ES 1021B – Materials Selection
Assignment (Clean Solution)
Question 1 – Materials Selection
a) Material Index Derivation
The deflection constraint is:
δ = (F L^3) / (4 E b h^3)
Rearranging for h:
h^3 = (F L^3) / (4 E b δ_max)
Mass of the beam:
m=ρbLh
Substituting h:
m ∝ ρ (1/E)^(1/3)
To minimize mass, maximize the material index:
M = E^(1/3) / ρ
b) Selection Chart Construction
Rearranging the material index:
M = E^(1/3) / ρ → E^(1/3) = Mρ
Taking logarithms:
log(E) = 3 log(ρ) + constant
Therefore:
- y-axis: Young’s modulus (E)
- x-axis: Density (ρ)
- Slope of selection line: 3
c) Material Selection
Candidate materials:
- Carbon fiber reinforced polymer (CFRP)
- Titanium alloys
- Aluminum alloys
- Magnesium alloys
- Steels
Ranking (best to worst):
1. CFRP
2. Titanium alloys
3. Aluminum alloys
4. Magnesium alloys
5. Steels
Materials are ranked based on the material index M = E^(1/3)/ρ. Materials with high
stiffness and low density perform best because they minimize mass while maintaining
stiffness requirements.
d) Recyclability Constraint
When recyclability is required, aluminum alloys are the best choice. They provide a strong
balance of stiffness and low density while being highly recyclable, unlike composite
materials such as CFRP.
Question 2 – Materials Selection
a) Material Index
Thermal stress:
σ = E α ΔT
To prevent yielding:
σ ≤ σ_y
Maximum temperature change:
ΔT_max = σ_y / (E α)
Material index to maximize:
M = σ_y / (E α)
b) Selection Chart
Rearranging:
σ_y = M (E α)
Taking logarithms:
log(σ_y) = log(E α) + constant
Therefore:
- y-axis: Yield strength (σ_y)
- x-axis: E × α
- Slope: 1
c) Best Material Class
Engineering ceramics outperform other material classes.
d) Explanation
Engineering ceramics have very low thermal expansion coefficients and high stiffness. This
reduces thermal stress under temperature changes, allowing them to withstand larger
temperature ranges without yielding.
Question 3 – Composite Material
a) Density
ρ_c = 0.45(1850) + 0.55(1200) = 1493 kg/m³
b) Young’s Modulus
Parallel:
E_parallel = 0.45(480) + 0.55(2.4) ≈ 217 GPa
Perpendicular:
1/E = (0.45/480) + (0.55/2.4)
E ≈ 4.35 GPa
c) Fracture Properties
i. Fracture stress:
σ = E ε = 217 × 0.021 ≈ 4.56 GPa
ii. Elastic energy:
U = (1/2) σ ε