## Engineering mechanics of materials |

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Page 213

t = y cos a — x sin a (5.1h) n = x cos a + y sin a

into the first of Eq. 5.1g leads to IK= \ (y cos a - x sin a)2 dA = cos2 a | y1 dA + sin2

a \ x2 dA - 2 sin a cos a | xy dA (5.1i) The mixed integral, j xy d/4, is known as ...

t = y cos a — x sin a (5.1h) n = x cos a + y sin a

**Substitution**of the first of Eq. 5.1hinto the first of Eq. 5.1g leads to IK= \ (y cos a - x sin a)2 dA = cos2 a | y1 dA + sin2

a \ x2 dA - 2 sin a cos a | xy dA (5.1i) The mixed integral, j xy d/4, is known as ...

Page 245

~m F2= "+ "j vdA (5.11c) 3. A shear force F3 representing the resultant of the

shear stresses xvx acting on surface efij. Since the shear stress xvx is assumed ...

**Substitution**of the value of ax as given by Eq. 5.10a into Eq. 5.1 la yields M +dM c~m F2= "+ "j vdA (5.11c) 3. A shear force F3 representing the resultant of the

shear stresses xvx acting on surface efij. Since the shear stress xvx is assumed ...

Page 274

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division by EIu yields the equation of the elastic curve as a function of x for the

given beam and loading. Thus D = £/;(l2* ~24* ~I**) (0^L) The equation for the

slope of ...

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**substitution**of these values for C, and C2 into the equation for v anddivision by EIu yields the equation of the elastic curve as a function of x for the

given beam and loading. Thus D = £/;(l2* ~24* ~I**) (0^L) The equation for the

slope of ...

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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