## Strength of Materials |

### From inside the book

Results 1-3 of 96

Page 116

500 lb 1000 lb 200

Fig . P - 428 . 427 . Beam loaded as shown in Fig . P - 427 . Ans . Max . M = 5000

ft - lb 428 . In the overhanging beam shown in Fig . P - 428 , determine P so that ...

500 lb 1000 lb 200

**lb**/**ft**5 ' 300**lb**/**ft**100**lb**/**ft**10 ' 10 ' 20 ' 1 R2 R Fig . P - 427 .Fig . P - 428 . 427 . Beam loaded as shown in Fig . P - 427 . Ans . Max . M = 5000

ft - lb 428 . In the overhanging beam shown in Fig . P - 428 , determine P so that ...

Page 212

30

120 x8 = 960 ' ft - lb M = 120x8 = 960 ft - lb TV = 120 lb M = - 720 ft - lb M = - 720 ft

- lb - 720 ft - lb - 120 x2 = - 240 ft - lb 11 30

...

30

**lb**/**ft**M = - 120 x 6 = - 720 ft - lb - = A1c - - - Bop - tbc 1V = 120 lb - 474 ( a )120 x8 = 960 ' ft - lb M = 120x8 = 960 ft - lb TV = 120 lb M = - 720 ft - lb M = - 720 ft

- lb - 720 ft - lb - 120 x2 = - 240 ft - lb 11 30

**lb**/**ft**4 M = - 120 x2 = - 240 ft - lb 30 lb...

Page 284

distributed load of w

+ B ) – 1 wL21 + 2a M3 = - 4 4 ( 1 + a ) ( a + B ) – 1 823 . A continuous beam

simply supported over three 10 - ft spans carries a concentrated load of 400 lb at

the ...

distributed load of w

**lb**/**ft**over the middle span . 1 + 2B Ans . M2 = - 4 ( 1 + a ) ( a+ B ) – 1 wL21 + 2a M3 = - 4 4 ( 1 + a ) ( a + B ) – 1 823 . A continuous beam

simply supported over three 10 - ft spans carries a concentrated load of 400 lb at

the ...

### 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