## Engineering mechanics of materials |

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Page 222

... m4 (P2)xy = 0 + 96(-4.67)(1.33) x 10"8 = -0.596 x 10"5 m4 The first term in Eq.

5.4f (i.e., Pxy) is zero for both rectangular areas Al and A2 because it represents

the product of inertia with respect to centroidal

... m4 (P2)xy = 0 + 96(-4.67)(1.33) x 10"8 = -0.596 x 10"5 m4 The first term in Eq.

5.4f (i.e., Pxy) is zero for both rectangular areas Al and A2 because it represents

the product of inertia with respect to centroidal

**axes**which are**axes**of symmetry.Page 224

Find the polar moments of inertia with respect to

FIGURE H5.4 5.5-5.7 The sections shown in Figures H5.5, H5.6. and H5.7 have

two perpendicular

...

Find the polar moments of inertia with respect to

**axes**through points C and O.FIGURE H5.4 5.5-5.7 The sections shown in Figures H5.5, H5.6. and H5.7 have

two perpendicular

**axes**of symmetry. For each of these three sections, determine...

Page 225

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

principal

symmetric or unsymmetric. Symmetric bending of beams is discussed in this

section and ...

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

principal

**axes**of inertia of the beam cross section, bending may be eithersymmetric or unsymmetric. Symmetric bending of beams is discussed in this

section and ...

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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