## Strength of Materials |

### From inside the book

Results 1-3 of 85

Page 170

This

it is assumed that the total vertical shear is carried by the web alone , the average

shearing stress in the web will be very close to the maximum shearing stress ...

This

**shows**that the flanges are almost ineffective in resisting the vertical shear . Ifit is assumed that the total vertical shear is carried by the web alone , the average

shearing stress in the web will be very close to the maximum shearing stress ...

Page 256

7 - 8d

long and perfectly restrained at the ends carries a uniformly distributed load over

part of its length as shown in Fig . 7 - 9a . Compute the end shears and end ...

7 - 8d

**shows**that the most dangerous moment is at the wall . 717 . A beam 12 ftlong and perfectly restrained at the ends carries a uniformly distributed load over

part of its length as shown in Fig . 7 - 9a . Compute the end shears and end ...

Page 257

Although we know from the downward concavity of the elastic curve that the

moment at A should be negative , we nevertheless

clockwise . Because of this deliberate error in the direction of the vector quantity

MA , our ...

Although we know from the downward concavity of the elastic curve that the

moment at A should be negative , we nevertheless

**show**it as positive , i . e . ,clockwise . Because of this deliberate error in the direction of the vector quantity

MA , our ...

### 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