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Mechanical

Shear stress on a rotated element

Shear stress on the rotated x′ face.

Ref: MEC-52
Rev: v1
Field: Mechanical
Result: τ_x′y′ [MPa]

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

τ_x′y′ is the shear stress on the rotated x′ face. With the tensor convention used here, a positive value acts in the +y′ direction on the +x′ face. The shear changes sign as the element turns and is exactly zero on the two principal planes, which is the definition of the principal directions. It reaches its largest size, the circle radius R, on planes 45° away from them. Do not read the sign from clockwise arrows on a Mohr diagram; use this tensor convention.

Shear stress on the rotated x′ face. It changes sign as the element turns and is exactly zero on the principal planes, which is how the principal directions are defined.

Calculator

Solve for

τ_x′y′Shear 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

Shear at the principal angle (should be zero)0 MPa
Largest shear in the plane, R50 MPa
Principal angle θ_p26.57 °

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.

Shear stress on the x′ face (MPa) current value −5.981 MPa

σ_xNormal stress in x (tension positive)Key driver

−114102

σ_yNormal stress in y (tension positive)

−88.3128

τ_xyShear stress on the x face in the y direction

−10149

θRotation of the x axis towards y

−5050

At 30° the shear is −5.981 MPa, nearly zero because 30° is close to the principal angle 26.57°. At exactly 26.57° it vanishes. It reaches +50 MPa at −18.43° and −50 MPa at 71.57°, the circle radius in size.
At θ = 0 the shear is τ_xy = 40 MPa and at θ = 90° it is −τ_xy. This is the complementary shear pairing of the original faces.

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	−114.2	102.3	216.5	+10 MPa	+72.4 %
σ_y Normal stress in y (tension positive)	−200 to 300 MPa	−88.25	128.3	216.5	+10 MPa	−72.4 %
τ_xy Shear stress on the x face in the y direction	−150 to 150 MPa	−101	49.02	150	+6 MPa	−50.2 %
θ Rotation of the x axis towards y	−90 to 90 °	−49.99	49.99	99.99	+3.6 °	+103 %

Sweep of Rotation of the x axis towards y

Rotation of the x axis towards y (°)	Shear stress on the x′ face (MPa)
−90	−40
−67.5	−7.071
−45	30
−22.5	49.5
0	40
22.5	7.071
45	−30
67.5	−49.5
90	−40

Sweep one variable

Vary

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

Variables and how each one contributes

τ_x′y′

Shear 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. Changes the shear by −½ sin 2θ times the change in σ_x, so a rise makes the shear more negative for angles between 0° and 90°. Raising it by 10 MPa (to 60 MPa) moves the result by +72.4 %, from −5.981 to −10.31 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. Changes the shear by +½ sin 2θ times the change in σ_y, the opposite of σ_x. Raising it by 10 MPa (to 0 MPa) moves the result by −72.4 %, from −5.981 to −1.651 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. Changes the shear by cos 2θ times the change in τ_xy: the full amount at θ = 0 and none at 45°. Raising it by 10 MPa (to 50 MPa) moves the result by −83.6 %, from −5.981 to −0.9808 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°. Swings the shear through zero at the principal angles and to its extreme values at 45° from them. Raising it by 10° (to 40°) moves the result by +278 %, from −5.981 to −22.6 MPa.

Worked example with the values above

Half-difference: (σ_x − σ_y)/2 = 30 MPa, and 2θ = 60°.
Terms: −30 × sin 60° + 40 × cos 60° = −25.98 + 20 MPa.
τ_x′y′ = −5.981 MPa.

Assumptions and limits

Plane stress at a point, tensor sign convention.
The angle is measured from x towards y, with the same convention as the normal stresses.

Common mistakes

Giving the shear the wrong sign after a rotation. Reverse the angle and the shear reverses sign.
Using the clockwise-positive convention from a textbook Mohr’s circle drawing with the tensor formulas.

Statistics and uncertainty

Near a principal direction the shear changes by about twice the circle radius per radian, so an alignment error of 1° (0.0175 rad) changes it by about 2R × 0.0175, roughly 3.5 % of R. Shear is therefore the quantity most sensitive to gauge or cut misalignment, which is why a shear reading at 45° is often used to check alignment.

Where it is used

Shear on a weld, glue line or inclined failure plane.
Verifying that the principal planes carry no shear.

Maths you need

Double-angle form: cos 2θ and sin 2θ
Complementary shear: equal shear on perpendicular faces

Related formulas

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