Strength of Materials |
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Page 62
... fibers . Slice ( 2 ) will rotate past slice ( 1 ) until the elastic fibers joining them are deformed enough to create a resisting torque which balances the applied torque . When this happens , slices ( 1 ) and ( 2 ) will act as a rigid ...
... fibers . Slice ( 2 ) will rotate past slice ( 1 ) until the elastic fibers joining them are deformed enough to create a resisting torque which balances the applied torque . When this happens , slices ( 1 ) and ( 2 ) will act as a rigid ...
Page 152
... fibers can be located at a greater distance from the neutral axis than the weaker fibers . The ideal treatment for such materials is to locate the cen- troidal or neutral axis in such a position that the ratio of the distances from it ...
... fibers can be located at a greater distance from the neutral axis than the weaker fibers . The ideal treatment for such materials is to locate the cen- troidal or neutral axis in such a position that the ratio of the distances from it ...
Page 495
... fibers and the decreased stress at the outer fibers , in comparison with the stresses computed from the flexure formula . To determine the shift in position of the neutral axis and to express the stress at any fiber in terms of the ...
... fibers and the decreased stress at the outer fibers , in comparison with the stresses computed from the flexure formula . To determine the shift in position of the neutral axis and to express the stress at any fiber in terms of the ...
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