## Engineering mechanics of materials |

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Page 55

Note that a

acts and it can be either tensile or compressive. A tensile

that tends to pull the material particles away from each other, while a

compressive ...

Note that a

**normal stress**acts in a direction perpendicular to the plane on which itacts and it can be either tensile or compressive. A tensile

**normal stress**is onethat tends to pull the material particles away from each other, while a

compressive ...

Page 85

However, for present purposes, it is only necessary to know that when a member

is subjected to a

direction of the applied stress (normal longitudinal strain), but also a smaller

normal ...

However, for present purposes, it is only necessary to know that when a member

is subjected to a

**normal stress**, it experiences not only a normal strain in thedirection of the applied stress (normal longitudinal strain), but also a smaller

normal ...

Page 117

the force normal to this area, Fu , is, in this case, equal to FB = P, the

, for example, the resultant force FB can be replaced by a stress system ax as

shown ...

the force normal to this area, Fu , is, in this case, equal to FB = P, the

**normal****stress**ax can be expressed by the equation ffx=7 = A=A (M) Thus, in Figure 3.1(b), for example, the resultant force FB can be replaced by a stress system ax as

shown ...

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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

### Other editions - View all

### 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