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10 FE practice problems: effective stress in soil, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

Effective stress is a core idea in the Geotechnical Engineering area of the FE Civil specification: total stress from the soil above, minus the pore water pressure, gives the stress the soil skeleton carries. These ten problems use layered profiles with a water table. Sketch the profile, add up unit weight times thickness layer by layer, and use the saturated unit weight below the water table.

How to use these problems

Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.

The problems

Problem 1 · FE Civil, Effective stress

A soil profile has 18 ft of sand over 16 ft of clay. The water table is 15 ft below the ground surface. The sand has a moist unit weight of 104 lb/ft³ above the water table and a saturated unit weight of 128 lb/ft³ below it. The clay has a saturated unit weight of 117 lb/ft³. Use γw = 62.4 lb/ft³. The vertical effective stress at the middle of the clay layer is most nearly:

  • A 2,190 lb/ft²
  • B 2,880 lb/ft²
  • C 1,510 lb/ft²
  • D 2,120 lb/ft²

Handbook: Civil Engineering, Geotechnical, Effective stress, FE Reference Handbook 10.6

Show the worked solution
Answer
A (2,190 lb/ft²)
Given
H1 = 18 ft, dw = 15 ft, H2 = 16 ft, γm = 104 lb/ft³, γsat,sand = 128 lb/ft³, γsat,clay = 117 lb/ft³, γw = 62.4 lb/ft³
Find
vertical effective stress σ' at the middle of the clay (lb/ft²)
Handbook
Civil Engineering, Geotechnical, Effective stress, page 269
Equation
σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)
Substitute
  1. Depth to the middle of the clay: z = H1 + H2/2 = 18 + 8 = 26 ft
  2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (104)(15) + (128)(3) + (117)(8) = 2,880 lb/ft²
  3. Pore-water pressure: u = γw(z - dw) = (62.4 lb/ft³)(11 ft) = 686.4 lb/ft²
  4. σ' = σ - u = 2,880 - 686.4 = 2,190 lb/ft²
Result
2,190 lb/ft², 3 significant figures
Check
with buoyant unit weights below the water table, γm·dw + (γsat,sand - γw)(H1 - dw) + (γsat,clay - γw)(H2/2) = 2,190 lb/ft², the same answer; units (lb/ft³)(ft) = lb/ft².
Why the others are wrong
  • B: is the total stress; the pore-water pressure u was not subtracted
  • C: used buoyant unit weights below the water table and then subtracted u again
  • D: used the moist unit weight for the whole sand layer, including the part below the water table

Problem 2 · FE Civil, Effective stress

A soil profile has 19 ft of sand over 18 ft of clay. The water table is 13 ft below the ground surface. The sand has a moist unit weight of 104 lb/ft³ above the water table and a saturated unit weight of 134 lb/ft³ below it. The clay has a saturated unit weight of 108 lb/ft³. Use γw = 62.4 lb/ft³. The vertical effective stress at the middle of the clay layer is most nearly:

  • A 1,260 lb/ft²
  • B 3,130 lb/ft²
  • C 1,380 lb/ft²
  • D 2,190 lb/ft²

Handbook: Civil Engineering, Geotechnical, Effective stress, FE Reference Handbook 10.6

Show the worked solution
Answer
D (2,190 lb/ft²)
Given
H1 = 19 ft, dw = 13 ft, H2 = 18 ft, γm = 104 lb/ft³, γsat,sand = 134 lb/ft³, γsat,clay = 108 lb/ft³, γw = 62.4 lb/ft³
Find
vertical effective stress σ' at the middle of the clay (lb/ft²)
Handbook
Civil Engineering, Geotechnical, Effective stress, page 269
Equation
σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)
Substitute
  1. Depth to the middle of the clay: z = H1 + H2/2 = 19 + 9 = 28 ft
  2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (104)(13) + (134)(6) + (108)(9) = 3,128 lb/ft²
  3. Pore-water pressure: u = γw(z - dw) = (62.4 lb/ft³)(15 ft) = 936.0 lb/ft²
  4. σ' = σ - u = 3,128 - 936.0 = 2,190 lb/ft²
Result
2,190 lb/ft², 3 significant figures
Check
with buoyant unit weights below the water table, γm·dw + (γsat,sand - γw)(H1 - dw) + (γsat,clay - γw)(H2/2) = 2,190 lb/ft², the same answer; units (lb/ft³)(ft) = lb/ft².
Why the others are wrong
  • A: used buoyant unit weights below the water table and then subtracted u again
  • B: is the total stress; the pore-water pressure u was not subtracted
  • C: measured the water head from the ground surface instead of from the water table

Problem 3 · FE Civil, Effective stress

A soil profile has 30 ft of sand over 24 ft of clay. The water table is 22 ft below the ground surface. The sand has a moist unit weight of 118 lb/ft³ above the water table and a saturated unit weight of 132 lb/ft³ below it. The clay has a saturated unit weight of 118 lb/ft³. Use γw = 62.4 lb/ft³. The vertical effective stress at the middle of the clay layer is most nearly:

  • A 2,570 lb/ft²
  • B 3,820 lb/ft²
  • C 4,130 lb/ft²
  • D 2,450 lb/ft²

Handbook: Civil Engineering, Geotechnical, Effective stress, FE Reference Handbook 10.6

Show the worked solution
Answer
B (3,820 lb/ft²)
Given
H1 = 30 ft, dw = 22 ft, H2 = 24 ft, γm = 118 lb/ft³, γsat,sand = 132 lb/ft³, γsat,clay = 118 lb/ft³, γw = 62.4 lb/ft³
Find
vertical effective stress σ' at the middle of the clay (lb/ft²)
Handbook
Civil Engineering, Geotechnical, Effective stress, page 269
Equation
σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)
Substitute
  1. Depth to the middle of the clay: z = H1 + H2/2 = 30 + 12 = 42 ft
  2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (118)(22) + (132)(8) + (118)(12) = 5,068 lb/ft²
  3. Pore-water pressure: u = γw(z - dw) = (62.4 lb/ft³)(20 ft) = 1,248 lb/ft²
  4. σ' = σ - u = 5,068 - 1,248 = 3,820 lb/ft²
Result
3,820 lb/ft², 3 significant figures
Check
with buoyant unit weights below the water table, γm·dw + (γsat,sand - γw)(H1 - dw) + (γsat,clay - γw)(H2/2) = 3,820 lb/ft², the same answer; units (lb/ft³)(ft) = lb/ft².
Why the others are wrong
  • A: used buoyant unit weights below the water table and then subtracted u again
  • C: used the saturated unit weight for the whole sand layer, including the part above the water table
  • D: measured the water head from the ground surface instead of from the water table

Problem 4 · FE Civil, Effective stress

A soil profile has 17 ft of sand over 6 ft of clay. The water table is 9 ft below the ground surface. The sand has a moist unit weight of 102 lb/ft³ above the water table and a saturated unit weight of 107 lb/ft³ below it. The clay has a saturated unit weight of 113 lb/ft³. Use γw = 62.4 lb/ft³. The vertical effective stress at the middle of the clay layer is most nearly:

  • A 1,430 lb/ft²
  • B 1,580 lb/ft²
  • C 1,390 lb/ft²
  • D 1,270 lb/ft²

Handbook: Civil Engineering, Geotechnical, Effective stress, FE Reference Handbook 10.6

Show the worked solution
Answer
A (1,430 lb/ft²)
Given
H1 = 17 ft, dw = 9 ft, H2 = 6 ft, γm = 102 lb/ft³, γsat,sand = 107 lb/ft³, γsat,clay = 113 lb/ft³, γw = 62.4 lb/ft³
Find
vertical effective stress σ' at the middle of the clay (lb/ft²)
Handbook
Civil Engineering, Geotechnical, Effective stress, page 269
Equation
σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)
Substitute
  1. Depth to the middle of the clay: z = H1 + H2/2 = 17 + 3 = 20 ft
  2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (102)(9) + (107)(8) + (113)(3) = 2,113 lb/ft²
  3. Pore-water pressure: u = γw(z - dw) = (62.4 lb/ft³)(11 ft) = 686.4 lb/ft²
  4. σ' = σ - u = 2,113 - 686.4 = 1,430 lb/ft²
Result
1,430 lb/ft², 3 significant figures
Check
with buoyant unit weights below the water table, γm·dw + (γsat,sand - γw)(H1 - dw) + (γsat,clay - γw)(H2/2) = 1,430 lb/ft², the same answer; units (lb/ft³)(ft) = lb/ft².
Why the others are wrong
  • B: computed the stress at the bottom of the clay layer instead of its middle
  • C: used the moist unit weight for the whole sand layer, including the part below the water table
  • D: computed the stress at the top of the clay layer instead of its middle

Problem 5 · FE Civil, Effective stress

A soil profile has 6.5 m of sand over 9.5 m of clay. The water table is 1.0 m below the ground surface. The sand has a moist unit weight of 16.1 kN/m³ above the water table and a saturated unit weight of 20.9 kN/m³ below it. The clay has a saturated unit weight of 17.4 kN/m³. Use γw = 9.81 kN/m³. The vertical effective stress at the middle of the clay layer is most nearly:

  • A 149 kPa
  • B 214 kPa
  • C 118 kPa
  • D 113 kPa

Handbook: Civil Engineering, Geotechnical, Effective stress, FE Reference Handbook 10.6

Show the worked solution
Answer
D (113 kPa)
Given
H1 = 6.5 m, dw = 1.0 m, H2 = 9.5 m, γm = 16.1 kN/m³, γsat,sand = 20.9 kN/m³, γsat,clay = 17.4 kN/m³, γw = 9.81 kN/m³
Find
vertical effective stress σ' at the middle of the clay (kPa)
Handbook
Civil Engineering, Geotechnical, Effective stress, page 269
Equation
σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)
Substitute
  1. Depth to the middle of the clay: z = H1 + H2/2 = 6.5 + 4.75 = 11.25 m
  2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (16.1)(1.0) + (20.9)(5.5) + (17.4)(4.75) = 213.7 kPa
  3. Pore-water pressure: u = γw(z - dw) = (9.81 kN/m³)(10.25 m) = 100.6 kPa
  4. σ' = σ - u = 213.7 - 100.6 = 113 kPa
Result
113 kPa, 3 significant figures
Check
with buoyant unit weights below the water table, γm·dw + (γsat,sand - γw)(H1 - dw) + (γsat,clay - γw)(H2/2) = 113 kPa, the same answer; units (kN/m³)(m) = kPa.
Why the others are wrong
  • A: computed the stress at the bottom of the clay layer instead of its middle
  • B: is the total stress; the pore-water pressure u was not subtracted
  • C: used the saturated unit weight for the whole sand layer, including the part above the water table

Problem 6 · FE Civil, Effective stress

A soil profile has 7.0 m of sand over 4.5 m of clay. The water table is 5.5 m below the ground surface. The sand has a moist unit weight of 17.2 kN/m³ above the water table and a saturated unit weight of 21.0 kN/m³ below it. The clay has a saturated unit weight of 17.1 kN/m³. Use γw = 9.81 kN/m³. The vertical effective stress at the middle of the clay layer is most nearly:

  • A 122 kPa
  • B 128 kPa
  • C 149 kPa
  • D 73.8 kPa

Handbook: Civil Engineering, Geotechnical, Effective stress, FE Reference Handbook 10.6

Show the worked solution
Answer
B (128 kPa)
Given
H1 = 7.0 m, dw = 5.5 m, H2 = 4.5 m, γm = 17.2 kN/m³, γsat,sand = 21.0 kN/m³, γsat,clay = 17.1 kN/m³, γw = 9.81 kN/m³
Find
vertical effective stress σ' at the middle of the clay (kPa)
Handbook
Civil Engineering, Geotechnical, Effective stress, page 269
Equation
σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)
Substitute
  1. Depth to the middle of the clay: z = H1 + H2/2 = 7.0 + 2.25 = 9.25 m
  2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (17.2)(5.5) + (21.0)(1.5) + (17.1)(2.25) = 164.6 kPa
  3. Pore-water pressure: u = γw(z - dw) = (9.81 kN/m³)(3.75 m) = 36.79 kPa
  4. σ' = σ - u = 164.6 - 36.79 = 128 kPa
Result
128 kPa, 3 significant figures
Check
with buoyant unit weights below the water table, γm·dw + (γsat,sand - γw)(H1 - dw) + (γsat,clay - γw)(H2/2) = 128 kPa, the same answer; units (kN/m³)(m) = kPa.
Why the others are wrong
  • A: used the moist unit weight for the whole sand layer, including the part below the water table
  • C: used the saturated unit weight for the whole sand layer, including the part above the water table
  • D: measured the water head from the ground surface instead of from the water table

Problem 7 · FE Civil, Effective stress

A soil profile has 7.5 m of sand over 5.5 m of clay. The water table is 3.5 m below the ground surface. The sand has a moist unit weight of 18.4 kN/m³ above the water table and a saturated unit weight of 19.4 kN/m³ below it. The clay has a saturated unit weight of 17.7 kN/m³. Use γw = 9.81 kN/m³. The vertical effective stress at the middle of the clay layer is most nearly:

  • A 191 kPa
  • B 128 kPa
  • C 124 kPa
  • D 120 kPa

Handbook: Civil Engineering, Geotechnical, Effective stress, FE Reference Handbook 10.6

Show the worked solution
Answer
C (124 kPa)
Given
H1 = 7.5 m, dw = 3.5 m, H2 = 5.5 m, γm = 18.4 kN/m³, γsat,sand = 19.4 kN/m³, γsat,clay = 17.7 kN/m³, γw = 9.81 kN/m³
Find
vertical effective stress σ' at the middle of the clay (kPa)
Handbook
Civil Engineering, Geotechnical, Effective stress, page 269
Equation
σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)
Substitute
  1. Depth to the middle of the clay: z = H1 + H2/2 = 7.5 + 2.75 = 10.25 m
  2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (18.4)(3.5) + (19.4)(4.0) + (17.7)(2.75) = 190.7 kPa
  3. Pore-water pressure: u = γw(z - dw) = (9.81 kN/m³)(6.75 m) = 66.22 kPa
  4. σ' = σ - u = 190.7 - 66.22 = 124 kPa
Result
124 kPa, 3 significant figures
Check
with buoyant unit weights below the water table, γm·dw + (γsat,sand - γw)(H1 - dw) + (γsat,clay - γw)(H2/2) = 124 kPa, the same answer; units (kN/m³)(m) = kPa.
Why the others are wrong
  • A: is the total stress; the pore-water pressure u was not subtracted
  • B: used the saturated unit weight for the whole sand layer, including the part above the water table
  • D: used the moist unit weight for the whole sand layer, including the part below the water table

Problem 8 · FE Civil, Effective stress

A soil profile has 3.5 m of sand over 3.5 m of clay. The water table is 2.0 m below the ground surface. The sand has a moist unit weight of 17.4 kN/m³ above the water table and a saturated unit weight of 18.9 kN/m³ below it. The clay has a saturated unit weight of 19.3 kN/m³. Use γw = 9.81 kN/m³. The vertical effective stress at the middle of the clay layer is most nearly:

  • A 96.9 kPa
  • B 48.4 kPa
  • C 33.2 kPa
  • D 65.0 kPa

Handbook: Civil Engineering, Geotechnical, Effective stress, FE Reference Handbook 10.6

Show the worked solution
Answer
D (65.0 kPa)
Given
H1 = 3.5 m, dw = 2.0 m, H2 = 3.5 m, γm = 17.4 kN/m³, γsat,sand = 18.9 kN/m³, γsat,clay = 19.3 kN/m³, γw = 9.81 kN/m³
Find
vertical effective stress σ' at the middle of the clay (kPa)
Handbook
Civil Engineering, Geotechnical, Effective stress, page 269
Equation
σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)
Substitute
  1. Depth to the middle of the clay: z = H1 + H2/2 = 3.5 + 1.75 = 5.25 m
  2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (17.4)(2.0) + (18.9)(1.5) + (19.3)(1.75) = 96.92 kPa
  3. Pore-water pressure: u = γw(z - dw) = (9.81 kN/m³)(3.25 m) = 31.88 kPa
  4. σ' = σ - u = 96.92 - 31.88 = 65.0 kPa
Result
65.0 kPa, 3 significant figures
Check
with buoyant unit weights below the water table, γm·dw + (γsat,sand - γw)(H1 - dw) + (γsat,clay - γw)(H2/2) = 65.0 kPa, the same answer; units (kN/m³)(m) = kPa.
Why the others are wrong
  • A: is the total stress; the pore-water pressure u was not subtracted
  • B: computed the stress at the top of the clay layer instead of its middle
  • C: used buoyant unit weights below the water table and then subtracted u again

Problem 9 · FE Civil, Effective stress

A soil profile has 17 ft of sand over 22 ft of clay. The water table is 11 ft below the ground surface. The sand has a moist unit weight of 113 lb/ft³ above the water table and a saturated unit weight of 133 lb/ft³ below it. The clay has a saturated unit weight of 112 lb/ft³. Use γw = 62.4 lb/ft³. The vertical effective stress at the middle of the clay layer is most nearly:

  • A 2,210 lb/ft²
  • B 2,090 lb/ft²
  • C 3,270 lb/ft²
  • D 1,150 lb/ft²

Handbook: Civil Engineering, Geotechnical, Effective stress, FE Reference Handbook 10.6

Show the worked solution
Answer
A (2,210 lb/ft²)
Given
H1 = 17 ft, dw = 11 ft, H2 = 22 ft, γm = 113 lb/ft³, γsat,sand = 133 lb/ft³, γsat,clay = 112 lb/ft³, γw = 62.4 lb/ft³
Find
vertical effective stress σ' at the middle of the clay (lb/ft²)
Handbook
Civil Engineering, Geotechnical, Effective stress, page 269
Equation
σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)
Substitute
  1. Depth to the middle of the clay: z = H1 + H2/2 = 17 + 11 = 28 ft
  2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (113)(11) + (133)(6) + (112)(11) = 3,273 lb/ft²
  3. Pore-water pressure: u = γw(z - dw) = (62.4 lb/ft³)(17 ft) = 1,061 lb/ft²
  4. σ' = σ - u = 3,273 - 1,061 = 2,210 lb/ft²
Result
2,210 lb/ft², 3 significant figures
Check
with buoyant unit weights below the water table, γm·dw + (γsat,sand - γw)(H1 - dw) + (γsat,clay - γw)(H2/2) = 2,210 lb/ft², the same answer; units (lb/ft³)(ft) = lb/ft².
Why the others are wrong
  • B: used the moist unit weight for the whole sand layer, including the part below the water table
  • C: is the total stress; the pore-water pressure u was not subtracted
  • D: used buoyant unit weights below the water table and then subtracted u again

Problem 10 · FE Civil, Effective stress

A soil profile has 6.5 m of sand over 8.5 m of clay. The water table is 4.0 m below the ground surface. The sand has a moist unit weight of 17.8 kN/m³ above the water table and a saturated unit weight of 20.7 kN/m³ below it. The clay has a saturated unit weight of 17.7 kN/m³. Use γw = 9.81 kN/m³. The vertical effective stress at the middle of the clay layer is most nearly:

  • A 98.4 kPa
  • B 132 kPa
  • C 198 kPa
  • D 92.7 kPa

Handbook: Civil Engineering, Geotechnical, Effective stress, FE Reference Handbook 10.6

Show the worked solution
Answer
B (132 kPa)
Given
H1 = 6.5 m, dw = 4.0 m, H2 = 8.5 m, γm = 17.8 kN/m³, γsat,sand = 20.7 kN/m³, γsat,clay = 17.7 kN/m³, γw = 9.81 kN/m³
Find
vertical effective stress σ' at the middle of the clay (kPa)
Handbook
Civil Engineering, Geotechnical, Effective stress, page 269
Equation
σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)
Substitute
  1. Depth to the middle of the clay: z = H1 + H2/2 = 6.5 + 4.25 = 10.75 m
  2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (17.8)(4.0) + (20.7)(2.5) + (17.7)(4.25) = 198.2 kPa
  3. Pore-water pressure: u = γw(z - dw) = (9.81 kN/m³)(6.75 m) = 66.22 kPa
  4. σ' = σ - u = 198.2 - 66.22 = 132 kPa
Result
132 kPa, 3 significant figures
Check
with buoyant unit weights below the water table, γm·dw + (γsat,sand - γw)(H1 - dw) + (γsat,clay - γw)(H2/2) = 132 kPa, the same answer; units (kN/m³)(m) = kPa.
Why the others are wrong
  • A: computed the stress at the top of the clay layer instead of its middle
  • C: is the total stress; the pore-water pressure u was not subtracted
  • D: measured the water head from the ground surface instead of from the water table

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

What is the difference between total and effective stress?

Total stress is the weight of everything above a point. Effective stress is total stress minus pore water pressure, and it controls strength and settlement.

Which unit weight do I use below the water table?

Use the saturated unit weight for soil below the water table when you compute total stress, then subtract the pore water pressure.

Where is effective stress in the handbook?

In the Civil Engineering chapter, under geotechnical topics; each solution gives the page.

Sources

  1. NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.