Engineering Mechanics of Materials |
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Page 378
... Solution of the preceding equation leads to the following value for diameter d : d = 3.09 in . ( Note that this solution may also be obtained by use of Eq . 7.13d . ) Example 7.7 Solve Example 7.6 using the maximum shear stress theory ...
... Solution of the preceding equation leads to the following value for diameter d : d = 3.09 in . ( Note that this solution may also be obtained by use of Eq . 7.13d . ) Example 7.7 Solve Example 7.6 using the maximum shear stress theory ...
Page 390
... Solution . A convenient way to consider variations of the dimensionless quan- tity Wb / 2kc2 is to imagine all ... solution , shown pictorially in Figure 8.2 ( c ) . Small - displacement - theory solutions are of great interest in the ...
... Solution . A convenient way to consider variations of the dimensionless quan- tity Wb / 2kc2 is to imagine all ... solution , shown pictorially in Figure 8.2 ( c ) . Small - displacement - theory solutions are of great interest in the ...
Page 439
... solution is obviously not valid . Under such conditions another type of solution based upon inelastic analysis is required . Such a solution is beyond the scope of this chapter . However , certain procedures based upon inelastic ...
... solution is obviously not valid . Under such conditions another type of solution based upon inelastic analysis is required . Such a solution is beyond the scope of this chapter . However , certain procedures based upon inelastic ...
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