## Engineering mechanics of materials |

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Page 225

FIGURE H5.21 5.3 Flexural Stresses due to Symmetric

definition, a beam is a long and slender member that is subjected to

action. Depending upon the position and orientation of the loads with respect to

the ...

FIGURE H5.21 5.3 Flexural Stresses due to Symmetric

**Bending**of Beams Bydefinition, a beam is a long and slender member that is subjected to

**bending**action. Depending upon the position and orientation of the loads with respect to

the ...

Page 258

In the particular case where <j> is zero (i.e., symmetric

from Eq. 5.14 that /S is also zero, which means that the neutral axis coincides

with the u principal centroidal axis of inertia, as was concluded earlier in

analyzing ...

In the particular case where <j> is zero (i.e., symmetric

**bending**), we concludefrom Eq. 5.14 that /S is also zero, which means that the neutral axis coincides

with the u principal centroidal axis of inertia, as was concluded earlier in

analyzing ...

Page 266

If these deflections become excessive, plaster cracking, which is expensive to

repair, may occur in buildings. Shafts acting in

in their bearings due to large deflections, resulting in excessive wear and

possible ...

If these deflections become excessive, plaster cracking, which is expensive to

repair, may occur in buildings. Shafts acting in

**bending**may become misalignedin their bearings due to large deflections, resulting in excessive wear and

possible ...

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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