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Composite Sections

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials
7 formulas
10 exam-style examples
~59 min
All Mechanics of Materials lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • The bending stresses in a beam composed of dissimilar materials (Material 1 and Material 2) where E1 > E2 are:
  • The composite section is transformed into a section composed of a single material. The centroid and then the moment of inertia
  • are found on the transformed section for use in the bending stress equations.

Core formulas for this FE topic

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

Worked exam-style examples

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

Example 1
Transformed area (composite sections) — solve for modular ratio — Composite Sections

A steel-reinforced timber beam is analyzed as one of the composite sections using a transformed area. Given modulus of material 2 (E_2) = 13,700 ksi; modulus of material 1 (base) (E_1) = 26,800 ksi, determine the modular ratio (n).

Given

  • modulusofmaterial2(E2)=13,700ksimodulus of material 2 (E_2) = 13,700 ksi
  • modulusofmaterial1(base)(E1)=26,800ksimodulus of material 1 (base) (E_1) = 26,800 ksi

Find

modular ratio (n)

Start with the thinking

  • The governing relation printed in this handbook section is Transformed area (composite sections).
  • Everything except n is given, so isolate n symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Composite sections made of two materials are analyzed using the modular ratio to transform one material into an equivalent area of the other.
Composite sections transformed areacentroidal axisb = 8h = 14

Figure 1 — schematic for Transformed area (composite sections) — solve for modular ratio — Composite Sections

Step-by-step solution

  1. Step 1 — State the governing relation:

    n=E2E1n = \dfrac{E_2}{E_1}
  2. Step 2 — Rearrange symbolically for n:

    n=E2E1n = \dfrac{E_2}{E_1}
  3. Step 3

    Listthegivens:modulusofmaterial2(E2)=13,700ksi,modulusofmaterial1(base)(E1)=26,800ksiList the givens: modulus of material 2 (E_2) = 13,700 ksi, modulus of material 1 (base) (E_1) = 26,800 ksi
  4. Step 4 — Substitute the given values:

    n=E2E1n = \dfrac{E_2}{E_1}
  5. Step 5 — Evaluate:

    n=0.5112n = 0.5112
  6. Step 6 — Check: returning n = 0.5112 to

    n=E2E1n = \dfrac{E_2}{E_1}

    reproduces the given quantities, and both sides carry the same units.

Answer:
n=0.5112n = 0.5112

Why the other options are there

  • 1.0224 — kept a factor of two that cancels in the correct rearrangement.
  • 0.2556 — dropped that same factor in the other direction.
  • 0.5623 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Composite Sections

Example 2
Transformed area (composite sections) — solve for modulus of material 2 — Composite Sections (2)

A flitch beam with steel plates bolted to wood is treated as composite sections. Given modulus of material 1 (base) (E_1) = 5,100 ksi; modular ratio (n) = 14.4500, determine the modulus of material 2 (E_2) in ksi.

Given

  • modulusofmaterial1(base)(E1)=5,100ksimodulus of material 1 (base) (E_1) = 5,100 ksi
  • modularratio(n)=14.4500modular ratio (n) = 14.4500

Find

modulus of material 2 (E_2), in ksi

Start with the thinking

  • The governing relation printed in this handbook section is Transformed area (composite sections).
  • Everything except E_2 is given, so isolate E_2 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Composite sections made of two materials are analyzed using the modular ratio to transform one material into an equivalent area of the other.
Composite sections transformed areacentroidal axisb = 8h = 14

Figure 2 — schematic for Transformed area (composite sections) — solve for modulus of material 2 — Composite Sections (2)

Step-by-step solution

  1. Step 1 — State the governing relation:

    n=E2E1n = \dfrac{E_2}{E_1}
  2. Step 2 — Rearrange symbolically for E_2:

    E2=nE1E_{2} = n E_1
  3. Step 3

    Listthegivens:modulusofmaterial1(base)(E1)=5,100ksi,modularratio(n)=14.4500List the givens: modulus of material 1 (base) (E_1) = 5,100 ksi, modular ratio (n) = 14.4500
  4. Step 4 — Substitute the given values:

    E2=14.4500E1E_{2} = 14.4500 E_1
  5. Step 5 — Evaluate:

    E2=73695 ksiE_{2} = 73695\ \text{ksi}
  6. Step 6 — Check: returning E_2 = 73,695 ksi to

    n=E2E1n = \dfrac{E_2}{E_1}

    reproduces the given quantities, and both sides carry the same units.

Answer:
E2=73695 ksiE_{2} = 73695\ \text{ksi}

Why the other options are there

  • 147,390 — kept a factor of two that cancels in the correct rearrangement.
  • 36,848 — dropped that same factor in the other direction.
  • 81,065 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Composite Sections

Example 3
Transformed area (composite sections) — solve for modulus of material 1 (base) — Composite Sections (3)

A composite steel-concrete deck is transformed into composite sections for stress analysis. Given modulus of material 2 (E_2) = 5,000 ksi; modular ratio (n) = 4.6500, determine the modulus of material 1 (base) (E_1) in ksi.

Given

  • modulusofmaterial2(E2)=5,000ksimodulus of material 2 (E_2) = 5,000 ksi
  • modularratio(n)=4.6500modular ratio (n) = 4.6500

Find

modulus of material 1 (base) (E_1), in ksi

Start with the thinking

  • The governing relation printed in this handbook section is Transformed area (composite sections).
  • Everything except E_1 is given, so isolate E_1 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Composite sections made of two materials are analyzed using the modular ratio to transform one material into an equivalent area of the other.
Composite sections transformed areacentroidal axisb = 8h = 14

Figure 3 — schematic for Transformed area (composite sections) — solve for modulus of material 1 (base) — Composite Sections (3)

Step-by-step solution

  1. Step 1 — State the governing relation:

    n=E2E1n = \dfrac{E_2}{E_1}
  2. Step 2 — Rearrange symbolically for E_1:

    E1=E2nE_{1} = \dfrac{E_2}{n}
  3. Step 3

    Listthegivens:modulusofmaterial2(E2)=5,000ksi,modularratio(n)=4.6500List the givens: modulus of material 2 (E_2) = 5,000 ksi, modular ratio (n) = 4.6500
  4. Step 4 — Substitute the given values:

    E1=E24.6500E_{1} = \dfrac{E_2}{4.6500}
  5. Step 5 — Evaluate:

    E1=1075 ksiE_{1} = 1075\ \text{ksi}
  6. Step 6 — Check: returning E_1 = 1,075 ksi to

    n=E2E1n = \dfrac{E_2}{E_1}

    reproduces the given quantities, and both sides carry the same units.

Answer:
E1=1075 ksiE_{1} = 1075\ \text{ksi}

Why the other options are there

  • 2,151 — kept a factor of two that cancels in the correct rearrangement.
  • 537.6 — dropped that same factor in the other direction.
  • 1,183 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Composite Sections

Example 4
Transformed area (composite sections) — solve for modular ratio (case 2) — Composite Sections (4)

A steel-reinforced timber beam is analyzed as one of the composite sections using a transformed area. Given modulus of material 2 (E_2) = 16,100 ksi; modulus of material 1 (base) (E_1) = 13,700 ksi, determine the modular ratio (n).

Given

  • modulusofmaterial2(E2)=16,100ksimodulus of material 2 (E_2) = 16,100 ksi
  • modulusofmaterial1(base)(E1)=13,700ksimodulus of material 1 (base) (E_1) = 13,700 ksi

Find

modular ratio (n)

Start with the thinking

  • The governing relation printed in this handbook section is Transformed area (composite sections).
  • Everything except n is given, so isolate n symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Composite sections made of two materials are analyzed using the modular ratio to transform one material into an equivalent area of the other.
Composite sections transformed areacentroidal axisb = 8h = 14

Figure 4 — schematic for Transformed area (composite sections) — solve for modular ratio (case 2) — Composite Sections (4)

Step-by-step solution

  1. Step 1 — State the governing relation:

    n=E2E1n = \dfrac{E_2}{E_1}
  2. Step 2 — Rearrange symbolically for n:

    n=E2E1n = \dfrac{E_2}{E_1}
  3. Step 3

    Listthegivens:modulusofmaterial2(E2)=16,100ksi,modulusofmaterial1(base)(E1)=13,700ksiList the givens: modulus of material 2 (E_2) = 16,100 ksi, modulus of material 1 (base) (E_1) = 13,700 ksi
  4. Step 4 — Substitute the given values:

    n=E2E1n = \dfrac{E_2}{E_1}
  5. Step 5 — Evaluate:

    n=1.1752n = 1.1752
  6. Step 6 — Check: returning n = 1.1752 to

    n=E2E1n = \dfrac{E_2}{E_1}

    reproduces the given quantities, and both sides carry the same units.

Answer:
n=1.1752n = 1.1752

Why the other options are there

  • 2.3504 — kept a factor of two that cancels in the correct rearrangement.
  • 0.5876 — dropped that same factor in the other direction.
  • 1.2927 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Composite Sections

Example 5
Transformed area (composite sections) — solve for modulus of material 2 (case 2) — Composite Sections (5)

A flitch beam with steel plates bolted to wood is treated as composite sections. Given modulus of material 1 (base) (E_1) = 14,100 ksi; modular ratio (n) = 4.6000, determine the modulus of material 2 (E_2) in ksi.

Given

  • modulusofmaterial1(base)(E1)=14,100ksimodulus of material 1 (base) (E_1) = 14,100 ksi
  • modularratio(n)=4.6000modular ratio (n) = 4.6000

Find

modulus of material 2 (E_2), in ksi

Start with the thinking

  • The governing relation printed in this handbook section is Transformed area (composite sections).
  • Everything except E_2 is given, so isolate E_2 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Composite sections made of two materials are analyzed using the modular ratio to transform one material into an equivalent area of the other.
Composite sections transformed areacentroidal axisb = 8h = 14

Figure 5 — schematic for Transformed area (composite sections) — solve for modulus of material 2 (case 2) — Composite Sections (5)

Step-by-step solution

  1. Step 1 — State the governing relation:

    n=E2E1n = \dfrac{E_2}{E_1}
  2. Step 2 — Rearrange symbolically for E_2:

    E2=nE1E_{2} = n E_1
  3. Step 3

    Listthegivens:modulusofmaterial1(base)(E1)=14,100ksi,modularratio(n)=4.6000List the givens: modulus of material 1 (base) (E_1) = 14,100 ksi, modular ratio (n) = 4.6000
  4. Step 4 — Substitute the given values:

    E2=4.6000E1E_{2} = 4.6000 E_1
  5. Step 5 — Evaluate:

    E2=64860 ksiE_{2} = 64860\ \text{ksi}
  6. Step 6 — Check: returning E_2 = 64,860 ksi to

    n=E2E1n = \dfrac{E_2}{E_1}

    reproduces the given quantities, and both sides carry the same units.

Answer:
E2=64860 ksiE_{2} = 64860\ \text{ksi}

Why the other options are there

  • 129,720 — kept a factor of two that cancels in the correct rearrangement.
  • 32,430 — dropped that same factor in the other direction.
  • 71,346 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Composite Sections

Example 6
Transformed area (composite sections) — solve for modulus of material 1 (base) (case 2) — Composite Sections (6)

A composite steel-concrete deck is transformed into composite sections for stress analysis. Given modulus of material 2 (E_2) = 13,700 ksi; modular ratio (n) = 1.8500, determine the modulus of material 1 (base) (E_1) in ksi.

Given

  • modulusofmaterial2(E2)=13,700ksimodulus of material 2 (E_2) = 13,700 ksi
  • modularratio(n)=1.8500modular ratio (n) = 1.8500

Find

modulus of material 1 (base) (E_1), in ksi

Start with the thinking

  • The governing relation printed in this handbook section is Transformed area (composite sections).
  • Everything except E_1 is given, so isolate E_1 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Composite sections made of two materials are analyzed using the modular ratio to transform one material into an equivalent area of the other.
Composite sections transformed areacentroidal axisb = 8h = 14

Figure 6 — schematic for Transformed area (composite sections) — solve for modulus of material 1 (base) (case 2) — Composite Sections (6)

Step-by-step solution

  1. Step 1 — State the governing relation:

    n=E2E1n = \dfrac{E_2}{E_1}
  2. Step 2 — Rearrange symbolically for E_1:

    E1=E2nE_{1} = \dfrac{E_2}{n}
  3. Step 3

    Listthegivens:modulusofmaterial2(E2)=13,700ksi,modularratio(n)=1.8500List the givens: modulus of material 2 (E_2) = 13,700 ksi, modular ratio (n) = 1.8500
  4. Step 4 — Substitute the given values:

    E1=E21.8500E_{1} = \dfrac{E_2}{1.8500}
  5. Step 5 — Evaluate:

    E1=7405 ksiE_{1} = 7405\ \text{ksi}
  6. Step 6 — Check: returning E_1 = 7,405 ksi to

    n=E2E1n = \dfrac{E_2}{E_1}

    reproduces the given quantities, and both sides carry the same units.

Answer:
E1=7405 ksiE_{1} = 7405\ \text{ksi}

Why the other options are there

  • 14,811 — kept a factor of two that cancels in the correct rearrangement.
  • 3,703 — dropped that same factor in the other direction.
  • 8,146 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Composite Sections

Example 7
Transformed area (composite sections) — solve for modular ratio (case 3) — Composite Sections (7)

A steel-reinforced timber beam is analyzed as one of the composite sections using a transformed area. Given modulus of material 2 (E_2) = 25,800 ksi; modulus of material 1 (base) (E_1) = 11,100 ksi, determine the modular ratio (n).

Given

  • modulusofmaterial2(E2)=25,800ksimodulus of material 2 (E_2) = 25,800 ksi
  • modulusofmaterial1(base)(E1)=11,100ksimodulus of material 1 (base) (E_1) = 11,100 ksi

Find

modular ratio (n)

Start with the thinking

  • The governing relation printed in this handbook section is Transformed area (composite sections).
  • Everything except n is given, so isolate n symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Composite sections made of two materials are analyzed using the modular ratio to transform one material into an equivalent area of the other.
Composite sections transformed areacentroidal axisb = 8h = 14

Figure 7 — schematic for Transformed area (composite sections) — solve for modular ratio (case 3) — Composite Sections (7)

Step-by-step solution

  1. Step 1 — State the governing relation:

    n=E2E1n = \dfrac{E_2}{E_1}
  2. Step 2 — Rearrange symbolically for n:

    n=E2E1n = \dfrac{E_2}{E_1}
  3. Step 3

    Listthegivens:modulusofmaterial2(E2)=25,800ksi,modulusofmaterial1(base)(E1)=11,100ksiList the givens: modulus of material 2 (E_2) = 25,800 ksi, modulus of material 1 (base) (E_1) = 11,100 ksi
  4. Step 4 — Substitute the given values:

    n=E2E1n = \dfrac{E_2}{E_1}
  5. Step 5 — Evaluate:

    n=2.3243n = 2.3243
  6. Step 6 — Check: returning n = 2.3243 to

    n=E2E1n = \dfrac{E_2}{E_1}

    reproduces the given quantities, and both sides carry the same units.

Answer:
n=2.3243n = 2.3243

Why the other options are there

  • 4.6486 — kept a factor of two that cancels in the correct rearrangement.
  • 1.1622 — dropped that same factor in the other direction.
  • 2.5568 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Composite Sections

Example 8
Transformed area (composite sections) — solve for modulus of material 2 (case 3) — Composite Sections (8)

A flitch beam with steel plates bolted to wood is treated as composite sections. Given modulus of material 1 (base) (E_1) = 7,900 ksi; modular ratio (n) = 3.3500, determine the modulus of material 2 (E_2) in ksi.

Given

  • modulusofmaterial1(base)(E1)=7,900ksimodulus of material 1 (base) (E_1) = 7,900 ksi
  • modularratio(n)=3.3500modular ratio (n) = 3.3500

Find

modulus of material 2 (E_2), in ksi

Start with the thinking

  • The governing relation printed in this handbook section is Transformed area (composite sections).
  • Everything except E_2 is given, so isolate E_2 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Composite sections made of two materials are analyzed using the modular ratio to transform one material into an equivalent area of the other.
Composite sections transformed areacentroidal axisb = 8h = 14

Figure 8 — schematic for Transformed area (composite sections) — solve for modulus of material 2 (case 3) — Composite Sections (8)

Step-by-step solution

  1. Step 1 — State the governing relation:

    n=E2E1n = \dfrac{E_2}{E_1}
  2. Step 2 — Rearrange symbolically for E_2:

    E2=nE1E_{2} = n E_1
  3. Step 3

    Listthegivens:modulusofmaterial1(base)(E1)=7,900ksi,modularratio(n)=3.3500List the givens: modulus of material 1 (base) (E_1) = 7,900 ksi, modular ratio (n) = 3.3500
  4. Step 4 — Substitute the given values:

    E2=3.3500E1E_{2} = 3.3500 E_1
  5. Step 5 — Evaluate:

    E2=26465 ksiE_{2} = 26465\ \text{ksi}
  6. Step 6 — Check: returning E_2 = 26,465 ksi to

    n=E2E1n = \dfrac{E_2}{E_1}

    reproduces the given quantities, and both sides carry the same units.

Answer:
E2=26465 ksiE_{2} = 26465\ \text{ksi}

Why the other options are there

  • 52,930 — kept a factor of two that cancels in the correct rearrangement.
  • 13,233 — dropped that same factor in the other direction.
  • 29,112 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Composite Sections

Example 9
Transformed area (composite sections) — solve for modulus of material 1 (base) (case 3) — Composite Sections (9)

A composite steel-concrete deck is transformed into composite sections for stress analysis. Given modulus of material 2 (E_2) = 14,500 ksi; modular ratio (n) = 7.3000, determine the modulus of material 1 (base) (E_1) in ksi.

Given

  • modulusofmaterial2(E2)=14,500ksimodulus of material 2 (E_2) = 14,500 ksi
  • modularratio(n)=7.3000modular ratio (n) = 7.3000

Find

modulus of material 1 (base) (E_1), in ksi

Start with the thinking

  • The governing relation printed in this handbook section is Transformed area (composite sections).
  • Everything except E_1 is given, so isolate E_1 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Composite sections made of two materials are analyzed using the modular ratio to transform one material into an equivalent area of the other.
Composite sections transformed areacentroidal axisb = 8h = 14

Figure 9 — schematic for Transformed area (composite sections) — solve for modulus of material 1 (base) (case 3) — Composite Sections (9)

Step-by-step solution

  1. Step 1 — State the governing relation:

    n=E2E1n = \dfrac{E_2}{E_1}
  2. Step 2 — Rearrange symbolically for E_1:

    E1=E2nE_{1} = \dfrac{E_2}{n}
  3. Step 3

    Listthegivens:modulusofmaterial2(E2)=14,500ksi,modularratio(n)=7.3000List the givens: modulus of material 2 (E_2) = 14,500 ksi, modular ratio (n) = 7.3000
  4. Step 4 — Substitute the given values:

    E1=E27.3000E_{1} = \dfrac{E_2}{7.3000}
  5. Step 5 — Evaluate:

    E1=1986 ksiE_{1} = 1986\ \text{ksi}
  6. Step 6 — Check: returning E_1 = 1,986 ksi to

    n=E2E1n = \dfrac{E_2}{E_1}

    reproduces the given quantities, and both sides carry the same units.

Answer:
E1=1986 ksiE_{1} = 1986\ \text{ksi}

Why the other options are there

  • 3,973 — kept a factor of two that cancels in the correct rearrangement.
  • 993.2 — dropped that same factor in the other direction.
  • 2,185 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Composite Sections

Example 10
Transformed area (composite sections) — solve for modular ratio (case 4) — Composite Sections (10)

A steel-reinforced timber beam is analyzed as one of the composite sections using a transformed area. Given modulus of material 2 (E_2) = 8,900 ksi; modulus of material 1 (base) (E_1) = 5,400 ksi, determine the modular ratio (n).

Given

  • modulusofmaterial2(E2)=8,900ksimodulus of material 2 (E_2) = 8,900 ksi
  • modulusofmaterial1(base)(E1)=5,400ksimodulus of material 1 (base) (E_1) = 5,400 ksi

Find

modular ratio (n)

Start with the thinking

  • The governing relation printed in this handbook section is Transformed area (composite sections).
  • Everything except n is given, so isolate n symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Composite sections made of two materials are analyzed using the modular ratio to transform one material into an equivalent area of the other.
Composite sections transformed areacentroidal axisb = 8h = 14

Figure 10 — schematic for Transformed area (composite sections) — solve for modular ratio (case 4) — Composite Sections (10)

Step-by-step solution

  1. Step 1 — State the governing relation:

    n=E2E1n = \dfrac{E_2}{E_1}
  2. Step 2 — Rearrange symbolically for n:

    n=E2E1n = \dfrac{E_2}{E_1}
  3. Step 3

    Listthegivens:modulusofmaterial2(E2)=8,900ksi,modulusofmaterial1(base)(E1)=5,400ksiList the givens: modulus of material 2 (E_2) = 8,900 ksi, modulus of material 1 (base) (E_1) = 5,400 ksi
  4. Step 4 — Substitute the given values:

    n=E2E1n = \dfrac{E_2}{E_1}
  5. Step 5 — Evaluate:

    n=1.6481n = 1.6481
  6. Step 6 — Check: returning n = 1.6481 to

    n=E2E1n = \dfrac{E_2}{E_1}

    reproduces the given quantities, and both sides carry the same units.

Answer:
n=1.6481n = 1.6481

Why the other options are there

  • 3.2963 — kept a factor of two that cancels in the correct rearrangement.
  • 0.8241 — dropped that same factor in the other direction.
  • 1.8130 — rounded an intermediate value before the final step.

Reference: FE Handbook — Mechanics of Materials: Composite Sections

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