In-Plane Principal Stresses

σ_{1, 2} = \frac{σ _{x}}{2} \pm (\frac{σ _{x}}{2})^{2} + τ_{x y}^{2} (σ_{y} = 0); σ_{1, 2} = \frac{σ _{x} + σ _{y}}{2} \pm (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}

Principal stresses are the eigenvalues of the plane-stress tensor; shear is zero on principal planes. When σy = 0 the general Mohr formula collapses to σx/2 ± sqrt((σx/2)²+τxy²). The in-plane τ_max is the Mohr radius.

$σ_{x}, σ_{y}$ (Pa): Normal stresses on x,y faces
$τ_{x y}$ (Pa): In-plane shear
$σ_{1, 2}$ (Pa): Algebraically max/min principal stress

Step-by-step solved example

σx = 80 MPa, σy = 0, τxy = 30 MPa. Find σ1 and σ2.

Mohr's circle: center at σavg, radius R. σ1,2 at the σ-axis intercepts.

1. Center and radius
σ_avg = 40 MPa. R = √(40²+30²) = 50 MPa.
$R = 50 MPa$
2. Principals
σ1 = 40+50 = 90 MPa, σ2 = 40−50 = −10 MPa.
$σ_{1} = 90, σ_{2} = - 10 MPa$

Answer: σ1 = 90 MPa, σ2 = −10 MPa

10 practice questions

0/10 correct

1.σx = 80 MPa, σy = 0, τ = 30 MPa. σ1 =

2.Mohr’s-circle center is at

3.Principal planes have

4.If σx = σy and τxy = 0, the principals are

5.In-plane τ_max equals the Mohr radius

6.σx = 50 MPa, σy = 0, τ = 0. Principals are

7.tan 2θp = 2τxy / (σx − σy) locates

8.Uniaxial tension σ (other stresses zero): principals are

9.σx = σy = 0, τxy = 40 MPa. σ1,2 =

10.Mohr’s-circle radius is also