NPTEL β ADVANCED FOUNDATION ENGINEERING-I
Module 4
(Lecture 16)
SHALLOW FOUNDATIONS: ALLOWABLE BEARING CAPACITY AND SETTLEMENT
Topics
1.1 STRIP FOUNDATION ON GRANULAR SOIL REINFORCED BY METALLIC STRIPS
Mode of Failure Location of Failure Surface Force Induced in Reinforcement Ties
1.2 FACTOR OF SAFETY OF TIES AGAINST BREAKING AND PULLOUT
1.3 DESIGN PROCEDURE FOR STRIP FOUNDATION ON REINFORCED EARTH
Determination of ππππ Determination of πΈπΈπΉπΉ Calculation of Tie Force Calculation of tie Resistance Due to Friction, πππ©π© Calculation of Tie Thickness to Resist Tie Breaking Calculation of Minimum Length of Ties
STRIP FOUNDATION ON GRANULAR SOIL REINFORCED BY METALLIC STRIPS
Mode of Failure
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The nature of bearing capacity failure of a shallow strip foundation resting on a compact and homogeneous soil mass was presented in figure 4.1a (from chapter 3). In contrast, if layers of reinforcing strips, or ties, are placed in the soil under a shallow strip foundation, the nature of failure in the soil mass will be like that shown in figure 4.36a, b, and c.
The type of failure in the soil mass shown in figure 4.36a generally occurs when the first layer of reinforcement is placed at a depth, d, greater than about 2
3π΅π΅ (B = width of the foundation). If the reinforcements in the first layer are strong and they are sufficiently concentrated, they may act as a rigid base at a limited depth. The bearing capacity of foundations in such cases can be evaluated by the theory presented by Mandel and Salencon (1972). Experimental laboratory results for the bearing capacity of shallow foundations resting on a sand layer with a rigid rough base at a limited depth have also been provided by Meyerhof (1974), Pfeifle, and Das (1979), and Das (1981).
The type of failure shown in figure 4.36b could occur if ππ/π΅π΅ is less than about 23 and the
numbers of layers of reinforcement, N, is less than about 2-3. In this type of failure, reinforcement tie pullout occurs.
The most beneficial effect of reinforced earth is obtained when ππ/π΅π΅less than about is 23
and the number of reinforcement layers is greater than 4 but no more than 6-7. In this case, the soil mass fails when the upper ties break (see figure 4.36c).
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Figure 4.36 Three modes of bearing capacity failure in reinforced earth (redrawn after Binquet and Lee, 1975b)
Location of Failure Surface
Figure 4.37 shows an idealized condition for development of the failure surface in soil for the condition shown in figure 4.36c. It consists of a central zone-zone I-immediately below the foundation that settles along with the foundation with the application of load. On each side of zone I, the soil is pushed outward and upward-this zone II. The points Aβ, Aβ, Aββ,β¦ and B;, Bβ, Bββ, β¦ which define the limiting lines between zones I and II, can be obtained by considering the shear stress distribution, πππ₯π₯π₯π₯ , in the soil caused by the foundation load. The term πππ₯π₯π₯π₯ refer to the shear stress developed at a depth z below the foundation at a distance x measured from the center line of the foundation. If integration of Boussinesqβs equation is performed, πππ₯π₯π₯π₯ is given by the relation
Figure 4.37 Failure mechanism under a foundation supported by reinforced earth [part (b) after Binquet and Lee, 1975b]
πππ₯π₯π₯π₯ = 4πππππ π π₯π₯π₯π₯ 2
ππ[(π₯π₯2+π₯π₯2βππ2)2+4ππ2π₯π₯2] [4.68]
Where q
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
ππ = half β width of the foundation = π΅π΅/2
π΅π΅ = width of foundation
πππ π = load per unit area on the foundation
The variation of πππ₯π₯π₯π₯ at any depth, z, is shown by the broken lines in figure 4.37a. Points Aβ and Bβ refer to the points at which the value of πππ₯π₯π₯π₯ is maximum at π₯π₯ = π₯π₯1.
Similarly, Aβ and Bβ refer to the points at which πππ₯π₯π₯π₯ is maximum at π₯π₯ = π₯π₯2. The distances π₯π₯ = ππππ at which the maximum value of πππ₯π₯π₯π₯ occurs take a nondimensional form and are shown in figure 4.37b.
Force Induced in Reinforcement Ties
Assumptions needed to obtain the tie force at any given depth are as follows:
1. Under the application of bearing pressure by the foundation, the reinforcing ties at points Aβ, Aβ, Aββ,β¦ and B;, Bβ, Bββ, β¦take the shape shown in figure 4.37c. That is, the ties take two right-angle turns on each side of zone I around two frictionless rollers.
2. For N reinforcing layers, the ratio of the load per unit area on the foundation supported by reinforced earth, πππ π , to the load per unit area on the foundation supported by unreinforced earth, ππππ , is constant irrespective of the settlement level, S (see figure 4.38). Benquet and Lee (1975a) proved this relation in laboratory experiments.
Figure 4.39a shows a continuous foundation supported by unreinforced soil and subjected to a load of ππππ per unit area. Similarly, figure 4.39b shows a continuous foundation supported by a reinforced soil layer (one layer of reinforcement, or ππ = 1) and subjected to a load of πππ π per unit area. (Due to symmetry only one-half of the foundation is shown in figure 4.39). In both cases-that is, in figure 4.39a and 4.39b-let the settlement equal ππππ . For one-half of each foundation under consideration, the following are the forces per unit length on a soil element of thickness βπ»π» located at a depth z.
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
Figure 4.38 Relationship between load per unit area and settlement for foundations resting on reinforced and unreinforced soil
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
Figure 4.39 Derivation of equation (87)
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
Figure 4.39a continued
Unreinforced Case πΉπΉ1 and πΉπΉ2 are the vertical forces and ππ1 is the shear force. Hence, for equilibrium,
πΉπΉ1 β πΉπΉ2 β ππ1 = 0 [4.69]
Reinforced Case Here, πΉπΉ3 and πΉπΉ4 are the vertical forces, ππ2 is the shear force, and ππ(ππ=1) is the tensile force developed in the reinforcement. The force ππ(ππ=1) is vertical because of the assumption made for the deformation of reinforcement as shown in figure 4.37c. So
πΉπΉ3 β πΉπΉ4 β ππ2 β ππ(ππ=1) = 0 [4.70]
If the foundation settlement, ππππ , is the same in both cases,
πΉπΉ2 = πΉπΉ4 [4.71]
Subtracting equation (69) from equation (70) and using the relationship given in equation (71), we obtain
ππ(ππ=1) = πΉπΉ3 β πΉπΉ1 β ππ2 + ππ1 [4.72]
Note that the force πΉπΉ1 is caused by the vertical stress, ππ, on the soil element under consideration as a result of the load ππππ on the foundation. Similarly, πΉπΉ3 is caused by the vertical stress imposed on the soil element as a result of the load πππ π . Hence
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
πΉπΉ1 = β« ππ(ππππ)πππ₯π₯π₯π₯ππ0 [4.73]
πΉπΉ3 = β« ππ(πππ π )πππ₯π₯π₯π₯ππ0 [4.74]
ππ1 = πππ₯π₯π₯π₯ (ππππ)βπ»π» [4.75]
ππ2 = πππ₯π₯π₯π₯ (πππ π )βπ»π» [4.76]
Where
ππ(ππππ) and ππ(πππ π ) are the vertical stresses at a depth z caused by the loads ππππ and πππ π on the foundation
πππ₯π₯π₯π₯ (ππππ) and πππ₯π₯π₯π₯ (πππ π ) are the shear stresses at a depth z and at a distance ππππ from the center line caused by the loads ππππ and πππ π
Integrating Boussinesqβs solution yields
ππ(ππππ) = πππππποΏ½π‘π‘π‘π‘π‘π‘β1 π₯π₯
π₯π₯βππβ π‘π‘π‘π‘π‘π‘β1 π₯π₯
π₯π₯+ππβ 2πππ₯π₯ (π₯π₯2βπ₯π₯2βππ2)
(π₯π₯2+π₯π₯2βππ2)2+4ππ2π₯π₯2οΏ½ [4.77]
ππ(πππ π ) = πππππποΏ½π‘π‘π‘π‘π‘π‘β1 π₯π₯
π₯π₯βππβ π‘π‘π‘π‘π‘π‘β1 π₯π₯
π₯π₯+ππβ 2πππ₯π₯ (π₯π₯2βπ₯π₯2βππ2)
(π₯π₯2+π₯π₯2βππ2)2+4ππ2π₯π₯2οΏ½ [4.78]
πππ₯π₯π₯π₯ (ππππ) = 4πππππππππππ₯π₯2
ππ[(ππππ2+π₯π₯2βππ2)2+4ππ2π₯π₯2] [4.79]
πππ₯π₯π₯π₯ (πππ π ) = 4πππππ π πππππ₯π₯2
ππ[(ππππ2+π₯π₯2βππ2)2+4ππ2π₯π₯2] [4.80]
Where
ππ = π΅π΅/2
The procedure for derivation of equations (77 to 80) is not presented here; for this information see a soil mechanics textbook (for example, Das, 1997. Proper substitution of equations (77 to 80) into equations (73 to 76) and simplification yields
πΉπΉ1 = π΄π΄1πππππ΅π΅ [4.81]
πΉπΉ3 = π΄π΄1πππ π π΅π΅ [4.82]
ππ1 = π΄π΄2ππππβπ»π» [4.83]
ππ2 = π΄π΄2πππ π βπ»π» [4.84]
Where
π΄π΄1 and π΄π΄2 = ππ(π₯π₯/π΅π΅)
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
The variations of π΄π΄1 and π΄π΄2 with nondimensional depth z are given in figure 4.40. Substituting equations (81)-(84) into equation (72) gives
Figure 4.40 Variation of π΄π΄1,π΄π΄2, and π΄π΄3 with π₯π₯/π΅π΅ (after Binquet and Lee, 1975b)
ππ(ππ=1) = π΄π΄1πππ π π΅π΅ β π΄π΄1πππππ΅π΅ β π΄π΄2πππ π βπ»π» +π΄π΄2ππππβπ»π» = π΄π΄1π΅π΅(πππ π β ππππ) β π΄π΄2βπ»π»(πππ π β ππππ) =ππππ οΏ½
πππ π ππππβ 1οΏ½ (π΄π΄1π΅π΅ β π΄π΄2βπ»π» [4.85]
Note that the derivation of equation (85) was based on the assumption that there is only one layer of reinforcement under the foundation shown in figure 4.39b. However, if there are N layers of reinforcement under the foundation with center-to-center spacing of βπ»π», as shown in figure 4.39c, the assumption can be made that
ππ(ππ) = ππ(ππ=1)
ππ [4.86]
Combining equations (85 and 86) gives
ππ(ππ) = 1πποΏ½ππππ οΏ½
πππ π ππππβ 1οΏ½ (π΄π΄1π΅π΅ β π΄π΄2βπ»π»οΏ½ [4.87]
The unit of ππ(ππ) in equation (87) is lb/ft (or kN) per unit length of foundation.
FACTOR OF SAFETY OF TIES AGAINST BREAKING AND PULLOUT
Once the tie forces that develop in each layer as the result of the foundation load are determined from equation (87), an engineer must determine whether the ties at any depth z will fail either by breaking or by pullout. The factor of safety against tie breaking at any depth z below the foundation can be calculated as
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
πΉπΉππ(π΅π΅) = π€π€π‘π‘π‘π‘ πππ¦π¦ππ(ππ)
[4.88]
Where
πΉπΉππ(π΅π΅) = factor of safety against the breakin
π€π€ = width of a single tie
π‘π‘ = thickness of each tie
π‘π‘ = number of ties per unit length of the foundation
πππ¦π¦ = yield or breaking strength of the tie material
The term π€π€π‘π‘ may be defined as the linear density ratio, LDR, so
πΉπΉππ(π΅π΅) = οΏ½ π‘π‘πππ¦π¦ππ(ππ)
οΏ½ (πΏπΏπΏπΏπ π ) [4.89]
The resistance against the tie being pulled out derives from the frictional resistance between the soil and the ties at any depth. From the fundamental principles of statics, we know that the frictional force per unit length of the foundation resisting tie pullout at any depth z (figure 4.41).
Figure 4.41 Variation of equation (91)
πΉπΉπ΅π΅ = 2 tanβ π’π’[normal force] [4.90]
=2 tanβ π’π’
βtwo sides of tie (i. e. , top and bottom)
β£β’β’β’β‘(πΏπΏπΏπΏπ π ) οΏ½ ππ(πππ π )πππ₯π₯ + (πΏπΏπΏπΏπ π )(πΎπΎ)(πΏπΏππ β ππππ)(π₯π₯ + πΏπΏππ)
πΏπΏππ
π₯π₯ππ β¦
β₯β₯β₯β€
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
Where
πΎπΎ = unit weight of soil
πΏπΏππ = depth of foundation
πππ’π’ = tie β soil friction angle
The relation for ππ(πππ π ) was defined in equation (78). The value of π₯π₯ = πΏπΏππ is generally assumed to be the distance at which ππ(πππ π ) equals to 0.1πππ π . The value of πΏπΏππ as a function of depth z is given in figure 4.42. Equation (90) may be simplified as
πΉπΉπ΅π΅ = 2 tanπππ¦π¦ (πΏπΏπΏπΏπ π ) οΏ½π΄π΄3π΅π΅ππππ οΏ½πππ π πππποΏ½ + πΎπΎ(πΏπΏππ β ππππ)(π₯π₯ + πΏπΏππ)οΏ½ [4.91]
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Figure 4.42 Variation of πΏπΏππ/π΅π΅ with π₯π₯/π΅π΅ (after Binquet and Lee, 1975b)
Where
π΄π΄3 is a nondimensionla quantity that may be expressed as a function of depth (π₯π₯/π΅π΅) (see figure 4.40)
The factor of safety against tie pullout, πΉπΉππ(ππ), is
πΉπΉππ(ππ) = πΉπΉπ΅π΅ππ(ππ)
[4.92]
DESIGN PROCEDURE FOR STRIP FOUNDATION ON REINFORCED EARTH
Following is a step-by-step procedure for the design of a strip foundation supported by granular soil reinforced by metallic strips:
1. Obtain the total load to be supported per unit length of the foundation. Also obtain the quantities a. Soil-friction angle, ππ b. Soil-tie friction angle, ππππ c. Factor of safety against bearing capacity failure d. Factor of safety against tie breaking, πΉπΉππ(π΅π΅) e. Factor of safety against tie pullout, πΉπΉππ(ππ)
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
f. Breaking strength of reinforcement ties, πππ¦π¦ g. Unit weight of soil, πΎπΎ h. Modulus of elasticity of soil, πΈπΈπ π i. Poissonβs ratio of soil, πππ π j. Allowable settlement of foundation, ππππ k. Depth of foundation, πΏπΏππ
2. Assume a width of foundation, B, and also d and N. the value of d should be less than 2
3π΅π΅. Also, the distance from the bottom of the foundation to the lowest layer of the reinforcement should be about 2B or less. Calculate Ξπ»π».
3. Assume a value of πΏπΏπΏπΏπ π 4. For width Bi (step 2) determine the ultimate bearing capacity, πππ’π’ , for unreinforced
soil [equation (3 from chapter 3); note: ππ = 0]. Determine πππ‘π‘ππππ (1): πππ‘π‘ππππ (1) = πππ’π’
πΉπΉππ against bearing capacity failure [4.93]
5. Calculate the allowable load, πππ‘π‘ππππ (2) based on the tolerable settlement, ππππ ,
assuming that the soil is not reinforced [equation (32a)]: ππππ = π΅π΅πππ‘π‘ππππ (2)
πΈπΈπ π (1 β πππ π 2)πΌπΌππ
For πΏπΏ/π΅π΅ = β, the value of πΌπΌππ may be taken as 2, or πππ‘π‘ππππ (2) = πΈπΈπ π ππππ
π΅π΅(1βπππ π 2)πΌπΌππ [4.94]
(The allowable load for a given settlement, ππππ , could have also been determined from equations that relate to standard penetration resistances).
6. Determine the lower of the two value of πππ‘π‘ππππ obtained from steps 4 and 5. The lower value of πππ‘π‘ππππ equals ππππ .
7. Calculate the magnitude of πππ π for the foundation supported by reinforced earth: πππ π = load on foundation per unit length
π΅π΅ [4.95]
8. Calculate the tie force, ππ(ππ), in each layer of reinforcement by using equation (87)
(note: unit of ππ(ππ) as kN/m of foundation). 9. Calculate the frictional resistance of ties for each layer per unit length of
foundation, πΉπΉπ΅π΅, by using equation (91). For each layer, determine whether πΉπΉπ΅π΅/ππ(ππ) β₯ πΉπΉππ(ππ). If πΉπΉπ΅π΅/ππ(ππ) < πΉπΉππ(ππ), the length of the reinforcing strips for a layer may be increased. That will increase the value of πΉπΉπ΅π΅ and thus πΉπΉππ(ππ), and so equation (91) must be rewritten as πΉπΉπ΅π΅ = 2 tanππππ (πΏπΏπΏπΏπ π ) οΏ½π΄π΄3π΅π΅ππππ οΏ½
πππ π πππποΏ½ + πΎπΎ(πΏπΏππ β ππππ)(π₯π₯ + πΏπΏππ)οΏ½ [4.96]
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Where πΏπΏ = the required length to obtain the desired value of FB
10. Use equation (89) to obtain the tie thickness for each layer. Some allowance should be made for the corrosion effect of the reinforcement during the life of the structure.
11. If he design is unsatisfactory, repeat steps 2-10.
The following example demonstrates the application of these steps.
Example 10
Design a strip foundation that will carry a load of 1.8 MN/m. Use the following parameters:
ππππππππ: πΎπΎ = 17.3 kN/m3; ππ = 35Β°; πΈπΈπ π = 3 Γ 104kN/m2; πππ π = 0.35
π π π π πππ‘π‘πππππππππ π π π π π π‘π‘π‘π‘ π‘π‘πππ π π π : πππ¦π¦2.5 Γ 105 kN/m2; ; ππππ = 28Β°; πΉπΉπ π (π π ) = 3; πΉπΉππ(ππ) = 2.5
πΉπΉπππ’π’π‘π‘πππ‘π‘π‘π‘πππππ‘π‘: πΏπΏf = 1m; Factor of safety against baring capacity failure = 3.
Tolerable settlement= ππππ = 25 mm; desired life of structure = 50 years
Solution
Let
π΅π΅ = 1 m
ππ = depth from the bottom of the foundation to the first reinforcing layer = 0.5
Ξπ»π» = 0.5 m
ππ = 5
πΏπΏπΏπΏπ π = 65%
If the reinforcing strips used are 75 mm wide, then
π€π€π‘π‘ = πΏπΏπΏπΏπ π
Or
π‘π‘ = πΏπΏπΏπΏπ π π€π€
= 0.650.075 m
= 8.67/m
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Hence each layer will contain 8.67 strips per meter length of the foundation.
Determination of ππππ
For an unreinforced foundation
πππ’π’ = πΎπΎπΏπΏππππππ + 12πΎπΎπ΅π΅πππΎπΎ
From table 4 (from chapter 3) for ππ = 35Β°,ππππ = 33.30 and πππΎπΎ48.03. Thus
πππ’π’ = (17.3)(1)(33.3) + 12
(17.3)(1)(48.03)
= 576.09 + 415.46 = 991.55 β 992 kN/m2
πππ‘π‘ππππ (1) = πππ’π’πΉπΉππ
= 9923
= 330.7 kN/m2
From equation (94)
πππ‘π‘ππππ (2) = (πΈπΈπ π )(ππππ)π΅π΅(1βπππ π 2)πΌπΌππ
= (30,000 kN /m2)(0.025m)(1m)(1β0.352)(2)
= 427.35 kN/m2
As πππ‘π‘ππππ (1) < πππ‘π‘ππππ (2), ππππ = πππ‘π‘ππππ (1) = 330.7 kN/m2
Determination of πΈπΈπΉπΉ
From equation (95),
πππ π = 1.8 MN /mπ΅π΅
= 1.8Γ103
1= 1.8 Γ 103kN/m2
Calculation of Tie Force
From equation (87),
ππ(ππ) = οΏ½πππππποΏ½ οΏ½πππ π
ππππβ 1οΏ½ (π΄π΄1π΅π΅ β π΄π΄2βπ»π»)
The tie forces for each layer are given in the following table.
Layer no οΏ½
πππππποΏ½ οΏ½πππ π ππππ
β 1οΏ½
π₯π₯(m) π₯π₯π΅π΅
π΄π΄1π΅π΅ π΄π΄2βπ»π» π΄π΄1π΅π΅β π΄π΄2βπ»π»
ππ(ππ)(kN/m)
1 293.7 0.5 0.5 0.35 0.125 0.225 66.08
2 293.7 1.0 1.0 0.34 0.09 0.25 73.43
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3 293.7 1.5 1.5 0.34 0.065 0.275 80.77
4 293.7 2.0 2.0 0.33 0.05 0.28 82.24
5 293.7 2.5 2.5 0.32 0.04 0.28 82.24
Note: π΄π΄1 is from figure 4.40; π΅π΅ = 1 m; βπ»π» = 0.5 m; π΄π΄2 is from figure 4.40; πππ’π’/ππππ = 1.8 Γ104/330.7 β 5.44
Calculation of tie Resistance Due to Friction, πππ©π©
Use equation (91):
πΉπΉπ΅π΅ = 2 tanππππ (πΏπΏπΏπΏπ π ) οΏ½π΄π΄3π΅π΅ππππ οΏ½πππ π πππποΏ½ + πΎπΎ(πΏπΏππ β ππππ)(π₯π₯ + πΏπΏππ)οΏ½
The following table shows the magnitude of πΉπΉπ΅π΅ for each layer:
Layer number
Quantity 1 2 3 4 5
2 tanππππ (πΏπΏπΏπΏπ π ) 0.691 0.691 0.691 0.691 0.691
π΄π΄3 0.125 0.14 0.15 0.15 0.15
π΄π΄3π΅π΅ππππ(πππ π /ππππ) 225.0 252.0 270.0 270.0 270.0
π₯π₯(π π ) 0.5 1.0 1.5 2.0 2.5
π₯π₯/π΅π΅ 0.5 1.0 1.5 2.0 2.5
πΏπΏππ(m) 1.55 2.6 3.4 3.85 4.2
ππππ(m) 0.55 0.8 1.1 1.4 1.65
πΏπΏππ β ππππ(m) 1.0 1.8 2.3 2.45 2.55
π₯π₯ + πΏπΏππ(m) 1.5 2.0 2.5 3.0 3.5
πΎπΎ(πΏπΏππ β ππππ)(π₯π₯ + πΏπΏππ (m) 25.95 62.28 99.48 127.16 154.4
πΉπΉπ΅π΅(kN/m) 173.4 217.2 255.1 274.4 292.3
πΉπΉππ(ππ) = πΉπΉπ΅π΅/ππ(ππ) 2.62 2.96 3.16 3.34 3.57
Note: π΄π΄3 is from figure 4.40; ππππ is from figure 4.37; πΏπΏππ is from figure 4.42; ππ(ππ) is from
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
the preceding table.
The minimum factor of safety is greater than the required values of πΉπΉπ π (ππ) which is 2.5.
Calculation of Tie Thickness to Resist Tie Breaking
From equation (89),
πΉπΉπ π (π΅π΅) = π‘π‘πππ¦π¦ππ(ππ)
(πΏπΏπΏπΏπ π )
Here, πππ¦π¦ = 2.5 Γ 105 kN/m2, πΏπΏπΏπΏπ π = 0.65 and πΉπΉπ π (π΅π΅) = 3, so
π‘π‘ = οΏ½ 3(2.5Γ105)(0.65)
οΏ½ ππ(ππ) = (1.846 Γ 105)ππ(ππ)
So, for layer 1
π‘π‘ = (1.846 Γ 10β5)(66.08) = 0.00122m = 1.22 mm
For layer 2
π‘π‘ = (1.846 Γ 10β5)(73.43) = 0.00136m = 1.36 mm
Similarly, for layer 3
π‘π‘ = 0.00149 = 1.49 mm
For layer 4
π‘π‘ = 1.52 mm
For layer 5
π‘π‘ = 1.52 mm
Thus in each layer ties with a thickness of 1.6 mm will be sufficient. However, if galvanized steel is used, the rate of corrosion is about 0.025 mm/yr, so t should be 1.6 + (0.025)(50) = 2.85 mm.
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
Calculation of Minimum Length of Ties
The minimum length of ties in each layer should equal 2πΏπΏππ . Following is the length of ties in each layer:
Layer no. Minimum length of the tie, 2πΏπΏππ (m)
1 3.1
2 5.2
3 6.8
4 7.7
5 8.4
Figure4. 43 is a diagram of the foundation with the ties. The design could be changed by varying π΅π΅,ππ,ππ, and βπ»π» to determine the most economical combination
Figure 4.43
Example 11
Refer to example 10. For the loading given, determine the width of the foundation that is needed for unreinforced earth. Note that the factor of safety against bearing capacity failure is 3 and that the tolerable settlement is 25 mm.
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
Solution
Bearing Capacity Consideration
For a continuous foundation,
πππ’π’ = πΎπΎπΏπΏππππππ + 12πΎπΎπ΅π΅πππΎπΎ
For ππ = 35Β°,ππππ = 33.3 and πππΎπΎ = 48.03, so
πππ‘π‘ππππ = πππ’π’πΉπΉππ
= 1πΉπΉπποΏ½πΎπΎπΏπΏππππππ + 1
2πΎπΎπ΅π΅πππΎπΎοΏ½
Or
πππ‘π‘ππππ = 13οΏ½(17.3)(1)(33.3) + 1
2(17.3)(π΅π΅)(48.03)οΏ½ = 192.03 + 138.5π΅π΅ [a]
However,
πππ‘π‘ππππ = 1.8Γ103kN(π΅π΅)(1)
[b]
Equating the right-hand sides of equations (a) and (b) yields
1800(π΅π΅)(1)
= 192.03 + 138.5π΅π΅
Solving the preceding equation gives π΅π΅ β 3 m, so with π΅π΅ = 3 m, ππall = 600 kN/m2.
Settlement Consideration
For a friction angle of ππ = 35Β°, the corrected average standard penetration number is about 10-15 (equation 11 from chapter 2). From equation (53) for the higher value, ππππππππ = 15,
πππ‘π‘ππππ = 11.98ππππππππ οΏ½3.28π΅π΅+1
3.28οΏ½
2οΏ½1 + 0.33πΏπΏππ
π΅π΅οΏ½
for a settlement of about 25 mm. now, we can make a few trials:
Assumed π΅π΅(m)(1) πππ‘π‘ππππ = 11.98ππππππππ οΏ½
3.28π΅π΅+13.28
οΏ½2οΏ½1 +
0.33πΏπΏπππ΅π΅
οΏ½ (kN/m2 (2)
ππ = (π΅π΅)(πππ‘π‘ππππ ) = Col. 1Γ Col. 2(kN/m)
6 209 1254
9 199 1791
Note: πΏπΏππ = 1 m
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
Required 1800 kN/m
For ππππππππ = 15, the width of the foundation should be 9m or more. Based on the consideration of bearing capacity failure and tolerable settlement, the latter criteria will control, so B is about 9 m.
Note: At first, the results of this calculation may show the use of reinforced earth for foundation construction to be desirable. However, several factors must be considered before a final decision is made. For example, reinforced earth needs overexcavation and backfilling. Hence, under many circumstances, proper material selection and compaction may make the construction of foundations on unreinforced soils more economical.
PROBLEMS
1. A flexible circular area is subjected to a uniformly distributed load of 3000 lb/ft2. The diameter of the loaded area is 9.5 ft. determine the stress increase in a soil mass at a point located 7.5 ft below the center of the loaded area.
2. Refer to figure 5, which shows a flexible rectangular area. Given: π΅π΅1 =1.2 m,π΅π΅2 = 3 m, πΏπΏ1 = 3 m, and πΏπΏ2 = 6 m. If the area is subjected to a uniform load of 110 kN/m2, determine the stress increase at a depth of 8 m located immediately below point O.
3. Repeat problem 2 with the following π΅π΅1 = 5 ft π΅π΅2 = 10 ft πΏπΏ1 = 7 ft πΏπΏ2 = 12 ft Uniform load on the flexible area = 2500 lb/ft2
4. Refer to figure P-1. Using the procedure outlined in section 5, determine the average stress increase in the clay layer below the center of the foundation due to the net foundation load of 900 kN.
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
Figure P-1
5. Figure P-2 shows an embankment load on a silty clay soil layer. Determine the stress increase at points π΄π΄,π΅π΅, and πΆπΆ whch are located at a depth of 15 ft below the ground surface.
Figure P-2
6. A flexible load area (figure P-3) is 2m Γ 3m in plan and carries a uniformly distributed load of 210 kN/m2. Estimate the elastic settlement below the center of the loaded area. Assume πΏπΏππ = 0 and π»π» = β.
Figure P-3
NPTEL β ADVANCED FOUNDATION ENGINEERING-I
7. A continuous foundation on a deposit of sand layer is shown in figure P-4 along with the variation of the modulus of elasticity of the soil (πΈπΈπ π ). Assuming πΎπΎ =115 lb/ft3 and πΆπΆ2 = 10 yr, calculate the elastic settlement of the foundation using the strain influence factor.
Figure P-4
8. Tow plate load tests with square plates were conducted in the field. At 1-in. settlement, the results were
Width of plane (in.)
Load (lb)
12 8,070
24 25,800
What size of square footing is required to carry a net load of 150,000 lb at a settlement of 1 in.?
9. The tie forces under a continuous foundation are given by equation (87). For the foundation described in problem 23, ππππ = 200 kN/m2 and πππ π /ππππ = 4.5. Determine the tie forces, ππ(ππ), in kN/m for each layer of reinforcement.