## Engineering mechanics of materials |

### From inside the book

Results 1-3 of 59

Page 141

If the stress decreases past this point, it is referred to as the upper yield point, in

contrast to the lower yield point represented by point D in Figure 3.12 and

beyond which the stress increases with further strain.

materials ...

If the stress decreases past this point, it is referred to as the upper yield point, in

contrast to the lower yield point represented by point D in Figure 3.12 and

beyond which the stress increases with further strain.

**Yield Strength**. Formaterials ...

Page 543

Consider that the top chord of this deck truss will be spliced at joint D and find the

number of A325 high-strength bolts ... Use an allowable bearing stress of 1.35

times the

...

Consider that the top chord of this deck truss will be spliced at joint D and find the

number of A325 high-strength bolts ... Use an allowable bearing stress of 1.35

times the

**yield stress**of the connected material and assume shear conditions as...

Page 609

shearing

a solid circular shaft to resist a service maximum torque of 2500 N-m applied at

the center of a 3-m-long shaft which is torsionally fixed at both ends. Use a load ...

shearing

**yield stress**is 300 MPa, select a pipe section for this shaft. 12.40 Designa solid circular shaft to resist a service maximum torque of 2500 N-m applied at

the center of a 3-m-long shaft which is torsionally fixed at both ends. Use a load ...

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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