## 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

Mv = K/v sin P (5.13e) From the geometry in Figure 5.17 and by using Eqs. 5.13d

and 5.13e, we obtain Equation 5.14 determines the orientation, from the u

principal centroidal axis of inertia, of the neutral axis for unsymmetric

terms ...

Mv = K/v sin P (5.13e) From the geometry in Figure 5.17 and by using Eqs. 5.13d

and 5.13e, we obtain Equation 5.14 determines the orientation, from the u

principal centroidal axis of inertia, of the neutral axis for unsymmetric

**bending**interms ...

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

Internal Forces in Members | 1 |

Hollow Circular ShaftsAngle of Twist and Shearing Stresses | 157 |

4i5 Power Transmission | 169 |

Copyright | |

7 other sections not shown

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### 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