Shallow Foundations Allowable Bearing Capacity & Settlement

NPTEL – 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 𝒒𝒒𝒐𝒐 Determination of 𝑸𝑸𝑹𝑹 Calculation of Tie Force Calculation of tie Resistance Due to Friction, 𝑭𝑭𝑩𝑩 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


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

3𝐵𝐵 (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 𝑑𝑑/𝐵𝐵 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 𝑑𝑑/𝐵𝐵less 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).


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 A’, A”, A’”,… and B;, B”, B’”, … which define the limiting lines between zones I and II, can be obtained by considering the shear stress distribution, 𝜏𝜏𝑥𝑥𝑥𝑥 , in the soil caused by the foundation load. The term 𝜏𝜏𝑥𝑥𝑥𝑥 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 Boussinesq’s equation is performed, 𝜏𝜏𝑥𝑥𝑥𝑥 is given by the relation

Figure 4.37 Failure mechanism under a foundation supported by reinforced earth [part (b) after Binquet and Lee, 1975b]

𝜏𝜏𝑥𝑥𝑥𝑥 = 4𝑏𝑏𝑞𝑞𝑅𝑅𝑥𝑥𝑥𝑥 2

𝜋𝜋[(𝑥𝑥2+𝑥𝑥2−𝑏𝑏2)2+4𝑏𝑏2𝑥𝑥2] [4.68]

Where q


𝑏𝑏 = half − width of the foundation = 𝐵𝐵/2

𝐵𝐵 = width of foundation

𝑞𝑞𝑅𝑅 = load per unit area on the foundation

The variation of 𝜏𝜏𝑥𝑥𝑥𝑥 at any depth, z, is shown by the broken lines in figure 4.37a. Points A’ and B’ refer to the points at which the value of 𝜏𝜏𝑥𝑥𝑥𝑥 is maximum at 𝑥𝑥 = 𝑥𝑥1.

Similarly, A” and B” refer to the points at which 𝜏𝜏𝑥𝑥𝑥𝑥 is maximum at 𝑥𝑥 = 𝑥𝑥2. The distances 𝑥𝑥 = 𝑋𝑋𝑜𝑜 at which the maximum value of 𝜏𝜏𝑥𝑥𝑥𝑥 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 A’, A”, A’”,… and B;, B”, B’”, …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, 𝑞𝑞𝑅𝑅 , to the load per unit area on the foundation supported by unreinforced earth, 𝑞𝑞𝑜𝑜 , 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 𝑞𝑞𝑜𝑜 per unit area. Similarly, figure 4.39b shows a continuous foundation supported by a reinforced soil layer (one layer of reinforcement, or 𝑁𝑁 = 1) and subjected to a load of 𝑞𝑞𝑅𝑅 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 𝑆𝑆𝑐𝑐 . For one-half of each foundation under consideration, the following are the forces per unit length on a soil element of thickness ∆𝐻𝐻 located at a depth z.


Figure 4.38 Relationship between load per unit area and settlement for foundations resting on reinforced and unreinforced soil


Figure 4.39 Derivation of equation (87)


Figure 4.39a continued

Unreinforced Case 𝐹𝐹1 and 𝐹𝐹2 are the vertical forces and 𝑆𝑆1 is the shear force. Hence, for equilibrium,

𝐹𝐹1 − 𝐹𝐹2 − 𝑆𝑆1 = 0 [4.69]

Reinforced Case Here, 𝐹𝐹3 and 𝐹𝐹4 are the vertical forces, 𝑆𝑆2 is the shear force, and 𝑇𝑇(𝑁𝑁=1) is the tensile force developed in the reinforcement. The force 𝑇𝑇(𝑁𝑁=1) is vertical because of the assumption made for the deformation of reinforcement as shown in figure 4.37c. So

𝐹𝐹3 − 𝐹𝐹4 − 𝑆𝑆2 − 𝑇𝑇(𝑁𝑁=1) = 0 [4.70]

If the foundation settlement, 𝑆𝑆𝑐𝑐 , is the same in both cases,

𝐹𝐹2 = 𝐹𝐹4 [4.71]

Subtracting equation (69) from equation (70) and using the relationship given in equation (71), we obtain

𝑇𝑇(𝑁𝑁=1) = 𝐹𝐹3 − 𝐹𝐹1 − 𝑆𝑆2 + 𝑆𝑆1 [4.72]

Note that the force 𝐹𝐹1 is caused by the vertical stress, 𝜎𝜎, on the soil element under consideration as a result of the load 𝑞𝑞𝑜𝑜 on the foundation. Similarly, 𝐹𝐹3 is caused by the vertical stress imposed on the soil element as a result of the load 𝑞𝑞𝑅𝑅 . Hence


𝐹𝐹1 = ∫ 𝜎𝜎(𝑞𝑞𝑜𝑜)𝑑𝑑𝑥𝑥𝑥𝑥𝑜𝑜0 [4.73]

𝐹𝐹3 = ∫ 𝜎𝜎(𝑞𝑞𝑅𝑅)𝑑𝑑𝑥𝑥𝑥𝑥𝑜𝑜0 [4.74]

𝑆𝑆1 = 𝜏𝜏𝑥𝑥𝑥𝑥 (𝑞𝑞𝑜𝑜)∆𝐻𝐻 [4.75]

𝑆𝑆2 = 𝜏𝜏𝑥𝑥𝑥𝑥 (𝑞𝑞𝑅𝑅)∆𝐻𝐻 [4.76]

Where

𝜎𝜎(𝑞𝑞𝑜𝑜) and 𝜎𝜎(𝑞𝑞𝑅𝑅) are the vertical stresses at a depth z caused by the loads 𝑞𝑞𝑜𝑜 and 𝑞𝑞𝑅𝑅 on the foundation

𝜏𝜏𝑥𝑥𝑥𝑥 (𝑞𝑞𝑜𝑜) and 𝜏𝜏𝑥𝑥𝑥𝑥 (𝑞𝑞𝑅𝑅) are the shear stresses at a depth z and at a distance 𝑋𝑋𝑜𝑜 from the center line caused by the loads 𝑞𝑞𝑜𝑜 and 𝑞𝑞𝑅𝑅

Integrating Boussinesq’s solution yields

𝜎𝜎(𝑞𝑞𝑜𝑜) = 𝑞𝑞𝑜𝑜𝜋𝜋�𝑡𝑡𝑡𝑡𝑡𝑡−1 𝑥𝑥

𝑥𝑥−𝑏𝑏− 𝑡𝑡𝑡𝑡𝑡𝑡−1 𝑥𝑥

𝑥𝑥+𝑏𝑏− 2𝑏𝑏𝑥𝑥 (𝑥𝑥2−𝑥𝑥2−𝑏𝑏2)

(𝑥𝑥2+𝑥𝑥2−𝑏𝑏2)2+4𝑏𝑏2𝑥𝑥2� [4.77]

𝜎𝜎(𝑞𝑞𝑅𝑅) = 𝑞𝑞𝑜𝑜𝜋𝜋�𝑡𝑡𝑡𝑡𝑡𝑡−1 𝑥𝑥

𝑥𝑥−𝑏𝑏− 𝑡𝑡𝑡𝑡𝑡𝑡−1 𝑥𝑥

𝑥𝑥+𝑏𝑏− 2𝑏𝑏𝑥𝑥 (𝑥𝑥2−𝑥𝑥2−𝑏𝑏2)

(𝑥𝑥2+𝑥𝑥2−𝑏𝑏2)2+4𝑏𝑏2𝑥𝑥2� [4.78]

𝜏𝜏𝑥𝑥𝑥𝑥 (𝑞𝑞𝑜𝑜) = 4𝑏𝑏𝑞𝑞𝑜𝑜𝑋𝑋𝑜𝑜𝑥𝑥2

𝜋𝜋[(𝑋𝑋𝑜𝑜2+𝑥𝑥2−𝑏𝑏2)2+4𝑏𝑏2𝑥𝑥2] [4.79]

𝜏𝜏𝑥𝑥𝑥𝑥 (𝑞𝑞𝑅𝑅) = 4𝑏𝑏𝑞𝑞𝑅𝑅𝑋𝑋𝑜𝑜𝑥𝑥2

𝜋𝜋[(𝑋𝑋𝑜𝑜2+𝑥𝑥2−𝑏𝑏2)2+4𝑏𝑏2𝑥𝑥2] [4.80]

Where

𝑏𝑏 = 𝐵𝐵/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

𝐹𝐹1 = 𝐴𝐴1𝑞𝑞𝑜𝑜𝐵𝐵 [4.81]

𝐹𝐹3 = 𝐴𝐴1𝑞𝑞𝑅𝑅𝐵𝐵 [4.82]

𝑆𝑆1 = 𝐴𝐴2𝑞𝑞𝑜𝑜∆𝐻𝐻 [4.83]

𝑆𝑆2 = 𝐴𝐴2𝑞𝑞𝑅𝑅∆𝐻𝐻 [4.84]

Where

𝐴𝐴1 and 𝐴𝐴2 = 𝑓𝑓(𝑥𝑥/𝐵𝐵)


The variations of 𝐴𝐴1 and 𝐴𝐴2 with nondimensional depth z are given in figure 4.40. Substituting equations (81)-(84) into equation (72) gives

Figure 4.40 Variation of 𝐴𝐴1,𝐴𝐴2, and 𝐴𝐴3 with 𝑥𝑥/𝐵𝐵 (after Binquet and Lee, 1975b)

𝑇𝑇(𝑁𝑁=1) = 𝐴𝐴1𝑞𝑞𝑅𝑅𝐵𝐵 − 𝐴𝐴1𝑞𝑞𝑜𝑜𝐵𝐵 − 𝐴𝐴2𝑞𝑞𝑅𝑅∆𝐻𝐻 +𝐴𝐴2𝑞𝑞𝑜𝑜∆𝐻𝐻 = 𝐴𝐴1𝐵𝐵(𝑞𝑞𝑅𝑅 − 𝑞𝑞𝑜𝑜) − 𝐴𝐴2∆𝐻𝐻(𝑞𝑞𝑅𝑅 − 𝑞𝑞𝑜𝑜) =𝑞𝑞𝑜𝑜 �

𝑞𝑞𝑅𝑅𝑞𝑞𝑜𝑜− 1� (𝐴𝐴1𝐵𝐵 − 𝐴𝐴2∆𝐻𝐻 [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 ∆𝐻𝐻, as shown in figure 4.39c, the assumption can be made that

𝑇𝑇(𝑁𝑁) = 𝑇𝑇(𝑁𝑁=1)

𝑁𝑁 [4.86]

Combining equations (85 and 86) gives

𝑇𝑇(𝑁𝑁) = 1𝑁𝑁�𝑞𝑞𝑜𝑜 �

𝑞𝑞𝑅𝑅𝑞𝑞𝑜𝑜− 1� (𝐴𝐴1𝐵𝐵 − 𝐴𝐴2∆𝐻𝐻� [4.87]

The unit of 𝑇𝑇(𝑁𝑁) 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


𝐹𝐹𝑆𝑆(𝐵𝐵) = 𝑤𝑤𝑡𝑡𝑡𝑡 𝑓𝑓𝑦𝑦𝑇𝑇(𝑁𝑁)

[4.88]

Where

𝐹𝐹𝑆𝑆(𝐵𝐵) = factor of safety against the breakin

𝑤𝑤 = width of a single tie

𝑡𝑡 = thickness of each tie

𝑡𝑡 = number of ties per unit length of the foundation

𝑓𝑓𝑦𝑦 = yield or breaking strength of the tie material

The term 𝑤𝑤𝑡𝑡 may be defined as the linear density ratio, LDR, so

𝐹𝐹𝑆𝑆(𝐵𝐵) = � 𝑡𝑡𝑓𝑓𝑦𝑦𝑇𝑇(𝑁𝑁)

� (𝐿𝐿𝐿𝐿𝑅𝑅) [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)

𝐹𝐹𝐵𝐵 = 2 tan∅𝑢𝑢[normal force] [4.90]

=2 tan∅𝑢𝑢

↑two sides of tie (i. e. , top and bottom)

⎣⎢⎢⎢⎡(𝐿𝐿𝐿𝐿𝑅𝑅) � 𝜎𝜎(𝑞𝑞𝑅𝑅)𝑑𝑑𝑥𝑥 + (𝐿𝐿𝐿𝐿𝑅𝑅)(𝛾𝛾)(𝐿𝐿𝑜𝑜 − 𝑋𝑋𝑜𝑜)(𝑥𝑥 + 𝐿𝐿𝑓𝑓)

𝐿𝐿𝑜𝑜

𝑥𝑥𝑜𝑜 ⎦

⎥⎥⎥⎤


Where

𝛾𝛾 = unit weight of soil

𝐿𝐿𝑓𝑓 = depth of foundation

𝜙𝜙𝑢𝑢 = tie − soil friction angle

The relation for 𝜎𝜎(𝑞𝑞𝑅𝑅) was defined in equation (78). The value of 𝑥𝑥 = 𝐿𝐿𝑜𝑜 is generally assumed to be the distance at which 𝜎𝜎(𝑞𝑞𝑅𝑅) equals to 0.1𝑞𝑞𝑅𝑅 . The value of 𝐿𝐿𝑜𝑜 as a function of depth z is given in figure 4.42. Equation (90) may be simplified as

𝐹𝐹𝐵𝐵 = 2 tan𝜙𝜙𝑦𝑦 (𝐿𝐿𝐿𝐿𝑅𝑅) �𝐴𝐴3𝐵𝐵𝑞𝑞𝑜𝑜 �𝑞𝑞𝑅𝑅𝑞𝑞𝑜𝑜� + 𝛾𝛾(𝐿𝐿𝑜𝑜 − 𝑋𝑋𝑜𝑜)(𝑥𝑥 + 𝐿𝐿𝑓𝑓)� [4.91]


Figure 4.42 Variation of 𝐿𝐿𝑜𝑜/𝐵𝐵 with 𝑥𝑥/𝐵𝐵 (after Binquet and Lee, 1975b)

Where

𝐴𝐴3 is a nondimensionla quantity that may be expressed as a function of depth (𝑥𝑥/𝐵𝐵) (see figure 4.40)

The factor of safety against tie pullout, 𝐹𝐹𝑆𝑆(𝑃𝑃), is

𝐹𝐹𝑆𝑆(𝑃𝑃) = 𝐹𝐹𝐵𝐵𝑇𝑇(𝑁𝑁)

[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, 𝜙𝜙 b. Soil-tie friction angle, 𝜙𝜙𝜇𝜇 c. Factor of safety against bearing capacity failure d. Factor of safety against tie breaking, 𝐹𝐹𝑆𝑆(𝐵𝐵) e. Factor of safety against tie pullout, 𝐹𝐹𝑆𝑆(𝑃𝑃)


f. Breaking strength of reinforcement ties, 𝑓𝑓𝑦𝑦 g. Unit weight of soil, 𝛾𝛾 h. Modulus of elasticity of soil, 𝐸𝐸𝑠𝑠 i. Poisson’s ratio of soil, 𝜇𝜇𝑠𝑠 j. Allowable settlement of foundation, 𝑆𝑆𝑐𝑐 k. Depth of foundation, 𝐿𝐿𝑓𝑓

2. Assume a width of foundation, B, and also d and N. the value of d should be less than 2

3𝐵𝐵. Also, the distance from the bottom of the foundation to the lowest layer of the reinforcement should be about 2B or less. Calculate Δ𝐻𝐻.

3. Assume a value of 𝐿𝐿𝐿𝐿𝑅𝑅 4. For width Bi (step 2) determine the ultimate bearing capacity, 𝑞𝑞𝑢𝑢 , for unreinforced

soil [equation (3 from chapter 3); note: 𝑐𝑐 = 0]. Determine 𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 (1): 𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 (1) = 𝑞𝑞𝑢𝑢

𝐹𝐹𝑆𝑆 against bearing capacity failure [4.93]

5. Calculate the allowable load, 𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 (2) based on the tolerable settlement, 𝑆𝑆𝑐𝑐 ,

assuming that the soil is not reinforced [equation (32a)]: 𝑆𝑆𝑐𝑐 = 𝐵𝐵𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 (2)

𝐸𝐸𝑠𝑠(1 − 𝜇𝜇𝑠𝑠2)𝛼𝛼𝑟𝑟

For 𝐿𝐿/𝐵𝐵 = ∞, the value of 𝛼𝛼𝑟𝑟 may be taken as 2, or 𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 (2) = 𝐸𝐸𝑠𝑠𝑆𝑆𝑐𝑐

𝐵𝐵(1−𝜇𝜇𝑠𝑠2)𝛼𝛼𝑟𝑟 [4.94]

(The allowable load for a given settlement, 𝑆𝑆𝑐𝑐 , could have also been determined from equations that relate to standard penetration resistances).

6. Determine the lower of the two value of 𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 obtained from steps 4 and 5. The lower value of 𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 equals 𝑞𝑞𝑜𝑜 .

7. Calculate the magnitude of 𝑞𝑞𝑅𝑅 for the foundation supported by reinforced earth: 𝑞𝑞𝑅𝑅 = load on foundation per unit length

𝐵𝐵 [4.95]

8. Calculate the tie force, 𝑇𝑇(𝑁𝑁), in each layer of reinforcement by using equation (87)

(note: unit of 𝑇𝑇(𝑁𝑁) as kN/m of foundation). 9. Calculate the frictional resistance of ties for each layer per unit length of

foundation, 𝐹𝐹𝐵𝐵, by using equation (91). For each layer, determine whether 𝐹𝐹𝐵𝐵/𝑇𝑇(𝑁𝑁) ≥ 𝐹𝐹𝑆𝑆(𝑃𝑃). If 𝐹𝐹𝐵𝐵/𝑇𝑇(𝑁𝑁) < 𝐹𝐹𝑆𝑆(𝑃𝑃), the length of the reinforcing strips for a layer may be increased. That will increase the value of 𝐹𝐹𝐵𝐵 and thus 𝐹𝐹𝑆𝑆(𝑃𝑃), and so equation (91) must be rewritten as 𝐹𝐹𝐵𝐵 = 2 tan𝜙𝜙𝜇𝜇 (𝐿𝐿𝐿𝐿𝑅𝑅) �𝐴𝐴3𝐵𝐵𝑞𝑞𝑜𝑜 �

𝑞𝑞𝑅𝑅𝑞𝑞𝑜𝑜� + 𝛾𝛾(𝐿𝐿𝑜𝑜 − 𝑋𝑋𝑜𝑜)(𝑥𝑥 + 𝐿𝐿𝑓𝑓)� [4.96]


Where 𝐿𝐿 = 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:

𝑆𝑆𝑜𝑜𝑆𝑆𝑎𝑎: 𝛾𝛾 = 17.3 kN/m3; 𝜙𝜙 = 35°; 𝐸𝐸𝑠𝑠 = 3 × 104kN/m2; 𝜇𝜇𝑠𝑠 = 0.35

𝑅𝑅𝑅𝑅𝑆𝑆𝑡𝑡𝑓𝑓𝑜𝑜𝑟𝑟𝑐𝑐𝑅𝑅𝑅𝑅𝑅𝑅𝑡𝑡𝑡𝑡 𝑡𝑡𝑆𝑆𝑅𝑅𝑠𝑠: 𝑓𝑓𝑦𝑦2.5 × 105 kN/m2; ; 𝜙𝜙𝜇𝜇 = 28°; 𝐹𝐹𝑠𝑠(𝑅𝑅) = 3; 𝐹𝐹𝑆𝑆(𝑃𝑃) = 2.5

𝐹𝐹𝑜𝑜𝑢𝑢𝑡𝑡𝑑𝑑𝑡𝑡𝑡𝑡𝑆𝑆𝑜𝑜𝑡𝑡: 𝐿𝐿f = 1m; Factor of safety against baring capacity failure = 3.

Tolerable settlement= 𝑆𝑆𝑐𝑐 = 25 mm; desired life of structure = 50 years

Solution

Let

𝐵𝐵 = 1 m

𝑑𝑑 = depth from the bottom of the foundation to the first reinforcing layer = 0.5

Δ𝐻𝐻 = 0.5 m

𝑁𝑁 = 5

𝐿𝐿𝐿𝐿𝑅𝑅 = 65%

If the reinforcing strips used are 75 mm wide, then

𝑤𝑤𝑡𝑡 = 𝐿𝐿𝐿𝐿𝑅𝑅

Or

𝑡𝑡 = 𝐿𝐿𝐿𝐿𝑅𝑅𝑤𝑤

= 0.650.075 m

= 8.67/m


Hence each layer will contain 8.67 strips per meter length of the foundation.

Determination of 𝒒𝒒𝒐𝒐

For an unreinforced foundation

𝑞𝑞𝑢𝑢 = 𝛾𝛾𝐿𝐿𝑓𝑓𝑁𝑁𝑞𝑞 + 12𝛾𝛾𝐵𝐵𝑁𝑁𝛾𝛾

From table 4 (from chapter 3) for 𝜙𝜙 = 35°,𝑁𝑁𝑞𝑞 = 33.30 and 𝑁𝑁𝛾𝛾48.03. Thus

𝑞𝑞𝑢𝑢 = (17.3)(1)(33.3) + 12

(17.3)(1)(48.03)

= 576.09 + 415.46 = 991.55 ≈ 992 kN/m2

𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 (1) = 𝑞𝑞𝑢𝑢𝐹𝐹𝑆𝑆

= 9923

= 330.7 kN/m2

From equation (94)

𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 (2) = (𝐸𝐸𝑠𝑠)(𝑆𝑆𝑐𝑐)𝐵𝐵(1−𝜇𝜇𝑠𝑠2)𝛼𝛼𝑟𝑟

= (30,000 kN /m2)(0.025m)(1m)(1−0.352)(2)

= 427.35 kN/m2

As 𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 (1) < 𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 (2), 𝑞𝑞𝑜𝑜 = 𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 (1) = 330.7 kN/m2

Determination of 𝑸𝑸𝑹𝑹

From equation (95),

𝑄𝑄𝑅𝑅 = 1.8 MN /m𝐵𝐵

= 1.8×103

1= 1.8 × 103kN/m2

Calculation of Tie Force

From equation (87),

𝑇𝑇(𝑁𝑁) = �𝑞𝑞𝑜𝑜𝑁𝑁� �𝑄𝑄𝑅𝑅

𝑞𝑞𝑜𝑜− 1� (𝐴𝐴1𝐵𝐵 − 𝐴𝐴2∆𝐻𝐻)

The tie forces for each layer are given in the following table.

Layer no �

𝑞𝑞𝑜𝑜𝑁𝑁� �𝑄𝑄𝑅𝑅𝑞𝑞𝑜𝑜

− 1�

𝑥𝑥(m) 𝑥𝑥𝐵𝐵

𝐴𝐴1𝐵𝐵 𝐴𝐴2∆𝐻𝐻 𝐴𝐴1𝐵𝐵− 𝐴𝐴2∆𝐻𝐻

𝑇𝑇(𝑁𝑁)(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


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: 𝐴𝐴1 is from figure 4.40; 𝐵𝐵 = 1 m; ∆𝐻𝐻 = 0.5 m; 𝐴𝐴2 is from figure 4.40; 𝑞𝑞𝑢𝑢/𝑞𝑞𝑜𝑜 = 1.8 ×104/330.7 ≈ 5.44

Calculation of tie Resistance Due to Friction, 𝑭𝑭𝑩𝑩

Use equation (91):

𝐹𝐹𝐵𝐵 = 2 tan𝜙𝜙𝜇𝜇 (𝐿𝐿𝐿𝐿𝑅𝑅) �𝐴𝐴3𝐵𝐵𝑞𝑞𝑜𝑜 �𝑞𝑞𝑅𝑅𝑞𝑞𝑜𝑜� + 𝛾𝛾(𝐿𝐿𝑜𝑜 − 𝑋𝑋𝑜𝑜)(𝑥𝑥 + 𝐿𝐿𝑓𝑓)�

The following table shows the magnitude of 𝐹𝐹𝐵𝐵 for each layer:

Layer number

Quantity 1 2 3 4 5

2 tan𝜙𝜙𝜇𝜇 (𝐿𝐿𝐿𝐿𝑅𝑅) 0.691 0.691 0.691 0.691 0.691

𝐴𝐴3 0.125 0.14 0.15 0.15 0.15

𝐴𝐴3𝐵𝐵𝑞𝑞𝑜𝑜(𝑞𝑞𝑅𝑅/𝑞𝑞𝑜𝑜) 225.0 252.0 270.0 270.0 270.0

𝑥𝑥(𝑅𝑅 ) 0.5 1.0 1.5 2.0 2.5

𝑥𝑥/𝐵𝐵 0.5 1.0 1.5 2.0 2.5

𝐿𝐿𝑜𝑜(m) 1.55 2.6 3.4 3.85 4.2

𝑋𝑋𝑜𝑜(m) 0.55 0.8 1.1 1.4 1.65

𝐿𝐿𝑜𝑜 − 𝑋𝑋𝑜𝑜(m) 1.0 1.8 2.3 2.45 2.55

𝑥𝑥 + 𝐿𝐿𝑓𝑓(m) 1.5 2.0 2.5 3.0 3.5

𝛾𝛾(𝐿𝐿𝑜𝑜 − 𝑋𝑋𝑜𝑜)(𝑥𝑥 + 𝐿𝐿𝑓𝑓 (m) 25.95 62.28 99.48 127.16 154.4

𝐹𝐹𝐵𝐵(kN/m) 173.4 217.2 255.1 274.4 292.3

𝐹𝐹𝑆𝑆(𝑃𝑃) = 𝐹𝐹𝐵𝐵/𝑇𝑇(𝑁𝑁) 2.62 2.96 3.16 3.34 3.57

Note: 𝐴𝐴3 is from figure 4.40; 𝑋𝑋𝑜𝑜 is from figure 4.37; 𝐿𝐿𝑜𝑜 is from figure 4.42; 𝑇𝑇(𝑁𝑁) is from


the preceding table.

The minimum factor of safety is greater than the required values of 𝐹𝐹𝑠𝑠(𝑃𝑃) which is 2.5.

Calculation of Tie Thickness to Resist Tie Breaking

From equation (89),

𝐹𝐹𝑠𝑠(𝐵𝐵) = 𝑡𝑡𝑓𝑓𝑦𝑦𝑇𝑇(𝑁𝑁)

(𝐿𝐿𝐿𝐿𝑅𝑅)

Here, 𝑓𝑓𝑦𝑦 = 2.5 × 105 kN/m2, 𝐿𝐿𝐿𝐿𝑅𝑅 = 0.65 and 𝐹𝐹𝑠𝑠(𝐵𝐵) = 3, so

𝑡𝑡 = � 3(2.5×105)(0.65)

� 𝑇𝑇(𝑁𝑁) = (1.846 × 105)𝑇𝑇(𝑁𝑁)

So, for layer 1

𝑡𝑡 = (1.846 × 10−5)(66.08) = 0.00122m = 1.22 mm

For layer 2

𝑡𝑡 = (1.846 × 10−5)(73.43) = 0.00136m = 1.36 mm

Similarly, for layer 3

𝑡𝑡 = 0.00149 = 1.49 mm

For layer 4

𝑡𝑡 = 1.52 mm

For layer 5

𝑡𝑡 = 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.


Calculation of Minimum Length of Ties

The minimum length of ties in each layer should equal 2𝐿𝐿𝑜𝑜 . Following is the length of ties in each layer:

Layer no. Minimum length of the tie, 2𝐿𝐿𝑜𝑜 (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 𝐵𝐵,𝑑𝑑,𝑁𝑁, and ∆𝐻𝐻 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.


Solution

Bearing Capacity Consideration

For a continuous foundation,

𝑞𝑞𝑢𝑢 = 𝛾𝛾𝐿𝐿𝑓𝑓𝑁𝑁𝑞𝑞 + 12𝛾𝛾𝐵𝐵𝑁𝑁𝛾𝛾

For 𝜙𝜙 = 35°,𝑁𝑁𝑞𝑞 = 33.3 and 𝑁𝑁𝛾𝛾 = 48.03, so

𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 = 𝑞𝑞𝑢𝑢𝐹𝐹𝑆𝑆

= 1𝐹𝐹𝑆𝑆�𝛾𝛾𝐿𝐿𝑓𝑓𝑁𝑁𝑞𝑞 + 1

2𝛾𝛾𝐵𝐵𝑁𝑁𝛾𝛾�

Or

𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 = 13�(17.3)(1)(33.3) + 1

2(17.3)(𝐵𝐵)(48.03)� = 192.03 + 138.5𝐵𝐵 [a]

However,

𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 = 1.8×103kN(𝐵𝐵)(1)

[b]

Equating the right-hand sides of equations (a) and (b) yields

1800(𝐵𝐵)(1)

= 192.03 + 138.5𝐵𝐵

Solving the preceding equation gives 𝐵𝐵 ≈ 3 m, so with 𝐵𝐵 = 3 m, 𝑞𝑞all = 600 kN/m2.

Settlement Consideration

For a friction angle of 𝜙𝜙 = 35°, the corrected average standard penetration number is about 10-15 (equation 11 from chapter 2). From equation (53) for the higher value, 𝑁𝑁𝑐𝑐𝑜𝑜𝑟𝑟 = 15,

𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 = 11.98𝑁𝑁𝑐𝑐𝑜𝑜𝑟𝑟 �3.28𝐵𝐵+1

3.28�

2�1 + 0.33𝐿𝐿𝑓𝑓

𝐵𝐵�

for a settlement of about 25 mm. now, we can make a few trials:

Assumed 𝐵𝐵(m)(1) 𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 = 11.98𝑁𝑁𝑐𝑐𝑜𝑜𝑟𝑟 �

3.28𝐵𝐵+13.28

�2�1 +

0.33𝐿𝐿𝑓𝑓𝐵𝐵

� (kN/m2 (2)

𝑄𝑄 = (𝐵𝐵)(𝑞𝑞𝑡𝑡𝑎𝑎𝑎𝑎 ) = Col. 1× Col. 2(kN/m)

6 209 1254

9 199 1791

Note: 𝐿𝐿𝑓𝑓 = 1 m


Required 1800 kN/m

For 𝑁𝑁𝑐𝑐𝑜𝑜𝑟𝑟 = 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: 𝐵𝐵1 =1.2 m,𝐵𝐵2 = 3 m, 𝐿𝐿1 = 3 m, and 𝐿𝐿2 = 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 𝐵𝐵1 = 5 ft 𝐵𝐵2 = 10 ft 𝐿𝐿1 = 7 ft 𝐿𝐿2 = 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.


Figure P-1

5. Figure P-2 shows an embankment load on a silty clay soil layer. Determine the stress increase at points 𝐴𝐴,𝐵𝐵, and 𝐶𝐶 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 × 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 𝐿𝐿𝑓𝑓 = 0 and 𝐻𝐻 = ∞.

Figure P-3


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 (𝐸𝐸𝑠𝑠). Assuming 𝛾𝛾 =115 lb/ft3 and 𝐶𝐶2 = 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, 𝑞𝑞𝑜𝑜 = 200 kN/m2 and 𝑞𝑞𝑅𝑅/𝑞𝑞𝑜𝑜 = 4.5. Determine the tie forces, 𝑇𝑇(𝑁𝑁), in kN/m for each layer of reinforcement.

Documents

Shallow Foundations Allowable Bearing Capacity & Settlement