Performance characteristics of materials

A designer specifies materials for an electric kettle: the element housing that heats water and the outer handle. Using physical properties, justify a different material for each.

1. 01Analyse the housing

   It must conduct heat efficiently into the water and resist corrosion - high thermal conductivity and corrosion resistance point to a metal such as stainless steel or aluminium.

2. 02Analyse the handle

   It must stay cool and safe to hold and insulate the user electrically - low thermal and electrical conductivity point to a polymer such as polypropylene or a thermoset.

3. 03Justify with the properties

   The metal's high thermal conductivity moves heat to the water; the polymer's low thermal and electrical conductivity protects the user - each material chosen for its physical response to heat and electricity.

Result: A conductive corrosion-resistant metal for the heated housing and an insulating polymer for the handle - each justified by its thermal and electrical conductivity, a physical-property decision.

Explain why a cold chisel is made from hardened high-carbon steel while a safety helmet shell is made from tough ABS, referring to the mechanical properties each product demands.

1. 01Analyse the chisel

   Its cutting edge must resist wear and indentation and stay sharp - it needs hardness, which hardened high-carbon steel provides; some brittleness is acceptable because it is not subject to sudden bending.

2. 02Analyse the helmet

   It must absorb a sudden impact without shattering to protect the head - it needs toughness (impact resistance), which ABS provides by deforming and spreading the energy rather than fracturing.

3. 03Contrast the properties

   Hardness suits the chisel's wear resistance at the edge; toughness suits the helmet's need to survive impact - the two products demand opposite ends of the hardness-toughness trade-off.

Result: The chisel uses hard high-carbon steel to keep a wear-resistant edge; the helmet uses tough ABS to absorb impact without shattering - a clear illustration of hardness versus toughness driving the material choice.

A steel tie rod of circular cross-section, diameter 10 mm, carries a tensile force of 12 kN. Calculate the tensile stress and, given the steel yields at 250 MPa, state the factor of safety.

1. 01Find the cross-sectional area

   For a circle, A = (pi/4) x d^2 = (pi/4) x (10 mm)^2 = 78.5 mm^2.

   $$A = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4}(10)^2 = 78.5\ \text{mm}^2$$
2. 02Convert the force

   12 kN = 12 000 N.

3. 03Calculate the stress

   Stress = F/A = 12 000 N / 78.5 mm^2 = 152.9 N/mm^2 = 152.9 MPa.

   $$\sigma = \dfrac{F}{A} = \dfrac{12000}{78.5} = 152.9\ \text{MPa}$$
4. 04Find the factor of safety

   Factor of safety = yield strength / working stress = 250 / 152.9 = 1.6, so the rod carries the load with a modest margin; a larger diameter would lower the stress and raise the factor of safety.

Result: The tensile stress is about 153 MPa; against a 250 MPa yield strength the factor of safety is about 1.6, so the rod is safe but not heavily over-engineered.

A manufacturer wants a baby's feeding spoon that visibly warns a parent when the food is too hot. Recommend a smart material and evaluate its suitability.

1. 01Match a stimulus to the need

   The warning must respond to temperature, so a thermochromic pigment - which changes colour above a set temperature - is the appropriate smart material.

2. 02Explain how it works

   Thermochromic pigment is moulded into or coated onto the spoon bowl; above the trigger temperature (set near the safe feeding limit) it changes colour, giving the parent an immediate visual warning without any electronics.

3. 03Evaluate

   The advantage is a self-contained, battery-free safety warning; the drawbacks are the pigment's cost, the need to verify the trigger temperature is reliable and food-safe, and possible fading of the effect over many wash cycles - which must be tested before launch.

Result: A thermochromic pigment that changes colour above the safe feeding temperature gives a simple, electronics-free hot-food warning; its suitability depends on a reliable, food-safe trigger temperature and durability across repeated washing, which testing must confirm.

A takeaway chain is considering replacing its oil-based polystyrene food trays with polylactic acid (PLA). Evaluate whether PLA is the more sustainable choice.

1. 01State the benefit

   PLA is made from renewable plant sugars rather than crude oil and can be composted, so if the trays reach industrial composting the material returns to natural substances and reduces reliance on fossil feedstock.

2. 02State the limitations

   PLA has lower heat resistance (it can soften with hot food), needs industrial composting to break down (it does little in landfill), and can contaminate the conventional plastics recycling stream if customers put it in the wrong bin.

3. 03Reach a judgement

   PLA is more sustainable only if the chain also provides and controls a composting route and communicates it clearly; without that infrastructure the environmental benefit is largely lost, so the decision depends on the disposal system, not the material alone.

Result: PLA is potentially more sustainable than polystyrene because it is bio-based and compostable, but only if industrial composting is actually available and the trays are correctly sorted - otherwise its lower heat resistance and recycling-contamination risk undermine the benefit.