## 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

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

...

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

...

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, F„ , 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 in ...

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

**normal****stress**ax can be expressed by the equation '.-h-r-i--A TM Thus, in Figure 3.1(b),for example, the resultant force FB can be replaced by a stress system ax as

shown in ...

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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