http://www.geotechnicalinfo.com/bearing_capacity_technical_guidance.html

**Bearing capacity** of soil is the value of the average contact pressure between the foundation and the soil which will produce shear failure in the soil. **Ultimate bearing capacity** is the theoretical maximum pressure which can be supported without failure. **Allowable bearing capacity** is what is used in geotechnical design, and is the ultimate bearing capacity divided by a factor of safety.

Theoretical (Ultimate) and allowable bearing capacity can be assessed for the following:

- Shallow Foundations
- strip footings
- square footings
- circular footings

- Deep foundations
- end bearing
- skin friction

For comprehensive examples of bearing capacity problems see:

- Bearing Capacity Examples

## Allowable Bearing Capacity

**Q _{a} = Qu ** Q

_{a}= Allowable bearing capacity (kN/m

^{2) }or (lb/ft

^{2})

**F.S.**

Where:

Q_{u} = ultimate bearing capacity (kN/m^{2}) or (lb/ft^{2}) *See below for theory

F.S. = Factor of Safety

## Ultimate Bearing Capacity for Shallow Foundations

__Terzaghi Ultimate Bearing Capacity Theory__

**Q _{u} = c N_{c} + g D N_{q} + 0.5 g B N_{g}**

= Ultimate bearing capacity equation for shallow strip footings, (kN/m

^{2}) (lb/ft

^{2})

**Q _{u} = 1.3 c N_{c} + g D N_{q} + 0.4 g B N_{g}**

= Ultimate bearing capacity equation for shallow square footings, (kN/m

^{2}) (lb/ft

^{2})

**Q _{u} = 1.3 c N_{c} + g D N_{q} + 0.3 g B N_{g}**

= Ultimate bearing capacity equation for shallow circular footings, (kN/m

^{2}) (lb/ft

^{2})

Where:

**c** = Cohesion of soil (kN/m^{2}) (lb/ft^{2}),

**g** = effective unit weight of soil (kN/m^{3}) (lb/ft^{3}), *see note below

**D** = depth of footing (m) (ft),

**B** = width of footing (m) (ft),

**N _{c}**=cotf(N

_{q}– 1), *see typical bearing capacity factors

**N**=e

_{q}^{2}(3p/4-f/2)tanf / [2 cos2(45+f/2)], *see typical bearing capacity factors

**N**

**=(1/2) tanf(k**

_{g}_{p}/cos

^{2}f – 1), *see typical bearing capacity factors

**e**= Napier’s constant = 2.718…,

**k**= passive pressure coefficient, and

_{p}**f**= angle of internal friction (degrees).

Notes:

Effective unit weight, g, is the unit weight of the soil for soils above the water table and capillary rise. For saturated soils, the effective unit weight is the unit weight of water, g_{w}, 9.81 kN/m^{3} (62.4 lb/ft^{3}), subtracted from the saturated unit weight of soil. Find more information in the foundations section.

__Meyerhof Bearing Capacity Theory Based on Standard Penetration Test Values__

**Q _{u} = 31.417(NB + ND)**

**(kN/m**

^{2}) (metric)**Q _{u} = NB + ND**

**(tons/ft**

^{2}) (standard)**10 10**

__For footing widths of 1.2 meters (4 feet) or less__

**Q _{a} = 11,970N (kN/m^{2}) (metric)**

**Q _{a} = 1.25N (tons/ft^{2}) (standard)**

**10**

__For footing widths of 3 meters (10 feet) or more__

**Q _{a} = 9,576N (kN/m^{2}) (metric)**

**Q _{a} = N (tons/ft^{2}) (standard)**

**10**

Where:

**N** = N value derived from Standard Penetration Test (SPT)

**D** = depth of footing (m) (ft), and

**B** = width of footing (m) (ft).

Note: All Meyerhof equations are for foundations bearing on clean sands. The first equation is for ultimate bearing capacity, while the second two are factored within the equation in order to provide an allowable bearing capacity. Linear interpolation can be performed for footing widths between 1.2 meters (4 feet) and 3 meters (10 feet). Meyerhof equations are based on limiting total settlement to 25 cm (1 inch), and differential settlement to 19 cm (3/4 inch).

## Ultimate Bearing Capacity for Deep Foundations (Pile)

**Q _{ult} = Q_{p} + Q_{f}**

Where:

**Q _{ult}** = Ultimate bearing capacity of pile, kN (lb)

**Q**= Theoretical bearing capacity for tip of foundation, or end bearing, kN (lb)

_{p}**Q**= Theoretical bearing capacity due to shaft friction, or adhesion between foundation shaft and soil, kN (lb)

_{f}

__End Bearing (Tip) Capacity of Pile Foundation__

**Q _{p} = A_{p}q_{p}**

Where:

**Q _{p}** = Theoretical bearing capacity for tip of foundation, or end bearing, kN (lb)

**A**= Effective area of the tip of the pile, m

_{p}^{2 }(ft

^{2})

For a circular closed end pile or circular plugged pile;

**A**m

_{p}= p(B/2)^{2}^{2}(ft

^{2})

**q**=

_{p}**g**

**DN**

_{q}= Theoretical unit tip-bearing capacity for cohesionless and silt soils, kN/m

^{2 }(lb/ft

^{2})

**q**=

_{p}**9c**

= Theoretical unit tip-bearing capacity for cohesive soils, kN/m

^{2 }(lb/ft

^{2})

**g**= effective unit weight of soil, kN/m

^{3}(lb/ft

^{3}), *See notes below

**D**= Effective depth of pile, m (ft), where

**D < D**,

_{c}**N**= Bearing capacity factor for piles,

_{q}**c**= cohesion of soil, kN/m

^{2}(lb/ft

^{2}),

**B**= diameter of pile, m (ft), and

**D**= critical depth for piles in loose silts or sands m (ft).

_{c}**D**

_{c}**= 10B**, for loose silts and sands

**D**

_{c}**= 15B**, for medium dense silts and sands

**D**

_{c}**= 20B**, for dense silts and sands

__Skin (Shaft) Friction Capacity of Pile Foundation__

**Q _{f} = A_{f}q_{f}** for one homogeneous layer of soil

**Q _{f} = pSq_{f}L** for multi-layers of soil

Where:

**Q _{f}** = Theoretical bearing capacity due to shaft friction, or adhesion between foundation shaft and soil, kN (lb)

**A**=

_{f}**pL**; Effective surface area of the pile shaft, m

^{2 }(ft

^{2})

**q**

_{f}**= ks tan d**= Theoretical unit friction capacity for cohesionless soils, kN/m

^{2 }(lb/ft

^{2})

**q**=

_{f}**c**= Theoretical unit friction capacity for silts, kN/m

_{A}+ ks tan d^{2 }(lb/ft

^{2})

**q**

_{f}**= aS**= Theoretical unit friction capacity for cohesive soils, kN/m

_{u}^{2 }(lb/ft

^{2})

**p**= perimeter of pile cross-section, m (ft)

for a circular pile;

**p = 2p(B/2)**

for a square pile;

**p = 4B**

**L**= Effective length of pile, m (ft) *See Notes below

**a**

**= 1 – 0.1(S**= adhesion factor, kN/m

_{uc})^{2}^{2}(ksf), where S

_{uc}< 48 kN/m

^{2}(1 ksf)

**a**=

**kN/m**

__1__[0.9 + 0.3(S_{uc}– 1)]^{2}, (ksf) where S

_{uc}> 48 kN/m

^{2}, (1 ksf)

**S**

_{uc}**S**

_{uc}**= 2c**= Unconfined compressive strength , kN/m

^{2}(lb/ft

^{2})

**c**= adhesion

_{A}=

**c**for rough concrete, rusty steel, corrugated metal

**0.8c**<

**c**<

_{A}**c**for smooth concrete

**0.5c**<

**c**<

_{A}**0.9c**for clean steel

**c**= cohesion of soil, kN/m

^{2}(lb/ft

^{2})

**d**= external friction angle of soil and wall contact (deg)

**f**= angle of internal friction (deg)

**s**

**= gD**= effective overburden pressure, kN/m

^{2}, (lb/ft

^{2})

**k**= lateral earth pressure coefficient for piles

**g**= effective unit weight of soil, kN/m

^{3}(lb/ft

^{3}) *See notes below

**B**= diameter or width of pile, m (ft)

**D**= Effective depth of pile, m (ft), where

**D < D**

_{c}**D**= critical depth for piles in loose silts or sands m (ft).

_{c}**D**

_{c}**= 10B**, for loose silts and sands

**D**

_{c}**= 15B**, for medium dense silts and sands

**D**

_{c}**= 20B**, for dense silts and sands

**S =**summation of differing soil layers (i.e. a

_{1}+ a

_{2}+ …. + a

_{n})

Notes: Determining effective length requires engineering judgment. The effective length can be the pile depth minus any disturbed surface soils, soft/ loose soils, or seasonal variation. The effective length may also be the length of a pile segment within a single soil layer of a multi layered soil. Effective unit weight, g, is the unit weight of the soil for soils above the water table and capillary rise. For saturated soils, the effective unit weight is the unit weight of water, g

_{w}, 9.81 kN/m

^{3}(62.4 lb/ft

^{3}), subtracted from the saturated unit weight of soil.

************

__Meyerhof Method for Determining qp and qf in Sand__

Theoretical unit tip-bearing capacity for driven piles in sand, when__ D __> 10:

B

** q _{p}** =

**4N**

_{c}**tons/ft**

^{2}standardTheoretical unit tip-bearing capacity for drilled piles in sand:

** q _{p}** =

**1.2N**

_{c}**tons/ft**

^{2}standardTheoretical unit friction-bearing capacity for driven piles in sand:

** q _{f}** =

**N**

__tons/ft__^{2}standard**50**

Theoretical unit friction-bearing capacity for drilled piles in sand:

** q _{f}** =

**N**

__tons/ft__^{2}standard**100**

Where:

**D** = pile embedment depth, ft

**B** = pile diameter, ft

**N _{c} = C_{n}(N)**

**C**

_{n}**= 0.77 log**

s

__20__s

**N**= N-Value from SPT test

**s = gD**= effective overburden stress at pile embedment depth, tons/ft

^{2}

= (

**g – g**)D = effective stress if below water table, tons/ft

_{w}^{2}

**g**= effective unit weight of soil, tons/ft

^{3}

**g**= 0.0312 tons/ft

_{w }^{3}= unit weight of water

## Examples for determining allowable bearing capacity

**Example #1: Determine allowable bearing capacity and width for a shallow strip footing on cohesionless silty sand and gravel soil. Loose soils were encountered in the upper 0.6 m (2 feet) of building subgrade. Footing must withstand a 144 kN/m ^{2} (3000 lb/ft^{2}) building pressure.**

__Given__

- bearing pressure from building = 144 kN/m
^{2 (}3000 lbs/ft^{2}) - unit weight of soil, g = 21 kN/m
^{3}(132 lbs/ft^{3}) *from soil testing, see typical g values - Cohesion, c = 0 *from soil testing, see typical c values
- angle of Internal Friction, f = 32 degrees *from soil testing, see typical f values
- footing depth, D = 0.6 m (2 ft) *because loose soils in upper soil strata

__Solution__

Try a minimal footing width, B = 0.3 m (B = 1 foot).

Use a factor of safety, F.S = 3. Three is typical for this type of application. See factor of safety for more information.

Determine bearing capacity factors N_{g}, N_{c} and N_{q}. See typical bearing capacity factors relating to the soils’ angle of internal friction.

- N
_{g}= 22 - N
_{c}= 35.5 - N
_{q}= 23.2

Solve for ultimate bearing capacity,

**Q _{u} = c N_{c} + g D N_{q} + 0.5 g B N_{g} ** *strip footing eq.

Q_{u} = 0(35.5) + 21 kN/m^{3}(0.6m)(23.2) + 0.5(21 kN/m^{3})(0.3 m)(22) **metric**

Q_{u} = 362 kN/m^{2}

Q_{u} = 0(35.5) + 132lbs/ft^{3}(2ft)(23.2) + 0.5(132lbs/ft^{3})(1ft)(22) **standard**

Q_{u} = 7577 lbs/ft^{2}

Solve for allowable bearing capacity,

**Q _{a} = Qu **

F.S.

Q_{a} =__ 362 kN/m__^{2} = 121 kN/m^{2}** not o.k. metric**

3** **

Q_{a} =__ 7577lbs/ft__^{2} = 2526 lbs/ft^{2}** not o.k. standard **

3** **

Since Q_{a} < required 144 kN/m^{2} (3000 lbs/ft^{2}) bearing pressure, increase footing width, B or foundation depth, D to increase bearing capacity.

Try footing width, B = 0.61 m (B = 2 ft).

Q_{u} = 0 + 21 kN/m^{3}(0.61 m)(23.2) + 0.5(21 kN/m^{3})(0.61 m)(22) **metric**

Q_{u} = 438 kN/m^{2}

Q_{u} = 0 + 132 lbs/ft^{3}(2 ft)(23.2) + 0.5(132 lbs/ft^{3})(2 ft)(22) **standard **

Q_{u} = 9029 lbs/ft^{2}

Q_{a} = __ 438 kN/m__^{2} = 146 kN/m^{2} **Q _{a} > 144 kN/m^{2 }o.k. metric**

3

Q_{a} = __ 9029 lbs/ft__^{2} = 3010 lbs/ft^{2} ** Q _{a} > 3000 lbs/ft^{2 }o.k. standard**

3

__Conclusion__

Footing shall be 0.61 meters (2 feet) wide at a depth of 0.61 meters (2 feet) below ground surface.Many engineers neglect the depth factor (i.e. D N_{q} = 0) for shallow foundations. This inherently increases the factor of safety. Some site conditions that may negatively effect the depth factor are foundations established at depths equal to or less than 0.3 meters (1 feet) below the ground surface, placement of foundations on fill, and disturbed/ fill soils located above or to the sides of foundations.

********************************

**Example #2: Determine allowable bearing capacity of a shallow, 0.3 meter (12-inch) square isolated footing bearing on saturated cohesive soil. The frost penetration depth is 0.61 meter (2 feet). Structural parameters require the foundation to withstand 4.4 kN (1000 lbs) of force on a 0.3 meter (12-inch) square column.**

__Given__

- bearing pressure from building column = 4.4 kN/ (0.3 m x 0.3 m) = 48.9 kN/m
^{2} - bearing pressure from building column = 1000 lbs/ (1 ft x 1 ft) = 1000 lbs/ft
^{2} - unit weight of saturated soil, g
_{sat}= 20.3 kN/m^{3}(129 lbs/ft^{3}) *see typical g values - unit weight of water, g
_{w}= 9.81 kN/m^{3}(62.4 lbs/ft^{3}) *constant - Cohesion, c = 21.1 kN/m
^{2}(440 lbs/ft^{2}) *from soil testing, see typical c values - angle of Internal Friction, f = 0 degrees *from soil testing, see typical f values
- footing width, B = 0.3 m (1 ft)

__Solution__

Try a footing depth, D = 0.61 meters (2 feet), because foundation should be below frost depth.

Use a factor of safety, F.S = 3. See factor of safety for more information.

Determine bearing capacity factors N_{g}, N_{c} and N_{q}. See typical bearing capacity factors relating to the soils’ angle of internal friction.

- N
_{g}= 0 - N
_{c}= 5.7 - N
_{q}= 1

Solve for ultimate bearing capacity,

**Q _{u }= 1.3c N_{c} + g D N_{q} + 0.4 g B N_{g}** *square footing eq.

Q_{u} =1.3(21.1kN/m^{2})5.7+(20.3kN/m^{3}-9.81kN/m^{3})(0.61m)1+0.4(20.3kN/m^{3}-9.81kN/m^{3})(0.3m)0

Q_{u} = 163 kN/m^{2 } **metric**

Q_{u} = 1.3(440lbs/ft^{2})(5.7) + (129lbs/ft^{3} – 62.4lbs/ft^{3})(2ft)(1) + 0.4(129lbs/ft^{3} – 62.4lbs/ft^{3})(1ft)(0)

Q_{u} = 3394 lbs/ft^{2 } ** standard**

Solve for allowable bearing capacity,

**Q _{a} = Qu **

F.S.

Q_{a} = __ 163 kN/m__^{2} = 54 kN/m^{2 } ** Q _{a} > 48.9 kN/m^{2}**

**o.k. metric**

3

Q

_{a}=

__3394lbs/ft__

^{2}= 1130 lbs/ft

^{2 }

**Q**

_{a}> 1000 lbs/ft^{2}**o.k. standard**

3

__Conclusion__

The 0.3 meter (12-inch) isolated square footing shall be 0.61 meters (2 feet) below the ground surface. Other considerations may be required for foundations bearing on moisture sensitive clays, especially for lightly loaded structures such as in this example. Sensitive clays could expand and contract, which could cause structural damage. Clay used as bearing soils may require mitigation such as heavier loads, subgrade removal and replacement below the foundation, or moisture control within the subgrade.

********************************

**Example #3: Determine allowable bearing capacity and width for a foundation using the Meyerhof Method. Soils consist of poorly graded sand. Footing must withstand a 144 kN/m ^{2}(1.5 tons/ft^{2}) building pressure.**

__Given__

- bearing pressure from building = 144 kN/m
^{2 (}1.5 tons/ft^{2}) - N Value, N = 10 at 0.3 m (1 ft) depth *from SPT soil testing
- N Value, N = 36 at 0.61 m (2 ft) depth *from SPT soil testing
- N Value, N = 50 at 1.5 m (5 ft) depth *from SPT soil testing

__Solution__

Try a minimal footing width, B = 0.3 m (B = 1 foot) at a depth, D = 0.61 meter (2 feet). Footings for single family residences are typically 0.3m (1 ft) to 0.61m (2ft) wide. This depth was selected because soil density greatly increases (i.e. higher N-value) at a depth of 0.61 m (2 ft).

Use a factor of safety, F.S = 3. Three is typical for this type of application. See factor of safety for more information.

__Solve for ultimate bearing capacity__

**Q _{u} = 31.417(NB + ND)**

**(kN/m**

^{2}) (metric)**Q _{u} = NB + ND**

**(tons/ft**

^{2}) (standard)**10 10**

Q_{u} = 31.417(36(0.3m) + 36(0.61m)) = 1029 kN/m^{2} **(metric)**

Q_{u} =__ 36(1 ft) __+ __ 36(2 ft) __ = 10.8 tons/ft^{2} ** (standard)**

10 10

__Solve for allowable bearing capacity__,

**Q _{a} = Qu **.

F.S

Q_{a} =__ 1029 kN/m__^{2} = 343 kN/m^{2}** Q _{a} > 144 kN/m^{2} o.k. (metric)**

3

Q

_{a}=

__10.8 tons/ft__

^{2}= 3.6 tons/ft

^{2}

**Q**

_{a}> 1.5 tons/ft^{2}o.k. (standard)3

__Conclusion__

Footing shall be 0.3 meters (1 feet) wide at a depth of 0.61 meters (2 feet) below the ground surface. A footing width of only 0.3 m (1 ft) is most likely insufficient for the structural engineer when designing the footing with the building pressure in this problem.

********************************

**Example #4: Determine allowable bearing capacity and diameter of a single driven pile. Pile must withstand a 66.7 kN (15 kips) vertical load.**

__Given__

- vertical column load = 66.7 kN (15 kips or 15,000 lb)
- homogeneous soils in upper 15.2 m (50 ft); silty soil
- Pile Information
- driven
- steel
- plugged end

__Solution__

Try a pile depth, **D = 1.5 meters (5 feet)**

Try pile diameter, **B = 0.61 m (2 ft)**

Use a factor of safety, F.S = 3. Smaller factors of safety are sometimes used if piles are load tested, or the engineer has sufficient experience with the regional soils.

**Determine ultimate end bearing of pile,**

**Q _{p} = A_{p}q_{p}**

**A**= p(B/2)

_{p}^{2}= p(0.61m/2)

^{2 }= 0.292 m

^{2}

**metric**

**A**= p(B/2)

_{p}^{2}= p(2ft/2)

^{2}= 3.14 ft

^{2 }

**standard**

**q _{p} = gDN_{q}**

g = 19.6 kN/m^{3} (125 lbs/ft^{3}); given soil unit weight

f = 30 degrees; given soil angle of internal friction

**B** = 0.61 m (2 ft); trial pile width

**D** = 1.5 m (5 ft); trial depth, may need to increase or decrease depending on capacity

check to see if** D < D _{c}**

**Dc**= 15B = 9.2 m (30 ft); critical depth for medium dense silts.

If D > D

_{c}, then use D

_{c}

**N**= 25; Meyerhof bearing capacity factor for driven piles, based on

_{q}**f**

**q _{p}** = 19.6 kN/m

^{3}(1.5 m)25 = 735 kN/m

^{2}

**metric**

**q**= 125 lb/ft

_{p}^{3}(5 ft)25 = 15,625 lb/ft

^{2}

**standard**

**Q**

_{p}**=**A

_{p}q

_{p }= (0.292 m

^{2})(735 kN/m

^{2}) =

**214.6 kN**

**metric**

**Q**= A

_{p}_{p}q

_{p }= (3.14 ft

^{2})(15,625 lb/ft

^{2}) =

**49,063 lb**

**standard**

**Determine ultimate friction capacity of pile,**

**Q _{f} = A_{f}q_{f}**

**A _{f}** =

**pL**

**p** = 2p(0.61m/2) = 1.92 m ** metric**

**p** = 2p(2 ft/2) = 6.28 ft ** standard**

**L** = D = 1.5 m (5 ft); length and depth used interchangeably. check **D _{c}** as above

**A _{f}** = 1.92 m(1.5 m) = 2.88 m

^{2}

**metric**

**A**= 6.28 ft(5 ft) = 31.4 ft

_{f}^{2}

**standard**

**q _{f} = c_{A} + ks tan d** =

**c**+

_{A}**kgD tan d**

**k** = 0.5; lateral earth pressure coefficient for piles, value chosen from Broms low density steel

**g **= 19.6 kN/m^{3 }(125 lb/ft^{3}); given effective soil unit weight. If water table, then g – g_{w}

**D = L** = 1.5 m (5 ft); pile length. Check to see if **D < D _{c}**

**Dc**= 15B = 9.2 m (30 ft); critical depth for medium dense silts. If D > D

_{c}, then use D

_{c}

d = 20 deg; external friction angle, equation chosen from Broms steel piles

**B**= 0.61 m (2 ft); selected pile diameter

**c**= 0.5c; for clean steel. See adhesion in pile theories above.

_{A}= 24 kN/m

^{2}(500 lb/ft

^{2})

**q _{f} **= 24 kN/m

^{2}+ 0.5(19.6 kN/m

^{3})(1.5m)tan 20 = 29.4 kN/m

^{2}

**metric**

**q**= 500 lb/ft

_{f}^{2}+ 0.5(125 lb/ft

^{3})(5ft)tan 20 = 614 lb/ft

^{2}

**standard**

**Q _{f} = **A

_{f}q

_{f}= 2.88 m

^{2}(29.4 kN/m

^{2})

**=**

**84.7 kN metric**

**Q**A

_{f}=_{f}q

_{f}= 31.4 ft

^{2}(614 lb/ft

^{2})

**=**

**19,280 lb**

**standard**

**Determine ultimate pile capacity,**

**Q _{ult} = Q_{p} + Q_{f}**

Q_{ult} = 214.6 kN + 84.7 kN = **299.3 kN metric**

Q_{ult} = 49,063 lb + 19,280 lb = **68,343 lb standard**

**Solve for allowable bearing capacity,**

**Q _{a} = Qult **

F.S.

Q_{a} = __ 299.3 kN __ = 99.8 kN; **Q _{a} > applied load (66.7 kN) o.k. metric**

3

Q

_{a}=

__68,343 lbs__= 22,781 lbs

**Q**

_{a}> applied load (15 kips) o.k. standard3

__Conclusion__

A 0.61 m (2 ft) steel pile shall be plugged and driven 1.5 m (5 feet) below the ground surface. Many engineers neglect the skin friction within the upper 1 to 5 feet of subgrade due to seasonal variations or soil disturbance. Seasonal variations may include freeze/ thaw or effects from water. The end bearing alone (neglect skin friction) is sufficient for this case. Typical methods for increasing the pile capacity are increasing the pile diameter or increasing the embedment depth of the pile.

********************************

**Example #5: Determine allowable bearing capacity and diameter of a single driven pile. Pile must withstand a 66.7 kN (15 kips) vertical load.**

__Given__

- vertical column load = 66.7 kN (15 kips or 15,000 lb)
- upper 1.5 m (5 ft) of soil is a medium dense gravelly sand
- soils below 1.5 m (5 ft) of soil is a stiff silty clay
- unit weight, g = 18.9 kN/m
^{3 }(120 lbs/ft^{3}) - cohesion, c = 47.9 kN/m
^{2}(1000 lb/ft^{2}) - angle of internal friction, f = 0 degrees

- unit weight, g = 18.9 kN/m
- Pile Information
- driven
- wood
- closed end

__Solution__

Try a pile depth, **D = 2.4 meters (8 feet)**

Try pile diameter, **B = 0.61 m (2 ft)**

Use a factor of safety, F.S = 3. Smaller factors of safety are sometimes used if piles are load tested, or the engineer has sufficient experience with the regional soils.

**Determine ultimate end bearing of pile,**

**Q _{p} = A_{p}q_{p}**

A

_{p}= p(B/2)

^{2}= p(0.61m/2)

^{2 }= 0.292 m

^{2}

**metric**

A

_{p}= p(B/2)

^{2}= p(2ft/2)

^{2}= 3.14 ft

^{2 }

**standard**

q

_{p}= 9c = 9(47.9 kN/m

^{2}) = 431.1 kN/m

^{2}

**metric**

q

_{p}= 9c = 9(1000 lb/ft

^{2}) = 9000 lb/ft

^{2}

**standard**

Q

_{p}

**=**A

_{p}q

_{p }= (0.292 m

^{2})(431.1 kN/m

^{2}) =

**125.9 kN**

**metric**

Q

_{p}= A

_{p}q

_{p }= (3.14 ft

^{2})(9000 lb/ft

^{2}) =

**28,260 lb**

**standard**

**Determine ultimate friction capacity of pile,**

**Q _{f} = pSq_{f}L**

p = 2p(0.61m/2) = 1.92 m ** metric**

p = 2p(2 ft/2) = 6.28 ft ** standard**

__upper 1.5 m (5 ft) of soil__

**q _{f}L = [ks tan d]L = [kgD tan d]L**

**k** = 1.5; lateral earth pressure coefficient for piles, value chosen from Broms low density timber

**g **= 19.6 kN/m^{3 }(125 lb/ft^{3}); given effective soil unit weight. If water table, then g – g_{w}

**D = L** = 1.5 m (5 ft); segment of pile within this soil strata. Check to see if **D < D _{c}**

**D**= 15B = 9.2 m (30 ft); critical depth for medium dense sands. This assumption is conservative, because the soil is gravelly, and this much soil unit weight for a sand would indicate dense soils. If D > D

_{c}_{c}, then use D

_{c}

d = f(2/3) = 20 deg; external friction angle, equation chosen from Broms timber piles

**B**= 0.61 m (2 ft); selected pile diameter

f = 30 deg; given soil angle of internal friction

q_{f}L = [1.5(19.6 kN/m^{3})(1.5m)tan (20)]1.5 m = **24.1 kN/m** **metric**

q_{f}L = [1.5(125 lb/ft^{3})(5ft)tan (20)]5 ft = **1706 lb/ft** **standard**

__soils below 1.5 m (5 ft) of subgrade__

**q _{f}L = aS_{u}**

**S _{uc}** = 2c = 95.8 kN/m

^{2 }(2000 lb/ft

^{2}); unconfined compressive strength

**c**= 47.9 kN/m

^{2 }(1000 lb/ft

^{2}); cohesion from soil testing (given)

**a**=

__1__[0.9 + 0.3(S

_{uc}– 1)] = 0.3; because S

_{uc}> 48 kN/m

^{2}, (1 ksf)

S

_{uc}

**L**= 0.91 m (3 ft); segment of pile within this soil strata

q_{f}L = [0.3(95.8 kN/m^{2})]0.91 m = **26.2 kN/m** ** metric**

q_{f}L = [0.3(2000 lb/ft^{2})]3 ft = **1800 lb/ft** ** standard**

__ultimate friction capacity of combined soil layers__

**Q _{f} = pSq_{f}L**

Q_{f} = 1.92 m(24.1 kN/m + 26.2 kN/m) = **96.6 kN metric **

Q_{f} = 6.28 ft(1706 lb/ft + 1800 lb/ft) = **22,018 lb standard **

**Determine ultimate pile capacity,**

**Q _{ult} = Q_{p} + Q_{f}**

Q_{ult} = 125.9 kN + 96.6 kN = **222.5 kN metric**

Q_{ult} = 28,260 lb + 22,018 lb = **50,278 lb standard**

**Solve for allowable bearing capacity,**

**Q _{a} = Qult **

F.S.

Q_{a} = __ 222.5 kN __ = 74.2 kN; **Q _{a} > applied load (66.7 kN) o.k. metric**

3

Q

_{a}=

__50,275 lbs__= 16,758 lbs

**Q**

_{a}> applied load (15 kips) o.k. standard3

__Conclusion__

Wood pile shall be driven 8 feet below the ground surface. Many engineers neglect the skin friction within the upper 1 to 5 feet of subgrade due to seasonal variations or soil disturbance. Seasonal variations may include freeze/ thaw or effects from water. Notice how the soil properties within the pile tip location is used in the end bearing calculations. End bearing should also consider the soil layer(s) directly beneath this layer. Engineering judgment or a change in design is warranted if subsequent soil layers are weaker than the soils within the vicinity of the pile tip. Typical methods for increasing the pile capacity are increasing the pile diameter or increasing the embedment depth of the pile.