Mechanical properties

Question 1

Define a mechanical property.

Answer 1

A characteristic describing how a material behaves under forces (loads), e.g., strength, stiffness, hardness, toughness, ductility.

Question 2

Strength vs stiffness—what’s the difference?

Answer 2

Strength is how much load a material can take before yielding or breaking; stiffness is how much it resists elastic deformation (slope of stress–strain).

Question 3

Define tensile strength (UTS).

Answer 3

The maximum stress a material can withstand in tension before fracture (from a tensile test).

Question 4

Define yield strength (or proof stress).

Answer 4

Stress at which noticeable plastic (permanent) deformation begins. Often taken at 0.2% strain for metals.

Question 5

Define compressive strength.

Answer 5

Maximum compressive stress before failure—important for brittle materials like ceramics and some polymers.

Question 6

Define shear strength.

Answer 6

Resistance to sliding failure along a plane within the material (shear loading).

Question 7

Define hardness.

Answer 7

Resistance to indentation or scratching. Measured by tests like Vickers (HV), Rockwell (HR), or Brinell (HB).

Question 8

Define toughness.

Answer 8

Ability to absorb energy before fracturing—area under the stress–strain curve; high toughness resists impact/shock.

Question 9

Define brittleness.

Answer 9

Tendency to fracture with little plastic deformation (low toughness).

Question 10

Define ductility.

Answer 10

Ability to plastically deform in tension without cracking; often reported as % elongation at break.

Question 11

Define malleability.

Answer 11

Ability to plastically deform in compression (e.g., rolling, pressing) without cracking.

Question 12

Define elasticity.

Answer 12

Ability to return to original shape after load is removed (reversible deformation).

Question 13

Define plasticity.

Answer 13

Permanent deformation remains after load is removed (irreversible).

Question 14

Define Young’s modulus (E).

Answer 14

Measure of stiffness: ratio of stress to strain in the elastic region (units: GPa). Higher E = stiffer.

Question 15

Define resilience.

Answer 15

Ability to absorb elastic energy and release it on unloading (area under elastic portion of stress–strain).

Question 16

Define fatigue.

Answer 16

Failure after many cycles of loading, even at stresses below static strength; often starts at stress raisers (notches).

Question 17

Define endurance limit (high‑cycle fatigue).

Answer 17

A stress level below which some materials (e.g., some steels) can endure a very large number of cycles without failing (approx. infinite life).

Question 18

Define creep (mechanical context).

Answer 18

Time‑dependent permanent deformation under constant load—accelerated at higher temperatures.

Question 19

What is fracture toughness (simple).

Answer 19

A measure of a material’s resistance to crack growth (e.g., K_IC). Higher value = more tolerant of flaws.

Question 20

What is hardness used for in workshop decisions?

Answer 20

Select tool materials and heat treatments; harder tools resist wear and indentation at the cutting edge.

Question 21

Name three common hardness tests.

Answer 21

Vickers (HV), Rockwell (HR), Brinell (HB).

Question 22

Impact tests—two examples and purpose.

Answer 22

Izod and Charpy; measure impact toughness (energy to break a notched specimen).

Question 23

Tensile test—two key results.

Answer 23

Yield strength/proof stress and ultimate tensile strength (UTS); also gives % elongation and Young’s modulus.

Question 24

Define stress and give formula.

Answer 24

Force per unit area. σ = F / A (Pa or N/m²).

Question 25

Define strain and give formula.

Answer 25

Change in length divided by original length. ε = ΔL / L₀ (no units).

Question 26

Give the formula linking E, stress and strain.

Answer 26

E = σ / ε in the elastic region.

Question 27

What is yield point vs UTS on a stress–strain curve?

Answer 27

Yield point marks start of plastic flow; UTS is the peak stress before necking/fracture.

Question 28

Why do notches reduce fatigue life?

Answer 28

They concentrate stress, making crack initiation easier under cyclic loading.

Question 29

Two design tips to improve fatigue performance.

Answer 29

Add fillets to reduce stress concentrations and improve surface finish to remove scratches.

Question 30

Why might a high‑strength steel still fail in impact?

Answer 30

If it is brittle (low toughness), it can crack suddenly under shock even though its static strength is high.

Question 31

What is work hardening?

Answer 31

Increase in strength/hardness due to plastic deformation (dislocation build‑up), often reduces ductility.

Question 32

How does heat treatment affect mechanical properties?

Answer 32

Processes like quenching/tempering change hardness, strength and toughness by altering microstructure.

Question 33

Simple calc: stress on a 10 mm diameter rod under 5 kN?

Answer 33

Area = π(0.01²)/4 ≈ 7.85e−5 m²; σ = 5000 / 7.85e−5 ≈ 63.7 MPa.

Question 34

Simple calc: a 200 mm bar extends 0.10 mm under load; what is strain?

Answer 34

ε = ΔL/L₀ = 0.10 / 200 = 0.0005 (0.05%).

Question 35

Lathe tool material—key mechanical property?

Answer 35

Hot hardness (retains hardness at high cutting temperatures) and wear resistance.

Question 36

Press forming sheet metal—two desirable properties.

Answer 36

High ductility and adequate yield strength to hold shape after forming.

Question 37

Structural frame—what property ratio matters?

Answer 37

Strength‑to‑weight (specific strength) and stiffness‑to‑weight (specific modulus).

Question 38

TIMWOOD link: how can poor toughness cause waste?

Answer 38

Leads to sudden fractures (Defects) and rework (Overprocessing); controlled by material choice and heat treatment.

Question 39

How does temperature affect mechanical properties? (general)

Answer 39

Rising temperature usually lowers strength and stiffness and can increase ductility; very low temperature can increase brittleness in some metals.

Question 40

What is a safety factor (FoS)?

Answer 40

Design margin = allowable stress / working stress; ensures reliability under variability and uncertainty.

Question 41

Give two reasons test specimens are standardised.

Answer 41

To compare materials fairly and ensure repeatable, traceable results across labs.

Question 42

Bend test—what does it indicate?

Answer 42

Material ductility and surface integrity (cracks or delamination when bent).

Question 43

Spring selection—two key properties.

Answer 43

High yield strength and resilience (high elastic energy storage without permanent set).