Strength of Materials |
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Page 67
... segment . The procedure we follow is exactly the same as that discussed in Art . 2-5 for statically indeterminate axially loaded members . Applying the conditions of static equilibrium and of geometric compatibility , we obtain first he ...
... segment . The procedure we follow is exactly the same as that discussed in Art . 2-5 for statically indeterminate axially loaded members . Applying the conditions of static equilibrium and of geometric compatibility , we obtain first he ...
Page 91
... segments . The free - body diagram of the left segment in Fig . 4-3b shows that the externally applied load is R1 . To maintain equilibrium in this segment of the beam , the fibers in the exploratory section a - a must supply the ...
... segments . The free - body diagram of the left segment in Fig . 4-3b shows that the externally applied load is R1 . To maintain equilibrium in this segment of the beam , the fibers in the exploratory section a - a must supply the ...
Page 104
... segment produces equilibrium of that segment . Thus in Fig . 4-18 , the segments to the left and right of section b - b in Fig . 4-17 are held in equilibrium by the shear P1 P2 R1 Fig . 4-18.- moment . w lb / ft M Mb V2 ( a ) V2 - ( b ) ...
... segment produces equilibrium of that segment . Thus in Fig . 4-18 , the segments to the left and right of section b - b in Fig . 4-17 are held in equilibrium by the shear P1 P2 R1 Fig . 4-18.- moment . w lb / ft M Mb V2 ( a ) V2 - ( b ) ...
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