Engineering Mechanics of Materials |
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Page 200
... cross section has lower stresses and twists less than a member with a thin - walled square cross section of equal perimeter . Comparative results for members of the closed cross section of Figure 4.23 ( a ) and the open cross sections ...
... cross section has lower stresses and twists less than a member with a thin - walled square cross section of equal perimeter . Comparative results for members of the closed cross section of Figure 4.23 ( a ) and the open cross sections ...
Page 241
... cross section 5 x 10 in . It carries a uniformly dis- tributed load whose intensity is 400 lb / ft . ( a ) If the ... cross - sectional area whose inside diameter is 0.15 m and outside diameter is 0.30 m . The beam is subjected to a ...
... cross section 5 x 10 in . It carries a uniformly dis- tributed load whose intensity is 400 lb / ft . ( a ) If the ... cross - sectional area whose inside diameter is 0.15 m and outside diameter is 0.30 m . The beam is subjected to a ...
Page 253
... cross section of the beam is shown in Figure H5.7 . Determine the vertical shear stress 5 ft from the right support and ( a ) In the web , 2 in . below the top of the section . ( b ) At the neutral axis for the section . 5.40 The cross ...
... cross section of the beam is shown in Figure H5.7 . Determine the vertical shear stress 5 ft from the right support and ( a ) In the web , 2 in . below the top of the section . ( b ) At the neutral axis for the section . 5.40 The cross ...
Contents
Internal Forces in Members | 1 |
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 torsional unit vertical yield zero