## Engineering mechanics of materials |

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Page 141

The proportional limit for a given material represents the value of

which the material no longer behaves in ...

The proportional limit for a given material represents the value of

**stress**beyondwhich the material no longer behaves in ...

**Yield**Point. The**yield**point, ay, is the**stress**at which the material continues to deform without further increase in the ...Page 488

Final design involves a consideration of numerous factors other than strength

and deformation characteristics. ... However, yielding can also occur at stresses

well below the yield point or

stresses ...

Final design involves a consideration of numerous factors other than strength

and deformation characteristics. ... However, yielding can also occur at stresses

well below the yield point or

**yield stress**if the material is subjected to thesestresses ...

Page 489

Thus, if failure is judged to be by inelastic action, the failure stress would be the

yield point or

failure by inelastic action, the design stresses (stresses to which the system is ...

Thus, if failure is judged to be by inelastic action, the failure stress would be the

yield point or

**yield stress**of the material. Therefore, for the system to be safe fromfailure by inelastic action, the design stresses (stresses to which the system is ...

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### Contents

Introduction | 1 |

Torsionally Loaded Members in Equilibrium | 14 |

Shear and Bending Moment in Beams | 23 |

Copyright | |

19 other sections not shown

### Other editions - View all

### Common terms and phrases

absolute maximum shear allowable stress aluminum angle of twist applied Assume axes 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 strains 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