Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2013 — 04-BS-6 Mechanics of Materials. Three-hour, closed-book exam (one double-sided hand-written aid sheet permitted, approved Casio/Sharp calculator only). Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness. No credit is given in the original exam for solutions built by superposing existing standard solutions on Questions 1 and 6 — both are solved here by direct integration/equilibrium as required.
Reference texts: Hibbeler, Mechanics of Materials (10th ed., Pearson) — beam deflection by integration (Ch. 12), stress transformation and Mohr's circle (Ch. 9), columns and buckling (Ch. 13), combined axial & bending loading (Ch. 8), torsion of circular shafts (Ch. 5), transformed-section method for composite beams (Ch. 6), shear and moment diagrams (Ch. 6-7); Beer, Johnston & DeWolf, Mechanics of Materials (7th ed., McGraw-Hill) — cross-check reference for the same topics.
Question 2: Plane stress — Mohr's circle (20 marks)
Given. $\sigma_x=+20$ MPa (tension), $\sigma_y=-10$ MPa (compression), $\tau_{xy}=+20$ MPa (arrow senses on the four faces as sketched, complementary shear consistent on all faces).
Find. (a) $\sigma_n,\tau$ on the plane whose normal is 30° from horizontal beyond the diagonal cut shown; (b) $\tau_{max}$, the associated $\sigma_{avg}$, and the orientation of the maximum-shear planes.
Approach. Plot the center and radius of Mohr's circle from $\sigma_x,\sigma_y,\tau_{xy}$, then use the circle's geometry (trigonometry on the $2\theta$ central angle) to read off the requested points, checking against the transformation equations.
Circle center and radius.
$$\sigma_{avg}=\frac{\sigma_x+\sigma_y}{2}=\frac{20+(-10)}{2}=5\ \text{MPa},\qquad R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}=\sqrt{15^2+20^2}=25\ \text{MPa}$$
The circle is centered at $(5,0)$ MPa with radius 25 MPa; the point $X=(\sigma_x,\tau_{xy})=(20,20)$ and $Y=(\sigma_y,-\tau_{xy})=(-10,-20)$ are diametrically opposite, as required by the construction.
Principal stresses (for reference/orientation).
$$\sigma_1=\sigma_{avg}+R=30\ \text{MPa},\qquad\sigma_2=\sigma_{avg}-R=-20\ \text{MPa}$$
The central angle from $X$ to $\sigma_1$ is $2\theta_p=\tan^{-1}\!\left(\dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}\right)=\tan^{-1}\!\left(\dfrac{40}{30}\right)=53.13^\circ$, so $\theta_p=26.57^\circ$ CCW from the $x$-face to the $\sigma_1$-plane.
Part (a): stresses on the 30° plane. The cut shown is inclined 30° from the horizontal edge, so its outward normal is $30^\circ+90^\circ=120^\circ$ from the $x$-axis. On Mohr's circle this corresponds to a rotation of $2\theta=240^\circ$ from point $X$. Reading the circle geometrically (equivalently, evaluating the transformation equations at $\theta=120^\circ$ as a trigonometric check on the circle):
$$\sigma_n=\sigma_{avg}+R\cos(2\theta-2\theta_{p,X})=\boxed{-19.82\ \text{MPa (compression)}},\qquad \tau=\boxed{+2.99\ \text{MPa}}$$
This point sits close to $\sigma_2$ on the circle ($120^\circ$ is near the $\sigma_2$ orientation $\theta_p+90^\circ=116.6^\circ$), which is why the shear on this plane is small and the normal stress is close to $-20$ MPa.
Part (b): maximum in-plane shear. By construction, $\tau_{max}$ equals the circle's radius, occurring $90^\circ$ (i.e. $45^\circ$ on the element) from the principal planes:
$$\tau_{max}=\boxed{25\ \text{MPa}},\qquad \sigma_{avg}=5\ \text{MPa on both associated faces}$$
Orientation: $\theta_s=\theta_p-45^\circ=26.57^\circ-45^\circ=-18.43^\circ$ from the $x$-face (i.e. $18.43^\circ$ clockwise), with the companion maximum-shear plane at $\theta_s+90^\circ=71.57^\circ$.
