## Engineering mechanics of materials |

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Page 165

Once again, the center of the circle lies at the origin of the coordinate system and

the radius of the circle

principal normal strains, kx and c3, each have magnitudes

the ...

Once again, the center of the circle lies at the origin of the coordinate system and

the radius of the circle

**equals**Tr/2JG. In the special case of pure torsion, theprincipal normal strains, kx and c3, each have magnitudes

**equal**to Tr/2JG andthe ...

Page 231

Equilibrium dictates that the resisting moment on section a-a of the beam be

the system of tensile and compressive stresses shown in section a-a in Figure 5.7

(c).

Equilibrium dictates that the resisting moment on section a-a of the beam be

**equal**in magnitude to the applied moment Mu . This resisting moment is due tothe system of tensile and compressive stresses shown in section a-a in Figure 5.7

(c).

Page 373

Thus the energy of distortion per unit volume stored in the material during the

uniaxial tension or compression test is given by the expression (7.11) which is

obtained from Eq. 7.10b by setting a2 and a3

Thus the energy of distortion per unit volume stored in the material during the

uniaxial tension or compression test is given by the expression (7.11) which is

obtained from Eq. 7.10b by setting a2 and a3

**equal**to zero and ct,**equal**to ct0 .### 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