## Engineering mechanics of materials |

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Page 499

Since this value exceeds the allowable

must be redesigned using Eq. 4.22 as follows: 9 L JG 2.00 x 10"4 = 20,000 (ttD4/

32)(10 x 106) D = 3.18 in. The shaft must be at least 3.18 in. in diameter to meet

the ...

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: 9 L JG 2.00 x 10"4 = 20,000 (ttD4/

32)(10 x 106) D = 3.18 in. The shaft must be at least 3.18 in. in diameter to meet

the ...

Page 501

Example 10.7 Design a solid shaft of rectangular cross section to resist a torque

of 1500 N-m. The shearing yield stress of the material is 300 MPa. Use a factor of

safety of Nf = 2 with respect to first yield in shear. An allowable

...

Example 10.7 Design a solid shaft of rectangular cross section to resist a torque

of 1500 N-m. The shearing yield stress of the material is 300 MPa. Use a factor of

safety of Nf = 2 with respect to first yield in shear. An allowable

**rotation**of 10.00°...

Page 503

Assume that the shafts have minimal stress concentrations and satisfactory

x diagram for this stepped shaft. The shaft is solid and is to be fabricated of a

material ...

Assume that the shafts have minimal stress concentrations and satisfactory

**rotation**characteristics. 10.14 Refer to Figure H10.14 and draw the torque versusx diagram for this stepped shaft. The shaft is solid and is to be fabricated of a

material ...

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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

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### 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