## Engineering mechanics of materials |

### From inside the book

Results 1-3 of 77

Page 147

m x 0.03 m and a gage length of 0.2 m is subjected to a tensile load of 60,000 N.

The 0.03-m dimension changed to 0.02999 m and the 0.2 m gage length ...

**Homework Problems**3.14 An aluminum bar with a rectangular cross section 0.01m x 0.03 m and a gage length of 0.2 m is subjected to a tensile load of 60,000 N.

The 0.03-m dimension changed to 0.02999 m and the 0.2 m gage length ...

Page 337

Section 6.8, Problems 6.57 to 6.64. Section 6.9, Problems 6.65 to 6.69. Section

6.10, Problems 6.70 to 6.76. 637 Refer to Figure H6.43 and use Castigliano's ...

**Homework Problems**Text sections and problems are to be associated as follows:Section 6.8, Problems 6.57 to 6.64. Section 6.9, Problems 6.65 to 6.69. Section

6.10, Problems 6.70 to 6.76. 637 Refer to Figure H6.43 and use Castigliano's ...

Page 685

load of intensity 2000 N/m and acting along the v principal centroidal axis. The

cross section for the beam is shown in Figure H14.23. Assume that £wood = 10 x

...

**Homework Problems**14.23 A simply supported beam 5 m long carries a uniformload of intensity 2000 N/m and acting along the v principal centroidal axis. The

cross section for the beam is shown in Figure H14.23. Assume that £wood = 10 x

...

### What people are saying - Write a review

We haven't found any reviews in the usual places.

### 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

### Other editions - View all

### 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