113_2009_HW5_sols

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online