Principal Stress, Mohr's Circle, and Combined Shear

Principal Stress, Mohr's Circle, and Combined Shear

Use the plane-stress relationships in the current NCEES PE Civil Reference Handbook supplied for the July 2026 test date. Stress transformation is sensitive to conventions. Before calculating, sketch the stress element, label x and y, take tensile normal stress as positive, and state the shear convention used by the selected handbook equation. Mixing one source's shear arrows with another source's Mohr-circle plotting convention can preserve the stress magnitudes while reversing the reported angle.

The plane-stress state

At a point, plane stress is described by σx, σy, and τxy. For the equations below, positive τxy acts in the positive y direction on the positive x face and in the positive x direction on the positive y face. Let the x' axis be rotated counterclockwise by physical angle θ from x. Then

σx' = (σx + σy)/2 + [(σx - σy)/2]cos(2θ) + τxy sin(2θ)

τx'y' = -[(σx - σy)/2]sin(2θ) + τxy cos(2θ).

The double angles arise from transforming an area orientation, not from a special material property. A plane physically rotated by θ corresponds to movement through 2θ around Mohr's circle. Depending on whether positive shear is plotted upward or downward, the plotted circle direction may oppose the physical rotation. State the convention rather than relying on visual memory.

A disciplined transformation workflow

A rotated stress element describes the same physical point under the same loading; only the faces used to report traction have changed. Use this sequence:

1. Draw positive faces and convert all stresses to one unit system. A negative normal value is compression under the tension-positive convention.
2. Identify whether the requested angle locates a plane, the plane's normal, or the rotated x' axis. A plane and its normal differ by 90 degrees.
3. Compute σavg and the half-difference (σx-σy)/2 before evaluating angles. These quantities expose many sign errors.
4. Use the full quadrant information for 2θ, then divide by two only once. List the perpendicular physical solution when orientation matters.
5. Substitute one candidate angle back into both transformation equations. Confirm the expected zero shear, maximum shear, or specified face stress.

The complementary shears on perpendicular faces have equal magnitude as required by moment equilibrium. The transformed normal stresses must sum to the original normal stresses, so σx' + σy' = σx + σy. If that trace changes, the transformation is inconsistent even if one ordinate looks reasonable.

Principal stresses and maximum shear

Define the circle center and radius as

C = σavg = (σx + σy)/2

R = sqrt{[(σx - σy)/2]^2 + τxy^2}.

The in-plane principal stresses are

σ1 = C + R and σ2 = C - R,

ordered so σ1 ≥ σ2. Shear is zero on principal planes. Their orientation satisfies

tan(2θp) = 2τxy/(σx - σy).

A one-argument arctangent can return the wrong quadrant. Use an atan2-style evaluation with numerator 2τxy and denominator σx - σy, or substitute the candidate angle back into the transformation equation and verify that τx'y' = 0. The two principal planes are 90 degrees apart physically and therefore 180 degrees apart on Mohr's circle.

The maximum in-plane shear magnitude is R. Its planes are 45 degrees from the principal planes, and the accompanying normal stress is C. Maximum shear does not generally occur on a zero-normal-stress plane.

Worked combined stress state

At a point, let

σx = 80 MPa, σy = 20 MPa, and τxy = +30 MPa.

First compute

C = (80 + 20)/2 = 50 MPa

and

R = sqrt{[(80 - 20)/2]^2 + 30^2}

= sqrt(30^2 + 30^2) = 42.43 MPa.

Thus

σ1 = 50 + 42.43 = 92.43 MPa,

σ2 = 50 - 42.43 = 7.57 MPa.

Both principal stresses are tensile, even though the original faces also carry shear. For orientation,

tan(2θp) = 2(30)/(80 - 20) = 1.

The quadrant-consistent solution associated with σ1 is 2θp = 45°, so θp = 22.5° counterclockwise. Substitute it:

σx' = 50 + 30cos45° + 30sin45° = 92.43 MPa,

τx'y' = -30sin45° + 30cos45° = 0.

This check identifies the rotated x' face as the σ1 plane rather than the perpendicular σ2 plane. The maximum in-plane shear planes are at θs = 22.5° + 45° = 67.5° and the perpendicular orientation. Substitution gives shear magnitude 42.43 MPa and normal stress 50 MPa.

Construct and read Mohr's circle

Plot the two perpendicular-face states at opposite ends of a diameter, commonly A = (σx, τxy) and B = (σy, -τxy) under the chosen plotting convention. The center lies on the normal-stress axis at 50 MPa. The horizontal intercepts are 92.43 and 7.57 MPa; the top and bottom points carry shear magnitude 42.43 MPa.

Do not read a plotted 45-degree arc as a 45-degree physical rotation. It represents 2θ = 45°, or θ = 22.5°. Also distinguish a plane from its normal: some prompts ask for the plane orientation while others ask for the direction of the principal stress axis, which can shift the verbal angle by 90 degrees.

In-plane versus absolute maximum shear

A plane-stress problem has third principal stress σ3 = 0 in the out-of-plane direction. The absolute maximum shear in three dimensions is half the largest separation among all three principal stresses:

τabsolute,max = [max(σ1,σ2,0) - min(σ1,σ2,0)]/2.

For the worked state, this is (92.43 - 0)/2 = 46.22 MPa, greater than the 42.43-MPa in-plane maximum. If σ1 and σ2 straddle zero, the in-plane radius may already be the absolute maximum. Read whether the question asks for maximum in-plane shear or absolute maximum shear.

Useful special cases provide quick checks. Equal biaxial normal stress with no shear gives a zero-radius circle. Pure shear, σx = σy = 0, gives principal stresses +|τ| and -|τ| on planes rotated 45 degrees, while the maximum shear magnitude remains |τ|. Finally, transformed stresses must preserve the invariants σx' + σy' = σx + σy and the same Mohr-circle radius. These checks catch sign and angle errors before they propagate into a combined-stress decision.