Question 4 of 8: Mohr's Circle for Stresses Parallel/Perpendicular to Wood Grain

Question 4: Mohr's Circle for Stresses Parallel/Perpendicular to Wood Grain (7+4+7+2 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: vertical stress $1.8\text{ MPa}$ COMPRESSIVE (both arrows point INTO the top/bottom faces; the printed element shows both normal stresses compressive). Horizontal stress $4.2\text{ MPa}$ COMPRESSIVE (both arrows point INTO the left/right faces). No shear stress is marked on the figure $\Rightarrow\tau_{xy}=0$. Grain at $\theta=20^{\circ}$ from vertical.

Find. (1) Normal stress perpendicular to grain, shear stress parallel to grain; (2) Mohr's circle; (3) maximum shear stress, its normal stress, and orientation; (4) why these directional stresses matter for wood.

Approach. With $\tau_{xy}=0$, the given $x$/$y$ faces are already the principal planes, so the circle is drawn directly from $\sigma_x,\sigma_y$; the grain-perpendicular plane is reached by rotating $2\theta=40^{\circ}$ around the circle from point $A(\sigma_x,0)$.

1. Circle centre and radius. $C=\dfrac{\sigma_x+\sigma_y}{2}=\dfrac{-4.2-1.8}{2}=\boxed{-3.0\text{ MPa}}$; $R=\left|\dfrac{\sigma_x-\sigma_y}{2}\right|=\left|\dfrac{-4.2+1.8}{2}\right|=\boxed{1.2\text{ MPa}}$. Since $\tau_{xy}=0$, points $A(-4.2,0)$ and $B(-1.8,0)$ are the circle's own left/right extremes — i.e. $\sigma_x,\sigma_y$ ARE the principal stresses: $\sigma_1=-1.8\text{ MPa}$, $\sigma_2=-4.2\text{ MPa}$.
2. Part (1) — rotate to the grain-normal plane. The normal to the grain lies $\theta=20^{\circ}$ from the $x$-axis (grain $20^{\circ}$ from vertical $\Rightarrow$ its normal is $20^{\circ}$ from horizontal); on Mohr's circle this is a rotation of $2\theta=40^{\circ}$ from point $A$: $\sigma_{\perp}=\sigma_x\cos^2\theta+\sigma_y\sin^2\theta=(-4.2)(0.9397)^2+(-1.8)(0.3420)^2=\boxed{-3.92\text{ MPa}}$ (compression); $\tau_{\parallel}=(\sigma_x-\sigma_y)\sin\theta\cos\theta=(-2.4)(0.3420)(0.9397)=\boxed{0.771\text{ MPa}}$ (magnitude; sense shown on the stress element).
3. Part (2) — plot. The grain-normal point sits at $(-3.92,\mp0.77)$ on the circle, $40^{\circ}$ around from $A$, as drawn.
4. Part (3) — maximum shear. $\tau_{max}=R=\boxed{1.2\text{ MPa}}$, occurring on planes at $45^{\circ}$ from the principal ($x,y$) directions, with corresponding normal stress equal to the centre, $\sigma=\boxed{-3.0\text{ MPa}}$ (both faces of the max-shear element).
5. Part (4) — why grain direction matters. Wood is a strongly anisotropic (orthotropic) material: its shear strength PARALLEL to the grain is far below its strength across the grain, so the member can fail in glue-line/grain shear at a stress well under the strength of the wood fibres themselves. Reporting stress only in the arbitrary $x,y$ directions would miss the actual failure plane the timber is weakest along.