## Engineering mechanics of materials |

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Page 163

I0k-ft V Uniformly distributed torque 2

FIGURE H4.7 4.8 A hollow shaft is subjected to the torques shown in Figure H4.8.

Plot the variation of torque versus a longitudinal coordinate measured along the

axis of ...

I0k-ft V Uniformly distributed torque 2

**k**-**ft**/ft V Knnnrj - A I D = 4 in. if -10 ft-FIGURE H4.7 4.8 A hollow shaft is subjected to the torques shown in Figure H4.8.

Plot the variation of torque versus a longitudinal coordinate measured along the

axis of ...

Page 747

Chapter 1 1.2 Maximum F = 25 lb at y = 8

1.6 Maximum F = 40 kN, x from 6 to 8 m 1.8 F = 102y (origin at bottom for y) 1.10

F = 15y (origin at bottom for y) 1.12 Maximum T = 10

Chapter 1 1.2 Maximum F = 25 lb at y = 8

**ft**1.4 Maximum F = 390 lb at y = 17**ft**1.6 Maximum F = 40 kN, x from 6 to 8 m 1.8 F = 102y (origin at bottom for y) 1.10

F = 15y (origin at bottom for y) 1.12 Maximum T = 10

**k**-in., between 0 and 10 in.Page 748

1.46 Maximum V = 2P, 4-

end 1.48 V = ( - wa/(2L))x2 ; M = (uy (6L))x3 - (w^)!2 1.50 V = - 1.25Q (0 < x < L);

= -0.25Q (L < x < 4L); maximum Af = 0.25QL, just to the right of x = 3L 1.52 ...

1.46 Maximum V = 2P, 4-

**ft**segment at left end; maximum M = 14.7 P, 8**ft**from leftend 1.48 V = ( - wa/(2L))x2 ; M = (uy (6L))x3 - (w^)!2 1.50 V = - 1.25Q (0 < x < L);

**K**= -0.25Q (L < x < 4L); maximum Af = 0.25QL, just to the right of x = 3L 1.52 ...

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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