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Recompression 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 — Recompression index

A normally consolidated clay layer is 23 ft thick with e₀ = 1.17 and Cc = 0.41. The initial effective stress at mid-depth is 1,738 psf and a fill adds Δσ = 1,166 psf. Find the primary settlement.

Given

  • H=23ftH = 23 ft
  • e0=1.17e_{0} = 1.17
  • Cc=0.41Cc = 0.41
  • σ0′=1,738psf\sigma'_{0} = 1,738 psf
  • Δσ=1,166psf\Delta\sigma = 1,166 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 = 1,166 psf)Clay (Cc = 0.41)

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

    (1,738+1,166)/1,738=1.671(1,738+1,166)/1,738 = 1.671

  3. Log term

    log10(1.671)=0.2229log_{10}(1.671) = 0.2229

  4. Coefficient

    CcH/(1+e0)=0.41(23)/(1+1.17)=4.346ftCcH/(1+e_{0}) = 0.41(23)/(1+1.17) = 4.346 ft

  5. Settlement

    S=4.346(0.2229)=0.969ft=11.63inS = 4.346(0.2229) = 0.969 ft = 11.63 in
Answer:

S ≈ 11.6 in. (0.97 ft)

Why the other options are there

  • 26.8 in. (natural log used)
  • 25.2 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Recompression index

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

A normally consolidated clay layer is 8 ft thick with e₀ = 1.16 and Cc = 0.42. The initial effective stress at mid-depth is 1,337 psf and a fill adds Δσ = 757.0 psf. Find the primary settlement.

Given

  • H=8ftH = 8 ft
  • e0=1.16e_{0} = 1.16
  • Cc=0.42Cc = 0.42
  • σ0′=1,337psf\sigma'_{0} = 1,337 psf
  • Δσ=757.0psf\Delta\sigma = 757.0 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 = 757.0 psf)Clay (Cc = 0.42)

Figure 2 — schematic for Primary consolidation settlement of a clay layer — Recompression 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

    (1,337+757.0)/1,337=1.566(1,337+757.0)/1,337 = 1.566

  3. Log term

    log10(1.566)=0.1948log_{10}(1.566) = 0.1948

  4. Coefficient

    CcH/(1+e0)=0.42(8)/(1+1.16)=1.556ftCcH/(1+e_{0}) = 0.42(8)/(1+1.16) = 1.556 ft

  5. Settlement

    S=1.556(0.1948)=0.303ft=3.64inS = 1.556(0.1948) = 0.303 ft = 3.64 in
Answer:

S ≈ 3.6 in. (0.30 ft)

Why the other options are there

  • 8.4 in. (natural log used)
  • 7.9 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Recompression index

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

A normally consolidated clay layer is 10 ft thick with e₀ = 0.85 and Cc = 0.24. The initial effective stress at mid-depth is 2,467 psf and a fill adds Δσ = 1,929 psf. Find the primary settlement.

Given

  • H=10ftH = 10 ft
  • e0=0.85e_{0} = 0.85
  • Cc=0.24Cc = 0.24
  • σ0′=2,467psf\sigma'_{0} = 2,467 psf
  • Δσ=1,929psf\Delta\sigma = 1,929 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 = 1,929 psf)Clay (Cc = 0.24)

Figure 3 — schematic for Primary consolidation settlement of a clay layer — Recompression 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

    (2,467+1,929)/2,467=1.782(2,467+1,929)/2,467 = 1.782

  3. Log term

    log10(1.782)=0.2509log_{10}(1.782) = 0.2509

  4. Coefficient

    CcH/(1+e0)=0.24(10)/(1+0.85)=1.297ftCcH/(1+e_{0}) = 0.24(10)/(1+0.85) = 1.297 ft

  5. Settlement

    S=1.297(0.2509)=0.325ft=3.91inS = 1.297(0.2509) = 0.325 ft = 3.91 in
Answer:

S ≈ 3.9 in. (0.33 ft)

Why the other options are there

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

Reference: FE Reference Handbook — Geotechnical → Recompression index

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

A normally consolidated clay layer is 19 ft thick with e₀ = 0.85 and Cc = 0.43. The initial effective stress at mid-depth is 1,840 psf and a fill adds Δσ = 1,045 psf. Find the primary settlement.

Given

  • H=19ftH = 19 ft
  • e0=0.85e_{0} = 0.85
  • Cc=0.43Cc = 0.43
  • σ0′=1,840psf\sigma'_{0} = 1,840 psf
  • Δσ=1,045psf\Delta\sigma = 1,045 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 = 1,045 psf)Clay (Cc = 0.43)

Figure 4 — schematic for Primary consolidation settlement of a clay layer — Recompression 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

    (1,840+1,045)/1,840=1.568(1,840+1,045)/1,840 = 1.568

  3. Log term

    log10(1.568)=0.1953log_{10}(1.568) = 0.1953

  4. Coefficient

    CcH/(1+e0)=0.43(19)/(1+0.85)=4.416ftCcH/(1+e_{0}) = 0.43(19)/(1+0.85) = 4.416 ft

  5. Settlement

    S=4.416(0.1953)=0.863ft=10.35inS = 4.416(0.1953) = 0.863 ft = 10.35 in
Answer:

S ≈ 10.4 in. (0.86 ft)

Why the other options are there

  • 23.8 in. (natural log used)
  • 19.1 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Recompression index

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

A normally consolidated clay layer is 19 ft thick with e₀ = 0.95 and Cc = 0.24. The initial effective stress at mid-depth is 1,297 psf and a fill adds Δσ = 856.0 psf. Find the primary settlement.

Given

  • H=19ftH = 19 ft
  • e0=0.95e_{0} = 0.95
  • Cc=0.24Cc = 0.24
  • σ0′=1,297psf\sigma'_{0} = 1,297 psf
  • Δσ=856.0psf\Delta\sigma = 856.0 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 = 856.0 psf)Clay (Cc = 0.24)

Figure 5 — schematic for Primary consolidation settlement of a clay layer — Recompression 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

    (1,297+856.0)/1,297=1.660(1,297+856.0)/1,297 = 1.660

  3. Log term

    log10(1.660)=0.2201log_{10}(1.660) = 0.2201

  4. Coefficient

    CcH/(1+e0)=0.24(19)/(1+0.95)=2.338ftCcH/(1+e_{0}) = 0.24(19)/(1+0.95) = 2.338 ft

  5. Settlement

    S=2.338(0.2201)=0.515ft=6.18inS = 2.338(0.2201) = 0.515 ft = 6.18 in
Answer:

S ≈ 6.2 in. (0.51 ft)

Why the other options are there

  • 14.2 in. (natural log used)
  • 12.0 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Recompression index

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

A normally consolidated clay layer is 12 ft thick with e₀ = 0.94 and Cc = 0.31. The initial effective stress at mid-depth is 1,251 psf and a fill adds Δσ = 1,310 psf. Find the primary settlement.

Given

  • H=12ftH = 12 ft
  • e0=0.94e_{0} = 0.94
  • Cc=0.31Cc = 0.31
  • σ0′=1,251psf\sigma'_{0} = 1,251 psf
  • Δσ=1,310psf\Delta\sigma = 1,310 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 = 1,310 psf)Clay (Cc = 0.31)

Figure 6 — schematic for Primary consolidation settlement of a clay layer — Recompression 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

    (1,251+1,310)/1,251=2.047(1,251+1,310)/1,251 = 2.047

  3. Log term

    log10(2.047)=0.3112log_{10}(2.047) = 0.3112

  4. Coefficient

    CcH/(1+e0)=0.31(12)/(1+0.94)=1.918ftCcH/(1+e_{0}) = 0.31(12)/(1+0.94) = 1.918 ft

  5. Settlement

    S=1.918(0.3112)=0.597ft=7.16inS = 1.918(0.3112) = 0.597 ft = 7.16 in
Answer:

S ≈ 7.2 in. (0.60 ft)

Why the other options are there

  • 16.5 in. (natural log used)
  • 13.9 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Recompression index

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

A normally consolidated clay layer is 18 ft thick with e₀ = 1.08 and Cc = 0.40. The initial effective stress at mid-depth is 2,061 psf and a fill adds Δσ = 1,442 psf. Find the primary settlement.

Given

  • H=18ftH = 18 ft
  • e0=1.08e_{0} = 1.08
  • Cc=0.40Cc = 0.40
  • σ0′=2,061psf\sigma'_{0} = 2,061 psf
  • Δσ=1,442psf\Delta\sigma = 1,442 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 = 1,442 psf)Clay (Cc = 0.40)

Figure 7 — schematic for Primary consolidation settlement of a clay layer — Recompression 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

    (2,061+1,442)/2,061=1.700(2,061+1,442)/2,061 = 1.700

  3. Log term

    log10(1.700)=0.2304log_{10}(1.700) = 0.2304

  4. Coefficient

    CcH/(1+e0)=0.40(18)/(1+1.08)=3.462ftCcH/(1+e_{0}) = 0.40(18)/(1+1.08) = 3.462 ft

  5. Settlement

    S=3.462(0.2304)=0.797ft=9.57inS = 3.462(0.2304) = 0.797 ft = 9.57 in
Answer:

S ≈ 9.6 in. (0.80 ft)

Why the other options are there

  • 22.0 in. (natural log used)
  • 19.9 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Recompression index

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

A normally consolidated clay layer is 19 ft thick with e₀ = 1.07 and Cc = 0.30. The initial effective stress at mid-depth is 2,542 psf and a fill adds Δσ = 1,256 psf. Find the primary settlement.

Given

  • H=19ftH = 19 ft
  • e0=1.07e_{0} = 1.07
  • Cc=0.30Cc = 0.30
  • σ0′=2,542psf\sigma'_{0} = 2,542 psf
  • Δσ=1,256psf\Delta\sigma = 1,256 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 = 1,256 psf)Clay (Cc = 0.30)

Figure 8 — schematic for Primary consolidation settlement of a clay layer — Recompression 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

    (2,542+1,256)/2,542=1.494(2,542+1,256)/2,542 = 1.494

  3. Log term

    log10(1.494)=0.1744log_{10}(1.494) = 0.1744

  4. Coefficient

    CcH/(1+e0)=0.30(19)/(1+1.07)=2.754ftCcH/(1+e_{0}) = 0.30(19)/(1+1.07) = 2.754 ft

  5. Settlement

    S=2.754(0.1744)=0.480ft=5.76inS = 2.754(0.1744) = 0.480 ft = 5.76 in
Answer:

S ≈ 5.8 in. (0.48 ft)

Why the other options are there

  • 13.3 in. (natural log used)
  • 11.9 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Recompression index

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

A normally consolidated clay layer is 20 ft thick with e₀ = 0.82 and Cc = 0.21. The initial effective stress at mid-depth is 2,545 psf and a fill adds Δσ = 1,654 psf. Find the primary settlement.

Given

  • H=20ftH = 20 ft
  • e0=0.82e_{0} = 0.82
  • Cc=0.21Cc = 0.21
  • σ0′=2,545psf\sigma'_{0} = 2,545 psf
  • Δσ=1,654psf\Delta\sigma = 1,654 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 = 1,654 psf)Clay (Cc = 0.21)

Figure 9 — schematic for Primary consolidation settlement of a clay layer — Recompression 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

    (2,545+1,654)/2,545=1.650(2,545+1,654)/2,545 = 1.650

  3. Log term

    log10(1.650)=0.2175log_{10}(1.650) = 0.2175

  4. Coefficient

    CcH/(1+e0)=0.21(20)/(1+0.82)=2.308ftCcH/(1+e_{0}) = 0.21(20)/(1+0.82) = 2.308 ft

  5. Settlement

    S=2.308(0.2175)=0.502ft=6.02inS = 2.308(0.2175) = 0.502 ft = 6.02 in
Answer:

S ≈ 6.0 in. (0.50 ft)

Why the other options are there

  • 13.9 in. (natural log used)
  • 11.0 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Recompression index

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

A normally consolidated clay layer is 11 ft thick with e₀ = 0.99 and Cc = 0.28. The initial effective stress at mid-depth is 1,254 psf and a fill adds Δσ = 1,883 psf. Find the primary settlement.

Given

  • H=11ftH = 11 ft
  • e0=0.99e_{0} = 0.99
  • Cc=0.28Cc = 0.28
  • σ0′=1,254psf\sigma'_{0} = 1,254 psf
  • Δσ=1,883psf\Delta\sigma = 1,883 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 = 1,883 psf)Clay (Cc = 0.28)

Figure 10 — schematic for Primary consolidation settlement of a clay layer — Recompression 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

    (1,254+1,883)/1,254=2.502(1,254+1,883)/1,254 = 2.502

  3. Log term

    log10(2.502)=0.3982log_{10}(2.502) = 0.3982

  4. Coefficient

    CcH/(1+e0)=0.28(11)/(1+0.99)=1.548ftCcH/(1+e_{0}) = 0.28(11)/(1+0.99) = 1.548 ft

  5. Settlement

    S=1.548(0.3982)=0.616ft=7.40inS = 1.548(0.3982) = 0.616 ft = 7.40 in
Answer:

S ≈ 7.4 in. (0.62 ft)

Why the other options are there

  • 17.0 in. (natural log used)
  • 14.7 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Recompression index

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