Engineering Mechanics of Materials |
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Page 653
... cylinder . This assumption is justified at least for transverse cross sections sufficiently removed from the two ends of the cylinder . Consider one part of a closed - ended cylindrical container subjected to both p1 and p2 and lying to ...
... cylinder . This assumption is justified at least for transverse cross sections sufficiently removed from the two ends of the cylinder . Consider one part of a closed - ended cylindrical container subjected to both p1 and p2 and lying to ...
Page 663
... cylinder , and ( c ) the normal and shearing stresses in the wall of the cylinder on a plane inclined to the longitudinal axis of the cylinder through a 30 ° angle . Solution ( a ) Consider a three - dimensional stress element on the ...
... cylinder , and ( c ) the normal and shearing stresses in the wall of the cylinder on a plane inclined to the longitudinal axis of the cylinder through a 30 ° angle . Solution ( a ) Consider a three - dimensional stress element on the ...
Page 666
... cylinder is made by shrinking a jacket onto a main cylinder as described in Example 14.3 . The following data are provided : E = 30 × 10o , = 20 in . , r2 = 30 in . , r3 = 40 in . , and Ar = 0.05 in . Determine the absolute maximum ...
... cylinder is made by shrinking a jacket onto a main cylinder as described in Example 14.3 . The following data are provided : E = 30 × 10o , = 20 in . , r2 = 30 in . , r3 = 40 in . , and Ar = 0.05 in . Determine the absolute maximum ...
Contents
Internal Forces in Members | 1 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
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acting allowable angle of twist applied Assume axes axis beam bending cantilever centroidal circle column components compressive Compute Consider constant construct coordinate cross section curve deflection deformation depicted in Figure Determine developed diameter direction discussed elastic element energy equal equation equilibrium Example expressed factor failure flexural force free-body diagram function given inertia joint length limit load material maximum shear stress method modulus moment moments neutral axis normal stress Note obtained plane plot positive principal stresses Problem properties quantity ratio reactions Refer to Figure relation represents resist respect rotation segment shaft shown in Figure slope Solution Solve static steel strain strength structural subjected Substitution supported surface tensile tension theory tion torque torsional unit vertical yield zero