## Engineering mechanics of materials |

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Page 213

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

h into the first of Eq. 5.1g leads to * — x sin a)2 dA = cos2 a | y 2 dA + sin2 a | x2

dA — 2 sin a cos a | xy dA (5. 1 i) The mixed integral, j xy dA, is known as the ...

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

**Substitution**of the first of Eq. 5.1h into the first of Eq. 5.1g leads to * — x sin a)2 dA = cos2 a | y 2 dA + sin2 a | x2

dA — 2 sin a cos a | xy dA (5. 1 i) The mixed integral, j xy dA, is known as the ...

Page 408

with one end free and the other end fixed will buckle elastically at a load that is

one-fourth as large as the load required to buckle it if both ends were pinned.

**Substitution**of this effective length into Eq. 8.21 yields a critical load: The columnwith one end free and the other end fixed will buckle elastically at a load that is

one-fourth as large as the load required to buckle it if both ends were pinned.

Page 576

2 2 + 4 or YA = \YB From the definition of unit strain, YA = eALA and Yb = ebLb

gives ŁA LA = yeB Lb or _ 1LB

— 3 ...

2 2 + 4 or YA = \YB From the definition of unit strain, YA = eALA and Yb = ebLb

**Substitution**of these expressions into the equation relating the displacementsgives ŁA LA = yeB Lb or _ 1LB

**Substitution**of LB = 25.0 and LA = 20.0 gives EA— 3 ...

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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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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