## Engineering mechanics of materials |

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Page 147

(

0.0270 4,800 0.0015 14,000 0.0330 6,400 0.0021 15,000 0.0400 8,000 0.0026

16,000 0.0490 9,600 0.0031 17,500 0.0670 11,200 0.0035 19,000 0.1050

12,000 ...

(

**lb) (in**.) 0 0 12,400 0.0170 1,600 0.0004 12,300 0.0200 3,200 0.0009 13,0000.0270 4,800 0.0015 14,000 0.0330 6,400 0.0021 15,000 0.0400 8,000 0.0026

16,000 0.0490 9,600 0.0031 17,500 0.0670 11,200 0.0035 19,000 0.1050

12,000 ...

Page 158

30,000

line Any cross section of the shaft (O 360,000

Angle of twist (d) FIGURE 4.6 Solution. Apply Eq. 4.23 for a hollow shaft to obtain

the ...

30,000

**lb**-ft 2 in. in. -60 in. (a) 30,000**lb**-ft r(**lb-in**.) (0,0) 3820 psi Typical radialline Any cross section of the shaft (O 360,000

**lb-in**. (b) (60, 0) • a- (in.) PinalAngle of twist (d) FIGURE 4.6 Solution. Apply Eq. 4.23 for a hollow shaft to obtain

the ...

Page 514

The maximum moment in the beam occurs at midspan and has a value equal to

5000

components, Mu and Mv as shown in Figure 10.7(b), in which Afu = Mv = 42,420

The maximum moment in the beam occurs at midspan and has a value equal to

5000

**lb**-ft = 60,000**lb-in**. This moment may be decomposed into its twocomponents, Mu and Mv as shown in Figure 10.7(b), in which Afu = Mv = 42,420

**lb-in**.### 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