## Engineering mechanics of materials |

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Page 498

Use a shearing yield stress of 16,000 psi with a factor of safety of Ny = 2 and an

allowable

be replaced by (a) a relatively brittle material, or (b)a thin-walled tube. Use G = 10

x ...

Use a shearing yield stress of 16,000 psi with a factor of safety of Ny = 2 and an

allowable

**rotation**of 2.00 x 10"4 rad/in. Comment on the design if steel were tobe replaced by (a) a relatively brittle material, or (b)a thin-walled tube. Use G = 10

x ...

Page 499

Since this value exceeds the allowable

must be redesigned using Eq. 4.22 as follows: 0 _ JT L~JG 2.00 x 10"4 = 20,000

(*D732)(10 x 106) D = 3.18 in. The shaft must be at least 3.18 in. in diameter to ...

Since this value exceeds the allowable

**rotation**of 2.00 x 10 4 rad/in., the shaftmust be redesigned using Eq. 4.22 as follows: 0 _ JT L~JG 2.00 x 10"4 = 20,000

(*D732)(10 x 106) D = 3.18 in. The shaft must be at least 3.18 in. in diameter to ...

Page 501

... ae 210 x 106 (30,560)D/2 Ne 3 D = 0.1305 m (7t/32)D4 Use Eq. 4.22 to check

the

77.5 x 109) Since 3.17° < 5.00°, the shaft has satisfactory

... ae 210 x 106 (30,560)D/2 Ne 3 D = 0.1305 m (7t/32)D4 Use Eq. 4.22 to check

the

**rotation**angle of this shaft 30,560(4) = 0.0554 or 3.17° JG (7t/32)(0.1305)4(77.5 x 109) Since 3.17° < 5.00°, the shaft has satisfactory

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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