22-Mec-B1 Advanced Machine Design · December 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, December 2018 — 07-Mec-B1 Advanced Machine Design. Three hours, open book, 100 marks. Part I (Problems 1 and 2) is compulsory; the candidate answers three of the four Part II problems (Problems 3–6). Any non-communicating calculator is permitted, assumptions must be stated, and tabulated values must be sourced. All six problems are worked here, so the set is complete as a study resource.
Reference texts. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed. (shaft fatigue §7, power screws §8-2, clutches and brakes §16, journal bearings §12); Norton, Machine Design: An Integrated Approach, 6th ed. (impact loading §3, brakes §16); Juvinall & Marshek, Fundamentals of Machine Component Design, 6th ed. (screws, clutches, brakes); Hibbeler, Mechanics of Materials, 10th ed. (beam bending, impact factors); CSA/ISO 14006 Eco-design management and ISO 14040 Life-cycle assessment for the green-design criteria.
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.
Yielding of a ductile metal is governed by the deviatoric (shape-changing) part of the stress state, not by the hydrostatic part. In uniaxial tension the specimen is free to contract laterally, so the only non-zero stress is the axial one and the whole of it is available to drive shear on the 45° planes. In plane-strain tension the specimen is constrained in one transverse direction — a wide, thin sheet gripped across its width, or the mid-thickness of a thick plate, where the surrounding material prevents the contraction that Poisson's ratio would otherwise produce.
That constraint requires a transverse stress. Setting the transverse plastic strain increment to zero in the Levy–Mises flow rule gives
$$\varepsilon_2 = 0 \;\Longrightarrow\; \sigma_2 = \tfrac{1}{2}\left(\sigma_1 + \sigma_3\right) = \tfrac{1}{2}\sigma_1 \quad (\sigma_3 = 0)$$so the state is no longer uniaxial: it is $\sigma_1$, $\sigma_1/2$, $0$. Substituting into the von Mises criterion,
$$\sigma_e = \sqrt{\tfrac{1}{2}\Big[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2\Big]} = \frac{\sqrt{3}}{2}\,\sigma_1 = S_y$$ $$\boxed{\;\sigma_{1,\text{yield}} = \frac{2}{\sqrt{3}}\,S_y = 1.155\,S_y\;}$$The added transverse tension raises the hydrostatic (mean) stress without adding a proportional amount of deviatoric stress, so a larger axial stress is needed before the deviatoric intensity reaches the yield surface. The material therefore carries about 15.5 % more axial load before it yields — the same triaxial-constraint effect that makes a notched bar appear stronger than a smooth one in a static tension test, and that produces the plane-strain "constraint factor" in fracture mechanics. The penalty is ductility: the constrained state suppresses plastic flow, so the material fails at a much smaller elongation and behaves in a more brittle manner.
Other acceptable criteria include minimal and recyclable packaging, low-impact manufacturing processes (dry machining, reduced scrap), and design for end-of-life take-back under extended-producer-responsibility programs. In Canadian practice these criteria are formalized by CSA/ISO 14006 (eco-design within an environmental management system) and quantified by ISO 14040/14044 life-cycle assessment.
In both torsion and bending, the stress carried by a fibre is proportional to its distance from the neutral axis or the shaft centreline, and its contribution to the section property is weighted by the square of that distance. Material close to the axis is therefore lightly stressed and contributes almost nothing to strength or stiffness while contributing fully to weight. Removing the core buys a large weight saving for a small loss of capacity. For a bore ratio $k = d_i/d_o$,
$$\frac{J_{\text{hollow}}}{J_{\text{solid}}} = 1-k^4, \qquad \frac{A_{\text{hollow}}}{A_{\text{solid}}} = 1-k^2$$At a typical $k = 0.6$ the shaft keeps $1-0.6^4 = 0.870$, i.e. 87.0 % of the torsional capacity, at $1-0.6^2 = 0.640$, i.e. 64.0 % of the weight — a
$$\boxed{\;\frac{0.8704}{0.6400} = 1.36 \;\;\text{i.e. a 36\% gain in strength-to-weight}\;}$$Secondary advantages follow from the lower mass: a higher critical (whirl) speed for the same stiffness, less rotating inertia so faster acceleration and braking, and a bore that can carry coolant, lubricant, wiring or a draw-bar. Hollow sections also cool and heat-treat more uniformly, avoiding the soft, untransformed core of a large solid forging.
Disadvantages. For the same torque the outside diameter must grow, so the shaft needs a larger envelope and correspondingly larger bearings, seals, couplings and gear bores. Manufacture is more expensive: deep-hole boring or trepanning, or a welded/drawn tube whose seam and wall-thickness tolerance must be controlled, and concentricity between bore and outside diameter is difficult to hold. The bore itself is a stress raiser, and thin walls are vulnerable to local denting, ovalisation and buckling under transverse or clamping loads. Keyways, splines, press fits and cross-holes are harder to machine and weaken a thin wall much more than a solid one, and hollow shafts are more difficult to join by welding or to straighten after heat treatment.
In a hydrodynamic journal bearing the journal runs eccentric to the bearing, and the minimum film thickness is set by the eccentricity ratio $\varepsilon = e/c_r$:
$$h_{\min} = c_r\,(1-\varepsilon)$$The eccentricity ratio is a function of the Sommerfeld (bearing characteristic) number
$$S = \left(\frac{r}{c_r}\right)^{2}\frac{\mu N}{P}$$where $\mu$ is the absolute viscosity, $N$ the journal speed in rev/s and $P$ the projected unit load. Raising the viscosity raises $S$; the Raimondi–Boyd design charts show $\varepsilon$ falling monotonically as $S$ rises, so a more viscous oil runs the journal more nearly concentric and gives a thicker minimum film. Physically, a thicker oil generates more pressure for a given wedge geometry and speed, so it can support the load on a larger clearance gap.
The relationship is monotonic but has strongly diminishing returns, and it is self-limiting: a higher viscosity also raises the friction power dissipated in the film, which raises the oil temperature, which drops the viscosity again. The design viscosity is therefore the value at the equilibrium operating temperature, not the value at the catalogue reference temperature, and the selection is a compromise between an adequate $h_{\min}$ (typically several times the combined surface roughness) and acceptable friction loss and oil temperature.
