## Strength of Materials |

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Page 258

Comparison with the

Eqs ( d ) and ( e ) used there are identical with Eqs . ( a ) and ( b ) here . On the

other hand , comparison with the superposition

that ...

Comparison with the

**solution**by double integration on page 250 discloses thatEqs ( d ) and ( e ) used there are identical with Eqs . ( a ) and ( b ) here . On the

other hand , comparison with the superposition

**solution**on page 252 disclosesthat ...

Page 308

procedure that was described in Prob . 874 for free ends on two spans . It is

inconvenient to treat a free end as fixed , carry moment over to it , and then

release it again .

**Solution**: There are two**solutions**. The first , in ( a ) , involves the sameprocedure that was described in Prob . 874 for free ends on two spans . It is

inconvenient to treat a free end as fixed , carry moment over to it , and then

release it again .

Page 309

... + 300 + 200 Carry - over + 150 2nd distribution - 50 - 100 + 20 + 13 Carry -

over - 50 | 3rd distribution - 7 + 30 + 20 Carry - over 4th distribution - 5 - 10 + 2 + 1

- 800 | + 800 - 1491 | + 1491 - 1234 + 1234

+ ...

... + 300 + 200 Carry - over + 150 2nd distribution - 50 - 100 + 20 + 13 Carry -

over - 50 | 3rd distribution - 7 + 30 + 20 Carry - over 4th distribution - 5 - 10 + 2 + 1

- 800 | + 800 - 1491 | + 1491 - 1234 + 1234

**Solution**( a ) Regular procedure - 33+ ...

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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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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