113_2009_HW5_sols

# 113_2009_HW5_sols - Problem 1 Plane stress 33 = 0 11 22 0...

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Problem 1 Plane stress: σ 33 = 0 σ 11 , σ 22 ≠ 0 which tells us it’s a biaxial stress field. σ 33 = 0 = E(ε 33 –υ(ε 11 22 )) ε 33 =υ(ε 11 22 ) ≠ 0 Maximum shear stress is at 45 degrees to σ 22 and σ 33 , thus material will yield nominally on this plane. As the normal stress at a free surface must be zero, if the specimen is thin enough, the stress through the thickness never reaches any appreciable value. Plane Strain: ε 33 = 0 ε 11 , ε 22 ≠ 0 which tells us it’s a triaxial stress field. ε 33 = 0 = E(σ 33 –υ(σ 11 + σ 22 )) σ 33 =υ(σ 11 + σ 22 ) ≠ 0 Maximum shear stress is 45 degrees to σ 11 and σ 33 thus the material will nominally yield on this plane. Triaxial stress field results in higher constraint in plane strain • In plane stress, as σ ij max goes to k, the shear yield 22 –0)/2 = k therefore σ 22 = 2k But in plane strain 22 σ 11 )/2 = k therefore σ 22 = 2k + σ 11 Measured tensile yield stress higher in plane strain then plain stress which reflect constraint. It is because of this constraint (triaxial stress state) that the plastic zone size in metallic materials is

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## This note was uploaded on 12/14/2009 for the course MSE 113 taught by Professor Ritchie during the Spring '09 term at Berkeley.

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113_2009_HW5_sols - Problem 1 Plane stress 33 = 0 11 22 0...

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