## Engineering mechanics of materials |

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Page 141

Elastic Limit. The elastic limit, ae , for a given material is the value of stress

beyond which the material experiences a ... The

constant of proportionality between stress and strain in Hooke's law as expressed

in Eq.

Elastic Limit. The elastic limit, ae , for a given material is the value of stress

beyond which the material experiences a ... The

**modulus of elasticity**, E, is theconstant of proportionality between stress and strain in Hooke's law as expressed

in Eq.

Page 147

(d) The maximum shearing stress in the bar. (e) The

steel bar 2 in. in diameter and 4 in. long is subjected to a compressive force. The

proportional limit for this steel is 40,000 psi, its

...

(d) The maximum shearing stress in the bar. (e) The

**modulus of elasticity**. 3.15 Asteel bar 2 in. in diameter and 4 in. long is subjected to a compressive force. The

proportional limit for this steel is 40,000 psi, its

**modulus of elasticity**30 x 106 psi,...

Page 208

For the data given below: (a) Find the shearing

equation of the shearing stress-strain line below the yield point. (c) Plot the stress

-strain diagram for shear. (d) Find the modulus of resilience. Circular cylindrical ...

For the data given below: (a) Find the shearing

**modulus of elasticity**. (b) Find theequation of the shearing stress-strain line below the yield point. (c) Plot the stress

-strain diagram for shear. (d) Find the modulus of resilience. Circular cylindrical ...

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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