Machine Design, Materials and Shop Practice
Machine Design
The shape of the exam
Machine Design items reward the candidate who can move cleanly from a free-body diagram, to a stress state, to a geometry decision. The working toolkit is: static failure theories (maximum shear stress and von Mises for ductile metals; maximum normal stress and Mohr-Coulomb for brittle materials), combined loading, stress concentration factors, fatigue assessment (endurance limit with its modifying factors, then Soderberg, Goodman or Gerber diagrams), and the standard routines for shafts, keys, belts, clutches, gears, springs, bolted and welded joints, and bearings.
Where candidates lose points
Watch four recurring traps. First, torsion: shear stress is $\tau = 16T/(\pi d^3)$ for a solid round shaft, so the diameter enters cubed - swapping radius for diameter corrupts the answer by a factor of eight. Second, the factor of safety is defined against a specific strength (yield, ultimate, endurance); quoting the wrong strength silently changes the result. Third, the Lewis form factor already absorbs the tooth geometry, so do not divide again by the circular pitch. Fourth, fits and tolerances are stated as deviations, not sizes - read whether a value is a hole or shaft limit before subtracting.
Worked example
A solid steel shaft transmits 20 kW at 1200 rpm with an allowable shear stress of 40 MPa. Torque:
$$T = \frac{9550,P(\text{kW})}{N} = \frac{9550 \times 20}{1200} = 159.2\ \text{N}\cdot\text{m}$$
$$d = \sqrt[3]{\frac{16T}{\pi \tau}} = \sqrt[3]{\frac{16 \times 159{,}200}{\pi \times 40}} \approx 27.3\ \text{mm}$$
Rule of thumb: check the units of torque before cubing anything - N·mm versus N·m is the single most common cause of a wrong shaft diameter in this exam.
Materials and Shop Practice
The shape of the exam
This domain mixes metallurgy with the physical acts of making parts. On the metallurgy side you must know the iron-carbon diagram (eutectoid at roughly 0.8% C, the ferrite-pearlite-cementite progression, the effect of carbon on hardness and ductility), the purpose of each heat treatment, and the difference between hardness, toughness and wear resistance. On the shop side you must handle cutting speeds and feeds, machining time, sheet-metal and welding symbols, casting allowances (shrinkage, draft, machining), forging, and measurement/tolerance practice.
Where candidates lose points
Annealing versus normalizing versus stress relieving: all soften or relax a part, but only annealing is the full-softening, furnace-cool slow cool; normalizing cools in still air and gives a finer, stronger structure. Quenching without tempering leaves a hard but brittle martensite - a favourite distractor. In machining problems, cutting speed is surface speed in m/min, not spindle speed in rpm; confusing them is the classic error. In welding, distinguish the electrode/process from the joint type and the position; in casting, remember that shrinkage allowance is added to the pattern while draft is a taper for withdrawal.
Worked example
Find the spindle speed for turning a 50 mm diameter bar at a cutting speed of 30 m/min:
$$N = \frac{1000V}{\pi D} = \frac{1000 \times 30}{\pi \times 50} \approx 191\ \text{rpm}$$
Rule of thumb: machining time = length of cut ÷ (feed per revolution × rpm), and always add approach and overtravel to the cut length before dividing.