## Strength of Materials |

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Page 375

91 ) = 310 , 000 in . - lb The safe bending moment is therefore 243 , 000 in . - lb .

Since the concrete governs , we conclude there is too much steel ; hence the

beam is over -

10 ...

91 ) = 310 , 000 in . - lb The safe bending moment is therefore 243 , 000 in . - lb .

Since the concrete governs , we conclude there is too much steel ; hence the

beam is over -

**reinforced**. PROBLEMS 1024 . In a**reinforced**concrete beam , b =10 ...

Page 376

The dimensions of a

sq in . , and n = 12 . If the allowable stresses are fc $ 800 psi and f . $ 18 , 000 psi

, determine the maximum bending moment that may be applied . In what state of

...

The dimensions of a

**reinforced**concrete beam are b = 12 in . , d = 18 in . , A , = 2sq in . , and n = 12 . If the allowable stresses are fc $ 800 psi and f . $ 18 , 000 psi

, determine the maximum bending moment that may be applied . In what state of

...

Page 379

Usually the available stock sizes of

area of steel , so the final design only closely approximates balanced

reach fo = 700 ...

Usually the available stock sizes of

**reinforcing**steel do not produce exactly thisarea of steel , so the final design only closely approximates balanced

**reinforcement**. PROBLEMS 1036 . A**reinforced**concrete beam is designed toreach fo = 700 ...

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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