Question 1 of 6: Short-answer fundamentals

Question 1: Short-answer fundamentals (10 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.

(a) The 0.2% offset yield strength (proof stress). Ductile metals such as aluminium, copper and many stainless and heat-treated steels do not show a sharp, well-defined yield point on the stress–strain curve. $\sigma_{0.2}$ is the stress that produces a permanent (plastic) strain of $0.2\% = 0.002$ after unloading. It is found graphically by drawing a line parallel to the elastic (Young’s-modulus) slope, offset by a strain of 0.002 along the strain axis; its intersection with the curve defines $\sigma_{0.2}$. This offset yield strength is the number used as $S_y$ in machine-design calculations for such materials.

(b) The Tresca (maximum-shear-stress) criterion. With principal stresses ordered $\sigma_1 \ge \sigma_2 \ge \sigma_3$, Tresca predicts yielding when $\sigma_1 - \sigma_3 = S_y$, while von Mises predicts yielding when $\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2 = S_y^2$ (for a plane-stress state $\sigma_3=0$). In biaxial stretching both in-plane stresses are tensile ($\sigma_1,\sigma_2 \gt 0,\ \sigma_3=0$), so Tresca uses $\sigma_1 - 0 = S_y$ and yields as soon as the larger principal stress reaches $S_y$. For any stress ratio between pure uniaxial and equibiaxial tension, Mises permits the load to rise higher (e.g. at $\sigma_2 = \tfrac{1}{2}\sigma_1$, Mises yields at $\sigma_1 = 1.15\,S_y$) before predicting yield. The Tresca hexagon is inscribed within the Mises ellipse, so Tresca always predicts yielding at the lower (more conservative) stress; the two coincide only at uniaxial and equibiaxial tension.

(c) Superior strength- and stiffness-to-weight. In both torsion and bending the stress varies linearly with radial distance from the neutral axis, $\tau = Tr/J$ and $\sigma = Mc/I$, so material clustered near the centre of a solid shaft is lightly stressed and contributes little to load capacity while adding full weight. Removing that low-stress core to form a hollow shaft barely reduces the section modulus and polar second moment ($J,\,I \propto d^4$, dominated by the outer fibres) yet cuts the mass in proportion to the removed area. The result is a much higher torque and bending capacity per unit weight (and higher critical speed), which is why drive shafts, axles and aircraft structures use tubes. The trade-offs are higher cost and reduced local buckling resistance of thin walls.

(d) Minimum film thickness follows directly from the eccentricity and clearance — the speed and viscosity are not required for the geometric film thickness.

Given. Journal bearing, $D = 45\text{ mm}$ ($r = 22.5\text{ mm}$), $L = 200\text{ mm}$, eccentricity ratio $\varepsilon = 0.55$, clearance ratio $c_r/r = 0.001$.

Find. The minimum oil-film thickness $h_0$.

Approach. Convert the clearance ratio to a radial clearance, then apply $h_0=c_r(1-\varepsilon)$.

1. Radial clearance from the clearance ratio. The clearance ratio is the radial clearance divided by the journal radius, so $$c_r = \left(\frac{c_r}{r}\right) r = 0.001 \times 22.5\text{ mm} = 0.0225\text{ mm}.$$ (The diametral-clearance reading $c_d/D=0.001$ gives the identical $c_r$, since $c_d/D=c_r/r$.)
2. Minimum film thickness. The eccentricity ratio is $\varepsilon = e/c_r$, and the film is thinnest where the journal is nearest the bearing: $$h_0 = c_r - e = c_r(1-\varepsilon) = 0.0225(1-0.55).$$ $$\boxed{h_0 = 0.0101\text{ mm} \approx 10.1\ \mu\text{m}.}$$