Strength of Materials |
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Page 215
... obtain 1 do de ρ = 1 or - ds dx ρ = d2y dx2 ( d ) In deriving the flexure formula in Art . 5–2 , we obtained on page 157 the relation 1 = M ρ ΕΙ Equating the values of 1 / p from Eqs . ( d ) and ( 5-1 ) , we have ( 5-1 ) d'y EI- = M dx2 ...
... obtain 1 do de ρ = 1 or - ds dx ρ = d2y dx2 ( d ) In deriving the flexure formula in Art . 5–2 , we obtained on page 157 the relation 1 = M ρ ΕΙ Equating the values of 1 / p from Eqs . ( d ) and ( 5-1 ) , we have ( 5-1 ) d'y EI- = M dx2 ...
Page 519
... obtain a method of determining deflections that is based on the principle of conservation of energy . This method will be shown to be extremely versatile . We begin by obtaining expressions for the strain energy U stored in a body under ...
... obtain a method of determining deflections that is based on the principle of conservation of energy . This method will be shown to be extremely versatile . We begin by obtaining expressions for the strain energy U stored in a body under ...
Page 627
... obtained from Art . A - 2 , which shows that the polar moment of inertia J , is the sum of the moments of inertia with respect to rectangular axes passing through the polar axis . Hence , since J , is a constant , we obtain as before J1 ...
... obtained from Art . A - 2 , which shows that the polar moment of inertia J , is the sum of the moments of inertia with respect to rectangular axes passing through the polar axis . Hence , since J , is a constant , we obtain as before J1 ...
Common terms and phrases
allowable stresses aluminum angle applied load assumed axial load beam carrying beam in Fig beam loaded beam shown bending moment bolt bronze cantilever beam centroid column compressive stress concentrated load continuous beam cross section deformation Determine the maximum diameter elastic curve end moments equal equilibrium equivalent exploratory section factor of safety fibers free-body diagram GN/m² Hence Hooke's law horizontal ILLUSTRATIVE PROBLEMS kN.m kN·m kN/m length load diagram loaded as shown M₁ M₂ maximum shearing stress midspan deflection mm² MN/m² Mohr's circle moment of inertia N/m² neutral axis obtain P₁ P₂ plane plate positive principal stresses proportional limit R₁ R₂ R2 Figure radius reaction resisting restrained beam resultant rivet segment shaft shear diagram shearing force shown in Fig simply supported beam Solution span statically indeterminate steel strain tangent drawn tensile stress three-moment equation torque torsional uniformly distributed load vertical shear weld ΕΙ