Strength of Materials |
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Page 31
... Hooke's law . Originally Hooke's law specified merely that stress was proportional to strain , but Thomas Young in 1807 introduced a constant of proportionality that came to be known as Young's modulus . Eventually this name was ...
... Hooke's law . Originally Hooke's law specified merely that stress was proportional to strain , but Thomas Young in 1807 introduced a constant of proportionality that came to be known as Young's modulus . Eventually this name was ...
Page 37
... a cone suspended from its base . Ans . 8wL2 ( D + d ) 6E ( D — d ) = - wL2d2 3ED ( Dd ) 2-4 . Poisson's Ratio . Biaxial and Triaxial Deformations Another Art . 2-3 ] 37 Hooke's Law . Axial Deformation 2-3 Hooke's Law Axial Deformation.
... a cone suspended from its base . Ans . 8wL2 ( D + d ) 6E ( D — d ) = - wL2d2 3ED ( Dd ) 2-4 . Poisson's Ratio . Biaxial and Triaxial Deformations Another Art . 2-3 ] 37 Hooke's Law . Axial Deformation 2-3 Hooke's Law Axial Deformation.
Page 128
... Hooke's law . 3. The moduli of elasticity for tension and compression are equal . 4. The beam is initially straight and of constant cross section . 5. The plane of loading must contain a principal axis of the beam cross section and the ...
... Hooke's law . 3. The moduli of elasticity for tension and compression are equal . 4. The beam is initially straight and of constant cross section . 5. The plane of loading must contain a principal axis of the beam cross section and 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