SME
Aerospace Structures · Problem type
Density and specific stiffness
Use this for “Carbon fibers (230 GPa, 1.8 g/cm³) fill 60% of an epoxy (3.5 GPa, 1.2 g/cm³) lamina. Find its density and its stiffness per unit mass.”
Diagram
Along: 230 × 0.6 + 3.5 × 0.4 = 139.4 GPa
Density: 1.8 × 0.6 + 1.2 × 0.4 = 1.56 g/cm³
Specific stiffness: 89.36 MN·m/kg
This is an example. Type your own values to start a new problem.
ρc (g/cm³)
=
ρf (g/cm³)
×
Vf
+
ρₘ (g/cm³)
×
Vₘ
Eₛₚ (MN·m/kg)
=
E₁ (GPa)
ρc (g/cm³)
Ef
Fiber modulus
example
Eₘ
Matrix modulus
example
Formulas
Vₘ = 1 − Vf
0.4 = 1 − 0.6
E₁
=
Ef
×
Vf
+
Eₘ
×
Vₘ
139.4 = 230 × 0.6 + 3.5 × 0.4
ρc = ρf × Vf + ρₘ × Vₘ
1.56 = 1.8 × 0.6 + 1.2 × 0.4
Eₛₚ
=
E₁
ρc
89.36
=
139.4
1.56
Assumptions
•
Fibers and matrix are bonded perfectly, with no voids.
•
Along the fibers both stretch the same (iso-strain), like springs side by side.
•
GPa ÷ (g/cm³) is MJ/kg, the same as MN·m/kg; aluminum is about 26 MJ/kg.
Step-by-step
WE KNOW
Ef = 230 GPa, Eₘ = 3.5 GPa, Vf = 0.6, ρf = 1.8 g/cm³, ρₘ = 1.2 g/cm³
FIND
matrix volume fraction (Vₘ), modulus along the fibers (E₁), composite density (ρc), specific stiffness (Eₛₚ)
Step 1 · Find matrix volume fraction (Vₘ)
Use Vₘ = 1 − Vf
With no voids, the matrix fills what the fibers leave.
Vₘ = 1 − 0.6
Vₘ = 0.4
Step 2 · Find modulus along the fibers (E₁)
Use E₁ = Ef × Vf + Eₘ × Vₘ
Along the fibers both stretch alike, so their stiffnesses add by share (springs side by side).
E₁
=
230
×
0.6
+
3.5
×
0.4
E₁
=
138
+
1.4
E₁
=
139.4
GPa
Step 3 · Find composite density (ρc)
Use ρc = ρf × Vf + ρₘ × Vₘ
Mass adds by volume share: each part’s density times its fraction.
ρc = 1.8 × 0.6 + 1.2 × 0.4
ρc = 1.08 + 0.48
ρc = 1.56 g/cm³
Step 4 · Find specific stiffness (Eₛₚ)
Use Eₛₚ = E₁ ÷ ρc
Stiffness
per
unit
mass:
divide
E₁
by
the
composite’s
density.
Eₛₚ
=
139.4
1.56
Eₛₚ = 89.36 MN·m/kg
Check
Put the numbers back in. Both sides should match.
0.4 = 1 − 0.6 ✓
139.4 = 230 × 0.6 + 3.5 × 0.4 ✓
1.56 = 1.8 × 0.6 + 1.2 × 0.4 ✓
89.36
=
139.4
1.56
✓
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