Engineering Mechanics of Materials |
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Page 337
... rotation at point A in terms of P , L , E , and I , by applying Castig- liano's second theorem . 6.59 Refer to Figure H6.45 and use Castigliano's second theorem to find the rotation at point A of this beam in terms of MA , L , E , and I ...
... rotation at point A in terms of P , L , E , and I , by applying Castig- liano's second theorem . 6.59 Refer to Figure H6.45 and use Castigliano's second theorem to find the rotation at point A of this beam in terms of MA , L , E , and I ...
Page 499
... rotation of 5.00 x 10-4 rad / in . Select the inside diameter equal to 0.6 of the outside diameter . This diametral ratio will assure that local buckling is not critical in this case and that shearing stress or rotation per unit length ...
... rotation of 5.00 x 10-4 rad / in . Select the inside diameter equal to 0.6 of the outside diameter . This diametral ratio will assure that local buckling is not critical in this case and that shearing stress or rotation per unit length ...
Page 501
... rotation angle of this shaft : TL 0 = = JG 30,560 ( 4 ) ( π / 32 ) ( 0.1305 ) * ( 77.5 × 109 ) = · 0.0554 or 3.17 ° Since 3.17 ° 5.00 ° , the shaft has satisfactory rotation characteristics . Choose : D = 0.1305 m Practical ...
... rotation angle of this shaft : TL 0 = = JG 30,560 ( 4 ) ( π / 32 ) ( 0.1305 ) * ( 77.5 × 109 ) = · 0.0554 or 3.17 ° Since 3.17 ° 5.00 ° , the shaft has satisfactory rotation characteristics . Choose : D = 0.1305 m Practical ...
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