Engineering Mechanics of Materials |
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Page 122
... tion . The reader is , once again , referred to Figure 3.4 ( d ) for the free - body diagram of that part of the member to the left of plane B - B . - ( c ) The solution for part ( a ) showed that σ1 = σ2 = 0 and that σ3 = − 10 ksi ...
... tion . The reader is , once again , referred to Figure 3.4 ( d ) for the free - body diagram of that part of the member to the left of plane B - B . - ( c ) The solution for part ( a ) showed that σ1 = σ2 = 0 and that σ3 = − 10 ksi ...
Page 242
... tion of the section in part ( a ) . 5.36 The cross section shown in Figure H5.21 is that for a simply supported beam 20 ft long which is to carry a uniform load of intensity 300 lb / ft acting ver- tically downward along the entire ...
... tion of the section in part ( a ) . 5.36 The cross section shown in Figure H5.21 is that for a simply supported beam 20 ft long which is to carry a uniform load of intensity 300 lb / ft acting ver- tically downward along the entire ...
Page 543
... tion . Consider a connection plate in . thick with a yield stress of 36 ksi . Assume that the combined thickness of the two angles of the connected member exceeds the plate thickness of 1⁄2 in . , which means that the bearing on the ...
... tion . Consider a connection plate in . thick with a yield stress of 36 ksi . Assume that the combined thickness of the two angles of the connected member exceeds the plate thickness of 1⁄2 in . , which means that the bearing on the ...
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