## Engineering mechanics of materials |

### From inside the book

Results 1-3 of 61

Page 290

Equations for Beam Slopes and Deflections of Selected Cases Case I. Simply

supported beam-uniformly distributed loading III ,3 - 24£/„ T + T J ...

Equations for Beam Slopes and Deflections of Selected Cases Case I. Simply

supported beam-uniformly distributed loading III ,3 - 24£/„ T + T J ...

**Cantilever****beam**-uniformly distributed loading "\ fl 11111 6£/» 8£/u Maximum at right end ...Page 297

633 Find the slope and deflection at the free end B of the cantilever shown in

Figure H6.33. Use Table 6.1 and the method of superposition. 2 kN/m upward

6kN FIGURE H6.33 6.34 A

633 Find the slope and deflection at the free end B of the cantilever shown in

Figure H6.33. Use Table 6.1 and the method of superposition. 2 kN/m upward

6kN FIGURE H6.33 6.34 A

**cantilever beam**is loaded as shown in Figure H6.34.Page 336

Although our discussion has focused on an end-loaded

approach is quite general. For example, the divisor of 3 in Eqs. 6.37 and 6.38

would be replaced by 48 if the problem involved computation of the center

deflection of ...

Although our discussion has focused on an end-loaded

**cantilever beam**, theapproach is quite general. For example, the divisor of 3 in Eqs. 6.37 and 6.38

would be replaced by 48 if the problem involved computation of the center

deflection of ...

### 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