## Strength of Materials |

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Page 10

An assembly used in drilling an oil well consists of three different sections of steel

tubing , each of constant

limited to 6000 psi and the lowest section has a

An assembly used in drilling an oil well consists of three different sections of steel

tubing , each of constant

**cross section**... If the maximum stress in each section islimited to 6000 psi and the lowest section has a

**cross**-**sectional**area of 2 in . ?Page 133

S = MC ( 5 – 2a ) If - is called the section modulus and denoted by Z , another

common variation of the flexure formula is M M Max . S = ; ( 5 – 2b ) 1 / c z ...

Various values of section modulus for common

5 – 1 .

S = MC ( 5 – 2a ) If - is called the section modulus and denoted by Z , another

common variation of the flexure formula is M M Max . S = ; ( 5 – 2b ) 1 / c z ...

Various values of section modulus for common

**cross sections**are listed in Table5 – 1 .

Page 499

Table 13 – 2 lists correction factors for various

resolved into two components : a shearing force V in the plane of the

Table 13 – 2 lists correction factors for various

**cross sections**. ... The force R isresolved into two components : a shearing force V in the plane of the

**cross****section**, and a normal force N perpendicular to the plane of the**cross section**.### 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