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Primary Bonds

Materials Science · FE Reference Handbook section

Materials Science
0 formulas
10 exam-style examples
~45 min
All Materials Science lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Composite modulus by the rule of mixtures — Primary Bonds

A fiber-reinforced polymer composite contains 55% fiber by volume. The fiber modulus is 66 GPa and the polymer matrix modulus is 3.0 GPa. Compute the longitudinal and transverse moduli of the composite material.

Given

  • Vf=0.55V_f = 0.55
  • Ef=66GPaE_f = 66 GPa
  • Em=3.0GPaE_m = 3.0 GPa

Find

E_longitudinal and E_transverse

Start with the thinking

  • Loaded along the fibers the strains are equal, giving the direct rule of mixtures.
  • Loaded across the fibers the stresses are equal, giving the inverse (harmonic) rule.

Step-by-step solution

  1. Formula

    E∥=VfEf+(1−Vf)EmE_∥ = V_f E_f + (1 - V_f) E_m

  2. Substituting

    E∥=0.55(66)+0.45(3.0)=37.65GPaE_∥ = 0.55(66) + 0.45(3.0) = 37.65 GPa

  3. Formula

    1/E⊥=Vf/Ef+(1−Vf)/Em1/E_⊥ = V_f/E_f + (1 - V_f)/E_m

  4. Substituting

    1/E⊥=0.55/66+0.45/3.0=0.158331/E_⊥ = 0.55/66 + 0.45/3.0 = 0.15833

  5. Evaluate

    E⊥=6.32GPaE_⊥ = 6.32 GPa
Answer:
E∥=37.7GPa,E⊥=6.32GPaE_∥ = 37.7 GPa, E_⊥ = 6.32 GPa

Why the other options are there

  • 34.5 GPa (simple average)
  • E_⊥ = 37.7 GPa (same as longitudinal)

Reference: FE Reference Handbook — Materials Science → Primary Bonds

Example 2
Composite modulus by the rule of mixtures — Primary Bonds (2)

A fiber-reinforced polymer composite contains 25% fiber by volume. The fiber modulus is 65 GPa and the polymer matrix modulus is 3.0 GPa. Compute the longitudinal and transverse moduli of the composite material.

Given

  • Vf=0.25V_f = 0.25
  • Ef=65GPaE_f = 65 GPa
  • Em=3.0GPaE_m = 3.0 GPa

Find

E_longitudinal and E_transverse

Start with the thinking

  • Loaded along the fibers the strains are equal, giving the direct rule of mixtures.
  • Loaded across the fibers the stresses are equal, giving the inverse (harmonic) rule.

Step-by-step solution

  1. Formula

    E∥=VfEf+(1−Vf)EmE_∥ = V_f E_f + (1 - V_f) E_m

  2. Substituting

    E∥=0.25(65)+0.75(3.0)=18.50GPaE_∥ = 0.25(65) + 0.75(3.0) = 18.50 GPa

  3. Formula

    1/E⊥=Vf/Ef+(1−Vf)/Em1/E_⊥ = V_f/E_f + (1 - V_f)/E_m

  4. Substituting

    1/E⊥=0.25/65+0.75/3.0=0.253851/E_⊥ = 0.25/65 + 0.75/3.0 = 0.25385

  5. Evaluate

    E⊥=3.94GPaE_⊥ = 3.94 GPa
Answer:
E∥=18.5GPa,E⊥=3.94GPaE_∥ = 18.5 GPa, E_⊥ = 3.94 GPa

Why the other options are there

  • 34.0 GPa (simple average)
  • E_⊥ = 18.5 GPa (same as longitudinal)

Reference: FE Reference Handbook — Materials Science → Primary Bonds

Example 3
Composite modulus by the rule of mixtures — Primary Bonds (3)

A fiber-reinforced polymer composite contains 35% fiber by volume. The fiber modulus is 71 GPa and the polymer matrix modulus is 3.0 GPa. Compute the longitudinal and transverse moduli of the composite material.

Given

  • Vf=0.35V_f = 0.35
  • Ef=71GPaE_f = 71 GPa
  • Em=3.0GPaE_m = 3.0 GPa

Find

E_longitudinal and E_transverse

Start with the thinking

  • Loaded along the fibers the strains are equal, giving the direct rule of mixtures.
  • Loaded across the fibers the stresses are equal, giving the inverse (harmonic) rule.

Step-by-step solution

  1. Formula

    E∥=VfEf+(1−Vf)EmE_∥ = V_f E_f + (1 - V_f) E_m

  2. Substituting

    E∥=0.35(71)+0.65(3.0)=26.80GPaE_∥ = 0.35(71) + 0.65(3.0) = 26.80 GPa

  3. Formula

    1/E⊥=Vf/Ef+(1−Vf)/Em1/E_⊥ = V_f/E_f + (1 - V_f)/E_m

  4. Substituting

    1/E⊥=0.35/71+0.65/3.0=0.221601/E_⊥ = 0.35/71 + 0.65/3.0 = 0.22160

  5. Evaluate

    E⊥=4.51GPaE_⊥ = 4.51 GPa
Answer:
E∥=26.8GPa,E⊥=4.51GPaE_∥ = 26.8 GPa, E_⊥ = 4.51 GPa

Why the other options are there

  • 37.0 GPa (simple average)
  • E_⊥ = 26.8 GPa (same as longitudinal)

Reference: FE Reference Handbook — Materials Science → Primary Bonds

Example 4
Composite modulus by the rule of mixtures — Primary Bonds (4)

A fiber-reinforced polymer composite contains 55% fiber by volume. The fiber modulus is 78 GPa and the polymer matrix modulus is 2.5 GPa. Compute the longitudinal and transverse moduli of the composite material.

Given

  • Vf=0.55V_f = 0.55
  • Ef=78GPaE_f = 78 GPa
  • Em=2.5GPaE_m = 2.5 GPa

Find

E_longitudinal and E_transverse

Start with the thinking

  • Loaded along the fibers the strains are equal, giving the direct rule of mixtures.
  • Loaded across the fibers the stresses are equal, giving the inverse (harmonic) rule.

Step-by-step solution

  1. Formula

    E∥=VfEf+(1−Vf)EmE_∥ = V_f E_f + (1 - V_f) E_m

  2. Substituting

    E∥=0.55(78)+0.45(2.5)=44.03GPaE_∥ = 0.55(78) + 0.45(2.5) = 44.03 GPa

  3. Formula

    1/E⊥=Vf/Ef+(1−Vf)/Em1/E_⊥ = V_f/E_f + (1 - V_f)/E_m

  4. Substituting

    1/E⊥=0.55/78+0.45/2.5=0.187051/E_⊥ = 0.55/78 + 0.45/2.5 = 0.18705

  5. Evaluate

    E⊥=5.35GPaE_⊥ = 5.35 GPa
Answer:
E∥=44.0GPa,E⊥=5.35GPaE_∥ = 44.0 GPa, E_⊥ = 5.35 GPa

Why the other options are there

  • 40.3 GPa (simple average)
  • E_⊥ = 44.0 GPa (same as longitudinal)

Reference: FE Reference Handbook — Materials Science → Primary Bonds

Example 5
Composite modulus by the rule of mixtures — Primary Bonds (5)

A fiber-reinforced polymer composite contains 50% fiber by volume. The fiber modulus is 72 GPa and the polymer matrix modulus is 3.0 GPa. Compute the longitudinal and transverse moduli of the composite material.

Given

  • Vf=0.50V_f = 0.50
  • Ef=72GPaE_f = 72 GPa
  • Em=3.0GPaE_m = 3.0 GPa

Find

E_longitudinal and E_transverse

Start with the thinking

  • Loaded along the fibers the strains are equal, giving the direct rule of mixtures.
  • Loaded across the fibers the stresses are equal, giving the inverse (harmonic) rule.

Step-by-step solution

  1. Formula

    E∥=VfEf+(1−Vf)EmE_∥ = V_f E_f + (1 - V_f) E_m

  2. Substituting

    E∥=0.50(72)+0.50(3.0)=37.50GPaE_∥ = 0.50(72) + 0.50(3.0) = 37.50 GPa

  3. Formula

    1/E⊥=Vf/Ef+(1−Vf)/Em1/E_⊥ = V_f/E_f + (1 - V_f)/E_m

  4. Substituting

    1/E⊥=0.50/72+0.50/3.0=0.173611/E_⊥ = 0.50/72 + 0.50/3.0 = 0.17361

  5. Evaluate

    E⊥=5.76GPaE_⊥ = 5.76 GPa
Answer:
E∥=37.5GPa,E⊥=5.76GPaE_∥ = 37.5 GPa, E_⊥ = 5.76 GPa

Why the other options are there

  • 37.5 GPa (simple average)
  • E_⊥ = 37.5 GPa (same as longitudinal)

Reference: FE Reference Handbook — Materials Science → Primary Bonds

Example 6
Composite modulus by the rule of mixtures — Primary Bonds (6)

A fiber-reinforced polymer composite contains 55% fiber by volume. The fiber modulus is 67 GPa and the polymer matrix modulus is 2.5 GPa. Compute the longitudinal and transverse moduli of the composite material.

Given

  • Vf=0.55V_f = 0.55
  • Ef=67GPaE_f = 67 GPa
  • Em=2.5GPaE_m = 2.5 GPa

Find

E_longitudinal and E_transverse

Start with the thinking

  • Loaded along the fibers the strains are equal, giving the direct rule of mixtures.
  • Loaded across the fibers the stresses are equal, giving the inverse (harmonic) rule.

Step-by-step solution

  1. Formula

    E∥=VfEf+(1−Vf)EmE_∥ = V_f E_f + (1 - V_f) E_m

  2. Substituting

    E∥=0.55(67)+0.45(2.5)=37.98GPaE_∥ = 0.55(67) + 0.45(2.5) = 37.98 GPa

  3. Formula

    1/E⊥=Vf/Ef+(1−Vf)/Em1/E_⊥ = V_f/E_f + (1 - V_f)/E_m

  4. Substituting

    1/E⊥=0.55/67+0.45/2.5=0.188211/E_⊥ = 0.55/67 + 0.45/2.5 = 0.18821

  5. Evaluate

    E⊥=5.31GPaE_⊥ = 5.31 GPa
Answer:
E∥=38.0GPa,E⊥=5.31GPaE_∥ = 38.0 GPa, E_⊥ = 5.31 GPa

Why the other options are there

  • 34.8 GPa (simple average)
  • E_⊥ = 38.0 GPa (same as longitudinal)

Reference: FE Reference Handbook — Materials Science → Primary Bonds

Example 7
Composite modulus by the rule of mixtures — Primary Bonds (7)

A fiber-reinforced polymer composite contains 35% fiber by volume. The fiber modulus is 71 GPa and the polymer matrix modulus is 3.5 GPa. Compute the longitudinal and transverse moduli of the composite material.

Given

  • Vf=0.35V_f = 0.35
  • Ef=71GPaE_f = 71 GPa
  • Em=3.5GPaE_m = 3.5 GPa

Find

E_longitudinal and E_transverse

Start with the thinking

  • Loaded along the fibers the strains are equal, giving the direct rule of mixtures.
  • Loaded across the fibers the stresses are equal, giving the inverse (harmonic) rule.

Step-by-step solution

  1. Formula

    E∥=VfEf+(1−Vf)EmE_∥ = V_f E_f + (1 - V_f) E_m

  2. Substituting

    E∥=0.35(71)+0.65(3.5)=27.12GPaE_∥ = 0.35(71) + 0.65(3.5) = 27.12 GPa

  3. Formula

    1/E⊥=Vf/Ef+(1−Vf)/Em1/E_⊥ = V_f/E_f + (1 - V_f)/E_m

  4. Substituting

    1/E⊥=0.35/71+0.65/3.5=0.190641/E_⊥ = 0.35/71 + 0.65/3.5 = 0.19064

  5. Evaluate

    E⊥=5.25GPaE_⊥ = 5.25 GPa
Answer:
E∥=27.1GPa,E⊥=5.25GPaE_∥ = 27.1 GPa, E_⊥ = 5.25 GPa

Why the other options are there

  • 37.3 GPa (simple average)
  • E_⊥ = 27.1 GPa (same as longitudinal)

Reference: FE Reference Handbook — Materials Science → Primary Bonds

Example 8
Composite modulus by the rule of mixtures — Primary Bonds (8)

A fiber-reinforced polymer composite contains 35% fiber by volume. The fiber modulus is 78 GPa and the polymer matrix modulus is 3.0 GPa. Compute the longitudinal and transverse moduli of the composite material.

Given

  • Vf=0.35V_f = 0.35
  • Ef=78GPaE_f = 78 GPa
  • Em=3.0GPaE_m = 3.0 GPa

Find

E_longitudinal and E_transverse

Start with the thinking

  • Loaded along the fibers the strains are equal, giving the direct rule of mixtures.
  • Loaded across the fibers the stresses are equal, giving the inverse (harmonic) rule.

Step-by-step solution

  1. Formula

    E∥=VfEf+(1−Vf)EmE_∥ = V_f E_f + (1 - V_f) E_m

  2. Substituting

    E∥=0.35(78)+0.65(3.0)=29.25GPaE_∥ = 0.35(78) + 0.65(3.0) = 29.25 GPa

  3. Formula

    1/E⊥=Vf/Ef+(1−Vf)/Em1/E_⊥ = V_f/E_f + (1 - V_f)/E_m

  4. Substituting

    1/E⊥=0.35/78+0.65/3.0=0.221151/E_⊥ = 0.35/78 + 0.65/3.0 = 0.22115

  5. Evaluate

    E⊥=4.52GPaE_⊥ = 4.52 GPa
Answer:
E∥=29.2GPa,E⊥=4.52GPaE_∥ = 29.2 GPa, E_⊥ = 4.52 GPa

Why the other options are there

  • 40.5 GPa (simple average)
  • E_⊥ = 29.2 GPa (same as longitudinal)

Reference: FE Reference Handbook — Materials Science → Primary Bonds

Example 9
Composite modulus by the rule of mixtures — Primary Bonds (9)

A fiber-reinforced polymer composite contains 35% fiber by volume. The fiber modulus is 44 GPa and the polymer matrix modulus is 4.5 GPa. Compute the longitudinal and transverse moduli of the composite material.

Given

  • Vf=0.35V_f = 0.35
  • Ef=44GPaE_f = 44 GPa
  • Em=4.5GPaE_m = 4.5 GPa

Find

E_longitudinal and E_transverse

Start with the thinking

  • Loaded along the fibers the strains are equal, giving the direct rule of mixtures.
  • Loaded across the fibers the stresses are equal, giving the inverse (harmonic) rule.

Step-by-step solution

  1. Formula

    E∥=VfEf+(1−Vf)EmE_∥ = V_f E_f + (1 - V_f) E_m

  2. Substituting

    E∥=0.35(44)+0.65(4.5)=18.33GPaE_∥ = 0.35(44) + 0.65(4.5) = 18.33 GPa

  3. Formula

    1/E⊥=Vf/Ef+(1−Vf)/Em1/E_⊥ = V_f/E_f + (1 - V_f)/E_m

  4. Substituting

    1/E⊥=0.35/44+0.65/4.5=0.152401/E_⊥ = 0.35/44 + 0.65/4.5 = 0.15240

  5. Evaluate

    E⊥=6.56GPaE_⊥ = 6.56 GPa
Answer:
E∥=18.3GPa,E⊥=6.56GPaE_∥ = 18.3 GPa, E_⊥ = 6.56 GPa

Why the other options are there

  • 24.3 GPa (simple average)
  • E_⊥ = 18.3 GPa (same as longitudinal)

Reference: FE Reference Handbook — Materials Science → Primary Bonds

Example 10
Composite modulus by the rule of mixtures — Primary Bonds (10)

A fiber-reinforced polymer composite contains 40% fiber by volume. The fiber modulus is 62 GPa and the polymer matrix modulus is 3.5 GPa. Compute the longitudinal and transverse moduli of the composite material.

Given

  • Vf=0.40V_f = 0.40
  • Ef=62GPaE_f = 62 GPa
  • Em=3.5GPaE_m = 3.5 GPa

Find

E_longitudinal and E_transverse

Start with the thinking

  • Loaded along the fibers the strains are equal, giving the direct rule of mixtures.
  • Loaded across the fibers the stresses are equal, giving the inverse (harmonic) rule.

Step-by-step solution

  1. Formula

    E∥=VfEf+(1−Vf)EmE_∥ = V_f E_f + (1 - V_f) E_m

  2. Substituting

    E∥=0.40(62)+0.60(3.5)=26.90GPaE_∥ = 0.40(62) + 0.60(3.5) = 26.90 GPa

  3. Formula

    1/E⊥=Vf/Ef+(1−Vf)/Em1/E_⊥ = V_f/E_f + (1 - V_f)/E_m

  4. Substituting

    1/E⊥=0.40/62+0.60/3.5=0.177881/E_⊥ = 0.40/62 + 0.60/3.5 = 0.17788

  5. Evaluate

    E⊥=5.62GPaE_⊥ = 5.62 GPa
Answer:
E∥=26.9GPa,E⊥=5.62GPaE_∥ = 26.9 GPa, E_⊥ = 5.62 GPa

Why the other options are there

  • 32.8 GPa (simple average)
  • E_⊥ = 26.9 GPa (same as longitudinal)

Reference: FE Reference Handbook — Materials Science → Primary Bonds

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