## Engineering mechanics of materials |

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Page 16

20

FIGURE 1.8 T4 = T, - T2 + T3 T4 = 20 - 30 + 25 TA = 15

= T, - T2 = 20 - 30 = 1 - 10

torque ...

20

**k**-**in**. Torque (**k**-**in**.) -20 in.- -30 in. -30 in.- A 30**k**-**in**. g 25**k**-**in**. 20 15 -10FIGURE 1.8 T4 = T, - T2 + T3 T4 = 20 - 30 + 25 TA = 15

**k**-**in**. TA = T,= | 20**k**-**in**. TB= T, - T2 = 20 - 30 = 1 - 10

**k**-**in**. Tc = T, - T2 + T3 = 20 - 30 + 25 =| 15**k**-**in**. Thetorque ...

Page 22

1.17 Draw the internal torque diagram for the shaft depicted in Figure H11. 17. 15

FIGURE H1.17 1.18 The shaft depicted in Figure H11. 18 is subjected to an

external ...

1.17 Draw the internal torque diagram for the shaft depicted in Figure H11. 17. 15

**k**-**in**. 30k-in. 40**k**-**in**. h ) ) ) ) fr- JmST 30 ^ 30 in-H^30 -I-25 in 'I /I B C D E FFIGURE H1.17 1.18 The shaft depicted in Figure H11. 18 is subjected to an

external ...

Page 747

Answers to Even-Numbered Problems (In some cases, partial answers are given.

) Chapter 1 1.2 Maximum F = 25 lb at y ... 1.14 Maximum T = 7.5 k-ft between 5

and 7 ft 1.16 Maximum T = 650

x ...

Answers to Even-Numbered Problems (In some cases, partial answers are given.

) Chapter 1 1.2 Maximum F = 25 lb at y ... 1.14 Maximum T = 7.5 k-ft between 5

and 7 ft 1.16 Maximum T = 650

**k**-**in**. at x = 90 in. 1.18 Maximum T = 127.3 N-m atx ...

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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