## Engineering mechanics of materialsThis book provides the students of various engineering disciplines with a clear and understandable treatment of the concepts of Mechanics of Materials or Strength of Materials. This subject is concerned with the behavior of deformable bodies when subjected to axial, torsional and flexural loads as well as combinations thereof. It is a 3rd, updated edition of the popular undergraduate level textbook useful for students of mechanical, structural, civil, aeronautical and other engineering disciplines. The book is supplied with problems and a solution manual will be available from the authors. |

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Page 169

333.3

Du = 0.40 m u-a 60

...

333.3

**kN**-**m**300**kN**-**m**40**kN**-**m**— D. = 0. 20 m | \ — m D, = 0. 1 2 m - const .T__Du = 0.40 m u-a 60

**kN**-**m**I m ) D0 = 0.24 m 0.75 m 1.20m -1.60m- d) Torque (**kN**-**m**) -60 x (m) -360 40**kN**-**m**(b) 360**kN**-**m**^\ [29.3 MN/m2 ) 60**kN**-**m**Do = 0.20m D...

Page 171

2.0 m 11 o 71 4 - 60,000 N-m ... 6kN-m Uniformly distributed torque 4

rrmr-. J B A D=0.1Om. r. —. 3 m- FIGURE P4.4 4.2 Determine the maximum

shearing stress in each segment (i.e., AB, BC, and CD) of the shaft shown in Fig.

P4.2.

2.0 m 11 o 71 4 - 60,000 N-m ... 6kN-m Uniformly distributed torque 4

**kN**-**m**/m y.rrmr-. J B A D=0.1Om. r. —. 3 m- FIGURE P4.4 4.2 Determine the maximum

shearing stress in each segment (i.e., AB, BC, and CD) of the shaft shown in Fig.

P4.2.

Page 378

340) after making the following changes: shaft diameter from 0.12 m to 0.10 m,

axial pull P from 650 kN to 800 kN, and torque T from 20

other information remains unchanged. 7.86 (Sec. 7.2) Solve Problem 7.7 (p.

340) after making the following changes: shaft diameter from 0.12 m to 0.10 m,

axial pull P from 650 kN to 800 kN, and torque T from 20

**kN**-**m**to 30**kN**-**m**. Allother information remains unchanged. 7.86 (Sec. 7.2) Solve Problem 7.7 (p.

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### Contents

Stress Strain and Their Relationships | 60 |

Stresses and Strains in Axially Loaded Members | 121 |

Torsional Stresses Strains and Rotations | 159 |

Copyright | |

14 other sections not shown

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### Common terms and phrases

absolute maximum shear aluminum angle of twist applied Assume axial force axially loaded beam shown bending cantilever beam Castigliano's second theorem column compressive constant coordinate cross section cross-sectional area cylinder deflection deformation depicted in Fig elastic curve equal equation equilibrium Euler EXAMPLE factor of safety FIGURE flexural stress FORTRAN free-body diagram function given by Eq k-ft k-in kN-m lb/ft length longitudinal material maximum in-plane shear maximum shear stress modulus of elasticity Mohr's circle neutral axis normal stress obtained perpendicular plane stress condition plot positive principal centroidal axis principal strains principal stresses radius ratio reactions Refer to Fig respect rotation shear force shear strain shown in Fig simply supported beam slope SOLUTION Solve Problem statically indeterminate steel stress concentration stress element subjected torque torsional uniform load vertical yield strength yield stress zero