## Strength of Materials |

### From inside the book

Results 1-3 of 89

Page 113

The concentrated load of 120 lb at E reduces the shear abruptly to zero . Before

we locate the positions of zero shear at F and G on the shear diagram , consider

the effect of narrowing the distance over which the reaction R2 is

The concentrated load of 120 lb at E reduces the shear abruptly to zero . Before

we locate the positions of zero shear at F and G on the shear diagram , consider

the effect of narrowing the distance over which the reaction R2 is

**distributed**.Page 303

Thus the unbalanced moment is

of B caused by these

moments of half the amount and of opposite sign . These carry - over moments

are ...

Thus the unbalanced moment is

**distributed**at the unlocked support . The rotationof B caused by these

**distributed**moments induces , at A and C , carry - overmoments of half the amount and of opposite sign . These carry - over moments

are ...

Page 308

To understand this modification , we shall show how the moment that is

the computations . This moment , denoted by M in Fig . 8 - 24 , carries over to A

as + * M if ...

To understand this modification , we shall show how the moment that is

**distributed**to the left of B in the first distribution of solution ( a ) is carried throughthe computations . This moment , denoted by M in Fig . 8 - 24 , carries over to A

as + * M if ...

### What people are saying - Write a review

We haven't found any reviews in the usual places.

### Contents

List of Symbols and Abbreviations | xvi |

SIMPLE STRAIN | 26 |

TORSION | 60 |

Copyright | |

18 other sections not shown

### Other editions - View all

### 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