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Mohr's circle and stress transformation

At a point of a solid the stresses depend on the orientation of the plane you look at: the same state can look like pure tension in one direction and shear in another. Transforming stresses means computing how they change when the element rotates, and Mohr's circle is the graphical way to do it. With it you find the principal stresses and the maximum shear, which are the ones compared with the strength of the material.

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Plane stress and sign convention

In a thin plate, on the surface of a part or in a beam, the stresses on the face perpendicular to $z$ are zero ( $σ_{z} = τ_{x z} = τ_{yz} = 0$ ). Three components remain in the $x$ – $y$ plane: $σ_{x}$ , $σ_{y}$ and $τ_{x y}$ . They are drawn on a very small square element around the point.

Element with the three stresses in their positive sense.

A value with the wrong sign is the most common mistake. Before calculating, draw the element with the arrows of your problem and compare: if the shear on the right face points down,

τ_{x y}

is negative.

Transformation equations

If the element rotates by an angle $θ$ (counter-clockwise positive), equilibrium of a wedge cut at that angle gives the stresses on the new axes $x^{'}$ – $y^{'}$ :

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

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

Two details matter: the double angle $2 θ$ always appears, and the sum $σ_{x^{'}} + σ_{y^{'}} = σ_{x} + σ_{y}$ does not change with the rotation (it is an invariant). That sum is the quickest check of a calculation.

Principal stresses and maximum shear

As the element rotates, $σ_{x^{'}}$ goes through a maximum and a minimum: the principal stresses $σ_{1} \geq σ_{2}$ . On those planes, the principal planes, the shear is zero. Differentiating $σ_{x^{'}}$ and setting it to zero:

tan 2 θ_{p} = \frac{2 τ _{x y}}{σ _{x} - σ _{y}} σ_{1, 2} = C \pm R C = \frac{σ _{x} + σ _{y}}{2} R = (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}

The tangent gives two angles 90° apart, one for each principal stress. To know which is which, substitute one into the $σ_{x^{'}}$ equation (or use the circle, which tells you without any algebra).

The maximum in-plane shear is $τ_{max} = R$ and acts on planes at 45° from the principal ones, where the normal stress is not zero but the average $C$ .

Mohr's circle

The transformation equations are the parametric equations of a circle in the $(σ, τ)$ plane, with center $C$ and radius $R$ . Each point of the circle is the state of one face of the element. To build it (with the simulator's convention):

Axes: $σ$ horizontal (positive to the right) and $τ$ vertical, positive downward.
Mark the center $C = (σ_{x} + σ_{y}) /2$ on the $σ$ axis.
Mark $X = (σ_{x}, τ_{x y})$ , the $x$ face, and draw the circle through $X$ centered at $C$ . The point $Y = (σ_{y}, - τ_{x y})$ lies at the opposite end of the diameter.
The crossings with the $σ$ axis are $σ_{1}$ and $σ_{2}$ ; the highest and lowest points, $\pm τ_{max}$ .
Rotating the element by $θ$ is rotating $2 θ$ on the circle, in the same sense: the angle from $X$ to $σ_{1}$ is $2 θ_{p}$ .

Circle of example 1. From point X to σ₁ there are 53.13° counter-clockwise, so the principal plane is θp = 26.57° counter-clockwise from the x axis.

Other textbooks draw

τ

positive upward; then the rotation on the circle goes in the opposite sense to that of the element. Both conventions give the same numbers: what matters is not to mix them.

State	Circle	What it shows
Uniaxial tension $σ$	passes through 0 and $σ$	the maximum shear is $σ /2$ , at 45°: that is why a ductile material yields on inclined planes
Pure shear $τ$ (torsion)	centered at the origin, radius $τ$	$σ_{1, 2} = \pm τ$ at 45°: chalk or cast iron break along a 45° helix
Equal biaxial $σ_{x} = σ_{y}$	a single point	the same stress in every direction and no shear

Absolute maximum shear (three circles)

Even when the state is plane, the element is a cube: the third principal stress is $σ_{3} = σ_{z} = 0$ . With the three principal stresses three circles are drawn, one for each pair, and the absolute maximum shear is the radius of the largest:

τ_{m \overset{a}{ˊ} x abs} = \frac{σ _{m \overset{a}{ˊ} x} - σ _{m \overset{ı}{ˊ} n}}{2} con σ_{m \overset{a}{ˊ} x}, σ_{m \overset{ı}{ˊ} n} entre σ_{1}, σ_{2} y 0

If $σ_{1}$ and $σ_{2}$ have opposite signs, the largest circle is the in-plane one and nothing changes. If they have the same sign (a pressure vessel, for example), the maximum shear acts out of the plane and is larger than $R$ : forgetting it leaves the Tresca check on the unsafe side.

Three circles of example 3: the in-plane one (blue) has radius 20, but the largest, between σ₃ = 0 and σ₁, has radius 40.

Common mistakes

Rotating $θ$ on the circle instead of $2 θ$ .
Marking $Y = (σ_{y}, τ_{x y})$ : the sign of the shear changes between the two faces.
Taking either solution of $tan 2 θ_{p}$ as the one for $σ_{1}$ without checking.
Assuming the normal stress is zero on the planes of maximum shear: it is $C$ .
Forgetting $σ_{3} = 0$ when $σ_{1}$ and $σ_{2}$ have the same sign.

Worked examples

Example 1Principal stresses and maximum shear

At a point,

σ_{x} = 50

MPa,

σ_{y} = - 10

MPa and

τ_{x y} = 40

MPa. Find the principal stresses, the orientation of their planes and the maximum in-plane shear.

Solution

Center and radius: $C = \frac{50 + ( - 10 )}{2} = 20$ MPa and $R = 3 0^{2} + 4 0^{2} = 50$ MPa.
Principal: $σ_{1} = 20 + 50 = 70$ MPa and $σ_{2} = 20 - 50 = - 30$ MPa.
$tan 2 θ_{p} = \frac{2 \cdot 40}{50 - ( - 10 )} = 1.333$ , so $2 θ_{p} = 53.13°$ and $θ_{p} = 26.57°$ . Check: $σ_{x^{'}} (26.57°) = 20 + 30 \cdot 0.6 + 40 \cdot 0.8 = 70$ MPa, it is $σ_{1}$ . The other plane, at $- 63.43°$ , carries $σ_{2}$ .
Maximum shear: $τ_{max} = R = 50$ MPa, at $θ_{s} = 26.57° - 45° = - 18.43°$ , with $σ = C = 20$ MPa on those faces.

σ_{1} = 70

MPa (

θ_{p} = 26.57°

σ_{2} = - 30

MPa (

- 63.43°

) and

τ_{max} = 50

MPa (

θ_{s} = - 18.43°

Solve it in Simuveliq

In Engineering → Mohr's circle, mode Element → circle.
In Differential element type σx = 50, σy = −10 and τxy = 40.
The Principal stress σ₁, Principal stress σ₂, Maximum in-plane shear and Center and radius cards give the values and angles. Below, the «Principal» and «Max shear» elements are drawn already rotated, and the «Step by step» repeats these calculations.

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Example 2Stresses on a rotated element

With

σ_{x} = 60

MPa,

σ_{y} = 20

MPa and

τ_{x y} = - 15

MPa, find the stresses on an element rotated 30° counter-clockwise.

Solution

$C = 40$ MPa and $\frac{σ _{x} - σ _{y}}{2} = 20$ MPa. With $2 θ = 60°$ : $cos 60° = 0.5$ and $sin 60° = 0.866$ .
$σ_{x^{'}} = 40 + 20 \cdot 0.5 + (- 15) \cdot 0.866 = 37.01$ MPa.
$τ_{x^{'} y^{'}} = - 20 \cdot 0.866 + (- 15) \cdot 0.5 = - 24.82$ MPa.
$σ_{y^{'}} = 60 + 20 - 37.01 = 42.99$ MPa. Invariant check: $37.01 + 42.99 = 80 = 60 + 20$ .

σ_{x^{'}} = 37.01

MPa,

σ_{y^{'}} = 42.99

MPa and

τ_{x^{'} y^{'}} = - 24.82

MPa.

Solve it in Simuveliq

Mode Element → circle with σx = 60, σy = 20 and τxy = −15.
In Rotated element, Angle θ (counter-clockwise +) = 30. Point X′ of the circle is 60° from X, and the «Rotated θ = 30°» element shows σx′ = 37, σy′ = 43 and τx′y′ = −24.8.
The last block of the «Step by step» has the same equations with four digits. Try dragging X′ around the circle and watch the element rotate.

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Example 3Pressure vessel: absolute maximum shear

In the wall of a cylindrical vessel,

σ_{x} = 80

MPa (hoop),

σ_{y} = 40

MPa (longitudinal) and

τ_{x y} = 0

. Find the maximum in-plane shear and the absolute one.

Solution

With no shear, the $x$ , $y$ axes are already principal: $σ_{1} = 80$ and $σ_{2} = 40$ MPa.
In the plane: $τ_{max} = R = \frac{80 - 40}{2} = 20$ MPa.
With $σ_{3} = 0$ (the outer surface is free): $τ_{abs max} = \frac{80 - 0}{2} = 40$ MPa, twice as much, and it acts on a plane at 45° out of the wall's plane.

In-plane

τ_{max} = 20

MPa; absolute

τ_{max} = 40

MPa.

Solve it in Simuveliq

Mode Element → circle with σx = 80, σy = 40 and τxy = 0.
Tick Include σz = 0 (three circles, absolute maximum τ): the three circles appear and the Absolute maximum shear card shows 40 MPa, next to Maximum in-plane shear with 20 MPa.

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Example 4From the circle to the element

It is known that

σ_{1} = 100

MPa,

σ_{2} = - 20

MPa and that the plane of

σ_{1}

is at

θ_{p} = 20°

(counter-clockwise) from the

x

axis. Find

σ_{x}

σ_{y}

and

τ_{x y}

Solution

$C = \frac{100 + ( - 20 )}{2} = 40$ MPa and $R = \frac{100 - ( - 20 )}{2} = 60$ MPa.
Point $X$ is $2 θ_{p} = 40°$ from $σ_{1}$ (the reverse of the rotation from $X$ to $σ_{1}$ ): $σ_{x} = C + R cos 40° = 40 + 45.96 = 85.96$ MPa and $τ_{x y} = R sin 40° = 38.57$ MPa.
$Y$ is at the other end of the diameter: $σ_{y} = C - R cos 40° = - 5.96$ MPa.
Check: $tan 2 θ_{p} = \frac{2 \cdot 38.57}{85.96 + 5.96} = 0.839$ , so $2 θ_{p} = 40°$ .

σ_{x} = 85.96

MPa,

σ_{y} = - 5.96

MPa and

τ_{x y} = 38.57

MPa.

Solve it in Simuveliq

Mode Circle → element; in Define by, «σ₁, σ₂ and θp»: σ₁ = 100, σ₂ = −20 and θp = 20.
The Result: the element that produces this circle card, right below the circle, gives σx = 85.96, σy = −5.963 and τxy = 38.57 MPa, with the element drawn.

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Mohr's circle and stress transformation

Plane stress and sign convention

Transformation equations

Principal stresses and maximum shear

Mohr's circle

Absolute maximum shear (three circles)

Common mistakes

Worked examples

Example 1Principal stresses and maximum shear

Solution

Solve it in Simuveliq

Example 2Stresses on a rotated element

Solution

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Example 3Pressure vessel: absolute maximum shear

Solution

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Example 4From the circle to the element

Solution

Solve it in Simuveliq

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