A 1.5 m wide strip footing sits 1.2 m deep in sand with γ = 18 kN/m³, Nq = 18.4 and Nγ = 22.4 (c = 0). Use FS = 3 to find the allowable bearing pressure.

Start with the thinking

• Cohesionless soil drops the cNc term.
• Net versus gross matters — this is a gross allowable pressure.

Step-by-step solution

1. Surcharge — q = γDf = 18(1.2) = 21.6 kPa

2. Surcharge term

   $$qNq = 21.6(18.4) = 397.4 kPa$$

3. Width term — 0.5γBNγ = 0.5(18)(1.5)(22.4) = 302.4 kPa

4. Ultimate

   $$q_ult = 397.4 + 302.4 = 699.8 kPa$$

5. Allowable

   $$q_all = 699.8/3$$

6. Result

   $$q_all = 233 kPa$$

Why the other options are there

• 700 kPa (FS omitted)
• 133 kPa (width term dropped)

Reference: FE Reference Handbook — Geotechnical — Bearing capacity

A geotechnical problem uses Terzaghi bearing capacity (strip, cohesion only). Given cohesion (c) = 1,590 psf; bearing capacity factor (Nc) = 12.0000; unit weight (gamma) = 108.0 pcf; footing depth (Df) = 10.5000 ft; bearing capacity factor (Nq) = 14.5000, determine the ultimate bearing capacity (qu) in psf.

Start with the thinking

• The governing relation printed in this handbook section is Terzaghi bearing capacity (strip, cohesion only).
• Everything except qu is given, so isolate qu 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.
• Geotechnical items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$q_u = c N_c + \gamma D_f N_q$$

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

3. Step 3 — List the givens: cohesion (c) = 1,590 psf, bearing capacity factor (Nc) = 12.0000, unit weight (gamma) = 108.0 pcf, footing depth (Df) = 10.5000 ft, bearing capacity factor (Nq) = 14.5000.

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

5. Step 5 — Evaluate:

   $$qu = 35523\ \text{psf}$$

6. Step 6 — Check: returning qu = 35,523 psf to

   $$q_u = c N_c + \gamma D_f N_q$$

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

Why the other options are there

• 71,046 — kept a factor of two that cancels in the correct rearrangement.
• 17,762 — dropped that same factor in the other direction.
• 39,075 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

A geotechnical problem uses Terzaghi bearing capacity (strip, cohesion only). Given bearing capacity factor (Nc) = 20.0000; unit weight (gamma) = 126.0 pcf; footing depth (Df) = 10.5000 ft; bearing capacity factor (Nq) = 10.0000; ultimate bearing capacity (qu) = 45,941 psf, determine the cohesion (c) in psf.

Start with the thinking

• The governing relation printed in this handbook section is Terzaghi bearing capacity (strip, cohesion only).
• Everything except c is given, so isolate 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.
• Geotechnical items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$q_u = c N_c + \gamma D_f N_q$$

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

3. Step 3 — List the givens: bearing capacity factor (Nc) = 20.0000, unit weight (gamma) = 126.0 pcf, footing depth (Df) = 10.5000 ft, bearing capacity factor (Nq) = 10.0000, ultimate bearing capacity (qu) = 45,941 psf.

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

5. Step 5 — Evaluate:

   $$c = 1636\ \text{psf}$$

6. Step 6 — Check: returning c = 1,636 psf to

   $$q_u = c N_c + \gamma D_f N_q$$

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

Why the other options are there

• 3,271 — kept a factor of two that cancels in the correct rearrangement.
• 817.8 — dropped that same factor in the other direction.
• 1,799 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

A geotechnical problem uses Terzaghi bearing capacity (strip, cohesion only). Given cohesion (c) = 590.0 psf; bearing capacity factor (Nc) = 5.0000; unit weight (gamma) = 102.0 pcf; bearing capacity factor (Nq) = 12.0000; ultimate bearing capacity (qu) = 14,180 psf, determine the footing depth (Df) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Terzaghi bearing capacity (strip, cohesion only).
• Everything except Df is given, so isolate Df 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.
• Geotechnical items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$q_u = c N_c + \gamma D_f N_q$$

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

3. Step 3 — List the givens: cohesion (c) = 590.0 psf, bearing capacity factor (Nc) = 5.0000, unit weight (gamma) = 102.0 pcf, bearing capacity factor (Nq) = 12.0000, ultimate bearing capacity (qu) = 14,180 psf.

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

5. Step 5 — Evaluate:

   $$Df = 9.1748\ \text{ft}$$

6. Step 6 — Check: returning Df = 9.1748 ft to

   $$q_u = c N_c + \gamma D_f N_q$$

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

Why the other options are there

• 18.3497 — kept a factor of two that cancels in the correct rearrangement.
• 4.5874 — dropped that same factor in the other direction.
• 10.0923 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

A geotechnical problem uses Terzaghi bearing capacity (strip, cohesion only). Given cohesion (c) = 640.0 psf; bearing capacity factor (Nc) = 11.5000; unit weight (gamma) = 127.0 pcf; footing depth (Df) = 6.5000 ft; bearing capacity factor (Nq) = 6.5000, determine the ultimate bearing capacity (qu) in psf.

Start with the thinking

• The governing relation printed in this handbook section is Terzaghi bearing capacity (strip, cohesion only).
• Everything except qu is given, so isolate qu 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.
• Geotechnical items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$q_u = c N_c + \gamma D_f N_q$$

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

3. Step 3 — List the givens: cohesion (c) = 640.0 psf, bearing capacity factor (Nc) = 11.5000, unit weight (gamma) = 127.0 pcf, footing depth (Df) = 6.5000 ft, bearing capacity factor (Nq) = 6.5000.

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

5. Step 5 — Evaluate:

   $$qu = 12726\ \text{psf}$$

6. Step 6 — Check: returning qu = 12,726 psf to

   $$q_u = c N_c + \gamma D_f N_q$$

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

Why the other options are there

• 25,452 — kept a factor of two that cancels in the correct rearrangement.
• 6,363 — dropped that same factor in the other direction.
• 13,998 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

A geotechnical problem uses Terzaghi bearing capacity (strip, cohesion only). Given bearing capacity factor (Nc) = 14.0000; unit weight (gamma) = 105.0 pcf; footing depth (Df) = 4.5000 ft; bearing capacity factor (Nq) = 13.5000; ultimate bearing capacity (qu) = 10,201 psf, determine the cohesion (c) in psf.

Start with the thinking

• The governing relation printed in this handbook section is Terzaghi bearing capacity (strip, cohesion only).
• Everything except c is given, so isolate 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.
• Geotechnical items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$q_u = c N_c + \gamma D_f N_q$$

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

3. Step 3 — List the givens: bearing capacity factor (Nc) = 14.0000, unit weight (gamma) = 105.0 pcf, footing depth (Df) = 4.5000 ft, bearing capacity factor (Nq) = 13.5000, ultimate bearing capacity (qu) = 10,201 psf.

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

5. Step 5 — Evaluate:

   $$c = 273.0\ \text{psf}$$

6. Step 6 — Check: returning c = 273.0 psf to

   $$q_u = c N_c + \gamma D_f N_q$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

A geotechnical problem uses Terzaghi bearing capacity (strip, cohesion only). Given cohesion (c) = 1,090 psf; bearing capacity factor (Nc) = 20.5000; unit weight (gamma) = 104.0 pcf; bearing capacity factor (Nq) = 14.5000; ultimate bearing capacity (qu) = 8,019 psf, determine the footing depth (Df) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Terzaghi bearing capacity (strip, cohesion only).
• Everything except Df is given, so isolate Df 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.
• Geotechnical items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$q_u = c N_c + \gamma D_f N_q$$

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

3. Step 3 — List the givens: cohesion (c) = 1,090 psf, bearing capacity factor (Nc) = 20.5000, unit weight (gamma) = 104.0 pcf, bearing capacity factor (Nq) = 14.5000, ultimate bearing capacity (qu) = 8,019 psf.

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

5. Step 5 — Evaluate:

   $$Df = -9.5000\ \text{ft}$$

6. Step 6 — Check: returning Df = -9.5000 ft to

   $$q_u = c N_c + \gamma D_f N_q$$

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

Why the other options are there

• -19.0000 — kept a factor of two that cancels in the correct rearrangement.
• -4.7500 — dropped that same factor in the other direction.
• -10.4500 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

A geotechnical problem uses Terzaghi bearing capacity (strip, cohesion only). Given cohesion (c) = 2,800 psf; bearing capacity factor (Nc) = 12.5000; unit weight (gamma) = 96.0000 pcf; footing depth (Df) = 2.5000 ft; bearing capacity factor (Nq) = 17.0000, determine the ultimate bearing capacity (qu) in psf.

Start with the thinking

• The governing relation printed in this handbook section is Terzaghi bearing capacity (strip, cohesion only).
• Everything except qu is given, so isolate qu 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.
• Geotechnical items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$q_u = c N_c + \gamma D_f N_q$$

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

3. Step 3 — List the givens: cohesion (c) = 2,800 psf, bearing capacity factor (Nc) = 12.5000, unit weight (gamma) = 96.0000 pcf, footing depth (Df) = 2.5000 ft, bearing capacity factor (Nq) = 17.0000.

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

5. Step 5 — Evaluate:

   $$qu = 39080\ \text{psf}$$

6. Step 6 — Check: returning qu = 39,080 psf to

   $$q_u = c N_c + \gamma D_f N_q$$

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

Why the other options are there

• 78,160 — kept a factor of two that cancels in the correct rearrangement.
• 19,540 — dropped that same factor in the other direction.
• 42,988 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

A geotechnical problem uses Terzaghi bearing capacity (strip, cohesion only). Given bearing capacity factor (Nc) = 19.0000; unit weight (gamma) = 128.0 pcf; footing depth (Df) = 5.0000 ft; bearing capacity factor (Nq) = 1.5000; ultimate bearing capacity (qu) = 14,667 psf, determine the cohesion (c) in psf.

Start with the thinking

• The governing relation printed in this handbook section is Terzaghi bearing capacity (strip, cohesion only).
• Everything except c is given, so isolate 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.
• Geotechnical items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$q_u = c N_c + \gamma D_f N_q$$

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

3. Step 3 — List the givens: bearing capacity factor (Nc) = 19.0000, unit weight (gamma) = 128.0 pcf, footing depth (Df) = 5.0000 ft, bearing capacity factor (Nq) = 1.5000, ultimate bearing capacity (qu) = 14,667 psf.

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

5. Step 5 — Evaluate:

   $$c = 721.4\ \text{psf}$$

6. Step 6 — Check: returning c = 721.4 psf to

   $$q_u = c N_c + \gamma D_f N_q$$

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

Why the other options are there

• 1,443 — kept a factor of two that cancels in the correct rearrangement.
• 360.7 — dropped that same factor in the other direction.
• 793.6 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity

A geotechnical problem uses Terzaghi bearing capacity (strip, cohesion only). Given cohesion (c) = 580.0 psf; bearing capacity factor (Nc) = 16.5000; unit weight (gamma) = 127.0 pcf; bearing capacity factor (Nq) = 16.5000; ultimate bearing capacity (qu) = 6,723 psf, determine the footing depth (Df) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Terzaghi bearing capacity (strip, cohesion only).
• Everything except Df is given, so isolate Df 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.
• Geotechnical items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$q_u = c N_c + \gamma D_f N_q$$

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

3. Step 3 — List the givens: cohesion (c) = 580.0 psf, bearing capacity factor (Nc) = 16.5000, unit weight (gamma) = 127.0 pcf, bearing capacity factor (Nq) = 16.5000, ultimate bearing capacity (qu) = 6,723 psf.

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

5. Step 5 — Evaluate:

   $$Df = -1.3586\ \text{ft}$$

6. Step 6 — Check: returning Df = -1.3586 ft to

   $$q_u = c N_c + \gamma D_f N_q$$

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

Why the other options are there

• -2.7173 — kept a factor of two that cancels in the correct rearrangement.
• -0.6793 — dropped that same factor in the other direction.
• -1.4945 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Geotechnical → Ultimate Bearing Capacity