MechanicalMachine DesignmediumFE ExamCourseworkPE / Advanced

Principal Stress from a Plane Stress State

A point in a machine component is under the following plane stress state:

$\sigma_x = 80\ \text{MPa}$ (tension)
$\sigma_y = -40\ \text{MPa}$ (compression)
$\tau_{xy} = 30\ \text{MPa}$

Find the maximum principal stress $\sigma_1$ .

Given

σx	80 MPa	Normal stress, x
σy	-40 MPa	Normal stress, y
τxy	30 MPa	Shear stress

Hint 1

Think of Mohr's circle: the principal stresses are the centre plus and minus the radius.

Hint 2

The centre is the average normal stress. The radius involves both the half-difference of normal stresses and the shear.

Hint 3

Be careful with the negative σy — it affects both the average and the difference term, and the difference is 120, not 40.

Worked solution — try the problem first

The principal stresses are the centre of Mohr's circle plus or minus its radius:

$\sigma_{1,2} = \underbrace{\frac{\sigma_x + \sigma_y}{2}}_{\text{centre}} \pm \underbrace{\sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}}_{\text{radius}}$

Centre: $\sigma_{avg} = \frac{80 + (-40)}{2} = \frac{40}{2} = 20\ \text{MPa}$

Radius. Note the difference $\sigma_x - \sigma_y = 80 - (-40) = 120$ , not 40 — the sign of $\sigma_y$ matters twice over here:

$R = \sqrt{\left(\frac{120}{2}\right)^2 + 30^2} = \sqrt{60^2 + 30^2} = \sqrt{3600 + 900} = \sqrt{4500} = 67.08\ \text{MPa}$

Principal stresses: $\sigma_1 = 20 + 67.08 = 87.08\ \text{MPa}$ $\sigma_2 = 20 - 67.08 = -47.08\ \text{MPa}$

$\boxed{\sigma_1 = 87.1\ \text{MPa}}$

Check. The sum of principal stresses must equal the sum of the original normal stresses (the first stress invariant): $87.08 + (-47.08) = 40 = 80 + (-40)$ ✓

Concepts:Principal stressesMohr's circlePlane stressStress transformation

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