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Compression index

Geotechnical · FE Reference Handbook section

Geotechnical
2 formulas
10 exam-style examples
~49 min
All Geotechnical lectures

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
Primary consolidation settlement of a clay layer — Compression index

A normally consolidated clay layer is 23 ft thick with e₀ = 0.87 and Cc = 0.37. The initial effective stress at mid-depth is 2,327 psf and a fill adds Δσ = 1,415 psf. Find the primary settlement.

Given

  • H=23ftH = 23 ft
  • e0=0.87e_{0} = 0.87
  • Cc=0.37Cc = 0.37
  • σ0′=2,327psf\sigma'_{0} = 2,327 psf
  • Δσ=1,415psf\Delta\sigma = 1,415 psf

Find

Primary consolidation settlement S

Start with the thinking

  • Normally consolidated → use the virgin compression branch only.
  • Stresses are evaluated at mid-depth of the compressible layer.
Fill (Δσ = 1,415 psf)Clay (Cc = 0.37)

Figure 1 — schematic for Primary consolidation settlement of a clay layer — Compression index

Step-by-step solution

  1. Settlement

    S=(CcH)/(1+e0)⋅log10[(σ0′+Δσ)/σ0′]S = (Cc H)/(1+e_{0}) \cdot log_{10}[(\sigma'_{0}+\Delta\sigma)/\sigma'_{0}]

  2. Stress ratio

    (2,327+1,415)/2,327=1.608(2,327+1,415)/2,327 = 1.608

  3. Log term

    log10(1.608)=0.2063log_{10}(1.608) = 0.2063

  4. Coefficient

    CcH/(1+e0)=0.37(23)/(1+0.87)=4.551ftCcH/(1+e_{0}) = 0.37(23)/(1+0.87) = 4.551 ft

  5. Settlement

    S=4.551(0.2063)=0.939ft=11.27inS = 4.551(0.2063) = 0.939 ft = 11.27 in
Answer:

S ≈ 11.3 in. (0.94 ft)

Why the other options are there

  • 25.9 in. (natural log used)
  • 21.1 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 2
Primary consolidation settlement of a clay layer — Compression index (2)

A normally consolidated clay layer is 19 ft thick with e₀ = 0.97 and Cc = 0.21. The initial effective stress at mid-depth is 2,616 psf and a fill adds Δσ = 1,946 psf. Find the primary settlement.

Given

  • H=19ftH = 19 ft
  • e0=0.97e_{0} = 0.97
  • Cc=0.21Cc = 0.21
  • σ0′=2,616psf\sigma'_{0} = 2,616 psf
  • Δσ=1,946psf\Delta\sigma = 1,946 psf

Find

Primary consolidation settlement S

Start with the thinking

  • Normally consolidated → use the virgin compression branch only.
  • Stresses are evaluated at mid-depth of the compressible layer.
Fill (Δσ = 1,946 psf)Clay (Cc = 0.21)

Figure 2 — schematic for Primary consolidation settlement of a clay layer — Compression index (2)

Step-by-step solution

  1. Settlement

    S=(CcH)/(1+e0)⋅log10[(σ0′+Δσ)/σ0′]S = (Cc H)/(1+e_{0}) \cdot log_{10}[(\sigma'_{0}+\Delta\sigma)/\sigma'_{0}]

  2. Stress ratio

    (2,616+1,946)/2,616=1.744(2,616+1,946)/2,616 = 1.744

  3. Log term

    log10(1.744)=0.2415log_{10}(1.744) = 0.2415

  4. Coefficient

    CcH/(1+e0)=0.21(19)/(1+0.97)=2.025ftCcH/(1+e_{0}) = 0.21(19)/(1+0.97) = 2.025 ft

  5. Settlement

    S=2.025(0.2415)=0.489ft=5.87inS = 2.025(0.2415) = 0.489 ft = 5.87 in
Answer:

S ≈ 5.9 in. (0.49 ft)

Why the other options are there

  • 13.5 in. (natural log used)
  • 11.6 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 3
Primary consolidation settlement of a clay layer — Compression index (3)

A normally consolidated clay layer is 21 ft thick with e₀ = 1.03 and Cc = 0.34. The initial effective stress at mid-depth is 2,927 psf and a fill adds Δσ = 922.0 psf. Find the primary settlement.

Given

  • H=21ftH = 21 ft
  • e0=1.03e_{0} = 1.03
  • Cc=0.34Cc = 0.34
  • σ0′=2,927psf\sigma'_{0} = 2,927 psf
  • Δσ=922.0psf\Delta\sigma = 922.0 psf

Find

Primary consolidation settlement S

Start with the thinking

  • Normally consolidated → use the virgin compression branch only.
  • Stresses are evaluated at mid-depth of the compressible layer.
Fill (Δσ = 922.0 psf)Clay (Cc = 0.34)

Figure 3 — schematic for Primary consolidation settlement of a clay layer — Compression index (3)

Step-by-step solution

  1. Settlement

    S=(CcH)/(1+e0)⋅log10[(σ0′+Δσ)/σ0′]S = (Cc H)/(1+e_{0}) \cdot log_{10}[(\sigma'_{0}+\Delta\sigma)/\sigma'_{0}]

  2. Stress ratio

    (2,927+922.0)/2,927=1.315(2,927+922.0)/2,927 = 1.315

  3. Log term

    log10(1.315)=0.1189log_{10}(1.315) = 0.1189

  4. Coefficient

    CcH/(1+e0)=0.34(21)/(1+1.03)=3.517ftCcH/(1+e_{0}) = 0.34(21)/(1+1.03) = 3.517 ft

  5. Settlement

    S=3.517(0.1189)=0.418ft=5.02inS = 3.517(0.1189) = 0.418 ft = 5.02 in
Answer:

S ≈ 5.0 in. (0.42 ft)

Why the other options are there

  • 11.6 in. (natural log used)
  • 10.2 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 4
Primary consolidation settlement of a clay layer — Compression index (4)

A normally consolidated clay layer is 17 ft thick with e₀ = 0.80 and Cc = 0.45. The initial effective stress at mid-depth is 2,261 psf and a fill adds Δσ = 827.0 psf. Find the primary settlement.

Given

  • H=17ftH = 17 ft
  • e0=0.80e_{0} = 0.80
  • Cc=0.45Cc = 0.45
  • σ0′=2,261psf\sigma'_{0} = 2,261 psf
  • Δσ=827.0psf\Delta\sigma = 827.0 psf

Find

Primary consolidation settlement S

Start with the thinking

  • Normally consolidated → use the virgin compression branch only.
  • Stresses are evaluated at mid-depth of the compressible layer.
Fill (Δσ = 827.0 psf)Clay (Cc = 0.45)

Figure 4 — schematic for Primary consolidation settlement of a clay layer — Compression index (4)

Step-by-step solution

  1. Settlement

    S=(CcH)/(1+e0)⋅log10[(σ0′+Δσ)/σ0′]S = (Cc H)/(1+e_{0}) \cdot log_{10}[(\sigma'_{0}+\Delta\sigma)/\sigma'_{0}]

  2. Stress ratio

    (2,261+827.0)/2,261=1.366(2,261+827.0)/2,261 = 1.366

  3. Log term

    log10(1.366)=0.1354log_{10}(1.366) = 0.1354

  4. Coefficient

    CcH/(1+e0)=0.45(17)/(1+0.80)=4.250ftCcH/(1+e_{0}) = 0.45(17)/(1+0.80) = 4.250 ft

  5. Settlement

    S=4.250(0.1354)=0.575ft=6.90inS = 4.250(0.1354) = 0.575 ft = 6.90 in
Answer:

S ≈ 6.9 in. (0.58 ft)

Why the other options are there

  • 15.9 in. (natural log used)
  • 12.4 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 5
Primary consolidation settlement of a clay layer — Compression index (5)

A normally consolidated clay layer is 25 ft thick with e₀ = 0.74 and Cc = 0.45. The initial effective stress at mid-depth is 2,405 psf and a fill adds Δσ = 1,084 psf. Find the primary settlement.

Given

  • H=25ftH = 25 ft
  • e0=0.74e_{0} = 0.74
  • Cc=0.45Cc = 0.45
  • σ0′=2,405psf\sigma'_{0} = 2,405 psf
  • Δσ=1,084psf\Delta\sigma = 1,084 psf

Find

Primary consolidation settlement S

Start with the thinking

  • Normally consolidated → use the virgin compression branch only.
  • Stresses are evaluated at mid-depth of the compressible layer.
Fill (Δσ = 1,084 psf)Clay (Cc = 0.45)

Figure 5 — schematic for Primary consolidation settlement of a clay layer — Compression index (5)

Step-by-step solution

  1. Settlement

    S=(CcH)/(1+e0)⋅log10[(σ0′+Δσ)/σ0′]S = (Cc H)/(1+e_{0}) \cdot log_{10}[(\sigma'_{0}+\Delta\sigma)/\sigma'_{0}]

  2. Stress ratio

    (2,405+1,084)/2,405=1.451(2,405+1,084)/2,405 = 1.451

  3. Log term

    log10(1.451)=0.1616log_{10}(1.451) = 0.1616

  4. Coefficient

    CcH/(1+e0)=0.45(25)/(1+0.74)=6.466ftCcH/(1+e_{0}) = 0.45(25)/(1+0.74) = 6.466 ft

  5. Settlement

    S=6.466(0.1616)=1.045ft=12.54inS = 6.466(0.1616) = 1.045 ft = 12.54 in
Answer:

S ≈ 12.5 in. (1.04 ft)

Why the other options are there

  • 28.9 in. (natural log used)
  • 21.8 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 6
Primary consolidation settlement of a clay layer — Compression index (6)

A normally consolidated clay layer is 17 ft thick with e₀ = 0.71 and Cc = 0.43. The initial effective stress at mid-depth is 2,580 psf and a fill adds Δσ = 641.0 psf. Find the primary settlement.

Given

  • H=17ftH = 17 ft
  • e0=0.71e_{0} = 0.71
  • Cc=0.43Cc = 0.43
  • σ0′=2,580psf\sigma'_{0} = 2,580 psf
  • Δσ=641.0psf\Delta\sigma = 641.0 psf

Find

Primary consolidation settlement S

Start with the thinking

  • Normally consolidated → use the virgin compression branch only.
  • Stresses are evaluated at mid-depth of the compressible layer.
Fill (Δσ = 641.0 psf)Clay (Cc = 0.43)

Figure 6 — schematic for Primary consolidation settlement of a clay layer — Compression index (6)

Step-by-step solution

  1. Settlement

    S=(CcH)/(1+e0)⋅log10[(σ0′+Δσ)/σ0′]S = (Cc H)/(1+e_{0}) \cdot log_{10}[(\sigma'_{0}+\Delta\sigma)/\sigma'_{0}]

  2. Stress ratio

    (2,580+641.0)/2,580=1.248(2,580+641.0)/2,580 = 1.248

  3. Log term

    log10(1.248)=0.0964log_{10}(1.248) = 0.0964

  4. Coefficient

    CcH/(1+e0)=0.43(17)/(1+0.71)=4.275ftCcH/(1+e_{0}) = 0.43(17)/(1+0.71) = 4.275 ft

  5. Settlement

    S=4.275(0.0964)=0.412ft=4.94inS = 4.275(0.0964) = 0.412 ft = 4.94 in
Answer:

S ≈ 4.9 in. (0.41 ft)

Why the other options are there

  • 11.4 in. (natural log used)
  • 8.5 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 7
Primary consolidation settlement of a clay layer — Compression index (7)

A normally consolidated clay layer is 15 ft thick with e₀ = 1.13 and Cc = 0.23. The initial effective stress at mid-depth is 2,829 psf and a fill adds Δσ = 1,385 psf. Find the primary settlement.

Given

  • H=15ftH = 15 ft
  • e0=1.13e_{0} = 1.13
  • Cc=0.23Cc = 0.23
  • σ0′=2,829psf\sigma'_{0} = 2,829 psf
  • Δσ=1,385psf\Delta\sigma = 1,385 psf

Find

Primary consolidation settlement S

Start with the thinking

  • Normally consolidated → use the virgin compression branch only.
  • Stresses are evaluated at mid-depth of the compressible layer.
Fill (Δσ = 1,385 psf)Clay (Cc = 0.23)

Figure 7 — schematic for Primary consolidation settlement of a clay layer — Compression index (7)

Step-by-step solution

  1. Settlement

    S=(CcH)/(1+e0)⋅log10[(σ0′+Δσ)/σ0′]S = (Cc H)/(1+e_{0}) \cdot log_{10}[(\sigma'_{0}+\Delta\sigma)/\sigma'_{0}]

  2. Stress ratio

    (2,829+1,385)/2,829=1.490(2,829+1,385)/2,829 = 1.490

  3. Log term

    log10(1.490)=0.1731log_{10}(1.490) = 0.1731

  4. Coefficient

    CcH/(1+e0)=0.23(15)/(1+1.13)=1.620ftCcH/(1+e_{0}) = 0.23(15)/(1+1.13) = 1.620 ft

  5. Settlement

    S=1.620(0.1731)=0.280ft=3.36inS = 1.620(0.1731) = 0.280 ft = 3.36 in
Answer:

S ≈ 3.4 in. (0.28 ft)

Why the other options are there

  • 7.7 in. (natural log used)
  • 7.2 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 8
Primary consolidation settlement of a clay layer — Compression index (8)

A normally consolidated clay layer is 9 ft thick with e₀ = 0.92 and Cc = 0.44. The initial effective stress at mid-depth is 2,547 psf and a fill adds Δσ = 1,141 psf. Find the primary settlement.

Given

  • H=9ftH = 9 ft
  • e0=0.92e_{0} = 0.92
  • Cc=0.44Cc = 0.44
  • σ0′=2,547psf\sigma'_{0} = 2,547 psf
  • Δσ=1,141psf\Delta\sigma = 1,141 psf

Find

Primary consolidation settlement S

Start with the thinking

  • Normally consolidated → use the virgin compression branch only.
  • Stresses are evaluated at mid-depth of the compressible layer.
Fill (Δσ = 1,141 psf)Clay (Cc = 0.44)

Figure 8 — schematic for Primary consolidation settlement of a clay layer — Compression index (8)

Step-by-step solution

  1. Settlement

    S=(CcH)/(1+e0)⋅log10[(σ0′+Δσ)/σ0′]S = (Cc H)/(1+e_{0}) \cdot log_{10}[(\sigma'_{0}+\Delta\sigma)/\sigma'_{0}]

  2. Stress ratio

    (2,547+1,141)/2,547=1.448(2,547+1,141)/2,547 = 1.448

  3. Log term

    log10(1.448)=0.1608log_{10}(1.448) = 0.1608

  4. Coefficient

    CcH/(1+e0)=0.44(9)/(1+0.92)=2.063ftCcH/(1+e_{0}) = 0.44(9)/(1+0.92) = 2.063 ft

  5. Settlement

    S=2.063(0.1608)=0.332ft=3.98inS = 2.063(0.1608) = 0.332 ft = 3.98 in
Answer:

S ≈ 4.0 in. (0.33 ft)

Why the other options are there

  • 9.2 in. (natural log used)
  • 7.6 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 9
Primary consolidation settlement of a clay layer — Compression index (9)

A normally consolidated clay layer is 16 ft thick with e₀ = 0.96 and Cc = 0.23. The initial effective stress at mid-depth is 1,316 psf and a fill adds Δσ = 1,895 psf. Find the primary settlement.

Given

  • H=16ftH = 16 ft
  • e0=0.96e_{0} = 0.96
  • Cc=0.23Cc = 0.23
  • σ0′=1,316psf\sigma'_{0} = 1,316 psf
  • Δσ=1,895psf\Delta\sigma = 1,895 psf

Find

Primary consolidation settlement S

Start with the thinking

  • Normally consolidated → use the virgin compression branch only.
  • Stresses are evaluated at mid-depth of the compressible layer.
Fill (Δσ = 1,895 psf)Clay (Cc = 0.23)

Figure 9 — schematic for Primary consolidation settlement of a clay layer — Compression index (9)

Step-by-step solution

  1. Settlement

    S=(CcH)/(1+e0)⋅log10[(σ0′+Δσ)/σ0′]S = (Cc H)/(1+e_{0}) \cdot log_{10}[(\sigma'_{0}+\Delta\sigma)/\sigma'_{0}]

  2. Stress ratio

    (1,316+1,895)/1,316=2.440(1,316+1,895)/1,316 = 2.440

  3. Log term

    log10(2.440)=0.3874log_{10}(2.440) = 0.3874

  4. Coefficient

    CcH/(1+e0)=0.23(16)/(1+0.96)=1.878ftCcH/(1+e_{0}) = 0.23(16)/(1+0.96) = 1.878 ft

  5. Settlement

    S=1.878(0.3874)=0.727ft=8.73inS = 1.878(0.3874) = 0.727 ft = 8.73 in
Answer:

S ≈ 8.7 in. (0.73 ft)

Why the other options are there

  • 20.1 in. (natural log used)
  • 17.1 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 10
Primary consolidation settlement of a clay layer — Compression index (10)

A normally consolidated clay layer is 13 ft thick with e₀ = 1.20 and Cc = 0.29. The initial effective stress at mid-depth is 1,872 psf and a fill adds Δσ = 753.0 psf. Find the primary settlement.

Given

  • H=13ftH = 13 ft
  • e0=1.20e_{0} = 1.20
  • Cc=0.29Cc = 0.29
  • σ0′=1,872psf\sigma'_{0} = 1,872 psf
  • Δσ=753.0psf\Delta\sigma = 753.0 psf

Find

Primary consolidation settlement S

Start with the thinking

  • Normally consolidated → use the virgin compression branch only.
  • Stresses are evaluated at mid-depth of the compressible layer.
Fill (Δσ = 753.0 psf)Clay (Cc = 0.29)

Figure 10 — schematic for Primary consolidation settlement of a clay layer — Compression index (10)

Step-by-step solution

  1. Settlement

    S=(CcH)/(1+e0)⋅log10[(σ0′+Δσ)/σ0′]S = (Cc H)/(1+e_{0}) \cdot log_{10}[(\sigma'_{0}+\Delta\sigma)/\sigma'_{0}]

  2. Stress ratio

    (1,872+753.0)/1,872=1.402(1,872+753.0)/1,872 = 1.402

  3. Log term

    log10(1.402)=0.1468log_{10}(1.402) = 0.1468

  4. Coefficient

    CcH/(1+e0)=0.29(13)/(1+1.20)=1.714ftCcH/(1+e_{0}) = 0.29(13)/(1+1.20) = 1.714 ft

  5. Settlement

    S=1.714(0.1468)=0.252ft=3.02inS = 1.714(0.1468) = 0.252 ft = 3.02 in
Answer:

S ≈ 3.0 in. (0.25 ft)

Why the other options are there

  • 7.0 in. (natural log used)
  • 6.6 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

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