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Mechanical

Normal stress on a rotated element, x′ face

Normal stress on a face turned by an angle θ from the x axis, found from the stresses on the original faces.

Ref: MEC-50
Rev: v1
Field: Mechanical
Result: σ_x′ [MPa]

σ_{x′} = C + \frac{σ_{x} - σ_{y}}{2} \cos 2 θ + τ_{xy} \sin 2 θ C = \frac{σ_{x} + σ_{y}}{2}

The element is turned through θ from the x axis towards the y axis, and σ_x′ is the normal stress on its new x′ face. Written about the circle centre C, the stress swings by (σ_x − σ_y)/2 · cos 2θ + τ_xy sin 2θ. The angle appears doubled, which is why a point on Mohr’s circle moves through 2θ when the element turns through θ. At θ = 0 the result is σ_x, and at θ = 90° it is σ_y. Tensor sign convention: positive τ_xy acts in the +y direction on the +x face.

Normal stress on a face turned by an angle θ from the x axis, found from the stresses on the original faces. The same state of stress gives a different normal stress on every plane through the point.

Calculator

Solve for

σ_x′Normal stress on the x′ faceComputed

MPa

σ_xNormal stress in x (tension positive)

MPa

−200300

σ_yNormal stress in y (tension positive)

MPa

−200300

τ_xyShear stress on the x face in the y direction

MPa

−150150

θRotation of the x axis towards y

−9090

Perpendicular face σ_y′−29.64 MPa
Shear stress τ_x′y′ on the same face−5.981 MPa
Sum σ_x′ + σ_y′ (equals σ_x + σ_y)40 MPa

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Normal stress on the x′ face (MPa) current value 69.64 MPa

σ_xNormal stress in x (tension positive)Key driver

−118257

τ_xyShear stress on the x face in the y direction

−94.9165

σ_yNormal stress in y (tension positive)

22.1147

θRotation of the x axis towards y

−29.869.8

At 30° the x′ face carries 69.64 MPa, against 50 MPa on the original x face. The most it can reach over all angles is the larger principal stress, 70 MPa, found at 26.57°; the least is −30 MPa, 90° further round.
The stress on the x′ face and the stress on the y′ face always add up to σ_x + σ_y = 40 MPa, so this entry and the next one together can be checked by addition.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (MPa)	Local step	Change in result
σ_x Normal stress in x (tension positive)	−200 to 300 MPa	−117.9	257.1	375	+10 MPa	+10.8 %
τ_xy Shear stress on the x face in the y direction	−150 to 150 MPa	−94.9	164.9	259.8	+6 MPa	+7.46 %
σ_y Normal stress in y (tension positive)	−200 to 300 MPa	22.14	147.1	125	+10 MPa	+3.59 %
θ Rotation of the x axis towards y	−90 to 90 °	−29.8	69.8	99.6	+3.6 °	−1.64 %

Sweep of Rotation of the x axis towards y

Rotation of the x axis towards y (°)	Normal stress on the x′ face (MPa)
−90	−10
−67.5	−29.5
−45	−20
−22.5	12.93
0	50
22.5	69.5
45	60
67.5	27.07
90	−10

Sweep one variable

Vary

Normal stress on the x′ face (MPa)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

σ_x′

Normal stress on the x′ face

Result · MPa

What the equation gives you. Everything else feeds it.

σ_x

Normal stress in x (tension positive)

MPa · typical −200 to 300

Normal stress on the x face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises σ_x′ by cos²θ times the change: the full amount at θ = 0 and none at θ = 90°. Raising it by 10 MPa (to 60 MPa) moves the result by +10.8 %, from 69.64 to 77.14 MPa.

σ_y

Normal stress in y (tension positive)

MPa · typical −200 to 300

Normal stress on the y face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises σ_x′ by sin²θ times the change: none at θ = 0 and the full amount at θ = 90°. Raising it by 10 MPa (to 0 MPa) moves the result by +3.59 %, from 69.64 to 72.14 MPa.

τ_xy

Shear stress on the x face in the y direction

MPa · typical −150 to 150

Shear stress on the x face acting in the +y direction, in MPa. The sign matters for directions and rotated stresses, so the slider runs from −150 to +150 MPa. Raises σ_x′ in proportion to sin 2θ, so it has most effect at 45° and none at 0° or 90°. Raising it by 10 MPa (to 50 MPa) moves the result by +12.4 %, from 69.64 to 78.3 MPa.

Rotation of the x axis towards y

° · typical −90 to 90

Angle through which the x axis is turned towards the y axis, in degrees. Positive is from x towards y, the opposite way for negative. The slider covers −90° to +90°, one half-turn, because the stresses repeat every 180°. Moves the face through the cycle of stresses; the slope here equals twice the shear on the face, 2τ_x′y′. Raising it by 10° (to 40°) moves the result by −7.24 %, from 69.64 to 64.6 MPa.

Worked example with the values above

Centre and half-difference: C = 20 MPa and (σ_x − σ_y)/2 = 30 MPa. The doubled angle is 2θ = 60°.
Swing: 30 × cos 60° + 40 × sin 60° = 30 × 0.5 + 40 × 0.866 = 49.64 MPa.
σ_x′ = 20 + (49.64) = 69.64 MPa.

Assumptions and limits

Plane stress at a point, with the stress components all in the same units and the same sign convention.
The angle is measured from x towards y. Reversing it, or using the opposite shear sign, changes the answer.

Common mistakes

Using θ instead of 2θ inside the sine and cosine.
Entering degrees into a calculator set to radians, or the other way round.

Statistics and uncertainty

The slope of σ_x′ with angle equals twice the shear stress on that face, so a gauge or cut misaligned by 1° changes the result by about 0.21 MPa at this angle. The slope is zero where the shear is zero, so aligning with a principal direction makes the normal stress least sensitive to angle error.

Where it is used

Resolving stress onto a weld line, glue line or strain-gauge direction.
Checking a principal-stress calculation by evaluating the stress at the principal angle.

Maths you need

Double-angle form: cos 2θ and sin 2θ
Mohr’s circle: centre and radius of the stress circle

Related formulas

Practise this formula