Stress transformation

A single point inside a loaded shaft, beam, or plate does not have "one" stress — it has infinitely many, one for every plane you could imagine slicing through it. The normal stress you measure on a cross-section perpendicular to the axis, σ=P/A\sigma = P/A, and the shear stress τ=Tc/J\tau = Tc/J in torsion are just two special readings of the same underlying stress state, taken on two particular planes. Rotate the cutting plane and the numbers change smoothly and predictably. Stress transformation is the equilibrium bookkeeping that predicts exactly how they change. Given the stress state (σx,σy,τxy)(\sigma_x, \sigma_y, \tau_{xy}) on an xx–yy element, it tells you the normal and shear stress on any plane rotated by angle θ\theta. This is not new physics — it is pure statics, resolved on a rotated free body — and it is the essential bridge to principal stresses, Mohr's circle, and the combined-loading problems that dominate the rest of this ME10001 unit. Only plane (2-D) stress is in scope here; the full 3-D stress tensor is left for later courses.

The core insight

Equilibrium of a wedge cut from a stressed element gives σx′\sigma_{x'} and τx′y′\tau_{x'y'} on any rotated plane directly from (σx,σy,τxy)(\sigma_x, \sigma_y, \tau_{xy}) and the angle θ\theta, with σx+σy\sigma_x+\sigma_y invariant under rotation.

The core idea

Take a small rectangular element with known stresses σx\sigma_x, σy\sigma_y on the faces perpendicular to xx and yy, and shear stress τxy\tau_{xy}, which by moment equilibrium is the same magnitude on all four faces. Cut the element along a plane whose outward normal makes angle θ\theta with the xx-axis, measured positive counter-clockwise, and write force equilibrium for the resulting triangular wedge. Two equations fall out of that single free-body diagram.

σx′=σx+σy2+σx−σy2cos⁡2θ+τxysin⁡2θ\sigma_{x'} = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\theta + \tau_{xy}\sin2\theta
τx′y′=−σx−σy2sin⁡2θ+τxycos⁡2θ\tau_{x'y'} = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta + \tau_{xy}\cos2\theta

where: σx′\sigma_{x'} is the normal stress on the rotated face, τx′y′\tau_{x'y'} is the shear stress on that same face, and θ\theta is the rotation angle from the original xx-axis to the new x′x'-axis. The normal stress on the perpendicular face, σy′\sigma_{y'}, is found by putting θ+90°\theta+90° into the first equation. A useful check falls straight out of adding σx′\sigma_{x'} and σy′\sigma_{y'}: the cos⁡2θ\cos2\theta terms cancel because cos⁡(2θ+180°)=−cos⁡2θ\cos(2\theta+180°) = -\cos2\theta, leaving

σx+σy=σx′+σy′\sigma_x + \sigma_y = \sigma_{x'} + \sigma_{y'}

where: this says the sum of normal stresses on any two perpendicular planes through a point is a constant, independent of how you orient the cutting plane — an invariant you can use to catch arithmetic errors in every transformation problem. Two things are worth internalising early. First, the sign convention: σ\sigma is positive in tension, τxy\tau_{xy} is positive when it tends to rotate the element clockwise, and θ\theta is positive counter-clockwise from xx to x′x' — mixing these up is the single biggest source of wrong answers. Second, these equations came purely from force equilibrium on a free body; no material law entered anywhere. That means they hold for any material, elastic or not, at any single point where the stress state is plane stress — the same equations you would use for steel apply just as well to a polymer or a soil element.

Figure
A square stress element with sigma x and sigma y on its faces and tau xy shown as shear arrows, next to a rotated wedge cut at angle theta showing the transformed stresses sigma x prime and tau x prime y prime on the inclined face
Wedge equilibrium: rotating the cutting plane by theta original element sigma y sigma x tau xy wedge cut at angle theta theta sigma x prime tau x prime y prime Equilibrium of the inclined face gives sigma x prime and tau x prime y prime in terms of theta. Trig functions inside the equations use 2 theta, not theta.
Derivation APlane-stress transformation from wedge equilibrium

The goal is to derive σx′\sigma_{x'} and τx′y′\tau_{x'y'} for a plane at angle θ\theta purely from force equilibrium of a triangular wedge cut from a stressed element carrying (σx,σy,τxy)(\sigma_x, \sigma_y, \tau_{xy}), with no material law used anywhere, so that the resulting equations apply to any linear-elastic or non-elastic material at that point.

Step 1. Isolate the wedge. Cut the element with an inclined plane whose normal makes angle θ\theta with xx, keeping the two original faces (areas Acos⁡θA\cos\theta and Asin⁡θA\sin\theta for a wedge of hypotenuse area AA) and the new inclined face of area AA. The stresses on the two original faces are known: σx\sigma_x, σy\sigma_y, and τxy\tau_{xy} on both.

Step 2. Resolve forces along x′x'. Multiply each stress by the area it acts on to get a force, then project every force onto the x′x'-direction using cos⁡θ\cos\theta or sin⁡θ\sin\theta as appropriate for that face's orientation. Summing and dividing back by AA turns the force balance into a stress balance, and using the double-angle identities cos⁡2θ=1+cos⁡2θ2\cos^2\theta = \frac{1+\cos2\theta}{2} and 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta = \sin2\theta collapses the result to

σx′=σx+σy2+σx−σy2cos⁡2θ+τxysin⁡2θ.\sigma_{x'} = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\theta + \tau_{xy}\sin2\theta.

Step 3. Resolve forces along y′y', perpendicular to the cut face, the direction that measures shear on that plane. The same bookkeeping, projected the other way, gives

τx′y′=−σx−σy2sin⁡2θ+τxycos⁡2θ.\tau_{x'y'} = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta + \tau_{xy}\cos2\theta.

Step 4. Get σy′\sigma_{y'} for free. Since the y′y'-face is simply the x′x'-face rotated a further 90°90°, substitute θ+90°\theta+90° into the Step 2 result; because cos⁡(2θ+180°)=−cos⁡2θ\cos(2\theta+180°)=-\cos2\theta and sin⁡(2θ+180°)=−sin⁡2θ\sin(2\theta+180°)=-\sin2\theta, this immediately gives σy′=σx+σy2−σx−σy2cos⁡2θ−τxysin⁡2θ\sigma_{y'} = \frac{\sigma_x+\sigma_y}{2} - \frac{\sigma_x-\sigma_y}{2}\cos2\theta - \tau_{xy}\sin2\theta, and adding it to the Step 2 result cancels the rotated terms, leaving the invariant σx+σy=σx′+σy′\sigma_x+\sigma_y=\sigma_{x'}+\sigma_{y'} — a self-consistency check built directly into the derivation.

Worked example

A stress element at a point in a machine part has σx=80 MPa\sigma_x = 80\text{ MPa}, σy=20 MPa\sigma_y = 20\text{ MPa}, τxy=30 MPa\tau_{xy} = 30\text{ MPa}. Find σx′\sigma_{x'} and τx′y′\tau_{x'y'} on a plane rotated θ=25°\theta = 25° counter-clockwise from the xx-axis, and confirm the result with the invariant.

Step 1. Compute the averages: σx+σy2=80+202=50 MPa\dfrac{\sigma_x+\sigma_y}{2} = \dfrac{80+20}{2} = 50\text{ MPa} and σx−σy2=80−202=30 MPa\dfrac{\sigma_x-\sigma_y}{2} = \dfrac{80-20}{2} = 30\text{ MPa}.

Step 2. Compute the angle terms. 2θ=50°2\theta = 50°, so cos⁡50°=0.6428\cos50° = 0.6428 and sin⁡50°=0.7660\sin50° = 0.7660.

Step 3. Apply the normal-stress equation. σx′=50+30(0.6428)+30(0.7660)=50+19.28+22.98=92.3 MPa\sigma_{x'} = 50 + 30(0.6428) + 30(0.7660) = 50 + 19.28 + 22.98 = 92.3\text{ MPa}.

Step 4. Apply the shear-stress equation. τx′y′=−30(0.7660)+30(0.6428)=−22.98+19.28=−3.7 MPa\tau_{x'y'} = -30(0.7660) + 30(0.6428) = -22.98 + 19.28 = -3.7\text{ MPa}.

Step 5. Check with the invariant. σy′=(σx+σy)−σx′=100−92.3=7.7 MPa\sigma_{y'} = (\sigma_x+\sigma_y) - \sigma_{x'} = 100 - 92.3 = 7.7\text{ MPa}. Substituting θ+90°=115°\theta+90°=115° directly into the σx′\sigma_{x'} formula gives the same 7.7 MPa7.7\text{ MPa}, confirming the arithmetic.

The rotated plane carries a normal stress of 92.3 MPa (tension) and a shear stress of 3.7 MPa.

Competitive stretch

Now push the same element to θ=45°\theta = 45° and θ=135°\theta = 135° for the pure-shear case σx=σy=0\sigma_x=\sigma_y=0, τxy=40 MPa\tau_{xy}=40\text{ MPa}, which strips the averages to zero and leaves σx′=τxysin⁡2θ\sigma_{x'} = \tau_{xy}\sin2\theta, τx′y′=τxycos⁡2θ\tau_{x'y'} = \tau_{xy}\cos2\theta. At θ=45°\theta=45°, 2θ=90°2\theta=90° gives σx′=40(1)=40 MPa\sigma_{x'} = 40(1) = 40\text{ MPa} tension with zero shear; at θ=135°\theta=135°, 2θ=270°2\theta=270° gives σx′=40sin⁡270°=−40 MPa\sigma_{x'} = 40\sin270° = -40\text{ MPa} compression, again with zero shear. A pure-shear state is therefore equivalent to equal tension and compression at 45°45° to the shear planes — the reasoning behind why torsion-test fractures in brittle shafts run along a 45°45° helix, and a result you will lean on constantly once Mohr's circle is introduced.

Common traps

  • Forgetting the factor of 2 on θ\theta. The angle inside the sine and cosine is 2θ2\theta, not θ\theta — a very common slip that produces wildly wrong numbers.
  • Sign convention confusion for τxy\tau_{xy}. Different texts define positive shear differently; be consistent with your course's convention and use the invariant σx+σy=σx′+σy′\sigma_x+\sigma_y=\sigma_{x'}+\sigma_{y'} as a check.
  • Mixing up the direction of θ\theta. It is measured from the xx-axis to the x′x'-axis; going the wrong way flips the sign of every sin⁡2θ\sin2\theta term.
  • Treating this as a 3-D problem. ME10001 restricts transformation to plane stress — do not attempt out-of-plane rotations or a full stress tensor here.

Ask the tutor: if you want the wedge free-body diagram walked through with force components drawn out explicitly, or extra practice with a different sign convention for τxy\tau_{xy}.

Why does the invariant sigma x plus sigma y equal sigma x prime plus sigma y prime even hold true, physically

It follows directly from the algebra of the transformation equations, but physically it reflects that the sum of normal stresses on any two perpendicular planes through a point is tied to the underlying stress state, not to how you chose to orient your cutting plane. Rotating your viewpoint cannot change the physical loading at that point, only how it is resolved onto different faces, so any two perpendicular directions must give the same total.

Do I need to memorise both the sigma x prime and tau x prime y prime formulas separately, or can I derive one from the other

You should be able to derive both from the single wedge free-body diagram shown in the derivation above, since that is what a stress-transformation exam question is really testing. In practice, memorising the pair is faster under time pressure, but always keep the wedge picture in mind so you can rebuild the sign convention if you forget which term gets a minus sign.

Why does a shaft under pure torsion often fracture along a helix instead of straight across

Pure torsion is a pure-shear stress state, and the worked stretch example above shows that pure shear on the original planes transforms into equal tension and compression on planes at 45 degrees to those original planes, with zero shear there. A brittle material fails in tension, so the fracture surface follows the 45-degree plane of maximum tensile stress, which traces out a helix along the length of a round shaft — this is exactly the transformation result applied to a real failure mode.

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Where this leads