How to work a plane-stress problem with Mohr’s circle
- Find the centre C = (σx + σy) ÷ 2 and the radius R = √(((σx − σy) ÷ 2)² + τxy²).
- Plot X at (σx, τxy) and Y at (σy, −τxy). The line XY is a diameter through C.
- Read the principal stresses where the circle crosses the σ axis: σ1 = C + R and σ2 = C − R.
- Measure 2θp1 from X to the σ1 end and halve it.
- The top and bottom of the circle give τmax = R, on planes 45° from the principal planes.
- For an element turned θ, turn the diameter 2θ: X′ and Y′ give σx′, τx′y′ and σy′.
An example from set EXA585
An element has σx = 80 MPa, σy = −10 MPa and τxy = −60 MPa. C = (80 − 10) ÷ 2 = 35 MPa and R = √(45² + 60²) = 75 MPa, so σ1 = 110 MPa and σ2 = −40 MPa. X (80, −60) is right of C, so 2θp1 = arctan(−60 ÷ 45) = −53.13° and θp1 = −26.57°, clockwise from x. τmax = 75 MPa at θs = −71.57°. Turned 30° counterclockwise, the element carries σx′ = 5.538 MPa, σy′ = 64.46 MPa and τx′y′ = −68.97 MPa. Open set EXA585 to work it yourself.
Questions
How do you draw Mohr’s circle for plane stress?
Mark the centre C = (σx + σy) ÷ 2 on the σ axis. Plot X at (σx, τxy) and Y at (σy, −τxy), the two ends of a diameter. The radius is R = √(((σx − σy) ÷ 2)² + τxy²). The circle crosses the σ axis at σ1 = C + R and σ2 = C − R.
Why do some courses plot τ positive downward?
With τ positive downward and X at (σx, τxy), turning the element counterclockwise by θ turns the circle counterclockwise by 2θ. Plotting τ upward gives a mirror image, so 2θ turns the other way. The answers are the same either way.
How do I know which angle is θp1?
tan 2θp has two answers 90° apart. On the circle, 2θp1 runs from X to the σ1 end. If σx > σy, X is right of C and 2θp1 is the arctangent itself. If σx < σy, add or take away 180°. Then halve it.
Why is the angle doubled on Mohr’s circle?
The transformation equations use sin 2θ and cos 2θ, so turning the element by θ turns the point on the circle by 2θ. Faces 90° apart on the element, like x and y, are opposite ends of a diameter.
What is the maximum in-plane shear stress?
It is the radius, τmax = R, at the top and bottom of the circle. Its planes are 45° from the principal planes, and the normal stress on them is C, not zero.
What is the absolute maximum shear stress in plane stress?
Count σ3 = 0 as a third principal stress. Take the largest and smallest of σ1, σ2 and 0 and halve the difference. When σ1 and σ2 have opposite signs that is R. When they have the same sign it is half the one with the larger size.
How do I give my class the same problems?
Send the Share with class link: everyone gets the same set. Print makes up to 30 different copies, each with its key.
Sources, checked 2 November 2026: David Roylance, Transformation of Stresses and Strains (MIT OpenCourseWare); Louie Yaw, Mohr’s Circle Rules and Comments (Walla Walla University); Purdue ME 323 homework notes.