## Engineering mechanics of materials |

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Page 55

In the special case where the

over the entire area A, then a = Fn/A and t = FJA. Note that a normal stress acts in

a direction perpendicular to the plane on which it acts and it can be either tensile

...

In the special case where the

**components**Fu and F, are uniformly distributedover the entire area A, then a = Fn/A and t = FJA. Note that a normal stress acts in

a direction perpendicular to the plane on which it acts and it can be either tensile

...

Page 482

Assume the cross-sectional area of the steel rods to be A, the modulus of

elasticity for steel to be E, and the moment of inertia for the beam to be /u , and

determine the force induced in the rods and the reaction

B. Use ...

Assume the cross-sectional area of the steel rods to be A, the modulus of

elasticity for steel to be E, and the moment of inertia for the beam to be /u , and

determine the force induced in the rods and the reaction

**components**at supportB. Use ...

Page 526

*10.7 Design of

1.9 will improve the reader's understanding of design of

such loadings. General review of Chapter 7 and intensive review of Section 7.6 is

also ...

*10.7 Design of

**Components**to Resist Combined Loadings Review of Section1.9 will improve the reader's understanding of design of

**components**to resistsuch loadings. General review of Chapter 7 and intensive review of Section 7.6 is

also ...

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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