Principal Stresses, Strains & Mohr's Circle

Principal stresses and Mohr's circle are near-certain GATE ME marks (~2) and feed directly into the failure theories. The two-dimensional stress-transformation formulas plus the circle geometry answer almost everything here.

Transformation & principals

For a 2D stress state (σx, σy, τxy), the principal stresses are:
σ₁,₂ = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²].

The maximum in-plane shear stress τmax = √[((σx − σy)/2)² + τxy²] = (σ₁ − σ₂)/2. Principal planes carry zero shear; the plane of max shear is 45° from the principal planes.

Orientation: tan 2θₚ = 2τxy/(σx − σy). The sum σx + σy = σ₁ + σ₂ is invariant (stress invariant).

Mohr's circle

Plot points (σx, −τxy) and (σy, +τxy); the centre = ((σx + σy)/2, 0); the radius = τmax = √[((σx − σy)/2)² + τxy²]. Principal stresses are where the circle cuts the σ-axis (τ = 0); the top/bottom give max shear. A physical rotation of angle θ corresponds to 2θ on the circle.

Special cases: pure shear → σ₁ = +τ, σ₂ = −τ (circle centred at origin); uniaxial tension → circle from 0 to σ, τmax = σ/2.

Exam Tricks & Tips

• 🎯 σx + σy is invariant — a fast sanity check: your two principal stresses must still sum to σx + σy.
• 🎯 τmax = (σ₁ − σ₂)/2 = radius of Mohr's circle — read it straight off the geometry.
• 🎯 Angles double on Mohr's circle: a 30° physical rotation moves 60° around the circle.
• 🎯 Pure shear gives principal stresses ±τ at 45° — the classic reason shafts fail on a 45° helix in torsion.
• 🎯 Absolute max shear may involve the third (zero) principal stress: for a 2D state with both principals positive, τabs = σ₁/2, not (σ₁−σ₂)/2.
• ❌ Common mistake: forgetting the sign convention for τxy when plotting Mohr's circle — mixing it up flips the rotation direction and misplaces the principal-plane angle.

Expected exam pattern

A 2-mark computation of principal stresses or max shear from (σx, σy, τxy), or reading a Mohr's-circle value. Pure-shear and the "absolute maximum shear includes the zero third stress" subtlety are recurring conceptual traps.

Quick recap

σ₁,₂ = (σx+σy)/2 ± R, with R = √[((σx−σy)/2)² + τxy²] = τmax = (σ₁−σ₂)/2. Mohr's centre ((σx+σy)/2, 0), radius R; angles double. σx+σy invariant. Pure shear → ±τ at 45°. Check absolute max shear against the zero third principal stress.

Cover the answer, recall, then check. 11 cards on 2D stress transformation for GATE ME.

Q1. Write the principal-stress formula for a 2D state.
A1. σ₁,₂ = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²].

Q2. Formula for maximum in-plane shear stress?
A2. τmax = √[((σx − σy)/2)² + τxy²] = (σ₁ − σ₂)/2 — the radius of Mohr's circle.

Q3. What is special about the principal planes?
A3. They carry the maximum/minimum normal stress and ZERO shear stress.

Q4. How far apart are the principal plane and the max-shear plane?
A4. 45° physically (which is 90° on Mohr's circle).

Q5. Give the centre and radius of Mohr's circle.
A5. Centre = ((σx + σy)/2, 0); radius = √[((σx − σy)/2)² + τxy²] = τmax.

Q6. What stress quantity is invariant under rotation?
A6. The sum of normal stresses: σx + σy = σ₁ + σ₂ (first stress invariant).

Q7. How do physical angles map onto Mohr's circle?
A7. A physical rotation of θ corresponds to a 2θ rotation on Mohr's circle.

Q8. Principal stresses for a pure-shear state τ?
A8. σ₁ = +τ and σ₂ = −τ, oriented at 45° to the shear planes. Mohr's circle is centred at the origin.

Q9. Where do principal stresses appear on Mohr's circle?
A9. Where the circle intersects the σ-axis (τ = 0) — the left and right extreme points.

Q10. For a 2D state with σ₁ > σ₂ > 0, what is the absolute maximum shear?
A10. τabs = (σ₁ − 0)/2 = σ₁/2, using the zero third principal stress — larger than the in-plane (σ₁ − σ₂)/2.

Q11. Orientation of the principal planes?
A11. tan 2θₚ = 2τxy/(σx − σy).

Real components rarely see stress along convenient axes — a shaft feels bending plus torsion, a pressure vessel feels biaxial tension. Principal-stress analysis finds the worst-case normal stress and the plane it acts on, which is exactly what failure theories need. Mohr's circle makes it visual and fast.

Core concept: at any point, rotating the reference axes changes the normal and shear stresses; the principal stresses are the maximum/minimum normal stresses (where shear is zero), and the maximum shear stress occurs 45° away.

Deep explanation

Beginner — the 2D transformation

Given σx, σy, τxy, the stresses on a plane at angle θ are found by transformation. The principal stresses are:
σ₁,₂ = (σx + σy)/2 ± √[ ((σx − σy)/2)² + τxy² ].
The average (σx+σy)/2 is the circle's centre; the square-root term is its radius R.

Intermediate — maximum shear and orientation

• Maximum in-plane shear stress: τ_max = R = √[ ((σx − σy)/2)² + τxy² ] = (σ₁ − σ₂)/2.
• Principal planes carry zero shear; the plane of maximum shear is 45° from the principal plane.
• Orientation: tan 2θ_p = 2τxy/(σx − σy).

Advanced — Mohr's circle

Plot points (σx, −τxy) and (σy, +τxy); the line joining them is a diameter. Centre C = ((σx+σy)/2, 0), radius R as above. Reading off:

• σ₁, σ₂ = C ± R (x-intercepts),
• τ_max = R (top of circle),
• angles on the circle are twice the physical angles (2θ).
  For 3D, remember the absolute maximum shear uses the largest minus smallest of all three principal stresses (including σ₃ = 0 for plane stress) — τ_abs,max = (σ_max − σ_min)/2.

Worked example

A point has σx = 80 MPa, σy = 20 MPa, τxy = 30 MPa. Find the principal stresses and maximum shear.
Centre = (80 + 20)/2 = 50 MPa. R = √[((80−20)/2)² + 30²] = √[30² + 30²] = √1800 = 42.4 MPa.
σ₁ = 50 + 42.4 = 92.4 MPa; σ₂ = 50 − 42.4 = 7.6 MPa.
τ_max = R = 42.4 MPa. (Note τ_max ≠ 30; it is the circle radius, not τxy.)

GATE relevance

Principal stresses and τ_max are directly examined and are the required input to failure theories (max shear stress / distortion energy) in Machine Design, and to combined bending-torsion shaft design. Mohr's circle is the fastest reliable route and appears as conceptual questions too.

Exam tricks & shortcuts

• Memorise σ₁,₂ = avg ± R and τ_max = R; you rarely need the full transformation equations.
• For a shaft under bending σ and torsion τ (σy = 0): σ₁,₂ = σ/2 ± √((σ/2)² + τ²).
• Mnemonic: "Centre is the average, radius is the worst shear."

Key formulas

• Principal stresses: σ₁,₂ = (σₓ+σ_y)/2 ± √[((σₓ−σ_y)/2)² + τ²].
• Max shear stress: τ_max = √[((σₓ−σ_y)/2)² + τ²] = (σ₁ − σ₂)/2.
• Principal plane angle: tan2θ_p = 2τ/(σₓ − σ_y).
• Mohr's circle: centre ((σₓ+σ_y)/2, 0), radius = τ_max.

Mohr's circle graphically gives principal stresses and maximum shear from a 2D stress state.

Worked Example

Problem: At a point in a loaded body the plane-stress state is σ_x = 80 MPa, σ_y = 20 MPa, and τ_xy = 30 MPa. Find the principal stresses, the maximum in-plane shear stress, and the orientation of the principal planes.

Solution:
Average (centre of Mohr's circle):
σ_avg = (σ_x + σ_y)/2 = (80 + 20)/2 = 50 MPa.

Radius of Mohr's circle:
R = √[((σ_x − σ_y)/2)² + τ_xy²] = √[(30)² + (30)²] = √(900 + 900) = √1800 ≈ 42.4 MPa.

Principal stresses (centre ± radius):
σ₁ = σ_avg + R = 50 + 42.4 = 92.4 MPa,
σ₂ = σ_avg − R = 50 − 42.4 = 7.6 MPa.

Maximum in-plane shear stress equals the radius:
τ_max = R ≈ 42.4 MPa.

Orientation of the principal planes:
tan(2θ_p) = 2τ_xy/(σ_x − σ_y) = 60/60 = 1 ⇒ 2θ_p = 45° ⇒ θ_p = 22.5°.

Answer: σ₁ = 92.4 MPa, σ₂ = 7.6 MPa, τ_max = 42.4 MPa, with principal planes at θ_p = 22.5°.