Mohr's Circle / Stress Transformation

Assumptions

• Plane stress: the out-of-plane stress σz = 0 and is the third principal stress for the absolute maximum shear check.
• Hibbeler sign convention: normal stresses tension-positive; positive τxy acts in the +y direction on the +x face. Angles are measured counterclockwise from the +x axis.

Solution steps

1. Center and radius of Mohr's circle

   The circle is centered on the normal-stress axis at the average normal stress; its radius comes from the half-difference of normal stresses and the shear.

   σavg = (σx + σy)/2; R = √(((σx − σy)/2)² + τxy²)

   σavg = 50 MPa; R = 50 MPa

2. Principal stresses

   The principal stresses sit at the two ends of the circle's horizontal diameter — the average stress plus and minus the radius. No shear acts on the principal planes.

   σ1 = σavg + R and σ2 = σavg − R

   σ1 = 100 MPa; σ2 = 0 MPa

   = 100 MPa

3. Principal and maximum-shear angles

   The half-angle arctangent has two solutions a right angle apart; the two-argument arctangent picks the rotation that carries the x-axis onto the σ1 direction (the other principal plane is 90° away). The maximum-shear planes bisect the principal planes.

   2θp = atan2(2τxy, σx − σy); θs = θp1 − 45°

   θp1 = 26.57 deg (σ1 direction, i.e. 2θp = 53.13 deg on the circle); θs = -18.43 deg or 71.57 deg

4. Maximum shear stresses

   The in-plane maximum shear equals the circle radius. For the absolute maximum, the out-of-plane principal stress σz = 0 joins the comparison: when σ1 and σ2 have the same sign, the largest circle is the one through zero.

   τmax(in-plane) = R; τabs = max(R, |σ1|/2, |σ2|/2)

   τmax(in-plane) = 50 MPa; τabs = 50 MPa

   = 50 MPa

Results

Assumptions

• Plane stress: the out-of-plane stress σz = 0 and is the third principal stress for the absolute maximum shear check.
• Hibbeler sign convention: normal stresses tension-positive; positive τxy acts in the +y direction on the +x face. Angles are measured counterclockwise from the +x axis.

Solution steps

1. Center and radius of Mohr's circle

   The circle is centered on the normal-stress axis at the average normal stress; its radius comes from the half-difference of normal stresses and the shear.

   σavg = (σx + σy)/2; R = √(((σx − σy)/2)² + τxy²)

   σavg = 15 ksi; R = 13 ksi

2. Principal stresses

   The principal stresses sit at the two ends of the circle's horizontal diameter — the average stress plus and minus the radius. No shear acts on the principal planes.

   σ1 = σavg + R and σ2 = σavg − R

   σ1 = 28 ksi; σ2 = 2 ksi

   = 28 ksi

3. Principal and maximum-shear angles

   The half-angle arctangent has two solutions a right angle apart; the two-argument arctangent picks the rotation that carries the x-axis onto the σ1 direction (the other principal plane is 90° away). The maximum-shear planes bisect the principal planes.

   2θp = atan2(2τxy, σx − σy); θs = θp1 − 45°

   θp1 = 33.69 deg (σ1 direction, i.e. 2θp = 67.38 deg on the circle); θs = -11.31 deg or 78.69 deg

4. Maximum shear stresses

   The in-plane maximum shear equals the circle radius. For the absolute maximum, the out-of-plane principal stress σz = 0 joins the comparison: when σ1 and σ2 have the same sign, the largest circle is the one through zero.

   τmax(in-plane) = R; τabs = max(R, |σ1|/2, |σ2|/2)

   τmax(in-plane) = 13 ksi; τabs = 14 ksi

   = 14 ksi

Results

Assumptions

• Plane stress: the out-of-plane stress σz = 0 and is the third principal stress for the absolute maximum shear check.
• Hibbeler sign convention: normal stresses tension-positive; positive τxy acts in the +y direction on the +x face. Angles are measured counterclockwise from the +x axis.

Solution steps

1. Center and radius of Mohr's circle

   The circle is centered on the normal-stress axis at the average normal stress; its radius comes from the half-difference of normal stresses and the shear.

   σavg = (σx + σy)/2; R = √(((σx − σy)/2)² + τxy²)

   σavg = 2 MPa; R = 11.66 MPa

2. Principal stresses

   The principal stresses sit at the two ends of the circle's horizontal diameter — the average stress plus and minus the radius. No shear acts on the principal planes.

   σ1 = σavg + R and σ2 = σavg − R

   σ1 = 13.66 MPa; σ2 = -9.662 MPa

   = 13.66 MPa

3. Principal and maximum-shear angles

   The half-angle arctangent has two solutions a right angle apart; the two-argument arctangent picks the rotation that carries the x-axis onto the σ1 direction (the other principal plane is 90° away). The maximum-shear planes bisect the principal planes.

   2θp = atan2(2τxy, σx − σy); θs = θp1 − 45°

   θp1 = -74.52 deg (σ1 direction, i.e. 2θp = -149 deg on the circle); θs = -119.5 deg or -29.52 deg

4. Maximum shear stresses

   The in-plane maximum shear equals the circle radius. For the absolute maximum, the out-of-plane principal stress σz = 0 joins the comparison: when σ1 and σ2 have the same sign, the largest circle is the one through zero.

   τmax(in-plane) = R; τabs = max(R, |σ1|/2, |σ2|/2)

   τmax(in-plane) = 11.66 MPa; τabs = 11.66 MPa

   = 11.66 MPa

5. Transformed stresses at the requested angle

   Rotating the element corresponds to walking twice the angle around Mohr's circle; the companion face carries the remainder of the normal-stress sum.

   σx' = σavg + ((σx − σy)/2)·cos2θ + τxy·sin2θ; σy' = σx + σy − σx'; τx'y' = −((σx − σy)/2)·sin2θ + τxy·cos2θ

   At θ = 30 deg: σx' = -8.196 MPa; σy' = 12.2 MPa; τx'y' = 5.66 MPa

Results