Fundamentals Of Structural Analysis Episode 2 Part 6 potx
... table: P L /2 L /2 w L L /2 L /2 L 0.2L0.2L 0.6L h h b a 0.2L0.8L h h b a 0.2L0.2L 0.3L c 0.2L0.8L h h b a h h 0.3L 100 kN 10 kN/m w P Other Topics by S. T. Mau 28 6 The analysis of composite ... 17 .63 0 42. 98 σ due to axial force (kN/cm 2 ) 0.0 72 0.0 72 0.088 0.088 0. 125 0. 124 σ due to moment (kN/cm 2 ) 0 0.0 06 0 0.013 0 0.0 32 Total σ (kN/cm 2 ) 0.0 72 0.0...
Ngày tải lên: 05/08/2014, 11:20
... 17 .63 0 42. 98 σ due to axial force (kN/cm 2 ) 0.0 72 0.0 72 0.088 0.088 0. 125 0. 124 σ due to moment (kN/cm 2 ) 0 0.0 06 0 0.013 0 0.0 32 Total σ (kN/cm 2 ) 0.0 72 0.078 0.088 0.101 0. 125 0.1 56 Error ... depth of the beam, h, is the same for both members. The fixed-end moments are obtained from the above table: P L /2 L /2 w L L /2 L /2 L 0.2L0.2L 0.6L h h b a 0....
Ngày tải lên: 05/08/2014, 09:20
... kN a b c d 2. 77 kN 3. 36 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 0 .69 ... nodal a b c 4 kN a b 2 kN 2 kN 2 kN-m 2 kN 2 kN 2 kN-m 2 kN-m + a b c 2 kN 2 kN-m 1 2 3 2 kN 2 kN-m x y 2 kN-m Beam and Frame Analysis: D...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 7 potx
... kN 0.5 kN 1.5 kN −1. 12 kN −1. 12 0 kN 2 kN 2. 24 −1 kN 0 kN 2. 24 0 kN 1 kN 1. 12 2 .24 −1. 12 −1 kN 1 kN 1 kN 1 4 kN 2 kN 1 kN Matrix Algebra Review by S. T. Mau 3 02 =1x(−3)−2x( 6) +3x(−3)=0 A matrix ... kN −5 kN 6 kN 6 kN 4 kN 5 kN 5 kN 1.8 kN 3 .2 kN 2. 4 kN −4 kN−3 kN 4 kN 1. 12 kN 2. 88 kN −4.8 kN −1.4 kN 5 kN 0.84 kN3.84 2. 4 4kN 5kN 4. 32 kN 4 .68 kN −5.4 kN −7....
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 6 doc
... right. 2m 3m 3m 2m 15 kN 13.5 kN-m M -6 kN/m 6 kN/m 1.5 kN M - 12 kN/m - 12 kN/m 30 kN-m 2m 6 kN 2m 6 kN 3m 3m 30 kN/m 2m 3m 3m 6 kN 2m 6 kN 5 kN 5 kN Beam and Frame Analysis: Force Method, Part ... forces. (2) Draw the shear diagram from left to right. Drawing shear diagram from left to 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...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 1 Part 6 pdf
... 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. ... 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 L inear...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 1 ppsx
... +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 ... reversed: M F ab = 2 kN-m M F bc = 2 kN-m 2 m 2 m 4 m a d b 4 kN c 2 m2 m 4 kN 1.33 1.33 1.33 2. 67 2. 67 I nflection point Beam and Frame Analysis: Displacement Meth...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 1 pdf
... reversed: M F ab = 2 kN-m M F bc = 2 kN-m 2 m 2 m 4 m a d b 4 kN c 2 m2 m 4 kN 1.33 1.33 1.33 2. 67 2. 67 I nflection point Beam and Frame Analysis: Displacement Method, Part I by S. T. Mau 190 Solution. (1) ... 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-m M F cb = 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m (d) Compute DF at b: 2 m 2 m 4 m a c b 3...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 2 pptx
... 0 .22 0.45 0 1 0 .23 0. 46 0.31 0 MEM M ab M ba M bc M be M eb M ab M ba M bc M be M eb EAM FEM − 62 . 5 62 . 5 − 62 . 5 62 . 5 DM 62 . 5 62 . 5 COM 31.3 31.3 DM −31.0 20 .6 − 42. 2 21 .6 −43 .2 29 .0 COM 0.0 21 .1 0.0 −14.5 SUM 0.0 ... Method, Part I by S. T. Mau 197 M dc = 0.0 + 0.0 = 0.0 kN-m M cd = − 62 . 8 + ( 72. 2) = 9.4 kN-m M cb = 20 .6 + (−43 .2) = 22 .6 kN-m M cf =...
Ngày tải lên: 05/08/2014, 09:20
Fundamentals of Structural Analysis Episode 2 Part 2 ppt
... 0 .22 0.45 0 1 0 .23 0. 46 0.31 0 MEM M ab M ba M bc M be M eb M ab M ba M bc M be M eb EAM FEM − 62 . 5 62 . 5 − 62 . 5 62 . 5 DM 62 . 5 62 . 5 COM 31.3 31.3 DM −31.0 20 .6 − 42. 2 21 .6 −43 .2 29 .0 COM 0.0 21 .1 0.0 −14.5 SUM 0.0 ... M F − 8 PL 8 PL − ab 2 PL − a 2 bPL − a(1-a)PL a(1-a)PL − (6 8a+3a 2 ) 12 22 wLa (4-3a) 12 23 wLa − 12 2 wL 12 2 wL − 20 2 wL 12...
Ngày tải lên: 05/08/2014, 09:20