Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2017 — 04-BS-6: Mechanics of Materials (3 hours, closed book, one hand-written aid sheet permitted). Any FIVE of the eight questions constitute a complete paper on the official exam; every question is answered below. A wide-flange (W-shape) section-property table is attached at the end of the official exam; every question below supplies its own built-up or standard section directly, so the table is not needed for any of the eight solutions.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (axial members with initial gaps, shear/moment diagrams, deflection by direct integration, transformation of stress via Mohr's circle, combined axial+bending on eccentrically-loaded columns, torsion of stepped shafts, Euler column buckling, unsymmetric-section flexure of brittle materials).
Question 4: Mohr's Circle for Plane Stress (20 marks)
Given. A plane-stress element with σx=−85 MPa, σy=−5 MPa (both compressive, arrows into the element), τxy=+20 MPa (standard positive-shear picture), and an inclined plane 15° from the horizontal.
Quantity
Value
σx
−85 MPa
σy
−5 MPa
τxy
+20 MPa
Inclined plane
15° from horizontal (normal at 75° from +x)
Find. Stresses on the 15° inclined plane and the maximum in-plane shear stress with orientation.
[Figure not reproduced: Stress element as printed: 5 MPa and 85 MPa normal stresses into every face (compressive), 20 MPa shear in the standard positive sense, and the 15° inclined cutting plane. See the official exam paper.]
Approach. Plot the center and radius of Mohr's circle from σx, σy, τxy, then read the principal stresses, the stress on the specified inclined plane, and the maximum shear directly off the circle geometry.
Plot the circle. With $\sigma_x=-85$, $\sigma_y=-5$, $\tau_{xy}=+20$ MPa (all read from the printed element: normal arrows point INTO the element on every face → both compressive; the shear arrows follow the standard "positive" picture, right face up / top face right):
$$\sigma_{avg}=\frac{\sigma_x+\sigma_y}{2}=\boxed{-45.0\text{ MPa}},\quad R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}=\boxed{44.72\text{ MPa}}$$
Principal stresses.
$$\sigma_1=\sigma_{avg}+R=\boxed{-0.279\text{ MPa}},\qquad \sigma_2=\sigma_{avg}-R=\boxed{-89.721\text{ MPa}}$$
at $2\theta_p=\arctan\!\left(\dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}\right)$, giving $\theta_p=76.72^\circ$ (measured CCW from the x-face to $\sigma_1$'s plane).
(a) Stress on the 15° inclined plane. The incline runs from the top-left corner down to the right edge, 15° below the horizontal top face — so its outward normal is 15° CLOCKWISE from the top face's own normal (+y, i.e. 90°), placing it at $\theta=90-15=75^\circ$ from the +x axis (NOT simply “15°” — the angle must be measured to the PLANE'S NORMAL, not to the face the figure happens to dimension from). Rotating the circle by $2\theta=150^\circ$ from the X-point:
$$\boxed{\sigma_n=-0.359\text{ MPa}},\qquad \boxed{\tau=2.679\text{ MPa}}$$
(the complementary face carries $\sigma_{n}'=-89.641$ MPa, by $\sigma_x+\sigma_y=\sigma_n+\sigma_n'$).
(b) Maximum in-plane shear. Directly off the circle,
$$\boxed{\tau_{max}=R=44.721\text{ MPa}}\ \text{at }\ \theta_s=\theta_p-45^\circ=31.72^\circ,\quad \sigma_{on\ these\ planes}=\sigma_{avg}=-45.0\text{ MPa}$$
Mohr's circle: center −45 MPa, radius 44.72 MPa, with the X and Y points and the principal stresses marked.
