## Engineering mechanics of materials |

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Page 141

The proportional limit for a given

which the

to strain. The proportional limit, ap, is represented by the ordinate to point A in ...

The proportional limit for a given

**material**represents the value of stress beyondwhich the

**material**no longer behaves in such a way that the stress is proportionalto strain. The proportional limit, ap, is represented by the ordinate to point A in ...

Page 680

In analyzing beams of two

equivalent cross section instead of dealing with the actual cross section of the

beam. This equivalent cross section is obtained by transforming either of the two

In analyzing beams of two

**materials**, it is very convenient to deal with anequivalent cross section instead of dealing with the actual cross section of the

beam. This equivalent cross section is obtained by transforming either of the two

**materials**...Page 684

In either case there is an abrupt change in the magnitude of the flexure stress at

the junction between

discussion, assume that EB < EA so that n is less than unity. It follows from Eq. (b)

...

In either case there is an abrupt change in the magnitude of the flexure stress at

the junction between

**material**A and**material**B, where vA = vB . For purposes ofdiscussion, assume that EB < EA so that n is less than unity. It follows from Eq. (b)

...

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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