## Strength of Materials |

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Page 164

It is this vertical

resisting vertical shear V , = SS , dA which balances the vertical shear V . Since it

is not feasible to determine Ss , directly , we have resorted to deriving the

numerically ...

It is this vertical

**shearing stress**Sso , shown in Fig . 5 – 23 , that forms theresisting vertical shear V , = SS , dA which balances the vertical shear V . Since it

is not feasible to determine Ss , directly , we have resorted to deriving the

numerically ...

Page 327

The notation used here defines a normal stress by means of a single subscript

corresponding to the face on which it acts . A face takes the name of the axis

normal to it , e . g . , the X face is perpendicular to the X axis . A

...

The notation used here defines a normal stress by means of a single subscript

corresponding to the face on which it acts . A face takes the name of the axis

normal to it , e . g . , the X face is perpendicular to the X axis . A

**shearing stress**is...

Page 341

The solid lines indicate the direction of the maximum compressive stresses , and

the dashed lines indicate the ... also act simultaneously on the shaft without

exceeding a resultant

stress S ...

The solid lines indicate the direction of the maximum compressive stresses , and

the dashed lines indicate the ... also act simultaneously on the shaft without

exceeding a resultant

**shearing stress**So = 12 , 000 psi or a resultant normalstress S ...

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### Contents

List of Symbols and Abbreviations | xvi |

SIMPLE STRAIN | 26 |

TORSION | 60 |

Copyright | |

18 other sections not shown

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### Common terms and phrases

acting actual allowable angle applied assumed axes axial axis beam beam shown bending bending moment cantilever carries caused centroid circle column compressive compressive stress compute concentrated concrete consider constant couple cross section deflection deformation Determine developed diameter direction distance distributed load effect elastic curve element equal equation equivalent expressed flange flexural stress force formula ft-lb given gives Hence horizontal ILLUSTRATIVE inertia joint lb/ft length limit load material maximum maximum shearing method midspan moments negative neutral axis normal obtain occurs plane plate positive principal Prob PROBLEMS produce radius reaction reduces reference reinforced relation resisting respect resultant rivet segment shaft shearing stress shown in Fig shows slope Solution Solve span steel strain strength supported Table tangent tensile thickness torsional uniformly varies vertical weight weld yield zero