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Lecture 3 mohr’s circle and theory of failure
 

Lecture 3 mohr’s circle and theory of failure

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    Lecture 3 mohr’s circle and theory of failure Lecture 3 mohr’s circle and theory of failure Presentation Transcript

    • Unit 1- Stress and StrainTopics Covered  Lecture -1 - Introduction, state of plane stress  Lecture -2 - Principle Stresses and Strains  Lecture -3 - Mohrs Stress Circle and Theory of Failure  Lecture -4- 3-D stress and strain, Equilibrium equations and impact loading  Lecture -5 - Generalized Hooks law and Castiglionos
    • Mohr Stress Circle We derived these two equations- These equations represent the equation of a circle σ1 + σ2 σ1 − σ2 σn = + cos2θ + τ sin2θ 2 2 σ1 − σ2 σt = sin2θ − τ cos2θ 2 € ⎛ σ1 + σ2 ⎞ 2 ⎛ σ1 − σ2 ⎞ 2 ⎜σn − ⎟ = ⎜ cos2θ + τ sin2θ ⎟ € ⎝ 2 ⎠ ⎝ 2 ⎠ 2 ⎛ σ1 − σ2 ⎞ 2 (σt ) = ⎜ ⎝ 2 sin2θ − τ cos2θ ⎟ ⎠€
    • Mohr Stress Circle ⎛ σ1 + σ2 ⎞ 2 ⎛ σ1 − σ2 ⎞ 2 ⎜σn − ⎟ = ⎜ cos2θ + τ sin2θ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ 2 ⎛ σ1 − σ2 ⎞ 2 (σt ) = ⎜ ⎝ 2 sin2θ − τ cos2θ ⎟ ⎠ Add above 2 equations. We will equation of circle.€ 2 2 ⎛ σ + σ ⎞ 1 2 ⎛ σ − σ ⎞ 2 1 2 2 ⎜σ − n ⎟ + σ = ⎜ t ⎟ + (τ ) € ⎝ 2 ⎠ ⎝ 2 ⎠ 2 2 ( x − a) y r 2 Equation of circle €
    • Mohr Stress Circle  Graphical method to determine stresses.   Body subjected to two mutually perpendicular principal stresses of unequal magnitude.   Body subjected to two mutually perpendicular principal stresses of unequal magnitude and unlike (one tensile and other compressive).   Body subjected to two mutually perpendicular principal stresses + simple shear stress.
    • Mohr Stress Circle   Body subjected to two mutually perpendicular principal stresses of unequal magnitude (σ1 - σ2 ) length AD = Normal stress on oblique plane σt = σn E length ED = Tangential stress on Oblique plane = σt 2θ B length AE = Resultant stress on Oblique plane θ σn O D σ1 2 2A C = σ +σt n € σ2 σ1 € €
    • Mohr Stress Circle   Body subjected to two mutually perpendicular principal stresses of unequal magnitude and unlike (one tensile and other compressive). (σ1+σ2 ) length AD = Normal stress on oblique plane σt E = σn length ED = Tangential stress on Oblique plane = σt 2θ length AE = Resultant stress on Oblique planeC θ B σn 2 2 A O D = σt + σn_ € + σ1 σ2 € €
    • Mohr Stress Circle   Body subjected to two mutually perpendicular principal stresses + simple shear stress. σt length AD = Normal stress on oblique plane = σn E length ED = Tangential stress on Oblique plane 2θ = σt length AE = Resultant stress on Oblique plane B M σ n L C D σ1 2 2A O = σ +σ t n € σ2 σ1 € €
    • Theories of failure  Maximum principal stress (Rankine theory)  Maximum principal strain (Saint Venant theory)  Maximum shear stress (Guest theory)  Maximum strain energy (Haigh theory)  Maximum shear strain energy (Mises & Henky theory)
    • 1. Maximum principal stress theory σ1,σ2 ,σ3 =principal stresses in 3 perpendicular directions * max(σ1,σ2 ,σ3 ) ≤ σ€ Maximum principal stress should be less than the max stress (yield stress) that material can bear in tension or compression. * σ = max tensile or compressive strength of material € σ* max principal stress= safety _ factor
    • 2. Maximum principal strain theory σ1,σ2 ,σ3 =principal stresses in 3 perpendicular directions σ1 υσ2 υσ3 σ2 υσ1 υσ3 σ3 υσ1 υσ2 e1 = − − e2 = − − e3 = − − E E E E E E E E E * σ*€ max(e1,e2 ,e3 ) ≤ e * e = E * σ = max tensile or compressive strength of material € Maximum principal strain should be less than the max strain (yield strain) that material € € can bear in tension or compression. € σ* € max principal stress= safety _ factor
    • 3. Maximum shear stress theory max shear stress =half the difference of max and min principal stresses 1 = (σ1 − σ3 ) 2 In simple tension the stressTo prevent failure max shear stress should be less that shear is existing in one directionstress in simple tension at elastic limit 1 * * = (σt − 0) σt max shear stress at elastic limit = max tensile of material € 2 (σ1 − σ 3 ) ≤ σ * tMaximum shear stress should be less than the max shear stress in simple tension (atelastic limit) that material can bear. € € σ*t allowable stress = € safety _ factor
    • 4. Maximum strain energy theory Strain energy per unit volume should be less than the strain energy per unit volume in simple tension (at elastic limit) that material can bear. 2 [σ 2 1 + σ + σ − 2υ (σ1σ2 + σ1σ3 + σ2σ3 ) ≤ (σ 2 2 2 3 ] * t ) σ*t max allowable stress= safety _ factor€ €
    • 5. Maximum shear strain energy theoryShear strain energy per unit volume should be less than the shear strain energy per unitvolume in simple tension (at elastic limit). 2 2 2 2 (σ1 − σ2 ) + (σ1 − σ3 ) + (σ2 − σ3 ) ≤ 2 * (σ * t ) σ*t max allowable stress= safety _ factor€ €
    • Important points  Brittle material -> Max principal stress   Brittle material do not fail in shear  Ductile material -> Max shear stress/max shear strain energy   Ductile material fail in shear because their yield strength is high.
    • Failure Theory  PROBLEM- The principal stresses at a point in an elastic material are 200 N/mm2 (tensile), 100 N/ mm2 (tensile) and 50 N/mm2 (compressive). If the stresses at the elastic limit in simple tension is 200 N/mm2, determine whether the failure of the material will occur according to different failure theory. (take Poissons ratio =0.3)   Max principal strain theory   Max shear stress theory   Max strain energy theory   Max shear strain energy theory