## Engineering mechanics of materials |

### From inside the book

Results 1-3 of 14

Page 659

Example 14.3 Frequently, in order to increase their load-carrying capacity,

composite

inner

outside ...

Example 14.3 Frequently, in order to increase their load-carrying capacity,

composite

**cylinders**are made by shrinking an outer**cylinder**or jacket onto aninner

**cylinder**. Initially, the inside radius of the jacket is a little smaller than theoutside ...

Page 660

The symbol r2 may be used to represent the outside radius of the inner

and the inside radius of the jacket because their difference Ar is very small.

Assume both jacket and the inner

When the ...

The symbol r2 may be used to represent the outside radius of the inner

**cylinder**and the inside radius of the jacket because their difference Ar is very small.

Assume both jacket and the inner

**cylinder**to be made of the same material.When the ...

Page 663

Example 14.4 A thin-walled

= 0.54 m is subjected to internal pressure pl = 2 MPa. Determine (a) the absolute

maximum shearing stress on the inner surface of the

Example 14.4 A thin-walled

**cylinder**with closed ends for which r, = 0.50 m and r2= 0.54 m is subjected to internal pressure pl = 2 MPa. Determine (a) the absolute

maximum shearing stress on the inner surface of the

**cylinder**, (b) the absolute ...### What people are saying - Write a review

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### Contents

Introduction | 1 |

Torsionally Loaded Members in Equilibrium | 14 |

Shear and Bending Moment in Beams | 23 |

Copyright | |

19 other sections not shown

### Other editions - View all

### Common terms and phrases

absolute maximum shear allowable stress aluminum angle of twist applied Assume axes 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 strains 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