Engineering Mechanics of Materials |
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Page 141
... stress beyond which the material no longer behaves in such a way that the stress is proportional to strain . The ... YIELD POINT . The yield point , o ,, is the stress at which the material continues to deform without further increase in ...
... stress beyond which the material no longer behaves in such a way that the stress is proportional to strain . The ... YIELD POINT . The yield point , o ,, is the stress at which the material continues to deform without further increase in ...
Page 488
... yielding . Yielding in a given material may occur at ordinary temperatures if the loads are sufficiently high to cause the stresses to exceed the yield point or yield stress of the material . However , yielding can also occur at stresses ...
... yielding . Yielding in a given material may occur at ordinary temperatures if the loads are sufficiently high to cause the stresses to exceed the yield point or yield stress of the material . However , yielding can also occur at stresses ...
Page 503
... stress concentrations . 10.9 Design a solid cylindrical shaft of circular cross section to resist an applied torque ... yield stress of 160 MPa . Use a factor of safety of N , 2 with respect to yield in shear and determine the diameters ...
... stress concentrations . 10.9 Design a solid cylindrical shaft of circular cross section to resist an applied torque ... yield stress of 160 MPa . Use a factor of safety of N , 2 with respect to yield in shear and determine the diameters ...
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 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 strain shown in Figure slope Solution Solve static statically indeterminate steel stress condition stress element T₁ tensile tension Tmax torque torsional V₁ yield strength yield stress zero σ₁