## Strength of Materials |

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Page 143

The explanation is that although the lightest beam is the cheapest on the basis of

the lightest one . 12 , 000 lb _ _ 15 _ ľ 5 ' R , = 3000 lb R2 = 9000 lb 3000 lb ...

The explanation is that although the lightest beam is the cheapest on the basis of

**weight**alone , frequently headroom clearances require a beam of less depth thanthe lightest one . 12 , 000 lb _ _ 15 _ ľ 5 ' R , = 3000 lb R2 = 9000 lb 3000 lb ...

Page 144

The

the maximum moment resulting from the combined live and dead loads . Hence

we compute the dead load moment Mp at x = 15 ft ( Fig . 5 – 8 ) . From the 27 lb /

ft ...

The

**weight**of the beam in this example is not sufficient to change the location ofthe maximum moment resulting from the combined live and dead loads . Hence

we compute the dead load moment Mp at x = 15 ft ( Fig . 5 – 8 ) . From the 27 lb /

ft ...

Page 146

In addition to section modulus and shape , the table lists the

encasement in lb / ft , and the product for checking lateral deflection when

necessary . The table is a modification of a similar table prepared by the

American ...

In addition to section modulus and shape , the table lists the

**weight**of a concreteencasement in lb / ft , and the product for checking lateral deflection when

necessary . The table is a modification of a similar table prepared by the

American ...

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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