Question 1 of 6: Short-answer set — proof stress, hollow shafts, plane-strain constraint, journal-bearing film (10 marks)

Question 1: Short-answer set — proof stress, hollow shafts, plane-strain constraint, journal-bearing film (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 meaning of σ_0.2 (2 marks)

Many engineering alloys — aluminium, copper, the austenitic stainless steels, quenched-and-tempered high-strength steels — have a stress–strain curve that bends over smoothly with no discontinuity, so there is no observable upper or lower yield point at which to record a yield strength. The convention is therefore to define a substitute: $\sigma_{0.2}$ is the 0.2 % offset proof stress, the stress at which a straight line drawn parallel to the initial elastic slope, but offset along the strain axis by a plastic strain of $0.002$, first intersects the measured curve.

Physically it is the stress that leaves behind a permanent set of two parts in a thousand once the load is removed, and it is the number a designer treats as $S_y$ for such materials. It is a defined, reproducible engineering quantity rather than a physical transition; because it is tied to a specific offset, the value must always be quoted with the offset used, and a material specified on a 0.1 % or 0.5 % offset will report a different figure from the same test data.

(b) Hollow versus solid shafts (3 marks)

Both the polar second moment $J=\int r^{2}\,dA$ and the rectangular second moment $I$ weight each element of area by the square of its distance from the neutral axis, whereas the mass weights it only by the area itself. Material near the centre of a solid shaft therefore contributes almost nothing to strength or stiffness while contributing fully to weight, and removing it is close to free. Concretely, replacing a solid 100 mm shaft with a hollow one of the same cross-sectional area (so the same mass per unit length) at $d_i/d_o = 0.6$ gives $d_o = 125$ mm, $d_i = 75$ mm and

so the same weight of steel carries 70 % more torque and is 113 % stiffer in bending. Beyond that ratio advantage, a hollow shaft has lower polar mass moment of inertia (quicker to accelerate, less gyroscopic reaction), a higher fundamental whirl speed for the same weight because stiffness rises faster than mass, a bore that can carry coolant, lubricant, wiring or a draw bar, and a section that can be produced by piercing or drawing rather than by machining away sound metal.

The disadvantages are mostly manufacturing and detail-design penalties. A hollow section costs more to make and to inspect; the bore must be concentric, because wall-thickness eccentricity puts the mass centre off the rotation axis and shows up directly as unbalance. Achieving equal strength requires a larger outside diameter, which drives up the size, cost and drag of every bearing, seal, coupling and housing that surrounds it. A thin wall is vulnerable where load is introduced locally: keyways, splines, set screws, press fits and interference collars can crush or locally buckle the wall, so the section usually has to be thickened at those stations. Torsional buckling of a very thin tube becomes a distinct failure mode that a solid bar simply does not have, welding and heat treatment distort a tube more, and the bore surface is difficult to finish or to inspect for the cracks that fatigue tends to start from.

(c) Why plane-strain tension raises the load-carrying capacity (3 marks)

The key point is that yielding is governed by the deviatoric (shape-changing) part of the stress state, not by the axial stress alone. In a plain uniaxial test the specimen is free to contract laterally and the transverse stresses stay zero, so yield occurs at $\sigma_x = S_y$. Under plane-strain tension one transverse direction is prevented from straining, $\varepsilon_z = 0$, and the constraint has to be paid for by a transverse stress. From Hooke’s law with $\varepsilon_z = 0$,

The state is now biaxial, $\left(\sigma_x,\ \nu\sigma_x,\ 0\right)$, and part of it is hydrostatic. Hydrostatic pressure changes volume but not shape, so it cannot drive plastic flow; only the differences between the principal stresses can. Adding $\sigma_z=\nu\sigma_x$ therefore increases the axial stress without increasing the deviatoric part in proportion, and the axial stress needed to reach the yield surface goes up. Taking the plastic value $\nu = 0.5$ (plastic flow is volume-conserving) and applying the distortion-energy criterion,

so the constrained material carries about 15.5 % more axial stress before it yields. This is the same $2/\sqrt{3}$ factor that appears as the “plane-strain flow stress” $2k$ in metal-forming analysis. Worth adding as a discriminator: the Tresca criterion, which depends only on $\sigma_1-\sigma_3$, predicts no increase at all here, because $\sigma_1-\sigma_3 = \sigma_x - 0$ is unchanged. The strengthening that is actually measured is therefore evidence in favour of the distortion-energy criterion for ductile metals. Note also that the same constraint that raises the yield stress raises the hydrostatic tension, and high hydrostatic tension promotes brittle cleavage — which is why thick, constrained sections and deep notches are dangerous even though they “test stronger”.

(d) Minimum film thickness and lubricant viscosity (2 marks)

In a full hydrodynamic journal bearing the shaft rides on a wedge of oil it drags into the converging clearance, and the film geometry is fixed by how far the journal centre has been pushed off the bearing centre. With radial clearance $c$ and eccentricity ratio $\varepsilon = e/c$,

so the question becomes how $\varepsilon$ depends on viscosity. That is set by the Sommerfeld (bearing characteristic) number

which is linear in the absolute viscosity $\mu$: doubling the viscosity doubles $S$ at fixed speed and unit load. On the Raimondi–Boyd charts a larger $S$ corresponds to a smaller $\varepsilon$ — a more viscous oil generates more pressure for the same wedge, so it holds the journal closer to the bearing centre. Hence the minimum film thickness increases with viscosity, steeply while the bearing is heavily loaded ($\varepsilon$ near unity) and then flattening out as $\varepsilon \to 0$. For a 50 mm journal with a clearance ratio $c/r = 0.001$ ($c = 0.025$ mm) at 1800 rev/min, moving from a lightly-loaded condition at $\varepsilon = 0.8$ to $\varepsilon = 0.5$ takes $h_{\min}$ from 5 $\mu$m to 12.5 $\mu$m — the difference between marginal and comfortable operation.

Two qualifications complete the answer. First, the benefit is self-limiting: a more viscous oil dissipates more friction power, the film runs hotter, and viscosity falls with temperature, so the equilibrium $\mu$ is lower than the bulk-supply value and must be found by iterating the thermal balance. Second, viscosity that is too high wastes power and can starve the inlet groove, so the design aim is the minimum viscosity that keeps $h_{\min}$ safely above the combined surface roughness of journal and bearing, not the highest viscosity available.