Strength of Materials |
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Page 185
... obtain 1 = do do 2 P ds dx 1 ď2y or = Ρ dx2 ( d ) In deriving the flexure formula in Art . 5-2 , we obtained on page 132 the relation 1 M = 1 ρ ΕΙ Equating the values of from Eqs . ( d ) and ( 5-1 ) , we have - ρ ΕΙ d2y dx2 = M ( 5-1 ) ...
... obtain 1 = do do 2 P ds dx 1 ď2y or = Ρ dx2 ( d ) In deriving the flexure formula in Art . 5-2 , we obtained on page 132 the relation 1 M = 1 ρ ΕΙ Equating the values of from Eqs . ( d ) and ( 5-1 ) , we have - ρ ΕΙ d2y dx2 = M ( 5-1 ) ...
Page 462
... obtain a method of determining deflections that is based on the principle of con- servation of energy . This method will be shown to be extremely versatile . We begin by obtaining expressions for the strain energy U stored in a body ...
... obtain a method of determining deflections that is based on the principle of con- servation of energy . This method will be shown to be extremely versatile . We begin by obtaining expressions for the strain energy U stored in a body ...
Page 517
... obtain the stress distribution shown . Superimposing the loadings and stress patterns of parts ( a ) and ( b ) , we obtain the unloaded bar with the residual stress distribution shown in part ( c ) . An interesting phenomenon of ...
... obtain the stress distribution shown . Superimposing the loadings and stress patterns of parts ( a ) and ( b ) , we obtain the unloaded bar with the residual stress distribution shown in part ( c ) . An interesting phenomenon of ...
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