Introduction to Mechanics of Deformable Solids |
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Page 72
1 Addition of hydrostatic pressure . recoverable changes , an increase in volume
due to increase in tension and a decrease due to increase in compression . Of
course , the stresses computed from the dead weights alone are not the actual ...
1 Addition of hydrostatic pressure . recoverable changes , an increase in volume
due to increase in tension and a decrease due to increase in compression . Of
course , the stresses computed from the dead weights alone are not the actual ...
Page 140
Pi = P2 = – 0oA Pz = + 10oA As the force in bar 4 cannot increase beyond 0oA ,
further increase in moment M must be carried solely by increases in the
magnitudes of the forces P1 , P2 , and Pz . The algebraic sum of the forces must
remain ...
Pi = P2 = – 0oA Pz = + 10oA As the force in bar 4 cannot increase beyond 0oA ,
further increase in moment M must be carried solely by increases in the
magnitudes of the forces P1 , P2 , and Pz . The algebraic sum of the forces must
remain ...
Page 417
Instead , the load is increased continually until failure is reached . Stability of path
must be examined , not just stability of existing configuration . Figure 16 . 6d
exhibits a combination of moment M and an accompanying minimum increase in
...
Instead , the load is increased continually until failure is reached . Stability of path
must be examined , not just stability of existing configuration . Figure 16 . 6d
exhibits a combination of moment M and an accompanying minimum increase in
...
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actual addition angle answer applied assemblage axes axial axis beam behavior bending circle circular column combined compatibility components compression compressive stress Consider constant creep cross section curve cylinder deflection deformation determined diameter direction displacement effect elastic equal equation equilibrium example Figure Find force given gives homogeneous idealization increase initial interior isotropic length limit linear linear-elastic load material maximum Maxwell modulus moment needed nonlinear normal obtained outer plane plastic positive pressure principal Prob problem produced pure radius range ratio relation replaced requires residual response result rotation shaft shear stress shell shown shows simple sketch solid solution steel strain stress-strain relations structural Suppose surface symmetry temperature tensile tension thickness thin-walled torsion tube twisting uniform unloading versus viscous yield zero