## Engineering mechanics of materials |

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Page 241

5.26 L = 15 ft, P = 30 kips, cross section

50 kips, cross section

5.27 if the load P acts in a plane that contains the u principal centroidal axis for ...

5.26 L = 15 ft, P = 30 kips, cross section

**shown in Figure**H5.6. 5.27 L = 25 ft, P =50 kips, cross section

**shown in Figure**H5.7. 5.28-5.30 Repeat Problems 5.25 to5.27 if the load P acts in a plane that contains the u principal centroidal axis for ...

Page 341

applicable to any arbitrary cross-sectional area as long as the material is not

stressed beyond the proportional limit. Consider, for example, the circular shaft

axial ...

applicable to any arbitrary cross-sectional area as long as the material is not

stressed beyond the proportional limit. Consider, for example, the circular shaft

**shown in Figure**7.1(a), which is subjected to the combined effects of the tensileaxial ...

Page 361

7.32 The following data apply to the crank shaft

, e = 0.30 m, d = 0.50 m, and the maximum tensile stress at point D is known to be

100 M Pa. Determine the diameter of the crank shaft. 7.33 Repeat Problem ...

7.32 The following data apply to the crank shaft

**shown in Figure**7.7(a): F = 12 kN, e = 0.30 m, d = 0.50 m, and the maximum tensile stress at point D is known to be

100 M Pa. Determine the diameter of the crank shaft. 7.33 Repeat Problem ...

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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