## Strength of Materials |

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Page 197

With this assumption , we obtain de = M , - 29 dx It is evident that

to the elastic curve at C and D in Fig . ... that will intersect a

curve at any other point A is the sum of the intercepts dt created by

With this assumption , we obtain de = M , - 29 dx It is evident that

**tangents**drawnto the elastic curve at C and D in Fig . ... that will intersect a

**tangent**drawn to thiscurve at any other point A is the sum of the intercepts dt created by

**tangents**to ...Page 199

Conversely , a computed positive value for deviation means that the point must

lie above the reference

Negative deviation ; B located above reference

Fig ...

Conversely , a computed positive value for deviation means that the point must

lie above the reference

**tangent**. А B / A ( a ) Positive deviation ; B located ( b )Negative deviation ; B located above reference

**tangent**below reference**tangent**Fig ...

Page 220

At the position of maximum deflection , the

horizontal . As shown in Fig . 6 – 24 , the change in slope between

this position B and at A ( i . e . , OAB ) is equal to the slope na at A , since for small

...

At the position of maximum deflection , the

**tangent**to the elastic curve will behorizontal . As shown in Fig . 6 – 24 , the change in slope between

**tangents**atthis position B and at A ( i . e . , OAB ) is equal to the slope na at A , since for small

...

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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