## Engineering mechanics of materialsThis book provides the students of various engineering disciplines with a clear and understandable treatment of the concepts of Mechanics of Materials or Strength of Materials. This subject is concerned with the behavior of deformable bodies when subjected to axial, torsional and flexural loads as well as combinations thereof. It is a 3rd, updated edition of the popular undergraduate level textbook useful for students of mechanical, structural, civil, aeronautical and other engineering disciplines. The book is supplied with problems and a solution manual will be available from the authors. |

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Results 1-3 of 26

Page 148

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 Fig. 3.14 and beyond

which the stress increases with further strain.

...

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 Fig. 3.14 and beyond

which the stress increases with further strain.

**Yield Strength**. For materials having...

Page 486

Thus, if failure is judged to be by inelastic action, the failure stress would be the

yield point or

failure by inelastic action, the design stresses (stresses to which the system is ...

Thus, if failure is judged to be by inelastic action, the failure stress would be the

yield point or

**yield stress**of the material. Therefore, for the system to be safe fromfailure by inelastic action, the design stresses (stresses to which the system is ...

Page 492

Assume that the concrete has an ultimate strength of 25 MPa and use a factor of

safety Nu = 3 based on failure by fracture. 10.2 A tension member is to be

fabricated from an aluminum alloy (£ = 10 x 106 psi) for which the

is 50 ...

Assume that the concrete has an ultimate strength of 25 MPa and use a factor of

safety Nu = 3 based on failure by fracture. 10.2 A tension member is to be

fabricated from an aluminum alloy (£ = 10 x 106 psi) for which the

**yield strength**is 50 ...

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### Contents

Stress Strain and Their Relationships | 60 |

Stresses and Strains in Axially Loaded Members | 121 |

Torsional Stresses Strains and Rotations | 159 |

Copyright | |

14 other sections not shown

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### Common terms and phrases

absolute maximum shear aluminum angle of twist applied Assume axial force axially loaded beam shown bending cantilever beam Castigliano's second theorem column compressive constant coordinate cross section cross-sectional area cylinder deflection deformation depicted in Fig elastic curve equal equation equilibrium Euler EXAMPLE factor of safety FIGURE flexural stress FORTRAN free-body diagram function given by Eq k-ft k-in kN-m lb/ft length longitudinal material maximum in-plane shear maximum shear stress modulus of elasticity Mohr's circle neutral axis normal stress obtained perpendicular plane stress condition plot positive principal centroidal axis principal strains principal stresses radius ratio reactions Refer to Fig respect rotation shear force shear strain shown in Fig simply supported beam slope SOLUTION Solve Problem statically indeterminate steel stress concentration stress element subjected torque torsional uniform load vertical yield strength yield stress zero