Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2016 — 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 section, so the table is not needed for any of the eight solutions.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (shear/moment diagrams, deflection by integration, combined axial+bending, transformation of stress via Mohr's circle, torsion of stepped shafts, axial members and pin connections, Euler column buckling).
Question 4: Mohr's Circle for Plane Stress (20 marks)
Given. 100 mm × 100 mm × 2 mm steel plate. Reading the edge forces off the figure (each face carries a normal and a shear force; complementary faces carry equal-magnitude, opposite-sense forces): 24 kN normal (tension, y-face), 4 kN normal (tension, x-face), 12 kN shear on every face, oriented so that τxy is negative by the standard sign convention.
Quantity
Value
Plate
100 mm × 100 mm × 2 mm (edge area = 200 mm²)
Normal force, y-faces
24 kN (tension)
Normal force, x-faces
4 kN (tension)
Shear force, all faces
12 kN
Find. Principal stresses σ1, σ2 and their plane orientation; maximum in-plane shear stress τmax, its associated normal stress, and its plane orientation — via Mohr's circle.
Approach. Convert the given edge forces to stresses by dividing by the 100×2 mm edge area, plot the stress state on a Mohr's circle (center, radius, points X and Y), then read the principal and maximum-shear states directly off the circle's geometry.
Stresses from the given forces. Each force acts on a 100×2 = 200 mm² edge:
$$\sigma_x=\frac{4000}{200}=20\text{ MPa (T)}, \quad \sigma_y=\frac{24000}{200}=120\text{ MPa (T)}, \quad \tau_{xy}=-\frac{12000}{200}=-60\text{ MPa}$$
Circle center and radius.
$$\sigma_{avg}=\frac{\sigma_x+\sigma_y}{2}=\frac{20+120}{2}=70\text{ MPa}$$
$$R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}}=\sqrt{(-50)^2+(-60)^2}=\boxed{78.1\text{ MPa}}$$
Principal stresses. Reading the circle's left/right extremes,
$$\sigma_{1,2}=\sigma_{avg}\pm R = 70 \pm 78.1$$
$$\boxed{\sigma_1=148.1\text{ MPa (T)}, \qquad \sigma_2=-8.1\text{ MPa (C)}}$$
The rotation angle from the X-face to the $\sigma_1$ plane, measured on the circle as $2\theta_p=\tan^{-1}\!\big(\tau_{xy}/[(\sigma_x-\sigma_y)/2]\big)$, gives $\boxed{\theta_p=-64.9^{\circ}}$ (physical angle, measured from the x-axis to the $\sigma_1$ direction, positive counterclockwise).
Maximum in-plane shear. The top of the circle gives $\tau_{max}=R=\boxed{78.1\text{ MPa}}$, with associated normal stress equal to the circle center, $\sigma'=\boxed{70\text{ MPa}}$ on both faces of that element. The maximum-shear planes sit $45^{\circ}$ from the principal planes: $\boxed{\theta_s=-199.9^{\circ}}$.
Given stress element (left) and the Mohr's circle construction (right): center C=70 MPa, radius R=78.1 MPa, with points X and Y, the principal points σ1/σ2, and the top-of-circle τmax point all marked.
Quantity
Value
σ1
148.1 MPa (T)
σ2
−8.1 MPa (C)
θp (to σ1)
-64.9°
τmax
78.1 MPa
σ on τmax planes
70 MPa
θs (to τmax plane)
-199.9°
Check: edge-force directions were read from the printed figure (arrowheads on all four edges confirmed complementary/consistent); both normal forces act in tension and the shear sense gives τxy<0 by the standard x-face/+y-direction sign convention.
