Wednesday, June 17, 2015

GEOTEXTILES


1. INTRODUCTION

According to the historical record, it is believed that the first applications of geotextiles were woven industrial fabrics used in 1950’s. One of the earliest documented cases was a waterfront structure built inFlorida in 1958. Then, the first nonwoven geotextile was developed in 1968 by the Rhone Poulence company in France. It was a comparatively thick needle-punched polyester, which was used in dam construction in France during 1970.
In fact, the geotextile is one of the members of the geosynthetic family. Those members include the following items

1.      Geogrids
  1. Geonets
  2. Geotextiles
  3. Geomembranes
  4. Geosynthetic Clay Liners
  5. Geopipe
  6. Geocomposites

What is geotextile?

As we know, the prefix of geotextile, geo, means earth and the ‘textile’ means fabric. Therefore, according to the definition of ASTM 4439, the geotextile is defined as follows:
"A permeable geosynthetic comprised solely of textiles. Geotextiles are used with foundation, soil, rock, earth, or any other geotechnical engineering-related material as an integral part of human-made project, structure, or system."
The ASAE (Society for Engineering in Agricultural, Food, and Biological Systems) defines a geotextile as a "fabric or synthetic material placed between the soil and a pipe, gabion, or retaining wall: to enhance water movement and retard soil movement, and as a blanket to add reinforcement and separation." A geotextile should consist of a stable network that retains its relative structure during handling, placement, and long-term service. Other terms that are used by the industry for similar materials and applications are geotextile cloth, agricultural fabric, and geosynthetic.

2. THE TYPES OF GEOTEXTILE

In general, the vast majority of geotextiles are made from polypropylene or polyester formed into fabrics as follows:

  • Woven monofilament

  • Woven multifilament

  • Woven slit-film monofilament

  • Woven slit-film multifilament

  • Nonwoven continuous filament heat bonded

  • Nonwoven continuous filament needle-punched

  • Nonwoven staple needle-punched

  • Nonwoven resin bonded

  • Other woven and nonwoven combinations

  • Knitted

  • 3. RAW MATERIAL OF GEOTEXTILE

The four main polymer families most widely used as the raw material for geotextiles are:

Polyester

Polyamide

Polypropylene

Polyethylene


The oldest of these is polyethylene, which was discovered in 1931 in the research laboratories of the ICI. Another group of polymers with a long production history is the polyamide family, the first of which was discovered in 1935.The next oldest of the four main polymer families relevant to geotextile manufacture is polyester which was first announced in 1941.The most recent polymer family relevant to geotextiles to be developed was polypropylene, which was discovered in 1954. The comparative properties of these four polymer are shown in very general items in Table 1.


Table 1
Polyester
Polyamide
Polypropylene
Polyethylene
Strength
H
M
L
L
Elastic modulus
H
M
L
L
Strain at failure
M
M
H
H
Creep
L
M
H
H
Unit weight
H
M
L
L
Cost
H
M
L
L
Resistance to:




U.V. light stabilized
H
M
H
H
unstabilized
H
M
M
L
Alkalis
L
H
H
H
Fungus, vermin
M
M
M
H
Fuel
M
M
L
L
Detergents
H
H
H
H
H: High; M: Medium; L: Low

4. THE BASIC PROPERTIES OF GEOTEXTILE[1]


The properties of polymer material are affected by its average molecular weight (MW ) and its statistical distribution. Increasing the average MW results in increasing:
  • tensile strength
  • elongation
  • impact strength
  • stress crack resistance
  • heat resistance

Narrowing the molecular weight distribution results in:

  • increased impact strength
  • decreased stress crack resistance
  • decreased processability

Increasing crystallinity results in:

  • increasing stiffness or hardness
  • increasing heat resistance
  • increasing tensile strength
  • increasing modulus
  • increasing chemical resistance
  • decreasing diffusive permeability
  • decreasing elongation or strain at failure
  • decreasing flexibility
  • decreasing impact strength
  • decreasing stress crack resistance

5. MARKET ACTIVITY
To say that the market activity of geosynthetics in the geotechnical, transportation, and environmental areas is strong is decidedly an understatement. To obtain an insight into the vitality of geosynthetics, note the curves in the graphs in fig 3a and 3b. The curves in Fig. 3a gives the estimated amount of geosynthetics used in North America over the years(geopipe is not shown ), while the curve in Fig 3b gives the estimated in-place expenditures of these products.




Used in the calculations were the data for 1995 (note that the values are in millions of square meters and millions of dollars ) [ 1 ]:
Geotextiles 500 Mm2 @ $ 0.9 / m2 = $ 450 M
Geogrids 40 Mm2 @ $ 2.50 / m2 = $ 100 M
Geonets 50 Mm2 @ $ 2.00 / m2 = $ 100 M
Geomembranes 75 Mm2 @ $ 10.00 / m2 = $ 750 M
Geosynthetic clay linears 50 Mm2 @ $ 2.5 / m2 = $ 125 M
Geocomposites 25 Mm2 @ $ 5.00 / m2 = $ 125 M
Geo-others 5 Mm2 @ $ 4.00 / m2 = $ 20 M
Total ( 1995 ) $ 1670 M

6. THE BASIC FUNCTION OF GEOTEXTILE

Geotextiles form one of the two largest groups of geosynthetics. Their rise in growth during the past fifteen years has been nothing short of awesome. They are indeed textiles in the traditional sense, but consist of synthetic fibers rather than natural ones such as cotton, wool, or silk. Thus biodegradation is not a problem. These synthetic fibers are made into a flexible, porous fabric by standard weaving machinery or are matted together in a random, or nonwoven, manner. Some are also knit. The major point is that they are porous to water flow across their manufactured plane and also within their plane, but to a widely varying degree. There are at least 80 specific applications area for geotextiles that have been developed; however, the fabric always performs at least one of five discrete functions:
1.     Separation 

Geotextiles function to prevent mutual mixing between 2 layers of soil having different particle sizes or different properties. Table 2 shows the required properties for separation:
 Table 2 The required properties for separation

Mechanical
Hydraulic
Long-term Performance
During installation
Impact resistance
Elongation at break
Apparent opening
size ( A.O.S.)
Thickness
UV resistance
During construction
Puncture resistance
Elongation at break
Apparent opening
size ( A.O.S.)
Thickness
Chemical stability
UV resistance
After completion of construction
Puncture resistance
Tear propagation resistance
Elongation at break
Apparent opening
size ( A.O.S.)
Thickness
Chemical stability
Resistance to decay
  1. Drainage :
The function of drainage is to gather water, which is not required functionally by the structure, such as rainwater or surplus water in the soil, and discharge it.









Table 3. The required properties for drainage:

Mechanical
Hydraulic
Long-term Performance
Permanent drainage function
Influence of normal overburden pressure
Permeability
Thickness
Apparent opening
size (A.O.S.)
Chemical properties of water and soil
Chemical stability
Decay resistance
Temporary drainage function
Influence of normal overburden pressure
Permeability
Thickness
Apparent opening
size (A.O.S.)

  1. Filtration :
Filtration involves the establishment of a stable interface between the drain and the surrounding soil. In all soils water flow will induce the movement of fine particles. Initially a portion of this fraction will be halted at the filter interface; some will be halted within the filter itself while the rest will pass into the drain. The geotextile provides an ideal interface for the creation of a reverse filter in the soil adjacent to the geotextile. The complex needle-punched structure of the geotextile provides for the retention of fine particles without reducing the permeability requirement of the drain.
Table 4. The required properties for Filtration:

 Mechanical filter stability
 Hydraulic filter stability
 Long-term performance
Permanent filter function
A.O.S.
Thickness
Geotextile permeability
Chemical properties of water and soil
Chemical stability
Decay resistance
Temporary filter function
A.O.S.
Thickness
Geotextile permeability


  1. Reinforcement
Due to their high soil fabric friction coefficient and high tensile strength, heavy grades of geotextiles are used to reinforce earth structures allowing the use of local fill material.

Table 5: The required properties for reinforcement:
Mechanical
Hydraulic
Long-term performance
Base failure
Shear strength of bonding system
Hydraulic boundary conditions
Chemical and decay resistance
Top failure
Tensile strength of geotextile
Geotextile/ soil friction
Hydraulic boundary conditions
Chemical and decay resistance
Slope failure
Tensile strength of geotextile
Geotextile/ soil friction

Creep of the geotextile/ soil system
Chemical and decay resistance

  1. Protection:
Erosion of earth embankments by wave action, currents and repeated drawdown is a constant problem requiring the use of non-erodable protection in the form of rock beaching or mattress structures. Beneath these is placed a layer of geotextile to prevent leaching of fine material. The geotextile is easily placed, even under water.




Table 6:. The required properties for protection

Mechanical
Long-term performance
Tunnel construction
Burst pressure resistance
Puncture resistance
Abrasion resistance
Chemically stable: pH=2-13
Decay resistance
Landfill and reservoir geomembrane construction
Puncture resistance
Burst pressure resistance
Friction coefficient
Chemically stable: pH=2-13
Decay resistance
Flat roof construction
Puncture resistance
Chemical compatibility

7. APPLICATIONS
Case ( I ) : 
 Wet soil conditions in animal feeding and high-traffic live-stock handling areas cause problems for both animals and producers, as well as the environment. Ruminating animals, such as beef, dairy, and sheep, often concentrate at stream crossings, in paddock lanes, and in feedlots and barnyards. In association with animal production, there will be concentrated farm vehicular and equipment traffic. When the animal and/or equipment traffic is excessively high, the vegetation is destroyed. During and after rainy weather, the soil in these areas turns to mud, creating an unhealthy environment for optimal livestock production, poor traction for farm equipment, and potentially poor surface water quality. Once these areas dry, they may provide rough and possibly hazardous footing for the animals.
After the vegetation in these concentrated areas is destroyed, the soil is bare and subject to erosion. In addition, once wet soil that has been trampled by livestock dries, it has a greatly reduced infiltration rate, and thus a much higher potential for producing runoff of soil and manure. Both of these conditions are conducive to creating a water quality problem. However, all of the conditions summarized above cause problems for producers as they try to properly manage the many operations for a profitable livestock production system.
The use of geotextile fabric in these high-traffic livestock areas can substantially reduce the occurrence of adverse conditions (see Figure 1). The installation of geotextile fabric combined with gravel can help provide a proper surface that animals, humans, vehicles, and equipment can travel on, and can also provide an erosion control benefit.
The purpose of this publication is to help producers, landowners, and agency and industry personnel who work with producers and landowners, understand the proper application, installation, and maintenance of geotextile fabric for agricultural applications. This publication provides an overview of a demonstration project (Using Geotextile Cloth in Livestock Operations to Reduce Nutrient and Sediment Loading in the Olentangy Watershed) on the use of geotextile fabric in high-traffic livestock areas. Some of the material provided is based on cooperative agency-industry-producer experiences from twelve project sites constructed in Morrow County, Ohio, during 1994.


Case ( II ):
The leading cause of pavement and roadway failure in the U.S. is contamination of the aggregate base and the resulting loss in aggregate strength. When aggregate is placed on a subgrade, the bottom layer becomes contaminated with soil. Over time, traffic loading and vibration punches pavement base aggregate into the soil and causes silt and clay to migrate upward. On wet sites, construction traffic causes pumping of weak subgrade soils into overlying aggregate. All of these conditions decrease the effective aggregate thickness destroying the road support and reducing roadway performance and life.





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CONCEPT Of Lugeon test

Lugeon test

The Lugeon test, sometimes call also Packer test, is an in-situ testing method widely used to estimate the avarage hydraulic conductivity of rock mass. The test is named after Maurice Lugeon (1933), a Swiss geologist who first formulated the test. Basically, the Lugeon test is a costant head permeability type test carried out in a isolated part of a borehole. The results provide information about hzdraulic condictivuty of the rock mass including the rock matrix and the discontinuities.

DESCRIPTION AND PROCEDURE of Lugeon test

The test is conducted in a portion of a borehole isolated by pneumatic packers. The water is injected into the isolated portion of the borehole using a slotted pipe which it self is bounded by the inflated packers. The packers can be inflated using a gas compressor on the surafce, and so they can isolate and seal that portion of  the borehole. A pressure transducer is also located in that portion to measure the pressure with a help of reading station on the surface.
First of all, a maximum test pressure (Pmax) is defined so that it does not exceed the in-situ minimum stress, thus avoiding hzdraulic fracturing. The test is carried out at five stages including increasing and decreasing pressure between zero and maximum pressure. At each stage, a constant pressure is applied for an interval of 10 miniutes while pumping water. Water pressure and flow rate are measured everz minute. The five loading and unloading stages form a pressure loop often with the following pressure intervals:
Stage
 Pressure
 1st    
 0.50 Pmax
 2nd
 0.75 Pmax
 3rd
 Pmax
 4th
 0.75 Pmax
 5th
 0.50 Pmax

Using the average values of water presure and flow rate measured at each stage, the average hydraulic conductivity of the rock mass can be determined. Following the empiricl original definition of the test, the hzdraulic conductiviy is experessed in terms of Lugeon Unit, being the conductivity required for a flow aret of 1 liter per minute per meter of the borehole interval under a constant pressure of 1 MPa. The Lugeon value for each test is therefore calculated as follows and then an average representative value is selected for the tested rock mass.
            Lugeon Value = (q / L) x (P0 / P)
 where
q - flow rate [lit/min]
L - Length of the borehole test interval [m]
P0 - reference pressure of 1 MPa [MPa]
P - Test pressure [MPa]
Considering a homogenous and isotropic condition, one Lugeon will be equal to 1.3e-7m/s. Contrary to the continuum media, the hzdraulic conductivity of the rock mass is very much influenced by the rock discontinuities. Therefore, the Lugeon value could represent not only the conductivity but also the rock jointing condition. Typical range of Lugeon values and the corresponding rock condition is indicated in th etable below [1]

 Lugeon Value
 Conductivity classification
 Rock discontinuity condition
 <1
 Very low
 Very tight
 1-5
 Low
 Tight
 5-15
 Moderate
 Few partly open
 15-50
 Medium
 Some open
 50-100
 High
 Many open
 >100
 Very high
 Open closely spaced or voids
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Bearing capacity of soil

Bearing capacity

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

Qa   =    Qu                                 Qa = Allowable bearing capacity  (kN/m2) or (lb/ft2)
              F.S.
Where:
Qu = ultimate bearing capacity (kN/m2) or (lb/ft2)                *See below for theory
F.S. = Factor of Safety                                    *See information on factor of safety

Ultimate Bearing Capacity for Shallow Foundations

Terzaghi Ultimate Bearing Capacity Theory

Qu = c Nc + g D Nq + 0.5 g B Ng
      = Ultimate bearing capacity equation for shallow strip footings, (kN/m2) (lb/ft2)
Qu = 1.3 c Nc + g D Nq + 0.4 g B Ng
      = Ultimate bearing capacity equation for shallow square footings, (kN/m2) (lb/ft2)
Qu = 1.3 c Nc + g D Nq + 0.3 g B Ng
      = Ultimate bearing capacity equation for shallow circular footings, (kN/m2) (lb/ft2)
Where:
c = Cohesion of soil (kN/m2) (lb/ft2),
g = effective unit weight of soil (kN/m3) (lb/ft3),  *see note below
D = depth of footing (m) (ft),
B = width of footing (m) (ft),
Nc=cotf(Nq – 1),                                             *see typical bearing capacity factors
Nq=e2(3p/4-f/2)tanf / [2 cos2(45+f/2)],         *see typical bearing capacity factors
N g=(1/2) tanf(kp /cos2 f - 1),                         *see typical bearing capacity factors
e = Napier's constant = 2.718...,
kp = passive pressure coefficient, and
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, gw, 9.81 kN/m3 (62.4 lb/ft3), 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
Qu = 31.417(NB + ND)      (kN/m2)                        (metric) 
Qu =   NB    +   ND            (tons/ft2)                       (standard)
            10           10
For footing widths of 1.2 meters (4 feet) or less
Qa =   11,970N               (kN/m2)                        (metric)

Qa =   1.25N                   (tons/ft2)                       (standard)
               10   
For footing widths of 3 meters (10 feet) or more
Qa =   9,576N                 (kN/m2)                        (metric)

Qa =   N                          (tons/ft2)                       (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)

Qult = Qp + Qf
Where:
Qult = Ultimate bearing capacity of pile, kN (lb)
Qp = Theoretical bearing capacity for tip of foundation, or end bearing, kN (lb)
Qf = Theoretical bearing capacity due to shaft friction, or adhesion between foundation shaft and soil, kN (lb)

End Bearing (Tip) Capacity of Pile Foundation
Qp = Apqp
Where:
Qp = Theoretical bearing capacity for tip of foundation, or end bearing, kN (lb)
Ap = Effective area of the tip of the pile, m2 (ft2)
         For a circular closed end pile or circular plugged pile; Ap = p(B/2)2 m2 (ft2)
qp = gDNq
     = Theoretical unit tip-bearing capacity for cohesionless and silt soils, kN/m2 (lb/ft2)
qp = 9c
    = Theoretical unit tip-bearing capacity for cohesive soils, kN/m2 (lb/ft2)
g = effective unit weight of soil, kN/m3 (lb/ft3),                                *See notes below
D = Effective depth of pile, m (ft), where D < Dc,
Nq = Bearing capacity factor for piles,
c = cohesion of soil, kN/m2 (lb/ft2),
B = diameter of pile, m (ft), and
Dc = critical depth for piles in loose silts or sands m (ft).
         Dc = 10B, for loose silts and sands
         Dc = 15B, for medium dense silts and sands
         Dc = 20B, for dense silts and sands
     
Skin (Shaft) Friction Capacity of Pile Foundation
Qf = Afqf       for one homogeneous layer of soil
Qf = pSqfL    for multi-layers of soil
Where:
Qf = Theoretical bearing capacity due to shaft friction, or adhesion between foundation shaft and soil, kN (lb)
Af = pL; Effective surface area of the pile shaft, m2 (ft2)
qf = ks tan d = Theoretical unit friction capacity for cohesionless soils, kN/m2 (lb/ft2)
qf = cA + ks tan d = Theoretical unit friction capacity for silts, kN/m2 (lb/ft2)
qf = aSu = Theoretical unit friction capacity for cohesive soils, kN/m2 (lb/ft2)
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(Suc)2 = adhesion factor, kN/m2 (ksf), where Suc < 48 kN/m2 (1 ksf)
a =    1    [0.9 + 0.3(Suc - 1)] kN/m2, (ksf) where Suc > 48 kN/m2, (1 ksf)
         Suc
Suc = 2c = Unconfined compressive strength , kN/m2 (lb/ft2)
cA = adhesion
     = c for rough concrete, rusty steel, corrugated metal
     0.8c < cA < c for smooth concrete
     0.5c < cA < 0.9c for clean steel
c = cohesion of soil, kN/m2 (lb/ft2)
d = external friction angle of soil and wall contact (deg)
f = angle of internal friction (deg)
s = gD = effective overburden pressure, kN/m2, (lb/ft2)
k = lateral earth pressure coefficient for piles
g = effective unit weight of soil, kN/m3 (lb/ft3)                    *See notes below
B = diameter or width of pile, m (ft)
D = Effective depth of pile, m (ft), where D < Dc
Dc = critical depth for piles in loose silts or sands m (ft).
         Dc = 10B, for loose silts and sands
         Dc = 15B, for medium dense silts and sands
         Dc = 20B, for dense silts and sands
S = summation of differing soil layers (i.e. a1 + a2 + .... + an)
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, gw, 9.81 kN/m3 (62.4 lb/ft3), 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
     qp = 4Nc  tons/ft2                      standard
Theoretical unit tip-bearing capacity for drilled piles in sand:
                                                                                      
     qp = 1.2Nc  tons/ft2                   standard
Theoretical unit friction-bearing capacity for driven piles in sand:
                                                                                      
     qf =  N   tons/ft2                         standard
             50
Theoretical unit friction-bearing capacity for drilled piles in sand:
                                                                                      
     qf =  N   tons/ft2                         standard
            100
Where:
D = pile embedment depth, ft
B = pile diameter, ft
Nc = Cn(N)
Cn = 0.77 log  20   
                        s
N = N-Value from SPT test   
s = gD = effective overburden stress at pile embedment depth,  tons/ft2
    = (g - gw)D = effective stress if below water table,  tons/ft2
g = effective unit weight of soil,  tons/ft3
gw = 0.0312 tons/ft3 = 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/m2 (3000 lb/ft2) building pressure.
Given
  • bearing pressure from building = 144 kN/m2 (3000 lbs/ft2)
  • unit weight of soil, g = 21 kN/m3 (132 lbs/ft3)  *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 Ng, Nc and Nq. See typical bearing capacity factors relating to the soils' angle of internal friction.
  • Ng = 22
  • Nc = 35.5
  • Nq = 23.2
Solve for ultimate bearing capacity,
Qu = c Nc + g D Nq + 0.5 g B Ng                                *strip footing eq.

Qu = 0(35.5) + 21 kN/m3(0.6m)(23.2) + 0.5(21 kN/m3)(0.3 m)(22)                 metric
Qu = 362 kN/m2
Qu = 0(35.5) + 132lbs/ft3(2ft)(23.2) + 0.5(132lbs/ft3)(1ft)(22)                          standard
Qu = 7577 lbs/ft2
Solve for allowable bearing capacity,
Qa =   Qu     
           F.S.
Qa =  362 kN/m2  = 121 kN/m2                                          not o.k.                 metric
               3   
Qa =  7577lbs/ft2  = 2526 lbs/ft2                                          not o.k.                 standard  
               3            
Since Qa < required 144 kN/m2 (3000 lbs/ft2) bearing pressure, increase footing width, B or foundation depth, D to increase bearing capacity.
Try footing width, B = 0.61 m (B = 2 ft).
Qu = 0 + 21 kN/m3(0.61 m)(23.2) + 0.5(21 kN/m3)(0.61 m)(22)                      metric
Qu = 438 kN/m2
Qu = 0 + 132 lbs/ft3(2 ft)(23.2) + 0.5(132 lbs/ft3)(2 ft)(22)                                standard 
Qu = 9029 lbs/ft2

Qa =   438 kN/m2   = 146 kN/m2          Qa > 144 kN/m2            o.k.                   metric
               3
Qa =   9029 lbs/ft2 = 3010 lbs/ft2           Qa > 3000 lbs/ft2           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 Nq = 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/m2
  • bearing pressure from building column = 1000 lbs/ (1 ft x 1 ft) = 1000 lbs/ft2
  • unit weight of saturated soil, gsat= 20.3 kN/m3 (129 lbs/ft3)            *see typical g values
  • unit weight of water, gw= 9.81 kN/m3 (62.4 lbs/ft3)                        *constant
  • Cohesion, c = 21.1 kN/m2 (440 lbs/ft2)                *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 Ng, Nc and Nq. See typical bearing capacity factors relating to the soils' angle of internal friction.
  • Ng = 0
  • Nc = 5.7
  • Nq = 1
Solve for ultimate bearing capacity,
Qu = 1.3c Nc + g D Nq + 0.4 g B Ng                                  *square footing eq.
Qu =1.3(21.1kN/m2)5.7+(20.3kN/m3-9.81kN/m3)(0.61m)1+0.4(20.3kN/m3-9.81kN/m3)(0.3m)0
Qu = 163 kN/m2                                                                                       metric
Qu = 1.3(440lbs/ft2)(5.7) + (129lbs/ft3 - 62.4lbs/ft3)(2ft)(1) + 0.4(129lbs/ft3 - 62.4lbs/ft3)(1ft)(0)
Qu = 3394 lbs/ft2                                                                                      standard
Solve for allowable bearing capacity,
Qa =   Qu     
           F.S.
Qa =   163 kN/m2   = 54 kN/m2             Qa > 48.9 kN/m2         o.k.          metric
               3        
Qa =    3394lbs/ft2   = 1130 lbs/ft2         Qa > 1000 lbs/ft2         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.
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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/m2 (1.5 tons/ft2) building pressure.
Given
  • bearing pressure from building = 144 kN/m2 (1.5 tons/ft2)
  • 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
Qu = 31.417(NB + ND)      (kN/m2)                          (metric) 
Qu =   NB    +   ND            (tons/ft2)                         (standard)
            10           10
Qu = 31.417(36(0.3m) + 36(0.61m)) = 1029 kN/m2    (metric) 
Qu =   36(1 ft)   +   36(2 ft)   = 10.8 tons/ft2                 (standard)
              10                10
Solve for allowable bearing capacity,
Qa =   Qu     
           F.S.
Qa =  1029 kN/m2  = 343 kN/m2    Qa > 144 kN/m2      o.k.    (metric)
               3   
Qa =  10.8 tons/ft2  = 3.6 tons/ft2     Qa > 1.5 tons/ft2      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.
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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
    • unit weight, g = 19.6 kN/m3 (125 lbs/ft3) *from soil testing, see typical g values
    • cohesion, c = 47.9 kN/m2 (1000 lb/ft2)   *from soil testing, see typical c values
    • angle of internal friction, f = 30 degrees   *from soil testing, see typical f values
  • 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,
Qp = Apqp
Ap = p(B/2)2 = p(0.61m/2)2 = 0.292 m2                                      metric
Ap = p(B/2)2 = p(2ft/2)2 = 3.14 ft2                                                        standard
qp = gDNq 
g = 19.6 kN/m3 (125 lbs/ft3); 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 < Dc
       Dc = 15B = 9.2 m (30 ft); critical depth for medium dense silts.
       If D > Dc, then use Dc
Nq = 25; Meyerhof bearing capacity factor for driven piles, based on f
qp = 19.6 kN/m3(1.5 m)25 = 735 kN/m2                                    metric
qp = 125 lb/ft3(5 ft)25 = 15,625 lb/ft2                                          standard

Qp = Apqp = (0.292 m2)(735 kN/m2) = 214.6 kN                     metric
Qp = Apqp = (3.14 ft2)(15,625 lb/ft2) = 49,063 lb                      standard

Determine ultimate friction capacity of pile,
Qf = Afqf
Af = 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 Dc as above
Af = 1.92 m(1.5 m) = 2.88 m2                                                      metric    
Af = 6.28 ft(5 ft) = 31.4 ft2                                                           standard
qf = cA + ks tan d = cA + kgD tan d
k = 0.5; lateral earth pressure coefficient for piles, value chosen from Broms low density steel
g = 19.6 kN/m3 (125 lb/ft3); given effective soil unit weight. If water table, then g - gw
D = L = 1.5 m (5 ft); pile length. Check to see if D < Dc
Dc = 15B = 9.2 m (30 ft); critical depth for medium dense silts. If D > Dc, then use Dc
d = 20 deg; external friction angle, equation chosen from Broms steel piles
B = 0.61 m (2 ft); selected pile diameter
cA = 0.5c; for clean steel. See adhesion in pile theories above.
     = 24 kN/m2 (500 lb/ft2) 
qf = 24 kN/m2 + 0.5(19.6 kN/m3)(1.5m)tan 20 = 29.4 kN/m2      metric
qf = 500 lb/ft2 + 0.5(125 lb/ft3)(5ft)tan 20 = 614 lb/ft2                  standard
Qf = Afqf = 2.88 m2(29.4 kN/m2) = 84.7 kN                               metric
Qf = Afqf = 31.4 ft2(614 lb/ft2) = 19,280 lb                                  standard

Determine ultimate pile capacity,
Qult = Qp + Qf
Qult = 214.6 kN + 84.7 kN = 299.3 kN                                       metric
Qult = 49,063 lb + 19,280 lb = 68,343 lb                                      standard

Solve for allowable bearing capacity,
Qa =  Qult      
           F.S.
Qa =     299.3 kN    = 99.8 kN;  Qa > applied load (66.7 kN)     o.k.     metric
                   3
Qa =     68,343 lbs    = 22,781 lbs  Qa > applied load (15 kips)   o.k.     standard
                   3

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.
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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
    • unit weight, g = 19.6 kN/m3 (125 lbs/ft3) *from soil testing, see typical g values
    • cohesion, c = 0                                        *from soil testing, see typical c values
    • angle of internal friction, f = 30 degrees   *from soil testing, see typical f values
  • soils below 1.5 m (5 ft) of soil is a stiff silty clay                     
    • unit weight, g = 18.9 kN/m3 (120 lbs/ft3)                    
    • cohesion, c = 47.9 kN/m2 (1000 lb/ft2)
    • angle of internal friction, f = 0 degrees
  • 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,
Qp = Apqp
Ap = p(B/2)2 = p(0.61m/2)2 = 0.292 m2                                      metric
Ap = p(B/2)2 = p(2ft/2)2 = 3.14 ft2                                                        standard

qp = 9c = 9(47.9 kN/m2) = 431.1 kN/m2                                     metric
qp = 9c = 9(1000 lb/ft2) = 9000 lb/ft2                                           standard

Qp = Apqp = (0.292 m2)(431.1 kN/m2) = 125.9 kN                    metric
Qp = Apqp = (3.14 ft2)(9000 lb/ft2) = 28,260 lb                           standard

Determine ultimate friction capacity of pile,
Qf = pSqfL
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
qfL = [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/m3 (125 lb/ft3); given effective soil unit weight. If water table, then g - gw
D = L = 1.5 m (5 ft); segment of pile within this soil strata. Check to see if D < Dc
Dc = 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 > Dc, then use Dc
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
qfL = [1.5(19.6 kN/m3)(1.5m)tan (20)]1.5 m = 24.1 kN/m            metric
qfL = [1.5(125 lb/ft3)(5ft)tan (20)]5 ft = 1706 lb/ft                         standard

soils below 1.5 m (5 ft) of subgrade
qfL = aSu

Suc = 2c = 95.8 kN/m2 (2000 lb/ft2); unconfined compressive strength 
c = 47.9 kN/m2 (1000 lb/ft2); cohesion from soil testing (given)
a =    1    [0.9 + 0.3(Suc - 1)] = 0.3; because Suc > 48 kN/m2, (1 ksf)
         Suc
L = 0.91 m (3 ft); segment of pile within this soil strata
qfL = [0.3(95.8 kN/m2)]0.91 m = 26.2 kN/m                                metric
qfL = [0.3(2000 lb/ft2)]3 ft = 1800 lb/ft                                         standard

ultimate friction capacity of combined soil layers
Qf = pSqfL
Qf = 1.92 m(24.1 kN/m + 26.2 kN/m) = 96.6 kN                         metric 
Qf = 6.28 ft(1706 lb/ft + 1800 lb/ft) = 22,018 lb                            standard  

Determine ultimate pile capacity,
Qult = Qp + Qf
Qult = 125.9 kN + 96.6 kN = 222.5 kN                                       metric
Qult = 28,260 lb + 22,018 lb = 50,278 lb                                      standard

Solve for allowable bearing capacity,
Qa =  Qult      
           F.S.
Qa =     222.5 kN    = 74.2 kN;  Qa > applied load (66.7 kN)     o.k.     metric
                   3
Qa =     50,275 lbs    = 16,758 lbs  Qa > applied load (15 kips)   o.k.     standard
                   3
                                                          

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.
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