## Strength of Materials |

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Page 160

Ferdinand Leon Singer. stresses Saand Sy multiplied by the area cdfg . Thus a

larger shear resistance must be developed over the

dce . Of course , the total compressive force C1 plus C2 acting over the area abgf

...

Ferdinand Leon Singer. stresses Saand Sy multiplied by the area cdfg . Thus a

larger shear resistance must be developed over the

**horizontal**layer at fg than atdce . Of course , the total compressive force C1 plus C2 acting over the area abgf

...

Page 162

Derivation of Formula for

sections ( 1 ) and ( 2 ) in a beam separated by the distance dx , as shown in Fig .

5 - 21 , and let the shaded part between them be isolated as a free body . Fig .

Derivation of Formula for

**Horizontal**Shearing Stress Consider two adjacentsections ( 1 ) and ( 2 ) in a beam separated by the distance dx , as shown in Fig .

5 - 21 , and let the shaded part between them be isolated as a free body . Fig .

Page 163

... the vertical shear ; so we obtain for dx the

Lovda = ħ ang = the ( 5 - 4 ) We have replaced the integral , y dA , which means

the sum of the moments of the differential areas dA about the neutral axis , by its ...

... the vertical shear ; so we obtain for dx the

**horizontal**shearing stress , S . = HoLovda = ħ ang = the ( 5 - 4 ) We have replaced the integral , y dA , which means

the sum of the moments of the differential areas dA about the neutral axis , by its ...

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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