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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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