Engineering Mechanics of Materials |
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Page 627
... solve Example 13.5 again . Compare your answers with the answers for Example 13.5 . 13.4 Solve Example 13.6 by assuming that the im- pacted beam is simply supported rather than fixed- ended . Compare your answers with the answers for ...
... solve Example 13.5 again . Compare your answers with the answers for Example 13.5 . 13.4 Solve Example 13.6 by assuming that the im- pacted beam is simply supported rather than fixed- ended . Compare your answers with the answers for ...
Page 647
... Solve Problem 13.34 for compressive forces and plot the Goodman straight line in the appro- priate quadrant . Assume the member is short enough that buckling is not a problem . 13.37 Solve Problem 13.35 for compressive forces and plot ...
... Solve Problem 13.34 for compressive forces and plot the Goodman straight line in the appro- priate quadrant . Assume the member is short enough that buckling is not a problem . 13.37 Solve Problem 13.35 for compressive forces and plot ...
Page 714
... solve for MA and V , by dealing with a 2 × 2 system . Use the method of two successive integrations to formulate the equations . Determine M , and VÅ using the equations of statics . 14.67 Refer to Figure 14.24 ( a ) and remove MB at ...
... solve for MA and V , by dealing with a 2 × 2 system . Use the method of two successive integrations to formulate the equations . Determine M , and VÅ using the equations of statics . 14.67 Refer to Figure 14.24 ( a ) and remove MB at ...
Contents
Internal Forces in Members | 1 |
Stress Strain and Their Relationships | 53 |
Stresses and Strains in Axially Loaded Members | 115 |
Copyright | |
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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