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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absolute maximum shear aluminum angle of twist applied Assume axes axial force axially loaded beam shown bending C₁ cantilever beam Castigliano's second theorem column components compressive Compute coordinate cross section cross-sectional area cylinder deflection deformation depicted in Figure Determine diameter elastic curve equal equation equilibrium Example factor of safety flexural stress free-body diagram Homework Problems k-ft k-in kN-m length longitudinal M₁ material maximum shear stress MN/m² modulus of elasticity Mohr's circle moment of inertia neutral axis normal stress obtained perpendicular plane stress plot principal centroidal axis principal stresses r₁ ratio Refer to Figure respect rotation section a-a segment shaft shear force shear strain shown in Figure slope Solution Solve static statically indeterminate steel stress condition stress element T₁ tensile tension Tmax torque torsional V₁ vertical yield strength yield stress zero