## Engineering mechanics of materials |

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Page 55

In the special case where the

the entire area A, then a = F„/A and t = F,/A. Note that a normal stress acts in a

direction perpendicular to the plane on which it acts and it can be either tensile or

...

In the special case where the

**components**F„ and F, are uniformly distributed overthe entire area A, then a = F„/A and t = F,/A. Note that a normal stress acts in a

direction perpendicular to the plane on which it acts and it can be either tensile or

...

Page 526

*10.7 Design of

1.9 will improve the reader's understanding of design of

such loadings. General review of Chapter 7 and intensive review of Section 7.6 is

also ...

*10.7 Design of

**Components**to Resist Combined Loadings Review of Section1.9 will improve the reader's understanding of design of

**components**to resistsuch loadings. General review of Chapter 7 and intensive review of Section 7.6 is

also ...

Page 573

Chapter I Analysis and Design for Inelastic Behavior 12-1 Introduction Increased

understanding of the behavior of materials and structural and machine

capacity.

Chapter I Analysis and Design for Inelastic Behavior 12-1 Introduction Increased

understanding of the behavior of materials and structural and machine

**components**has led to a change in design emphasis from stress levels to loadcapacity.

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### 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