Strength of Materials |
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Page 110
... lb which is reduced to zero in the interval BE at the rate of 200 lb / ft . Hence BE d = 1400 7 ft . = = 200 As a preliminary to computing the bending moments , we determine the areas of the shear diagram marked A1 , A2 , A3 , and A4 ...
... lb which is reduced to zero in the interval BE at the rate of 200 lb / ft . Hence BE d = 1400 7 ft . = = 200 As a preliminary to computing the bending moments , we determine the areas of the shear diagram marked A1 , A2 , A3 , and A4 ...
Page 114
... lb and acts at the centroid of the triangular area , i.e. , at of 6 ft from F. Thus we obtain = [ M ( ZM ) L ] MF = 240 ( 6 ) - 240 ( ) = 960 ft - lb Similarly , the moment at B where x = 9 ft is found to be [ M = ( ΣM ) z ] MB = 240 ...
... lb and acts at the centroid of the triangular area , i.e. , at of 6 ft from F. Thus we obtain = [ M ( ZM ) L ] MF = 240 ( 6 ) - 240 ( ) = 960 ft - lb Similarly , the moment at B where x = 9 ft is found to be [ M = ( ΣM ) z ] MB = 240 ...
Page 212
Ferdinand Leon Singer. 30 lb / ft M = -120x6 = -720 ft - lb SA = tA / C B 8B = B / C A · 4'- -4'- V = 120 lb ( a ) 8 ' 120x8 = 960 ft - lb M = 120x8 = 960 ft - lb Tv = 120 lb M = 8 ' -720 ft - lb M = -720 ft - lb 18 ' ( b ) -720 ft - lb ...
Ferdinand Leon Singer. 30 lb / ft M = -120x6 = -720 ft - lb SA = tA / C B 8B = B / C A · 4'- -4'- V = 120 lb ( a ) 8 ' 120x8 = 960 ft - lb M = 120x8 = 960 ft - lb Tv = 120 lb M = 8 ' -720 ft - lb M = -720 ft - lb 18 ' ( b ) -720 ft - lb ...
Contents
APPENDIX B TABLES | 2 |
Shearing Stress | 10 |
SIMPLE STRAIN | 26 |
Copyright | |
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acting actual allowable angle applied assumed axial axis beam beam shown bending bolt carries caused centroid circle column compressive compressive stress compute concentrated concrete consider constant couple cross section deflection deformation determine developed diagram diameter direction distance effect elastic curve element equal equation equivalent expressed flexural stress force formula ft-lb given gives Hence horizontal ILLUSTRATIVE inertia joint lb/ft length limit load material maximum method midspan moments negative neutral axis normal obtain occurs plane plate positive Prob PROBLEMS produce R₁ radius reaction relation resisting respect resultant rivet segment shaft shearing stress shown in Fig shows simply slope Solution Solve span spring steel strain strength supported Table tangent tensile thickness torque torsional uniformly varies vertical wall yield zero ΕΙ