## Engineering mechanics of materials |

### From inside the book

Results 1-3 of 73

Page 160

400

29.3 MN/m' 60

0.40m 0,= 0.12 m /->„ = 0.24 m O, = 0.l2m Shearing stresses along typical radial ...

400

**kN**-**m**300**kN**-**m**(a) Torque (**kN**-**m**+40 -60 X (m) -360 40**kN**-**m**(b) 360**kN**-**m**29.3 MN/m' 60

**kN**-**m**AB 0„ = 0.20 m Dj = 0.12 m BC 28.9 MN/nr 23.6 MN/m2 O„ =0.40m 0,= 0.12 m /->„ = 0.24 m O, = 0.l2m Shearing stresses along typical radial ...

Page 163

Calculate the maximum shearing stress in this shaft. Use G = 80.0 GPa.

Determine the angle of twist of end D with respect to end A of the shaft. 20

Da = 0.10 m. 0, = 0.0ft m 20

H4.8 -2m ...

Calculate the maximum shearing stress in this shaft. Use G = 80.0 GPa.

Determine the angle of twist of end D with respect to end A of the shaft. 20

**kN**-**m**Da = 0.10 m. 0, = 0.0ft m 20

**kN**-**m**1 ) ) JJ 20**kN**-**m*** 20**kN**-**m**/ 2 m 2m- FIGUREH4.8 -2m ...

Page 296

10

Find the deflection of the simply supported beam at the center of the span due to

the loadings shown in Figure H6.27. Use the method of superposition and obtain

...

10

**kN**-**m**lOkN-m £/u = 6.00 X 103kN-m 5) L = 10m V/9//J FIGURE H6.26 6.27Find the deflection of the simply supported beam at the center of the span due to

the loadings shown in Figure H6.27. Use the method of superposition and obtain

...

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### Contents

Internal Forces in Members | 1 |

Hollow Circular ShaftsAngle of Twist and Shearing Stresses | 157 |

4i5 Power Transmission | 169 |

Copyright | |

7 other sections not shown

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### Common terms and phrases

absolute maximum shear allowable stress aluminum angle of twist applied Assume 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 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