Normal Stress on a Rotated Plane

Equation: the normal stress on a plane whose normal is rotated θ from the x-axis is σθ = (σx + σy)/2 + [(σx − σy)/2]·cos2θ + τxy·sin2θ, with θ positive counterclockwise. Values: σx = 60 MPa and σy = 20 MPa are both positive because the arrows point outward (tension). τxy = +20 MPa is positive because on the right (+x) face it points up (+y). θ = +30° because the element is rotated counterclockwise, so 2θ = 60°. Step 1: average stress, (σx + σy)/2 = (60 + 20)/2 = 40 MPa. Step 2: half the difference, (σx − σy)/2 = (60 − 20)/2 = 20 MPa. With cos60° = 0.5, this term is 20 × 0.5 = 10 MPa. Step 3: shear term, τxy·sin2θ = 20 × sin60° = 20 × 0.866 = 17.3 MPa. Step 4: σθ = 40 + 10 + 17.3 = 67.3 MPa. Taking θ = −30° (clockwise) flips the sign of the shear term and gives 32.7 MPa. Dropping the cos2θ term gives 57.3 MPa, and 40 MPa is only the average stress.