## Strength of Materials |

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Page 283

Ferdinand Leon Singer. 816 .

length of the other span . 817 . See Fig . P - 817 . 818 . In Prob . 817 , determine

the changed value of the applied couple that will cause M2 to become zero .

Ferdinand Leon Singer. 816 .

**Solve**Prob . 815 if one span is three - fourths thelength of the other span . 817 . See Fig . P - 817 . 818 . In Prob . 817 , determine

the changed value of the applied couple that will cause M2 to become zero .

Page 310

8 - 25 . Here , instead of starting with an adjusted FEM at A and B , the

unbalanced moment at all supports is first distributed at each support and then

carried over as shown . PROBLEMS By means of the moment - distribution

method ,

8 - 25 . Here , instead of starting with an adjusted FEM at A and B , the

unbalanced moment at all supports is first distributed at each support and then

carried over as shown . PROBLEMS By means of the moment - distribution

method ,

**solve**...Page 376

remaining unchanged . 1033 .

inertia of the transformed section , and then applying the flexure formula

according to the ...

**Solve**Prob . 1031 if the bending moment is 50 , 000 ft - lb , all other dataremaining unchanged . 1033 .

**Solve**Prob . 1027 by computing the moment ofinertia of the transformed section , and then applying the flexure formula

according to the ...

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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