## Engineering mechanics of materials |

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Page 264

Homework Problems In the following problems, assume the loads to be so

placed that they produce no twisting action. 5.60 A simply supported beam 20 ft

long carries a

...

Homework Problems In the following problems, assume the loads to be so

placed that they produce no twisting action. 5.60 A simply supported beam 20 ft

long carries a

**uniform load**of intensity 1200 lb/ft along its entire length. The cross...

Page 604

-20 ft- 10 ft - FIGURE H12.23 12.24 A two-span continuous beam is loaded with

proportional loads as shown in Figure H 12.24. Determine the ... (b) The

-20 ft- 10 ft - FIGURE H12.23 12.24 A two-span continuous beam is loaded with

proportional loads as shown in Figure H 12.24. Determine the ... (b) The

**uniform****load**associated with collapse of the simply supported beam. (c) The ratio of the ...Page 695

14.41 Refer to Example 14.11 and determine the total ultimate

may be applied to a simply supported beam of 20 ft length with this cross section.

Subtract the

14.41 Refer to Example 14.11 and determine the total ultimate

**uniform load**thatmay be applied to a simply supported beam of 20 ft length with this cross section.

Subtract the

**uniform loading**due to the weight of the beam and thus determine ...### What people are saying - Write a review

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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

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### 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