Mohr's Circle: Principal Stresses and Maximum Shear: 04-BS-6, May 2018, Q4

Question 4: Mohr's Circle: Principal Stresses and Maximum Shear (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Plane-stress element: top face normal stress $70\text{ MPa}$ compressive ($\sigma_y=-70\text{ MPa}$); right face normal stress $10\text{ MPa}$ tensile ($\sigma_x=10\text{ MPa}$); shear on the right face $30\text{ MPa}$ ($\tau_{xy}=30\text{ MPa}$).

Find. The principal stresses $\sigma_1,\sigma_2$ and their plane orientation; the maximum in-plane shear stress $\tau_{\max}$, its associated normal stress, and its plane orientation.

Plane-stress element: 70 MPa compression (top), 10 MPa tension (right), 30 MPa shear (right face).

Mohr's circle: centre -30 MPa, radius 50 MPa, principal stresses +20/-80 MPa.

Approach. Plot the two known stress states as diametrically-opposite points on the circle, read the centre and radius from the circle's geometry, then read off the principal and maximum-shear states directly.

Plot the circle. Centre $C=\dfrac{\sigma_x+\sigma_y}{2}=\dfrac{10-70}{2}=-30\text{ MPa}$. Radius $R=\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}=\sqrt{40^2+30^2}=\boxed{50\text{ MPa}}$, giving the point $X(\sigma_x,\tau_{xy})=(10,30)$ and $Y(\sigma_y,-\tau_{xy})=(-70,-30)$, which lie diametrically opposite on the circle as required.
Principal stresses. $\sigma_1=C+R=-30+50=\boxed{20\text{ MPa}}$; $\sigma_2=C-R=-30-50=\boxed{-80\text{ MPa}}$.
Principal plane orientation. $\tan(2\theta_p)=\dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}=\dfrac{60}{80}=0.75\Rightarrow2\theta_p=36.87^{\circ}\Rightarrow\theta_p=\boxed{18.43^{\circ}}$, measured CCW on the circle from point X. Substituting back into the transformation formula at this angle reproduces $\sigma_{x^{\prime}}=20\text{ MPa}=\sigma_1$, confirming the rotation sense.
Maximum in-plane shear. $\tau_{\max}=R=\boxed{50\text{ MPa}}$, on planes at $45^{\circ}$ to the principal planes (i.e. at $\theta_p+45^{\circ}=63.43^{\circ}$ from the x-axis); the associated normal stress on those planes is the circle centre, $\sigma_{\text{avg}}=\boxed{-30\text{ MPa}}$ on each face.

Final results
Quantity	Value
σ₁	20 MPa
σ₂	-80 MPa
θ_p (to σ₁)	18.43° CCW from the x-face
τ_max	50 MPa
σ on max-shear planes	-30 MPa
Max-shear plane angle	63.43° from the x-axis

Topic: Mohr's circle for plane stress

Key relations: Centre $C=(\sigma_x+\sigma_y)/2$; radius $R=\sqrt{((\sigma_x-\sigma_y)/2)^2+\tau_{xy}^2}$; $\sigma_{1,2}=C\pm R$; $\tau_{\max}=R$ on planes $45^{\circ}$ from the principal planes.

Why this works: Mohr's circle is a graphical restatement of the stress-transformation equations: every point on the circle represents the normal/shear stress pair on SOME plane through the point, so the extreme points on the horizontal axis (zero shear) are automatically the principal stresses, and the topmost/bottommost points are automatically the maximum shear.

Common pitfall: Mixing up the sign convention for plotting the shear coordinate of the X and Y points (they must be plotted with OPPOSITE signs of $\tau$, since they sit at opposite ends of a diameter) is the most common construction error; verifying that $X$ and $Y$ are diametrically opposite is a cheap self-check.

Source: Hibbeler, Mechanics of Materials, 10th ed., §9.4 Mohr's Circle: Plane Stress

Question 4 of 8: Mohr's Circle: Principal Stresses and Maximum Shear

Question 4: Mohr's Circle: Principal Stresses and Maximum Shear (20 marks)