Engineering Mechanics of Materials |
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Page 165
... equals Tr / 2JG . In the special case of pure torsion , the principal normal strains , & , and ε3 , each have magnitudes equal to Tr / 2JG and the maximum and minimum shearing strains have magnitudes equal to Tr / JG . Mohr's circle for ...
... equals Tr / 2JG . In the special case of pure torsion , the principal normal strains , & , and ε3 , each have magnitudes equal to Tr / 2JG and the maximum and minimum shearing strains have magnitudes equal to Tr / JG . Mohr's circle for ...
Page 231
... equal in magnitude to the applied moment M. This resisting moment is due to the system of tensile and compressive stresses shown in section a - a in Figure 5.7 ( c ) . The neutral axis is , of course , unstressed and , as mentioned ...
... equal in magnitude to the applied moment M. This resisting moment is due to the system of tensile and compressive stresses shown in section a - a in Figure 5.7 ( c ) . The neutral axis is , of course , unstressed and , as mentioned ...
Page 514
... equal - legged angle with the load acting perpendicularly to one of its sides , as shown in Figure 10.7 ( b ) . The material has a yield strength in tension or compression of 36,000 psi . Select a suitable equal - legged angle section ...
... equal - legged angle with the load acting perpendicularly to one of its sides , as shown in Figure 10.7 ( b ) . The material has a yield strength in tension or compression of 36,000 psi . Select a suitable equal - legged angle section ...
Contents
T Shear and Moment at Specified Sections | 24 |
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 unit yield zero