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# Lecture 3 mohr’s circle and theory of failure

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### Lecture 3 mohr’s circle and theory of failure

1. 1. 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
2. 2. 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 ⎟ ⎠€
3. 3. 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 €
4. 4. 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.
5. 5. 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 € €
6. 6. 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 € €
7. 7. 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 € €
8. 8. 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)
9. 9. 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
10. 10. 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
11. 11. 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
12. 12. 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€ €
13. 13. 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€ €
14. 14. 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.
15. 15. 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