## Strength of Materials |

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Page xxiv

A more exact definition of stress is

the differential area over which it acts ... Applying pass through the centroid C .

the conditions of equilibrium , we

...

A more exact definition of stress is

**obtained**by dividing the differential load dp bythe differential area over which it acts ... Applying pass through the centroid C .

the conditions of equilibrium , we

**obtain**( EZ = 0 ) P = SdP = SS dA ( EM , = 0 ] Pb...

Page 94

Applying the definitions of vertical shear and bending moment , and noting that

they apply only to external loads , we

= 630 – 200x Mab = 630x – ( 200x ) = 630x – 100c ( 6 ) ( 200 x ) lb 1000 lb - 2 .

Applying the definitions of vertical shear and bending moment , and noting that

they apply only to external loads , we

**obtain**( a ) ( V = ( EY ) . ] ( M = ( EM ) , VAB= 630 – 200x Mab = 630x – ( 200x ) = 630x – 100c ( 6 ) ( 200 x ) lb 1000 lb - 2 .

Page 462

from which we

Of these several theories of failure , experimental work shows best agreement

with the Mises yield theory when applied to ductile materials . For such materials

...

from which we

**obtain**28 , p2 = ( S – S2 ) 2 + ( S , – Sz ) ? + ( S3 – S , ) Summary .Of these several theories of failure , experimental work shows best agreement

with the Mises yield theory when applied to ductile materials . For such materials

...

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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