## Engineering mechanics of materials |

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Page 55

In the special case where the components F„ and F, are uniformly distributed over

the entire area A, then a = F„/A and t = F,/A. Note that a normal stress acts in a

direction

...

In the special case where the components F„ and F, are uniformly distributed over

the entire area A, then a = F„/A and t = F,/A. Note that a normal stress acts in a

direction

**perpendicular**to the plane on which it acts and it can be either tensile or...

Page 231

Thus any stress element at a distance v above the neutral axis, such as the one

shown in Figure 5.7(c) on the surface of the beam, has normal tensile stresses av

acting on its two planes that are

Thus any stress element at a distance v above the neutral axis, such as the one

shown in Figure 5.7(c) on the surface of the beam, has normal tensile stresses av

acting on its two planes that are

**perpendicular**to the axis of the beam. Its other ...Page 309

Tangential deviations are measured

from points on the tangent drawn to the elastic curve. (Point B also lies on the

initial position given by the straight line ABC, since the support at B does not

settle.) ...

Tangential deviations are measured

**perpendicular**to the undeformed beam axisfrom points on the tangent drawn to the elastic curve. (Point B also lies on the

initial position given by the straight line ABC, since the support at B does not

settle.) ...

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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