Combined Stresses, Principal Stresses and Mohr's Circle

NCEES FE Civil spec: Mechanics of Materials, D. Combined stresses, principal stresses, and Mohr's circle. FE Reference Handbook: Mechanics of Materials chapter, plane stress and Mohr's circle.

Real members often carry several loads at once: a shaft in bending and torsion, or a beam web with bending and shear. Combined stress problems add the stresses on one element, then find the planes where normal stress is greatest (principal stresses) and where shear is greatest. Mohr's circle is the graphical shortcut, and the FE Civil exam tests it every year.

What the NCEES spec covers

Section D covers combined stresses, stress transformation in plane stress, principal stresses and planes, maximum in-plane shear stress, and Mohr's circle.

Key concepts

Superposition. Add normal stresses that act in the same direction (for example, axial plus bending). Shear stresses from torsion and transverse shear add if they act on the same face in the same direction.
Principal planes carry the maximum and minimum normal stress and have zero shear stress.
Maximum in-plane shear occurs on planes 45° from the principal planes and equals the circle radius.
Mohr's circle plots each plane as a point (σ, τ). Its center is the average normal stress and its radius combines half the normal stress difference with τxy. Rotating θ on the element is 2θ on the circle.
Absolute maximum shear. If both principal stresses have the same sign, the absolute maximum shear is half the larger one, not the in-plane value.

Core formulas

σ_{1, 2} = \frac{σ _{x} + σ _{y}}{2} \pm (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}

τ_{m a x, in-plane} = (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2} tan 2 θ_{p} = \frac{2 τ _{x y}}{σ _{x} - σ _{y}}

σ_{θ} = \frac{σ _{x} + σ _{y}}{2} + \frac{σ _{x} - σ _{y}}{2} cos 2 θ + τ_{x y} sin 2 θ

Where it is in the FE Reference Handbook

The Mechanics of Materials chapter covers plane stress, principal stresses and Mohr's circle, including the sign convention it uses for plotting shear. Check that convention before plotting, because textbooks differ on whether positive shear is plotted up or down.

Worked example

An element has $σ_{x} = 90 MPa$ , $σ_{y} = - 30 MPa$ and $τ_{x y} = 40 MPa$ . Find the principal stresses, their orientation and the maximum in-plane shear stress.

Step 1: center and radius.

C = \frac{90 + ( - 30 )}{2} = 30 MPa R = 6 0^{2} + 4 0^{2} = 72.1 MPa

Step 2: principal stresses.

σ_{1} = 30 + 72.1 = 102.1 MPa σ_{2} = 30 - 72.1 = - 42.1 MPa

Step 3: orientation.

tan 2 θ_{p} = \frac{2 ( 40 )}{90 - ( - 30 )} = 0.667 \Rightarrow θ_{p} = 16. 8^{\circ}

Plug θ = 16.8° into the transformation equation: 30 + 60 cos 33.7° + 40 sin 33.7° = 102.1 MPa, so σ₁ acts on the plane rotated 16.8° counterclockwise from x.

Step 4: maximum shear. τmax = R = 72.1 MPa, on planes at 16.8° + 45° = 61.8° (and −28.2°). Because σ₁ and σ₂ have opposite signs, this is also the absolute maximum shear.

Common mistakes

Dropping the sign of a compressive σy, which changes both the center and the radius.
Reporting 2θ as the element rotation. Halve the angle measured on the circle.
Plotting τxy with the wrong sign convention and getting the rotation direction backwards.
Forgetting the absolute maximum shear check when both principal stresses have the same sign.
Adding stresses that act on different planes or in different directions.

Plot any stress state with the Mohr's circle calculator.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

At a point, σx = 80 MPa, σy = 20 MPa and τxy = 30 MPa. The principal stresses are most nearly: See the worked solution or practice the Stress & Strain deck
At a point, σx = 80 MPa, σy = 20 MPa and τxy = 30 MPa. The maximum in-plane shear stress is most nearly: See the worked solution or practice the Stress & Strain deck
At a point, σx = 80 MPa, σy = 20 MPa and τxy = 30 MPa. The angle from the x-axis to the principal plane is most nearly: See the worked solution or practice the Stress & Strain deck
An element is subjected only to a positive shear stress τxy = 40 MPa. The principal stresses are: See the worked solution or practice the Stress & Strain deck
For σx = 60 MPa, σy = 20 MPa and τxy = 20 MPa, the normal stress on a plane rotated 30° counterclockwise from the x-axis is most nearly: See the worked solution or practice the Stress & Strain deck
At a point in plane stress, σ_x = 80 MPa, σ_y = 20 MPa, and τ_xy = 0. What are the two in-plane principal stresses? See the worked solution or practice the Stress & Strain deck

Related Calculators and Examples

Frequently Asked Questions

How do I find principal stresses?

Take the average normal stress (sigma_x + sigma_y) / 2 and add and subtract the radius sqrt(((sigma_x - sigma_y) / 2)^2 + tau_xy^2). These are the maximum and minimum normal stresses, acting on planes with zero shear.

What does the radius of Mohr's circle represent?

The maximum in-plane shear stress. It occurs on planes rotated 45 degrees from the principal planes.

Why is the angle on Mohr's circle doubled?

The stress transformation equations depend on 2 theta, so a rotation of theta on the physical element corresponds to 2 theta around the circle.

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