## Strength of Materials |

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Page 110

The shear diagram passes through zero at B , where x = 3 ft , and also at E . The

position of E is determined from the fact that the shear at the right of B is 1400 lb

which is reduced to zero in the interval BE at the rate of 200 lb / ft .

...

The shear diagram passes through zero at B , where x = 3 ft , and also at E . The

position of E is determined from the fact that the shear at the right of B is 1400 lb

which is reduced to zero in the interval BE at the rate of 200 lb / ft .

**Hence**BE = d...

Page 203

, we obtain diagram ( d ) . The shaded area of diagram ( d ) is evidently equal to

the area of the conventional moment diagram ( e ) obtained by plotting the ...

**Hence**, if the triangular area is revolved about the line of zero moment as an axis, we obtain diagram ( d ) . The shaded area of diagram ( d ) is evidently equal to

the area of the conventional moment diagram ( e ) obtained by plotting the ...

Page 481

any section along the beam are equal respectively to the product of the vertical

shears V , and V , in the flanges and the distance to the section .

**Hence**, applying p = 7 gives = * = ( c ) But the bending moments M1 and My atany section along the beam are equal respectively to the product of the vertical

shears V , and V , in the flanges and the distance to the section .

**Hence**Eq . ( c ) ...### What people are saying - Write a review

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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