SME
Materials Science & Material Properties · Problem type
Load share and longitudinal strength
Use this for “Carbon fibers (230 GPa) make up 60% of an epoxy (3.5 GPa). What share of an axial load do the fibers carry, and what is the strength if the fibers break at 3500 MPa while the epoxy carries 40 MPa?”
Diagram
This is an example. Type your own values to start a new problem.
Ff ÷ F (%)
=
100
×
Ef (GPa)
Ef (GPa)
+
Eₘ (GPa)
(1
−
)
σcu (MPa)
=
σfu (MPa)
×
Vf
+
σₘ′ (MPa)
×
(1
−
Vf
)
Formulas
Ff
F
=
100
×
Ef
×
Vf
Ef × Vf + Eₘ × (1 − Vf)
99
=
100
×
230
×
0.6
230 × 0.6 + 3.5 × (1 − 0.6)
σcu = σfu × Vf + σₘ′ × (1 − Vf)
2,116 = 3,500 × 0.6 + 40 × (1 − 0.6)
Assumptions
•
Isostrain along the fibers, so each part’s load is its modulus times its fraction, times the strain.
•
The fibers break first and the composite fails with them (Vf above the critical fraction).
•
σₘ′ is the matrix stress at the fibers’ breaking strain, not the matrix’s own strength.
Step-by-step
WE KNOW
Vf = 0.6, Ef = 230 GPa, Eₘ = 3.5 GPa, σfu = 3,500 MPa, σₘ′ = 40 MPa
FIND
fiber load share (Ff ÷ F), composite strength (σcu)
Step 1 · Find fiber load share (Ff ÷ F)
Use Ff ÷ F = 100 × Ef × Vf ÷ (Ef × Vf + Eₘ × (1 − Vf))
At one strain each part’s load goes as its modulus times its fraction; the fibers’ part over the sum, times 100.
Ff
F
=
100
×
230
×
0.6
230 × 0.6 + 3.5 × (1 − 0.6)
Ff
F
=
100
×
138
138 + 1.4
Ff
F
=
100
×
138
139.4
Ff
F
=
99%
Step 2 · Find composite strength (σcu)
Use σcu = σfu × Vf + σₘ′ × (1 − Vf)
When the fibers break, each part carries its stress times its fraction of the section.
σcu = 3,500 × 0.6 + 40 × (1 − 0.6)
σcu = 3,500 × 0.6 + 40 × 0.4
σcu = 2,100 + 16
σcu = 2,116 MPa
Check
Put the numbers back in. Both sides should match.
99
=
100
×
230
×
0.6
230 × 0.6 + 3.5 × (1 − 0.6)
✓
2,116 = 3,500 × 0.6 + 40 × (1 − 0.6) ✓
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