Question 3 of 8: Mohr's Circle for a Welded Axial Joint

Question 3: Mohr's Circle for a Welded Axial Joint (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.

Find. (a) σ, τ on the weld; (b) the maximum in-plane shear stress and its associated normal stress; (c) alternative solution methods.

Approach. This is a UNIAXIAL stress state ($\sigma_y=\tau_{xy}=0$) since the plate carries only the axial load. Plot the two known points on Mohr's circle, draw the circle through them, then read every requested quantity as a point on that circle located by trigonometry.

1. Axial stress and circle parameters. $$\sigma_x = \frac{P}{A} = \frac{960{,}000}{320(25)} = \frac{960{,}000}{8000} = 120\ \text{MPa}\ (\sigma_y=0,\ \tau_{xy}=0)$$ Point A (x-face) plots at (120, 0); point B (y-face) plots at (0, 0) — the circle passes through the origin, centred at $$C = \frac{\sigma_x+\sigma_y}{2} = 60\ \text{MPa}, \qquad R = \frac{\sigma_x-\sigma_y}{2} = \boxed{60\ \text{MPa}}$$
2. (a) Locate the weld plane on the circle. The weld LINE makes 20° with the horizontal (load) axis, so its outward NORMAL is rotated $\theta_n = 90^\circ-20^\circ = 70^\circ$ from the x-axis. On Mohr's circle this is a rotation of $2\theta_n=140^\circ$ from point A, read by trigonometry: $$\sigma_n = C + R\cos(2\theta_n) = 60+60\cos(140^\circ) = \boxed{14.0\ \text{MPa (tension)}}$$ $$\tau_n = R\sin(2\theta_n) = 60\sin(140^\circ) = \boxed{38.6\ \text{MPa}}$$ (cross-check with the transformation equations at $\theta_n=70^\circ$, as the question permits only for verification, reproduces the same two numbers).
3. (b) Maximum in-plane shear. The top and bottom of the circle give $$\boxed{\tau_{max} = R = 60\ \text{MPa}}, \qquad \sigma_{avg} = C = \boxed{60\ \text{MPa (on both faces of that element)}}$$ occurring on planes 45° from the load axis (since the circle's centre sits at $\sigma_x/2$, the max-shear planes always bisect the axial and transverse directions for a uniaxial state).
4. (c) Alternative methods. (i) The stress transformation equations, evaluated directly at $\theta=70^\circ$ (used only as the permitted check above). (ii) A direct wedge (free-body) method: cut a triangular wedge along the weld line, resolve the 960 kN axial force into components normal and tangential to the cut face, and divide each by the weld's own inclined area $A/\sin(20^\circ)$ — this reproduces the same $\sigma_n,\tau_n$ without any transformation formula at all.

Final Results.