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 Mu -h / 2 F1 = Iu v dA v1 ( 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 Mu -h / 2 F1 = Iu v dA v1 ( 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 Iu = tan I ( 5.13f ) ( 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 Iu = tan I ( 5.13f ) ( 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 - − - [ ( 01 − 02 ) 2 + ( 02 − 03 ) 2 + ( 03 − 01 ) 2 ] 6E = 1 + μ 0 20 ( 7.12a ) 3E which , when simplified , reduces to - -- ( 01 − 02 ) ...
... given by Eq . 7.10b reaches the critical value of the material given by Eq . 7.11 . Thus 1 + μ - − - [ ( 01 − 02 ) 2 + ( 02 − 03 ) 2 + ( 03 − 01 ) 2 ] 6E = 1 + μ 0 20 ( 7.12a ) 3E which , when simplified , reduces to - -- ( 01 − 02 ) ...
Contents
Internal Forces in Members | 1 |
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 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 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 shaft shear strain shown in Figure slope Solution Solve static statically indeterminate steel stress at point stress condition stress element T₁ tensile tension Tmax torque torsional V₁ yield strength yield stress zero σ₁