Horizontal Stress Profiles and Forces

Geotechnical · FE Reference Handbook section

Geotechnical

10 formulas

10 exam-style examples

~60 min

All Geotechnical lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Active forces on retaining wall per unit wall length (as shown):
Passive forces on retaining wall per unit wall length (similar to the active forces shown):
At rest forces on wall per unit length of wall

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

Horizontal Stress Profiles and Forces — solve for horizontal effective stress — Horizontal Stress Profiles and Forces

horizontal stress profiles and forces behind a basement wall Given lateral earth pressure coefficient (K) = 1.6300; vertical effective stress (sigma_v) = 50.0000 kPa, determine the horizontal effective stress (sigma_h) in kPa.

Given

$lateral earth pressure coefficient (K) = 1.6300$
$vertical effective stress (sigma_v) = 50.0000 kPa$

Find

horizontal effective stress (sigma_h), in kPa

Start with the thinking

The governing relation printed in this handbook section is Horizontal Stress Profiles and Forces.
Everything except sigma_h is given, so isolate sigma_h 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.
Horizontal stress profiles and forces are used to determine lateral pressure acting on a retaining structure.

Figure 1 — schematic for Horizontal Stress Profiles and Forces — solve for horizontal effective stress — Horizontal Stress Profiles and Forces

Step-by-step solution

Step 1 — State the governing relation:
$\sigma'_h = K \sigma'_v$
Step 2 — Rearrange the relation so that sigma_h stands alone on the left-hand side.
Step 3 — List the givens: lateral earth pressure coefficient (K) = 1.6300, vertical effective stress (sigma_v) = 50.0000 kPa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma_{h} = 81.5000\ \text{kPa}$
Step 6 — Check: returning sigma_h = 81.5000 kPa to
$\sigma'_h = K \sigma'_v$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma_{h} = 81.5000\ \text{kPa}

Why the other options are there

163.0 — kept a factor of two that cancels in the correct rearrangement.
40.7500 — dropped that same factor in the other direction.
89.6500 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Stress Profiles and Forces

Example 2

Horizontal Stress Profiles and Forces — solve for lateral earth pressure coefficient — Horizontal Stress Profiles and Forces (2)

horizontal stress profiles and forces on a sheet-pile bulkhead Given vertical effective stress (sigma_v) = 39.0000 kPa; horizontal effective stress (sigma_h) = 58.0000 kPa, determine the lateral earth pressure coefficient (K).

Given

$vertical effective stress (sigma_v) = 39.0000 kPa$
$horizontal effective stress (sigma_h) = 58.0000 kPa$

Find

lateral earth pressure coefficient (K)

Start with the thinking

The governing relation printed in this handbook section is Horizontal Stress Profiles and Forces.
Everything except K is given, so isolate K 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.
Horizontal stress profiles and forces are used to determine lateral pressure acting on a retaining structure.

Figure 2 — schematic for Horizontal Stress Profiles and Forces — solve for lateral earth pressure coefficient — Horizontal Stress Profiles and Forces (2)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma'_h = K \sigma'_v$
Step 2 — Rearrange the relation so that K stands alone on the left-hand side.
Step 3 — List the givens: vertical effective stress (sigma_v) = 39.0000 kPa, horizontal effective stress (sigma_h) = 58.0000 kPa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$K = 1.4872$
Step 6 — Check: returning K = 1.4872 to
$\sigma'_h = K \sigma'_v$
reproduces the given quantities, and both sides carry the same units.

Answer:

K = 1.4872

Why the other options are there

2.9744 — kept a factor of two that cancels in the correct rearrangement.
0.7436 — dropped that same factor in the other direction.
1.6359 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Stress Profiles and Forces

Example 3

Horizontal Stress Profiles and Forces — solve for vertical effective stress — Horizontal Stress Profiles and Forces (3)

horizontal stress profiles and forces for lateral earth pressure design Given lateral earth pressure coefficient (K) = 1.8800; horizontal effective stress (sigma_h) = 60.0000 kPa, determine the vertical effective stress (sigma_v) in kPa.

Given

$lateral earth pressure coefficient (K) = 1.8800$
$horizontal effective stress (sigma_h) = 60.0000 kPa$

Find

vertical effective stress (sigma_v), in kPa

Start with the thinking

The governing relation printed in this handbook section is Horizontal Stress Profiles and Forces.
Everything except sigma_v is given, so isolate sigma_v 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.
Horizontal stress profiles and forces are used to determine lateral pressure acting on a retaining structure.

Figure 3 — schematic for Horizontal Stress Profiles and Forces — solve for vertical effective stress — Horizontal Stress Profiles and Forces (3)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma'_h = K \sigma'_v$
Step 2 — Rearrange the relation so that sigma_v stands alone on the left-hand side.
Step 3 — List the givens: lateral earth pressure coefficient (K) = 1.8800, horizontal effective stress (sigma_h) = 60.0000 kPa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma_{v} = 31.9149\ \text{kPa}$
Step 6 — Check: returning sigma_v = 31.9149 kPa to
$\sigma'_h = K \sigma'_v$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma_{v} = 31.9149\ \text{kPa}

Why the other options are there

63.8298 — kept a factor of two that cancels in the correct rearrangement.
15.9574 — dropped that same factor in the other direction.
35.1064 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Stress Profiles and Forces

Example 4

Horizontal Stress Profiles and Forces — solve for horizontal effective stress (case 2) — Horizontal Stress Profiles and Forces (4)

horizontal stress profiles and forces behind a basement wall Given lateral earth pressure coefficient (K) = 2.8400; vertical effective stress (sigma_v) = 186.0 kPa, determine the horizontal effective stress (sigma_h) in kPa.

Given

$lateral earth pressure coefficient (K) = 2.8400$
$vertical effective stress (sigma_v) = 186.0 kPa$

Find

horizontal effective stress (sigma_h), in kPa

Start with the thinking

The governing relation printed in this handbook section is Horizontal Stress Profiles and Forces.
Everything except sigma_h is given, so isolate sigma_h 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.
Horizontal stress profiles and forces are used to determine lateral pressure acting on a retaining structure.

Figure 4 — schematic for Horizontal Stress Profiles and Forces — solve for horizontal effective stress (case 2) — Horizontal Stress Profiles and Forces (4)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma'_h = K \sigma'_v$
Step 2 — Rearrange the relation so that sigma_h stands alone on the left-hand side.
Step 3 — List the givens: lateral earth pressure coefficient (K) = 2.8400, vertical effective stress (sigma_v) = 186.0 kPa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma_{h} = 528.2\ \text{kPa}$
Step 6 — Check: returning sigma_h = 528.2 kPa to
$\sigma'_h = K \sigma'_v$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma_{h} = 528.2\ \text{kPa}

Why the other options are there

1,056 — kept a factor of two that cancels in the correct rearrangement.
264.1 — dropped that same factor in the other direction.
581.1 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Stress Profiles and Forces

Example 5

Horizontal Stress Profiles and Forces — solve for lateral earth pressure coefficient (case 2) — Horizontal Stress Profiles and Forces (5)

horizontal stress profiles and forces on a sheet-pile bulkhead Given vertical effective stress (sigma_v) = 15.0000 kPa; horizontal effective stress (sigma_h) = 6.0000 kPa, determine the lateral earth pressure coefficient (K).

Given

$vertical effective stress (sigma_v) = 15.0000 kPa$
$horizontal effective stress (sigma_h) = 6.0000 kPa$

Find

lateral earth pressure coefficient (K)

Start with the thinking

The governing relation printed in this handbook section is Horizontal Stress Profiles and Forces.
Everything except K is given, so isolate K 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.
Horizontal stress profiles and forces are used to determine lateral pressure acting on a retaining structure.

Figure 5 — schematic for Horizontal Stress Profiles and Forces — solve for lateral earth pressure coefficient (case 2) — Horizontal Stress Profiles and Forces (5)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma'_h = K \sigma'_v$
Step 2 — Rearrange the relation so that K stands alone on the left-hand side.
Step 3 — List the givens: vertical effective stress (sigma_v) = 15.0000 kPa, horizontal effective stress (sigma_h) = 6.0000 kPa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$K = 0.4000$
Step 6 — Check: returning K = 0.4000 to
$\sigma'_h = K \sigma'_v$
reproduces the given quantities, and both sides carry the same units.

Answer:

K = 0.4000

Why the other options are there

0.8000 — kept a factor of two that cancels in the correct rearrangement.
0.2000 — dropped that same factor in the other direction.
0.4400 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Stress Profiles and Forces

Example 6

Horizontal Stress Profiles and Forces — solve for vertical effective stress (case 2) — Horizontal Stress Profiles and Forces (6)

horizontal stress profiles and forces for lateral earth pressure design Given lateral earth pressure coefficient (K) = 1.2400; horizontal effective stress (sigma_h) = 378.0 kPa, determine the vertical effective stress (sigma_v) in kPa.

Given

$lateral earth pressure coefficient (K) = 1.2400$
$horizontal effective stress (sigma_h) = 378.0 kPa$

Find

vertical effective stress (sigma_v), in kPa

Start with the thinking

The governing relation printed in this handbook section is Horizontal Stress Profiles and Forces.
Everything except sigma_v is given, so isolate sigma_v 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.
Horizontal stress profiles and forces are used to determine lateral pressure acting on a retaining structure.

Figure 6 — schematic for Horizontal Stress Profiles and Forces — solve for vertical effective stress (case 2) — Horizontal Stress Profiles and Forces (6)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma'_h = K \sigma'_v$
Step 2 — Rearrange the relation so that sigma_v stands alone on the left-hand side.
Step 3 — List the givens: lateral earth pressure coefficient (K) = 1.2400, horizontal effective stress (sigma_h) = 378.0 kPa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma_{v} = 304.8\ \text{kPa}$
Step 6 — Check: returning sigma_v = 304.8 kPa to
$\sigma'_h = K \sigma'_v$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma_{v} = 304.8\ \text{kPa}

Why the other options are there

609.7 — kept a factor of two that cancels in the correct rearrangement.
152.4 — dropped that same factor in the other direction.
335.3 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Stress Profiles and Forces

Example 7

Horizontal Stress Profiles and Forces — solve for horizontal effective stress (case 3) — Horizontal Stress Profiles and Forces (7)

horizontal stress profiles and forces behind a basement wall Given lateral earth pressure coefficient (K) = 0.3100; vertical effective stress (sigma_v) = 229.0 kPa, determine the horizontal effective stress (sigma_h) in kPa.

Given

$lateral earth pressure coefficient (K) = 0.3100$
$vertical effective stress (sigma_v) = 229.0 kPa$

Find

horizontal effective stress (sigma_h), in kPa

Start with the thinking

The governing relation printed in this handbook section is Horizontal Stress Profiles and Forces.
Everything except sigma_h is given, so isolate sigma_h 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.
Horizontal stress profiles and forces are used to determine lateral pressure acting on a retaining structure.

Figure 7 — schematic for Horizontal Stress Profiles and Forces — solve for horizontal effective stress (case 3) — Horizontal Stress Profiles and Forces (7)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma'_h = K \sigma'_v$
Step 2 — Rearrange the relation so that sigma_h stands alone on the left-hand side.
Step 3 — List the givens: lateral earth pressure coefficient (K) = 0.3100, vertical effective stress (sigma_v) = 229.0 kPa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma_{h} = 70.9900\ \text{kPa}$
Step 6 — Check: returning sigma_h = 70.9900 kPa to
$\sigma'_h = K \sigma'_v$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma_{h} = 70.9900\ \text{kPa}

Why the other options are there

142.0 — kept a factor of two that cancels in the correct rearrangement.
35.4950 — dropped that same factor in the other direction.
78.0890 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Stress Profiles and Forces

Example 8

Horizontal Stress Profiles and Forces — solve for lateral earth pressure coefficient (case 3) — Horizontal Stress Profiles and Forces (8)

horizontal stress profiles and forces on a sheet-pile bulkhead Given vertical effective stress (sigma_v) = 203.0 kPa; horizontal effective stress (sigma_h) = 68.0000 kPa, determine the lateral earth pressure coefficient (K).

Given

$vertical effective stress (sigma_v) = 203.0 kPa$
$horizontal effective stress (sigma_h) = 68.0000 kPa$

Find

lateral earth pressure coefficient (K)

Start with the thinking

The governing relation printed in this handbook section is Horizontal Stress Profiles and Forces.
Everything except K is given, so isolate K 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.
Horizontal stress profiles and forces are used to determine lateral pressure acting on a retaining structure.

Figure 8 — schematic for Horizontal Stress Profiles and Forces — solve for lateral earth pressure coefficient (case 3) — Horizontal Stress Profiles and Forces (8)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma'_h = K \sigma'_v$
Step 2 — Rearrange the relation so that K stands alone on the left-hand side.
Step 3 — List the givens: vertical effective stress (sigma_v) = 203.0 kPa, horizontal effective stress (sigma_h) = 68.0000 kPa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$K = 0.3350$
Step 6 — Check: returning K = 0.3350 to
$\sigma'_h = K \sigma'_v$
reproduces the given quantities, and both sides carry the same units.

Answer:

K = 0.3350

Why the other options are there

0.6700 — kept a factor of two that cancels in the correct rearrangement.
0.1675 — dropped that same factor in the other direction.
0.3685 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Stress Profiles and Forces

Example 9

Horizontal Stress Profiles and Forces — solve for vertical effective stress (case 3) — Horizontal Stress Profiles and Forces (9)

horizontal stress profiles and forces for lateral earth pressure design Given lateral earth pressure coefficient (K) = 1.3500; horizontal effective stress (sigma_h) = 126.0 kPa, determine the vertical effective stress (sigma_v) in kPa.

Given

$lateral earth pressure coefficient (K) = 1.3500$
$horizontal effective stress (sigma_h) = 126.0 kPa$

Find

vertical effective stress (sigma_v), in kPa

Start with the thinking

The governing relation printed in this handbook section is Horizontal Stress Profiles and Forces.
Everything except sigma_v is given, so isolate sigma_v 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.
Horizontal stress profiles and forces are used to determine lateral pressure acting on a retaining structure.

Figure 9 — schematic for Horizontal Stress Profiles and Forces — solve for vertical effective stress (case 3) — Horizontal Stress Profiles and Forces (9)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma'_h = K \sigma'_v$
Step 2 — Rearrange the relation so that sigma_v stands alone on the left-hand side.
Step 3 — List the givens: lateral earth pressure coefficient (K) = 1.3500, horizontal effective stress (sigma_h) = 126.0 kPa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma_{v} = 93.3333\ \text{kPa}$
Step 6 — Check: returning sigma_v = 93.3333 kPa to
$\sigma'_h = K \sigma'_v$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma_{v} = 93.3333\ \text{kPa}

Why the other options are there

186.7 — kept a factor of two that cancels in the correct rearrangement.
46.6667 — dropped that same factor in the other direction.
102.7 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Stress Profiles and Forces

Example 10

Horizontal Stress Profiles and Forces — solve for horizontal effective stress (case 4) — Horizontal Stress Profiles and Forces (10)

horizontal stress profiles and forces behind a basement wall Given lateral earth pressure coefficient (K) = 1.8500; vertical effective stress (sigma_v) = 69.0000 kPa, determine the horizontal effective stress (sigma_h) in kPa.

Given

$lateral earth pressure coefficient (K) = 1.8500$
$vertical effective stress (sigma_v) = 69.0000 kPa$

Find

horizontal effective stress (sigma_h), in kPa

Start with the thinking

The governing relation printed in this handbook section is Horizontal Stress Profiles and Forces.
Everything except sigma_h is given, so isolate sigma_h 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.
Horizontal stress profiles and forces are used to determine lateral pressure acting on a retaining structure.

Figure 10 — schematic for Horizontal Stress Profiles and Forces — solve for horizontal effective stress (case 4) — Horizontal Stress Profiles and Forces (10)

Step-by-step solution

Step 1 — State the governing relation:
$\sigma'_h = K \sigma'_v$
Step 2 — Rearrange the relation so that sigma_h stands alone on the left-hand side.
Step 3 — List the givens: lateral earth pressure coefficient (K) = 1.8500, vertical effective stress (sigma_v) = 69.0000 kPa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\sigma_{h} = 127.7\ \text{kPa}$
Step 6 — Check: returning sigma_h = 127.7 kPa to
$\sigma'_h = K \sigma'_v$
reproduces the given quantities, and both sides carry the same units.

Answer:

\sigma_{h} = 127.7\ \text{kPa}

Why the other options are there

255.3 — kept a factor of two that cancels in the correct rearrangement.
63.8250 — dropped that same factor in the other direction.
140.4 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Stress Profiles and Forces