## Engineering mechanics of materials |

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Page 338

6.61 Find the deflection of point A of the beam shown in Figure H6.47 using

you are ready to express the deflection in terms of w, a, E, and Iu . 6.62 Refer to

Figure ...

6.61 Find the deflection of point A of the beam shown in Figure H6.47 using

**Castigliano's second theorem**. Avoid replacing P by wa until the final step whenyou are ready to express the deflection in terms of w, a, E, and Iu . 6.62 Refer to

Figure ...

Page 482

*9.7 Statically Indeterminate Members Under Flexural Loads —

used to satisfy the principle of consistent deformations and therefore to

supplement ...

*9.7 Statically Indeterminate Members Under Flexural Loads —

**Castigliano's****Second Theorem Castigliano's second theorem**as expressed in Eq. 6.29 may beused to satisfy the principle of consistent deformations and therefore to

supplement ...

Page 483

dA~ElJ0 MudAvUX From the free-body diagram of Figure 9.6(c), Iw Therefore, Mu

= Avx-"^x3 0 L dA„ (c) (e) Substitution of Eqs. (d) and (e) into Eq. (c) leads to the

following expression: Using

dA~ElJ0 MudAvUX From the free-body diagram of Figure 9.6(c), Iw Therefore, Mu

= Avx-"^x3 0 L dA„ (c) (e) Substitution of Eqs. (d) and (e) into Eq. (c) leads to the

following expression: Using

**Castigliano's second theorem**as expressed in Eq.### What people are saying - Write a review

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### Contents

Internal Forces in Members | 1 |

Hollow Circular ShaftsAngle of Twist and Shearing Stresses | 157 |

4i5 Power Transmission | 169 |

Copyright | |

7 other sections not shown

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### Common terms and phrases

absolute maximum shear allowable stress aluminum angle of twist applied Assume axial force axially loaded beam shown bending cantilever beam Castigliano's second theorem circular column components compressive Compute constant construct coordinate cross section cross-sectional area cylinder deflection deformation depicted in Figure Determine diameter differential elastic curve equal equation equilibrium factor of safety flexural stress free-body diagram function given by Eq Homework Problems k-ft k-in kN-m length longitudinal material maximum shear stress modulus of elasticity Mohr's circle neutral axis normal stress obtained perpendicular plane stress plot principal centroidal axis principal stresses radius Refer to Figure respect rotation section a-a segment shear center shear force shear strain shown in Figure slope Solution Solve static statically indeterminate steel stress element subjected Substitution tension tion torque torsional uniform load vertical yield strength yield stress zero