Engineering Mechanics of Materials |
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Page 338
... Castigliano's second theorem . Avoid replacing P by wa until the final step when you are ready to express the deflection in terms of w , a , E , and I. 6.62 Refer to Figure H6.48 and use Castigliano's second theorem to express the ...
... Castigliano's second theorem . Avoid replacing P by wa until the final step when you are ready to express the deflection in terms of w , a , E , and I. 6.62 Refer to Figure H6.48 and use Castigliano's second theorem to express the ...
Page 482
... Castigliano's Second Theorem Castigliano's second theorem as expressed in Eq . 6.29 may be used to satisfy the principle of consistent deformations and therefore to supplement the equations of equilibrium as they apply to a given ...
... Castigliano's Second Theorem Castigliano's second theorem as expressed in Eq . 6.29 may be used to satisfy the principle of consistent deformations and therefore to supplement the equations of equilibrium as they apply to a given ...
Page 486
... Castigliano's second theorem . 9.70 Solve Problem 9.33 by Castigliano's second theorem . 9.71 Solve Problem 9.34 by Castigliano's second theorem . 9.72 Solve Problem 9.35 by Castigliano's second theorem . 9.73 Solve Problem 9.36 by ...
... Castigliano's second theorem . 9.70 Solve Problem 9.33 by Castigliano's second theorem . 9.71 Solve Problem 9.34 by Castigliano's second theorem . 9.72 Solve Problem 9.35 by Castigliano's second theorem . 9.73 Solve Problem 9.36 by ...
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