Engineering Mechanics of SolidsPopov (civil engineering, U. Cal., Berkeley) has written this textbook for undergraduate students. Traditional topics are supplemented by an exposure to several newly-emerging disciplines, such as the probabilistic basis for structural analysis, and matrix methods. Annotation copyright Book News, In |
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... material begins and continues to yield . No such stress points can lie outside the hexagon because one of the three yield criteria equa- tions given before for perfectly plastic material would be violated . The stress points falling ...
... material begins and continues to yield . No such stress points can lie outside the hexagon because one of the three yield criteria equa- tions given before for perfectly plastic material would be violated . The stress points falling ...
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... materials is the same , and tentatively assuming elastic response of both materials , one has the following : From equilibrium : Pa Ps = P1 or P2 + From compatibility : △ a = As or Ɛa = Es From material properties : Ea Ɛa = σa / Ea and ...
... materials is the same , and tentatively assuming elastic response of both materials , one has the following : From equilibrium : Pa Ps = P1 or P2 + From compatibility : △ a = As or Ɛa = Es From material properties : Ea Ɛa = σa / Ea and ...
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... material , substitution into Eq . 2-22 of the value of the stress at the proportional limit gives an index of the material's ability to store or absorb energy without permanent deformation . The quantity so found is called the modulus ...
... material , substitution into Eq . 2-22 of the value of the stress at the proportional limit gives an index of the material's ability to store or absorb energy without permanent deformation . The quantity so found is called the modulus ...
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A₁ allowable stress aluminum analysis angle applied force applied load assumed axes axial force axially loaded bar beam bending moment bending stress bolts buckling C₁ caused centroid column compression concentrated force considered constant cross section cross-sectional area cylinder deflection deformations Determine diameter direction elastic curve elastic modulus element equal equations equilibrium example figure flange flexure formula given by Eq Hence Hooke's law horizontal in² indeterminate problems inertia infinitesimal internal kips length linearly elastic material maximum shear stress mm² modulus Mohr's circle moment of inertia neutral axis normal stress obtained P₁ plane plastic principal stresses problem procedure reactions rectangular rotation segment shaft shear center shear stress shown in Fig solution statically indeterminate steel strain energy stress distribution stress-strain stresses acting tensile tensile stress Tmax torque torsion tube vertical yield zero