Failure_and_Fracture

Failure_and_Fracture - Failure Theory (7.6, 7.7, 7.10) MAE...

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Failure Theory (7.6, 7.7, 7.10) MAE 316 – Strength of Mechanical Components C. Heeter Tran Failure Theory 1
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tatic Loading Static Loading ` Failure theories r a given stress state ( use properties from a simple tension ` For a given stress state ( σ 1 , σ 2 σ 3 ) use properties from a simple tension test (S y , S u ) to assess the strength. y diameter = d F T x A y 3 16 d T J Tr π τ = = x A 2 4 d F A F σ = = Failure Theory 2
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tatic Loading Static Loading ` For what values of T and F will the material fail if the yield rength is S strength is S y ? ` If T = 0 load) (yield 4 4 2 2 y y y x S d F S d F π σ = = = ` For T and F non-zero, how do we calculate an equivalent or effective stress to assess the strength? Failure Theory 3
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aximum Normal Stress Theory .10) Maximum Normal Stress Theory (7.10) ` Failure occurs when maximum principal stress exceeds the ultimate strength. ` Primarily applies to brittle materials ` Principal stresses: σ 1 , σ 2 σ 3 σ y τ xy ` Define σ I σ II σ III where ` σ I = max ( σ 1 σ 2 σ 3 ) y x σ x ` σ III = min ( σ 1 σ 2 σ 3 ) ` σ II is the value in between Failure Theory 4
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aximum Normal Stress Theory .10) Maximum Normal Stress Theory (7.10) ` For failure 0 (where ultimate strength in tension) ` σ I = S ut if σ I > 0 (where S ut is ultimate strength in tension) ` σ III = -S uc if σ III < 0 (where S uc is ultimate strength in compression) ` Case I: Uni-axial tension (bar, σ x = σ o = P/A, σ y = τ xy = 0) ` σ 1 = σ o , σ 2 = 0 & σ 3 = 0 (plane stress) 0 0 ` σ I = σ o , σ II = 0, σ III = 0 ` Failure when σ o = S ut 9 Failure Theory 5
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aximum Normal Stress Theory .10) Maximum Normal Stress Theory (7.10) ` Case 2: Pure torsion (shaft, τ xy = Tr/J, σ x = σ y = 0) + 0 (plane stress) ` σ 1 = + τ xy , σ 2 = - τ xy & σ 3 = 0 (plane stress) ` σ I = + τ xy , σ II = 0, σ III = - τ xy ` Failure when τ y = S t or τ y = S c 8 xy ut xy uc ` Does not agree with experimental data. ` Experimental data would show that failure occurs when 6 S τ 0.6 S ut . Failure Theory 6
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aximum Shear Stress Theory .6) Maximum Shear Stress Theory (7.6) ` Tresca yield criterion ` Failure occurs when the maximum shear stress exceeds the yield strength (max shear stress in a tension test is S y /2).
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Failure_and_Fracture - Failure Theory (7.6, 7.7, 7.10) MAE...

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