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Mechanical

Principal angle in plane stress

Angle from the x axis to the direction of the larger in-plane principal stress, positive towards y.

Ref: MEC-53
Rev: v1
Field: Mechanical
Result: θ_p [°]

θ_{p} = \frac{1}{2} atan2 (2 τ_{xy}, σ_{x} - σ_{y})

θ_p = ½ atan2(2τ_xy, σ_x − σ_y) is the angle from the x axis, positive towards y, to the direction of the larger in-plane principal stress σ_a. The smaller one, σ_b, acts 90° further round. The two-argument arctangent is needed because it keeps track of the quadrant: the plain arctangent of the ratio would point to the smaller stress whenever σ_x is less than σ_y. The angle is undefined when the circle has zero radius (σ_x = σ_y and τ_xy = 0), because every direction is then principal. The result lies between −90° and +90°.

Angle from the x axis to the direction of the larger in-plane principal stress, positive towards y. On the planes at this angle and 90° beyond it the shear stress is zero.

Calculator

Solve for

θ_pPrincipal angleComputed

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

Angle of the smaller principal stress, θ_p + 90°116.6 °
Larger in-plane principal stress σ_a70 MPa
Shear on the plane at θ_p (should be zero)0 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.

Principal angle (°) current value 26.57 °

τ_xyShear stress on the x face in the y directionKey driver

−39.339.3

σ_yNormal stress in y (tension positive)

8.8781.1

σ_xNormal stress in x (tension positive)

7.2478.6

The lecture state gives 26.57° (half the angle whose tangent is 80/60). At that angle the shear on the face is 0 MPa and the normal stress is the larger principal stress, 70 MPa.
Reversing the sign of the shear to −40 MPa reverses the angle to −26.57°, but leaves the principal stresses unchanged. This is why the shear needs a negative range for directions, but not for the principal values or the yield criteria.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (°)	Local step	Change in result
τ_xy Shear stress on the x face in the y direction	−150 to 150 MPa	−39.35	39.35	78.69	+6 MPa	+1.88 °
σ_y Normal stress in y (tension positive)	−200 to 300 MPa	8.872	81.13	72.26	+10 MPa	+2.43 °
σ_x Normal stress in x (tension positive)	−200 to 300 MPa	7.235	78.58	71.35	+10 MPa	−2.16 °

Sweep of Shear stress on the x face in the y direction

Shear stress on the x face in the y direction (MPa)	Principal angle (°)
−150	−39.35
−112.5	−37.53
−75	−34.1
−37.5	−25.67
0	0
37.5	25.67
75	34.1
112.5	37.53
150	39.35

Sweep one variable

Vary

Principal angle (°)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

θ_p

Principal angle

Result · °

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. Turns the principal direction towards the x axis when it is raised above σ_y, because it increases the difference σ_x − σ_y. Raising it by 10 MPa (to 60 MPa) moves the result by −8.12 %, from 26.57 to 24.41 °.

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. Turns the principal direction away from the x axis when it is raised, because it reduces the difference σ_x − σ_y. Raising it by 10 MPa (to 0 MPa) moves the result by +9.16 %, from 26.57 to 29 °.

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. Turns the principal direction towards 45°; reversing the sign of the shear reverses the angle. Raising it by 10 MPa (to 50 MPa) moves the result by +11.1 %, from 26.57 to 29.52 °.

Worked example with the values above

Arguments: 2τ_xy = 80 MPa and σ_x − σ_y = 60 MPa.
atan2(80, 60) = 53.13°.
Halve it: θ_p = 26.57° from the x axis towards y.

Assumptions and limits

Plane stress, with the angle measured from the x axis towards the y axis.
A stress state whose circle has a non-zero radius.

Common mistakes

Using ½ arctan(2τ_xy/(σ_x − σ_y)), which can point to the smaller principal stress when σ_x is less than σ_y.
Entering the arguments the wrong way round in a spreadsheet, where the order is reversed compared with the mathematical convention.

Statistics and uncertainty

The angle is sensitive to small errors when the two normal stresses are nearly equal and the shear is small. Raising σ_x by only 1 MPa here moves the angle by 0.23°, but near σ_x = σ_y with a small shear the same 1 MPa could move it by many degrees. A principal direction from a near-circular stress state should not be trusted.

Where it is used

Orienting a strain-gauge rosette or a reinforcement along the principal direction.
Setting the rotation at which to evaluate the principal stresses in a rotated element.

Maths you need

atan2: the angle of a point, with its quadrant resolved
Double-angle form: the circle angle is twice the element angle

Related formulas

Practise this formula