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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absolute maximum shear aluminum angle of twist applied Assume axes axial force axially loaded beam shown bending C₁ cantilever beam Castigliano's second theorem column components compressive Compute coordinate cross section cross-sectional area cylinder deflection deformation depicted in Figure Determine diameter elastic curve equal equation equilibrium Example factor of safety flexural stress free-body diagram Homework Problems k-ft k-in kN-m length longitudinal M₁ material maximum shear stress MN/m² modulus of elasticity Mohr's circle moment of inertia neutral axis normal stress obtained perpendicular plane stress plot principal centroidal axis principal stresses r₁ ratio Refer to Figure respect rotation section a-a segment shaft shear force shear strain shown in Figure slope Solution Solve static statically indeterminate steel stress condition stress element T₁ tensile tension Tmax torque torsional V₁ vertical yield strength yield stress zero