## Strength of Materials |

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Page 164

For equilibrium of this

requires an equal balancing shearing stress on the top face . The forces causing

these shearing stresses ( Fig . 5 – 23c ) form | Vr = S , DA a counterclockwise

couple ...

For equilibrium of this

**element**, the shearing stress Son on the bottom facerequires an equal balancing shearing stress on the top face . The forces causing

these shearing stresses ( Fig . 5 – 23c ) form | Vr = S , DA a counterclockwise

couple ...

Page 325

Stress at a point varies with inclination of plane through the point . only to a

normal stress , but the

shearing stress caused by the N and T components of the equilibrant E . Thus at

the ...

Stress at a point varies with inclination of plane through the point . only to a

normal stress , but the

**element**in Fig . 9 – 9c is subjected to both normal andshearing stress caused by the N and T components of the equilibrant E . Thus at

the ...

Page 338

Solve the preceding problem if the stress in part ( c ) of the figure is changed to

10 , 000 psi compression . 941 . The principal stresses on an

, and Sz . Assuming S , > Sy > Sz , show that the maximum shearing stress on

any ...

Solve the preceding problem if the stress in part ( c ) of the figure is changed to

10 , 000 psi compression . 941 . The principal stresses on an

**element**are S . , Sy, and Sz . Assuming S , > Sy > Sz , show that the maximum shearing stress on

any ...

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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