124/3337–11FE Reference Handbook 10.4 · Mechanics of Materials · Mohr / Failure

Handbook formula

Maximum Shear (Tresca) Stress

Tresca: yield when τ_max ≥ Sy/2. In plane stress, remember σ3 might be 0 — the largest of |σ1|, |σ2|, |σ1−σ2| decides the Mohr diameter.

In-plane or absolute max shear
Mohr radius

Step-by-step solved example

σx=40, σy=−20, τ=0 MPa. τ_max (absolute, plane stress)?

στσ₂σ₁C
τ_max is the Mohr radius (absolute max may use σ=0).
  1. 1. Principals

    σ1=40, σ2=−20, σ3=0.

  2. 2. Max shear

    (σ1−σ2)/2=30 MPa (larger than 20 or 10).

Answer: τ_max = 30 MPa

10 practice questions

0/10 correct

1.In-plane Mohr radius is

2.Tresca yield in uniaxial tension at Sy means τ_max =

3.Pure shear τ yields (Tresca) when τ =

4.Center of Mohr’s circle is

5.If σx=σy and τ=0, in-plane τ_max is

6.Principal orientation tan 2θp =

7.Absolute max shear in plane stress can use σ=0 as a principal

8.σx=100, σy=0, τ=0: R =

9.Sign convention on FE Mohr: +τ

10.Tresca hexagon vs von Mises circle in π-plane: Tresca is