Engineering Mechanics of Materials |
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Page 245
... given by Eq . 5.10a . Thus substituting Eq . 5.10a into Eq . 5.11a yields x Fi = M u Iu V1 -h / 2 v dA ( 5.11b ) 2. A normal force F2 representing the resultant of the normal stresses on plane efgh produced by the moment M + dM . This ...
... given by Eq . 5.10a . Thus substituting Eq . 5.10a into Eq . 5.11a yields x Fi = M u Iu V1 -h / 2 v dA ( 5.11b ) 2. A normal force F2 representing the resultant of the normal stresses on plane efgh produced by the moment M + dM . This ...
Page 258
... Eq . 5.13f for tan ẞ leads to tan ß = I ( 5.13f ) Iu tan I ( 5.14 ) Equation ... given cross section will be the algebraic sum of the stresses produced by M ... given by Eq . 5.10a . Also , the moment M ,, acting alone , would produce ...
... Eq . 5.13f for tan ẞ leads to tan ß = I ( 5.13f ) Iu tan I ( 5.14 ) Equation ... given cross section will be the algebraic sum of the stresses produced by M ... given by Eq . 5.10a . Also , the moment M ,, acting alone , would produce ...
Page 373
... given by Eq . 7.10b reaches the critical value of the material given by Eq . 7.11 . Thus 1 μ 3E [ ( 0 , − 02 ) 2 + ( 02 − 03 ) 2 + ( 03 − 01 ) 2 ] 03 ) 2 + ( 03 − 01 ) 2 ] = 1 + H 2 1 + μ 6E which , when simplified , reduces to ...
... given by Eq . 7.10b reaches the critical value of the material given by Eq . 7.11 . Thus 1 μ 3E [ ( 0 , − 02 ) 2 + ( 02 − 03 ) 2 + ( 03 − 01 ) 2 ] 03 ) 2 + ( 03 − 01 ) 2 ] = 1 + H 2 1 + μ 6E which , when simplified , reduces to ...
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