: 05/08/2014, 09:20