## Strength of Materials |

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Page 268

Select the lightest WF beam that will support these loads without exceeding a

30 X 106 psi . Ans . 5 = 0 . 187 in . 749 . A timber beam 6 in . wide by 12 in . deep

...

Select the lightest WF beam that will support these loads without exceeding a

**flexural stress**of 18 , 000 psi . Compute the midspan deflection of this beam if E =30 X 106 psi . Ans . 5 = 0 . 187 in . 749 . A timber beam 6 in . wide by 12 in . deep

...

Page 314

Similarly , at a point B in the same section , also at a distance y from the neutral

axis but above it , the resultant stress is the difference between the axial and

stress ...

Similarly , at a point B in the same section , also at a distance y from the neutral

axis but above it , the resultant stress is the difference between the axial and

**flexural stresses**. If tensile stress is denoted by a positive sign and compressivestress ...

Page 387

If the eccentricity is small and the member short , the lateral deflection is

negligible , - Unavoidable or accidental and the

comeccentricity pared with the direct compressive stress . A long member ,

however , is quite ...

If the eccentricity is small and the member short , the lateral deflection is

negligible , - Unavoidable or accidental and the

**flexural stress**is insignificantcomeccentricity pared with the direct compressive stress . A long member ,

however , is quite ...

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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