Combined stress and Mohr's circle

Stresses at an angle

• A bracket point carries $\sigma_x = 100$, $\sigma_y = 40$, $\tau_{xy} = 25$ MPa - but a weld or gauge plane sits at an angle. Mohr's circle 莫尔圆 draws the whole state: centre $\sigma_{avg} = (\sigma_x + \sigma_y)/2$, radius $R = \sqrt{((\sigma_x - \sigma_y)/2)^2 + \tau_{xy}^2}$.

The principals

• For the point above: centre $\mathbf{70}$, $R = \sqrt{30^2 + 25^2} = \sqrt{1525} = \mathbf{39.1}$ MPa, so the principal stresses 主应力 $\sigma_1 = 70 + 39.1 = \mathbf{109.1}$ and $\sigma_2 = 70 - 39.1 = \mathbf{30.9}$ MPa sit at $\theta_p = \tan^{-1}(50/60)/2 = \mathbf{19.9°}$.
• The circle turns $2\theta$ as the element turns $\theta$ - halve carefully.

Which shear is maximum

• Both principals positive: the absolute-maximum shear is $\sigma_1/2 \approx \mathbf{54.5}$ MPa.
• Swap the second normal stress to $-40$ MPa: centre 40, $R = \sqrt{80^2 + 30^2} = \mathbf{85.4}$, so $\sigma_1 = \mathbf{125.4}$ and $\sigma_2 = \mathbf{-45.4}$ MPa - now the principals straddle zero and the absolute-maximum shear IS the radius, $\mathbf{85.4}$ MPa ✓.

Three rotation slips

Recap