the factor of safety against seepage liquefaction beneath a levee toe Given critical hydraulic gradient (i_c) = 0.9000; actual (exit) hydraulic gradient (i) = 1.0000, determine the factor of safety (FS).

Start with the thinking

• The governing relation printed in this handbook section is Factor of safety against seepage liquefaction.
• Everything except FS is given, so isolate FS 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.
• The factor of safety against seepage liquefaction compares the critical gradient to the actual exit gradient beneath a structure.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$FS = \dfrac{i_c}{i}$$

2. Step 2 — Rearrange the relation so that FS stands alone on the left-hand side.

3. Step 3 — List the givens: critical hydraulic gradient (i_c) = 0.9000, actual (exit) hydraulic gradient (i) = 1.0000.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$FS = 0.9000$$

6. Step 6 — Check: returning FS = 0.9000 to

   $$FS = \dfrac{i_c}{i}$$

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

Why the other options are there

• 1.8000 — kept a factor of two that cancels in the correct rearrangement.
• 0.4500 — dropped that same factor in the other direction.
• 0.9900 — rounded an intermediate value before the final step.

Reference: FE Handbook — Factor of safety against seepage liquefaction

the factor of safety against seepage liquefaction near a sheet-pile cofferdam Given critical hydraulic gradient (i_c) = 1.1800; factor of safety (FS) = 2.8100, determine the actual (exit) hydraulic gradient (i).

Start with the thinking

• The governing relation printed in this handbook section is Factor of safety against seepage liquefaction.
• Everything except i is given, so isolate i 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.
• The factor of safety against seepage liquefaction compares the critical gradient to the actual exit gradient beneath a structure.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$FS = \dfrac{i_c}{i}$$

2. Step 2 — Rearrange the relation so that i stands alone on the left-hand side.

3. Step 3 — List the givens: critical hydraulic gradient (i_c) = 1.1800, factor of safety (FS) = 2.8100.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$i = 0.4199$$

6. Step 6 — Check: returning i = 0.4199 to

   $$FS = \dfrac{i_c}{i}$$

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

Why the other options are there

• 0.8399 — kept a factor of two that cancels in the correct rearrangement.
• 0.2100 — dropped that same factor in the other direction.
• 0.4619 — rounded an intermediate value before the final step.

Reference: FE Handbook — Factor of safety against seepage liquefaction

the factor of safety against seepage liquefaction under a dam's downstream slope Given actual (exit) hydraulic gradient (i) = 0.9600; factor of safety (FS) = 5.7000, determine the critical hydraulic gradient (i_c).

Start with the thinking

• The governing relation printed in this handbook section is Factor of safety against seepage liquefaction.
• Everything except i_c is given, so isolate i_c 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.
• The factor of safety against seepage liquefaction compares the critical gradient to the actual exit gradient beneath a structure.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$FS = \dfrac{i_c}{i}$$

2. Step 2 — Rearrange the relation so that i_c stands alone on the left-hand side.

3. Step 3 — List the givens: actual (exit) hydraulic gradient (i) = 0.9600, factor of safety (FS) = 5.7000.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$i_{c} = 5.4720$$

6. Step 6 — Check: returning i_c = 5.4720 to

   $$FS = \dfrac{i_c}{i}$$

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

Why the other options are there

• 10.9440 — kept a factor of two that cancels in the correct rearrangement.
• 2.7360 — dropped that same factor in the other direction.
• 6.0192 — rounded an intermediate value before the final step.

Reference: FE Handbook — Factor of safety against seepage liquefaction

the factor of safety against seepage liquefaction beneath a levee toe Given critical hydraulic gradient (i_c) = 1.1800; actual (exit) hydraulic gradient (i) = 1.2000, determine the factor of safety (FS).

Start with the thinking

• The governing relation printed in this handbook section is Factor of safety against seepage liquefaction.
• Everything except FS is given, so isolate FS 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.
• The factor of safety against seepage liquefaction compares the critical gradient to the actual exit gradient beneath a structure.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$FS = \dfrac{i_c}{i}$$

2. Step 2 — Rearrange the relation so that FS stands alone on the left-hand side.

3. Step 3 — List the givens: critical hydraulic gradient (i_c) = 1.1800, actual (exit) hydraulic gradient (i) = 1.2000.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$FS = 0.9833$$

6. Step 6 — Check: returning FS = 0.9833 to

   $$FS = \dfrac{i_c}{i}$$

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

Why the other options are there

• 1.9667 — kept a factor of two that cancels in the correct rearrangement.
• 0.4917 — dropped that same factor in the other direction.
• 1.0817 — rounded an intermediate value before the final step.

Reference: FE Handbook — Factor of safety against seepage liquefaction

the factor of safety against seepage liquefaction near a sheet-pile cofferdam Given critical hydraulic gradient (i_c) = 1.1500; factor of safety (FS) = 1.2000, determine the actual (exit) hydraulic gradient (i).

Start with the thinking

• The governing relation printed in this handbook section is Factor of safety against seepage liquefaction.
• Everything except i is given, so isolate i 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.
• The factor of safety against seepage liquefaction compares the critical gradient to the actual exit gradient beneath a structure.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$FS = \dfrac{i_c}{i}$$

2. Step 2 — Rearrange the relation so that i stands alone on the left-hand side.

3. Step 3 — List the givens: critical hydraulic gradient (i_c) = 1.1500, factor of safety (FS) = 1.2000.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$i = 0.9583$$

6. Step 6 — Check: returning i = 0.9583 to

   $$FS = \dfrac{i_c}{i}$$

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

Why the other options are there

• 1.9167 — kept a factor of two that cancels in the correct rearrangement.
• 0.4792 — dropped that same factor in the other direction.
• 1.0542 — rounded an intermediate value before the final step.

Reference: FE Handbook — Factor of safety against seepage liquefaction

the factor of safety against seepage liquefaction under a dam's downstream slope Given actual (exit) hydraulic gradient (i) = 0.1400; factor of safety (FS) = 4.0300, determine the critical hydraulic gradient (i_c).

Start with the thinking

• The governing relation printed in this handbook section is Factor of safety against seepage liquefaction.
• Everything except i_c is given, so isolate i_c 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.
• The factor of safety against seepage liquefaction compares the critical gradient to the actual exit gradient beneath a structure.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$FS = \dfrac{i_c}{i}$$

2. Step 2 — Rearrange the relation so that i_c stands alone on the left-hand side.

3. Step 3 — List the givens: actual (exit) hydraulic gradient (i) = 0.1400, factor of safety (FS) = 4.0300.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$i_{c} = 0.5642$$

6. Step 6 — Check: returning i_c = 0.5642 to

   $$FS = \dfrac{i_c}{i}$$

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

Why the other options are there

• 1.1284 — kept a factor of two that cancels in the correct rearrangement.
• 0.2821 — dropped that same factor in the other direction.
• 0.6206 — rounded an intermediate value before the final step.

Reference: FE Handbook — Factor of safety against seepage liquefaction

the factor of safety against seepage liquefaction beneath a levee toe Given critical hydraulic gradient (i_c) = 0.8900; actual (exit) hydraulic gradient (i) = 0.1700, determine the factor of safety (FS).

Start with the thinking

• The governing relation printed in this handbook section is Factor of safety against seepage liquefaction.
• Everything except FS is given, so isolate FS 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.
• The factor of safety against seepage liquefaction compares the critical gradient to the actual exit gradient beneath a structure.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$FS = \dfrac{i_c}{i}$$

2. Step 2 — Rearrange the relation so that FS stands alone on the left-hand side.

3. Step 3 — List the givens: critical hydraulic gradient (i_c) = 0.8900, actual (exit) hydraulic gradient (i) = 0.1700.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$FS = 5.2353$$

6. Step 6 — Check: returning FS = 5.2353 to

   $$FS = \dfrac{i_c}{i}$$

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

Why the other options are there

• 10.4706 — kept a factor of two that cancels in the correct rearrangement.
• 2.6176 — dropped that same factor in the other direction.
• 5.7588 — rounded an intermediate value before the final step.

Reference: FE Handbook — Factor of safety against seepage liquefaction

the factor of safety against seepage liquefaction near a sheet-pile cofferdam Given critical hydraulic gradient (i_c) = 1.0800; factor of safety (FS) = 5.8300, determine the actual (exit) hydraulic gradient (i).

Start with the thinking

• The governing relation printed in this handbook section is Factor of safety against seepage liquefaction.
• Everything except i is given, so isolate i 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.
• The factor of safety against seepage liquefaction compares the critical gradient to the actual exit gradient beneath a structure.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$FS = \dfrac{i_c}{i}$$

2. Step 2 — Rearrange the relation so that i stands alone on the left-hand side.

3. Step 3 — List the givens: critical hydraulic gradient (i_c) = 1.0800, factor of safety (FS) = 5.8300.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$i = 0.1852$$

6. Step 6 — Check: returning i = 0.1852 to

   $$FS = \dfrac{i_c}{i}$$

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

Why the other options are there

• 0.3705 — kept a factor of two that cancels in the correct rearrangement.
• 0.0926 — dropped that same factor in the other direction.
• 0.2038 — rounded an intermediate value before the final step.

Reference: FE Handbook — Factor of safety against seepage liquefaction

the factor of safety against seepage liquefaction under a dam's downstream slope Given actual (exit) hydraulic gradient (i) = 1.2100; factor of safety (FS) = 2.4900, determine the critical hydraulic gradient (i_c).

Start with the thinking

• The governing relation printed in this handbook section is Factor of safety against seepage liquefaction.
• Everything except i_c is given, so isolate i_c 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.
• The factor of safety against seepage liquefaction compares the critical gradient to the actual exit gradient beneath a structure.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$FS = \dfrac{i_c}{i}$$

2. Step 2 — Rearrange the relation so that i_c stands alone on the left-hand side.

3. Step 3 — List the givens: actual (exit) hydraulic gradient (i) = 1.2100, factor of safety (FS) = 2.4900.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$i_{c} = 3.0129$$

6. Step 6 — Check: returning i_c = 3.0129 to

   $$FS = \dfrac{i_c}{i}$$

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

Why the other options are there

• 6.0258 — kept a factor of two that cancels in the correct rearrangement.
• 1.5065 — dropped that same factor in the other direction.
• 3.3142 — rounded an intermediate value before the final step.

Reference: FE Handbook — Factor of safety against seepage liquefaction

the factor of safety against seepage liquefaction beneath a levee toe Given critical hydraulic gradient (i_c) = 1.0400; actual (exit) hydraulic gradient (i) = 0.6300, determine the factor of safety (FS).

Start with the thinking

• The governing relation printed in this handbook section is Factor of safety against seepage liquefaction.
• Everything except FS is given, so isolate FS 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.
• The factor of safety against seepage liquefaction compares the critical gradient to the actual exit gradient beneath a structure.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$FS = \dfrac{i_c}{i}$$

2. Step 2 — Rearrange the relation so that FS stands alone on the left-hand side.

3. Step 3 — List the givens: critical hydraulic gradient (i_c) = 1.0400, actual (exit) hydraulic gradient (i) = 0.6300.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$FS = 1.6508$$

6. Step 6 — Check: returning FS = 1.6508 to

   $$FS = \dfrac{i_c}{i}$$

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

Why the other options are there

• 3.3016 — kept a factor of two that cancels in the correct rearrangement.
• 0.8254 — dropped that same factor in the other direction.
• 1.8159 — rounded an intermediate value before the final step.

Reference: FE Handbook — Factor of safety against seepage liquefaction