## Engineering mechanics of materials |

### From inside the book

Results 1-3 of 81

Page 274

Back substitution of these values for C, and C2 into the

division by EIu yields the

given beam and loading. Thus D = £/;(l2* ~24* ~I**) (0^L) The

slope of ...

Back substitution of these values for C, and C2 into the

**equation**for v anddivision by EIu yields the

**equation**of the elastic curve as a function of x for thegiven beam and loading. Thus D = £/;(l2* ~24* ~I**) (0^L) The

**equation**for theslope of ...

Page 322

Equating these two

term jdPt dvt is a second-order differential and without approximation may be

dropped from the

Equating these two

**equations**leads to U + If dpi = \ dpi dv' + dP' V' + U oPi 2 Theterm jdPt dvt is a second-order differential and without approximation may be

dropped from the

**equation**. This yields the mathematical form of Castigliancfs ...Page 697

By applying the method of two successive integrations, this single moment

of integration. These two constants of integration require only two boundary

conditions ...

By applying the method of two successive integrations, this single moment

**equation**yields one slope and one deflection**equation**, containing two constantsof integration. These two constants of integration require only two boundary

conditions ...

### What people are saying - Write a review

We haven't found any reviews in the usual places.

### Contents

Introduction | 1 |

Torsionally Loaded Members in Equilibrium | 14 |

Shear and Bending Moment in Beams | 23 |

Copyright | |

19 other sections not shown

### Other editions - View all

### Common terms and phrases

absolute maximum shear allowable stress aluminum angle of twist applied Assume axes axial force axially loaded beam shown bending cantilever beam Castigliano's second theorem circular column components compressive Compute constant construct coordinate cross section cross-sectional area cylinder deflection deformation depicted in Figure Determine diameter differential elastic curve equal equation equilibrium factor of safety flexural stress free-body diagram function given by Eq Homework Problems k-ft k-in kN-m length longitudinal material maximum shear stress modulus of elasticity Mohr's circle neutral axis normal stress obtained perpendicular plane stress plot principal centroidal axis principal strains principal stresses radius Refer to Figure respect rotation section a-a segment shear center shear force shear strain shown in Figure slope solution Solve static statically indeterminate steel stress element subjected Substitution tension tion torque torsional uniform load vertical yield strength yield stress zero