--------------- | ------------------------------ | --------------------------------- | | Centre | C = (σx + σy)/2 | On the σ-axis | | Radius | R = √[((σx − σy)/2)² + τxy²] | Distance from centre to any point | | σ₁ (max principal) | C + R | Rightmost point on circle | | σ₂ (min principal) | C - R | Leftmost point on circle | | τmax (max shear) | R | Top and bottom of circle | | Principal angle θp | (1/2) arctan(2τxy / (σx − σy)) | Physical rotation angle | | Angle on circle | 2θ (double angle) | Circle angle = 2 × physical angle |

Yield Criteria for Structural Steel

Von Mises (distortion energy) criterion

The von Mises equivalent stress for plane stress (σz = 0):

σvm = √[σx² - σx·σy + σy² + 3τxy²]

Yield check: σvm ≤ Fy

This is the most commonly used yield criterion for structural steel. AISC 360 uses von Mises implicitly in combined-stress interaction formulas. For pure shear (σx = σy = 0): σvm = √3 · τ = 1.732τ, so shear yield stress τy = Fy/√3 ≈ 0.577Fy.

Tresca (maximum shear stress) criterion

τmax = (σ₁ - σ₂)/2 ≤ Fy/2

Yield check: σ₁ - σ₂ ≤ Fy

Tresca is more conservative than von Mises. For pure shear: τy = Fy/2 = 0.50Fy, which is lower than the von Mises value of 0.577Fy. AISC 360 uses a shear yield of 0.6Fy (approximating von Mises).

Comparison of yield criteria

Worked Example — Beam Web Stress State

Problem: A W18x46 beam (A992 steel, Fy = 50 ksi) carries a factored moment Mu = 250 kip-ft and factored shear Vu = 80 kips at a critical section near a support. Evaluate the principal stresses and maximum shear in the beam web at mid-depth.

Step 1 — Stress components at mid-depth of web

W18x46 properties: d = 18.06 in, tw = 0.305 in, Ix = 712 in⁴

Bending stress at mid-depth (y = 0 from neutral axis):
  σx = M·y/I = 0 (neutral axis)
  → Bending stress is zero at the neutral axis

Shear stress at neutral axis (maximum):
  For W-shapes, VQ/(I·tw) at the NA ≈ Vu/(d·tw) for preliminary estimate
  τxy ≈ Vu/(d·tw) = 80,000/(18.06 × 0.305) = 14,525 psi ≈ 14.5 ksi

Step 2 — Principal stresses

At the neutral axis: σx = 0, σy = 0, τxy = 14.5 ksi
σavg = 0
R = τxy = 14.5 ksi

σ₁ = 0 + 14.5 = 14.5 ksi (tension)
σ₂ = 0 - 14.5 = -14.5 ksi (compression)
τmax = 14.5 ksi

θp = (1/2) arctan(2×14.5 / 0) = (1/2)(90°) = 45°

The principal stresses at the neutral axis are equal and opposite (pure shear state), oriented at 45° to the beam axis.

Step 3 — Von Mises check

σvm = √[0 - 0 + 0 + 3×(14.5)²] = √(630.75) = 25.1 ksi
Fy = 50 ksi
σvm / Fy = 25.1/50 = 0.50 → OK (well below yield)

Step 4 — AISC shear check comparison

AISC nominal shear capacity (Chapter G):
  Cv = 1.0 (web not slender for W18x46)
  Vn = 0.6Fy·Aw = 0.6 × 50 × (18.06 × 0.305) = 165.2 kips
  φVn = 1.0 × 165.2 = 165.2 kips

  Vu / φVn = 80/165.2 = 0.48 → OK

The AISC shear check (0.6Fy × Aw) implicitly uses a von Mises shear yield of approximately 0.6Fy, which is consistent with the principal stress analysis.

Common pitfalls and how to avoid confusion

• Sign convention: Tension is positive on Mohr's circle. A common error is plotting compressive stresses (negative σ) on the wrong side of the origin, which flips the entire circle. Always confirm: positive σ is to the right.
• Double angle: Mohr's circle uses 2θ, so a 15° physical rotation corresponds to 30° on the circle. Forgetting the factor of 2 is the most common mistake in graphical construction.
• Axis confusion: major/minor axis properties, sign conventions, and local coordinate systems can flip a result. Confirm whether the input stress state uses x-y or some other local convention before computing.
• Plane stress assumption: Mohr's circle as presented here applies to plane stress (σz = 0). Thick pressure vessels, tunnel linings, and triaxial stress states require 3D Mohr's circles with all three principal stresses.
• Yield criterion mismatch: AISC uses approximately von Mises (0.6Fy shear yield). If you check principal stress against Tresca (0.5Fy), you will get a more conservative but potentially misleading result.

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Frequently Asked Questions

What are principal stresses and how do you find them from Mohr's circle? Principal stresses are the normal stresses acting on planes where shear stress is zero — they represent the maximum and minimum normal stress values for a given stress state. On Mohr's circle, the principal stresses σ₁ and σ₂ are the two points where the circle intersects the horizontal axis (τ = 0). Algebraically, σ₁,₂ = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²]. The principal planes are oriented at an angle θp = (1/2)arctan(2τxy / (σx − σy)) from the original x-face.

What is maximum shear stress and where does it occur on Mohr's circle? The maximum in-plane shear stress is the radius of Mohr's circle: τmax = √[((σx − σy)/2)² + τxy²]. It occurs on planes rotated 45° from the principal planes. At the planes of maximum shear stress, there is also a normal stress equal to the average stress (σx + σy)/2 — these planes are not stress-free. For structural steel yield criteria under combined loading, τmax is compared to the shear yield stress τy = Fy/√3 (von Mises) or Fy/2 (Tresca).

How do you transform stresses to a plane rotated by angle θ? The stress transformation equations give the normal stress σθ and shear stress τθ on a plane at angle θ from the x-axis: σθ = (σx + σy)/2 + (σx − σy)/2·cos(2θ) + τxy·sin(2θ), and τθ = −(σx − σy)/2·sin(2θ) + τxy·cos(2θ). On Mohr's circle this corresponds to rotating the point representing the x-face by 2θ around the circle centre. Note that the angle on the circle is twice the physical rotation angle of the plane.

What is the difference between plane stress and plane strain? Plane stress assumes that all stress components in the out-of-plane direction are zero (σz = τxz = τyz = 0) — this applies to thin plates and webs where the through-thickness stress is negligible. Plane strain assumes that out-of-plane strains are zero (εz = 0), which applies to thick sections or long prismatic members where the geometry constrains expansion. In plane strain, an out-of-plane stress σz = ν(σx + σy) is induced even though no force is applied in that direction. Mohr's circle and stress transformation apply directly to both states for the in-plane components.

When is principal stress analysis needed in structural steel design? Principal stress analysis is most relevant when a steel element is subject to combined normal and shear stresses simultaneously — for example, in beam webs near supports where high shear and bending moment coexist (the critical section for combined stress), in gusset plates with eccentric load paths, or in welded connections where throat stresses combine normal and shear components. AISC 360 checks for combined loading in webs (Section G) and weld throat (Section J2) implicitly use principal stress concepts; Mohr's circle makes these limits intuitive and helps identify the orientation of critical planes for detailing.

What are the principal stresses for σx = 10 ksi, σy = 4 ksi, τxy = 3 ksi? Using the principal stress formula: σavg = (10 + 4)/2 = 7 ksi; R = √[((10 − 4)/2)² + 3²] = √[9 + 9] = √18 = 4.24 ksi. Therefore σ₁ = 7 + 4.24 = 11.24 ksi ≈ 11.2 ksi, σ₂ = 7 − 4.24 = 2.76 ksi ≈ 2.8 ksi, and τmax = 4.24 ksi. The principal plane angle is θp = (1/2) arctan(2 × 3 / (10 − 4)) = (1/2) arctan(1.0) = 22.5°. To verify: on the principal planes, the shear stress is zero and the Mohr's circle diameter equals σ₁ − σ₂ = 8.48 ksi = 2R ✓.

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