... +7.50 DM −4.69 −2. 81 COM −2.35 1. 41 DM +0. 71 +0.70 COM +0.36 0.35 DM −0.22 −0 .14 COM −0 .11 −0.07 DM +0.04 +0.03 COM +0.02 +0.02 DM −0. 01 −0. 01 COM 0.00 0.00 SUM −2.46 −4.92 +4.92 +14 .27 +15 .73 +7.87 In ... distributed loads −2.46 4.92 -14 .27 15 .73 -7.87 I nflection point Beam and Frame Analysis: Displacement Method, Part I by S. T. Mau 17 9 M bc = 2 1 ( M ba + M bc...
Ngày tải lên: 05/08/2014, 09:20
... −2. 81 COM −2.35 1. 41 DM +0. 71 +0.70 COM +0.36 0.35 DM −0.22 −0 .14 COM −0 .11 −0.07 DM +0.04 +0.03 COM +0.02 +0.02 DM −0. 01 −0. 01 COM 0.00 0.00 SUM −2.46 −4.92 +4.92 +14 .27 +15 .73 +7.87 In the ... moments of the frame shown. 0.4 1. 2 −3.8 3.8 −0.4 1. 2 I nflection point P P /2 P /2 P /2 P /2 = + 17 5 Beam and Frame Analysis: Displacement Method Part I 1. Introduction Th...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals Of Structural Analysis Episode 2 Part 1 pps
... M ab M ba M bc M cb M cd M dc EAM 30 DM +15 +15 COM +7.50 +7.50 DM −4.69 −2. 81 COM −2.35 1. 41 DM +0. 71 +0.70 COM +0.36 0.35 DM −0.22 −0 .14 COM −0 .11 −0.07 DM +0.04 +0.03 COM +0.02 +0.02 DM −0. 01 −0. 01 COM 0.00 0.00 SUM ... kN-m Beam and Frame Analysis: Displacement Method, Part I by S. T. Mau 18 2 DF ba : DF bc = 4EK ab : 4EK bc = 4( L EI ) ab : 4( L EI ) bc = 10 1 :...
Ngày tải lên: 05/08/2014, 11:20
Fundamentals of Structural Analysis Episode 2 Part 2 pptx
... kN a b c d 2.36 kN-m 1. 27 kN-m 2.27 kN 1. 73 kN b 1. 27 kN-m 0.37 kN-m 0. 81 kN 0. 81 kN 1. 73 kN 1. 73 kN 0.37 kN-m 0 .17 kN-m 0.28 kN 0.28 kN 0. 81 kN 0. 81 kN 0. 81 kN 0. 81 kN 1. 73 kN 0.28 kN R eaction = 2. 01 kN 2 ... from the Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau 213 M ab = (4EK) ab θ a + (2EK) ab θ b (11 ) M ba = (4EK) ab θ b + (2EK...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 2 ppt
... kN a b c d 2.36 kN-m 1. 27 kN-m 2.27 kN 1. 73 kN b 1. 27 kN-m 0.37 kN-m 0. 81 kN 0. 81 kN 1. 73 kN 1. 73 kN 0.37 kN-m 0 .17 kN-m 0.28 kN 0.28 kN 0. 81 kN 0. 81 kN 0. 81 kN 0. 81 kN 1. 73 kN 0.28 kN R eaction = 2. 01 kN 2 ... and Frame Analysis: Displacement Method, Part II by S. T. Mau 213 M ab = (4EK) ab θ a + (2EK) ab θ b (11 ) M ba = (4EK) ab θ b + (2EK) ab θ a (11...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 3 potx
... kN-m M ab = −37 /11 kN-m = −3.36 kN-m M bc = −3 /11 kN-m = −0.27 kN-m M cb = 18 /11 kN-m = 1. 64 kN-m M cd = 18 /11 kN-m = 1. 64 kN-m M dc = −9 /11 kN-m = −0.82 kN-m From the member-end moments, ... kN-m 1. 23 kN 0.27 kN-m 0.27 kN-m 1. 64 kN-m 0.69 kN 0.69 kN c b 1. 64 kN-m 0.82 kN-m 1. 23 kN 1. 23 kN 1. 23 kN 0.69 kN 0.69 kN 0.69 kN 0.69 kN 0.69 kN 1. 23 kN c 1. 23 kN 1. 23 kN...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 3 pps
... diagram. a 11 .43 kN-m 10 .52 kN 10 .52 kN 11 .59 kN-m 4.60 kN 4.60 kN b 11 .43 kN-m 11 .22 kN-m 5.66 kN 5.66 kN 5.66 kN 4.60 kN 10 kN 10 .52 kN b b c Beam and Frame Analysis: Displacement Method, Part II ... kN c b 1. 64 kN-m 0.82 kN-m 1. 23 kN 1. 23 kN 1. 23 kN 0.69 kN 0.69 kN 0.69 kN 0.69 kN 0.69 kN 1. 23 kN c 1. 23 kN 1. 23 kN 0.69 kN b 1. 23 kN 1. 23 kN Beam and Fram...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 4 ppsx
... (k G ) 1 = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 60,000022,500-30,000022,500 01x10001x10-0 22,500- 011 ,25022,500- 011 ,250- 30,0000 22,500-60,000022,500 0 1x10-001x100 22,500 011 ,250-22,500 011 ,250 99 99 Member 2: (k G ) 2 = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 44 99 44 99 12 x1090,00006x1090,000-0 90,00090,000-090,00090,000-0 002x10002x10- 6x1090,000 012 x1090,000-0 90,000-90,00...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 4 doc
... (k G ) 1 = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 60,000022,500-30,000022,500 01x10001x10-0 22,500- 011 ,25022,500- 011 ,250- 30,0000 22,500-60,000022,500 0 1x10-001x100 22,500 011 ,250-22,500 011 ,250 99 99 Member 2: (k G ) 2 = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 44 99 44 99 12 x1090,00006x1090,000-0 90,00090,000-090,00090,000-0 002x10002x10- 6x1090,000 012 x1090,000-0 90,000-90,00...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 5 pptx
... =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 0.625 0. 41 F CJ 10 kN/m 6 m 9 m 1. 18 1. 82 1. 25 1. 25 0.625 0. 41 F CJ 6 m 9 m 1. 18 1. 82 Influence Lines by S. ... 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 6 m 1. 25 1. 25 0.625 0. 41 F CJ 2...
Ngày tải lên: 05/08/2014, 09:20