Engineering Mechanics of Materials |
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Page 469
... Example 9.6 . The determination of the reaction components B , and MR is identical to what was done in Example 9.6 . B Example 9.10 Solve Example 9.8 by superposition . Solution . For convenience , the beam of Example 9.8 is repeated in ...
... Example 9.6 . The determination of the reaction components B , and MR is identical to what was done in Example 9.6 . B Example 9.10 Solve Example 9.8 by superposition . Solution . For convenience , the beam of Example 9.8 is repeated in ...
Page 535
... Example 10.16 and redesign the member as a hollow circular cylindrical one such that the inside diameter is 0.8 of the outside diameter . 10.42 Refer to Example 10.16 , increase the axial force to P = 200 kN and the torque to T = 40 kN ...
... Example 10.16 and redesign the member as a hollow circular cylindrical one such that the inside diameter is 0.8 of the outside diameter . 10.42 Refer to Example 10.16 , increase the axial force to P = 200 kN and the torque to T = 40 kN ...
Page 560
... Example 11.5 and Figure 11.9 and design members CE and CD . As stated in Example 11.5 , these two members are flexible steel rods . Assume an allowable stress of 200 MPa and an allow- able deformation of 4 x 10-2 m . Let E = 200 GN / m2 ...
... Example 11.5 and Figure 11.9 and design members CE and CD . As stated in Example 11.5 , these two members are flexible steel rods . Assume an allowable stress of 200 MPa and an allow- able deformation of 4 x 10-2 m . Let E = 200 GN / m2 ...
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