## Engineering mechanics of materials |

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Page 238

Therefore, the top fibers are in tension and the maximum tensile fiber stress is

10.186x10-' =2129MPa Position D: At this location, the bending moment is

positive.

Therefore, the top fibers are in tension and the maximum tensile fiber stress is

**given by Eq**. 5.10a, with v = 9.64 x 10" 2 m. Thus , v 22,500(9.64) x 10 2 (g^ =10.186x10-' =2129MPa Position D: At this location, the bending moment is

positive.

Page 245

This resultant force is given by -hi 2 F,= l axdA (5.11a) where dA is an element of

area in plane ijkl at a distance v below the neutral axis and the stress ax is

This resultant force is given by -hi 2 F,= l axdA (5.11a) where dA is an element of

area in plane ijkl at a distance v below the neutral axis and the stress ax is

**given****by Eq**. 5.10a. Thus substituting Eq. 5.10a into Eq. 5.11a yields F1=_?j vdA ...Page 373

Thus the energy of distortion per unit volume stored in the material during the

uniaxial tension or compression test is

from

Thus the energy of distortion per unit volume stored in the material during the

uniaxial tension or compression test is

**given**by the expression which is obtainedfrom

**Eq**. 7.10b by setting a2 and a3 equal to zero and equal to a0. The quantity ...### 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

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### 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