Lecture 19

NPTEL - ADVANCED FOUNDATION ENGINEERING-1

Module 5

(Lecture 19)

MAT FOUNDATIONS

Topics

1.1 STRUCTURAL DESIGN OF MAT FOUNDATIONS Conventional Rigid Method

1.2 Approximate Flexible Method

Foundations on Sandy Soils Foundations on Clays

1.3 Example

1.4 PROBLEMS

STRUCTURAL DESIGN OF MAT FOUNDATIONS

The structural design of mat foundations can be carried out by two conventional methods: the conventional rigid method and the approximate flexible method. Finite difference and finite element methods can also be used, but this section covers only the basic concepts of the first two design methods.

Conventional Rigid Method

The conventional rigid method of mat foundation design can be explained step by step with reference to figure 5.8.

1. Figure 5.8a shows mat dimensions of and columns loads of 1,2,3, Calculate the total column load as = 1 + 2 + 3 [5.24]


2. Determine the pressure on the soil, q, below the mat at points ,,,, , by using the equation =

[5.25]

Where

= = (1/12)3 = moment of inertia about the axis = (1/12)3 = moment of inertia about the axis = moment of the column loads about the axis = = moment of the column loads about the axis = The load eccentricities, and , in the and directions can be determined by using ( , ) coordinates: = 11+22+33+

[5.26]

And

= 2 [5.27] Similarly

= 11+22+33+

[5.28]

And

= 2 [5.29] 3. Compare the values of the soil pressures determined in step 2 with the net

allowable soil pressure to determine whether all (net ). 4. Divide the mat into several strips in x and y directions (see figure 5.8a). Let the

width of any strip be 1. 5. Draw the shear, V, and the moment, M, diagrams for each individual strip (in the x

and y directions). For example, the average soil pressure of the bottom strip in the x direction of figure 5.8a is

+2 [5.30]


Where and = soil pressures at poins and as determined from step 2.

The total soil reaction is equal to 1. Now obtain the total column load on the strip as 1 + 2 + 3 + 4. The sum of the column loads on the strip will not equal 1 because the shear between the adjacent strips has not been taken into account. For this reason, the soil reaction and the column loads need to be adjusted, or Average load = 1+(1+2+3+4)2 [5.31]

Now, the modified average soil reaction becomes

(modified ) = average load1 [5.32]

And the column load modification factor is

= average load1+2+3+4 [5.33]

So, the modified column loads are 1,2,3, and 4. This modified loading on the strip under consideration is shown in figure 5.8b. The shear and the moment diagram for this strip can now be drawn. This procedure is repeated for all strips in the x and y directions.

6. Determine the effective depth of the mat d by checking for diagonal tension shear near various columns. According to ACI Code 318-95. American Concrete Institute, 1995), for the critical section, = [(0.34)] [5.34] Where = factored column load (MN), or (column load) (load factor) = reduction factor = 0.85 = compressive strength of concrete at 28 days (MN/m2)

The units of and in equation (34) are in meters. In English units, equation (34) may be expressed as

= (4 ) [5.35]


Where

is in lb, and are in in. , and is in lb/in2 The expression for in terms of , which depends on the location of the column with respect to the plan of the mat, can be obtained from figure 5.8c.

7. From the moment diagrams of all strips in one direction (x or y), obtain the maximum positive and negative moments per unit width (that is, = /1).

8. Determine the areas of steep per unit width for positive and negative reinforcement in the x and y directions. = ()(load factor) = 2 [5.36] And = 0.85 [5.37] Where = area of steel per unit width = yield stress of reinforcement in tension = factored moment = 0.9 = reduction factor

Examples 5 and 6 illustrate the use of the conventional rigid method of mat foundation design.

Approximate Flexible Method

In the conventional rigid method of design, the mat is assumed to be infinitely rigid. Also, the soil pressure is distributed in a straight line, and the centroid of the soil pressure is coincidental with the line of action of the resultant column loads (see figure 5.9). In the approximate flexible method of design, the soil is assumed to be equivalent to infinite number of elastic springs, as shown in figure 5.9b. It is sometimes referred to as the Winkler foundation. The elastic constant of these assumed springs is referred to as the coefficient of subgrade reaction k.


Figure 5.9 (a) Principles of design conventional rigid method; (b) principles of approximate flexible method; (c) derivation of equation (42) for beams on elastic foundation

To understand the fundamental concepts behind flexible foundation design, consider a beam of width 1 having infinite length, as shown in figure 5.9c. The beam is subjected to a single concentrated load Q. from the fundamental of mechanics of materials,

= 2 2 [5.38]

Where

= moment at any section = modulus of elasticity of foundation material = moment of inertia of the cross section of the beam = 11213 (see figure 5.9c) However

= shear force =


And

= = soil reaction

Hence

2 2 = [5.39]

Combining equations (38 and 39) yields

4 4 = [5.40]

However, the soil reaction is

= Where

= deflection = 1 = coefficient of subgrade reaction (kN/m3 or lb/in3) So

4 4 = 1 [5.41]

Solution of equation (41) yields

= ( cos + " sin ) [5.42]

Where and " are constants and = 144 [5.43] The unit of the term as defined by the preceding equation is (length)1. This parameter is very important in determining whether a mat foundation should be designed by conventional rigid method or approximate flexible method. According to the American Concrete Institute Committee 336 (1988), mats should be designed by the conventional rigid method if the spacing of columns in a strip is less than 1.75/. If the spacing of columns is larger than 1.75/, the approximate flexible method may be used.


To perform the analysis for the structural design of a flexible mat, you must know the principles of evaluating the coefficient of subgrade reaction, k. before proceeding with the discussion of the approximate flexible design method, let as discuss this coefficient in more detail.

If a foundation of width B (figure 5.10) is subjected to a load per unit area of q, it will undergo a settlement, . The coefficient of subgrade modulus, k, can be defined as

= [5.44]

Figure 5.10 Definition of coefficient of subgrade reaction,

The unit of is kN/m3 (or lb/in3). The value of the coefficient of subgrade reaction is not a constant for a given soil. T depends on several factors, such as the length, , and width, , of the foundation and also the depth of embedment of the foundation. Terzaghi (1955) made a comprehensive study of the parameters affecting the coefficient of subgrade reaction. It indicated that the value of the coefficient of subgrade reaction decreases with the width of the foundation. In the field, load tests can be carried out by means of square plate measuring 1 ft 1 ft (0.3 m 0.3 m) and values of k can be calculated. The value of k can be related to large foundations measuring in the following ways.

Foundations on Sandy Soils

= 0.3 +0.32 2 [5.45] Where

0.3 and = coefficients of subgrade reaction of foundation measuring 0.3m 0.3m and (m) (m), respectively (units is kN/m3 In English units, equation (45) may be expressed as

= 1 +12 2 [5.46]


Where

1 and = coefficient of subgrade reaction of foundation measuring 1 ft a ft and (ft) , respectively (units is lb/in3) Foundations on Clays

(kN/m3) = 0.3(kN/m3) 0.3 (m) (m) [5.47] The definition of k in equation (47) is the same as in equation (45).

In English units,

(lb/in3) = 1(lb/in3) 1 (ft) (ft) [5.48]

The definitions of and 1 are the same as in equation (46). For rectangular foundations having dimensions of (for similar soil and q), = ()1+0.51.5 [5.49] Where

= coefficient of subgrade modulus of the rectangular foundation ( ) () =coefficient of subgrade modulus of a square foundation having dimension of Equation (49) indicates that the value of k of a very long foundation with a width B is approximately 0.67(). The modulus of elasticity of granular soils increases with depth. Because the settlement of a foundation depends on the modulus of elasticity, the value of k increases as the depth of the foundation increases.

Following are some typical ranges of value for the coefficient of subgrade reaction 1 for sandy and clayey soils.

Sand (dry or moist)

Loose: 29 92lb/in3(8 25MN/m3) Medium: 91 460lb/in3(25 125MN/m3) Dense: 460 1380lb/in3(125 375MN/m3)


Sand (saturated)

Loose: 38 55lb/in3(10 15MN/m3) Medium: 128 147lb/in3(35 40MN/m3) Dense: 478 552lb/in3(130 150MN/m3) Clay Stiff: 44 92lb/in3(12 25MN/m3 Very stiff: 92 184lb/in3(25 50MN/m3 Hard: > 184lb/in3(> 50/m3 Scott (1981) proposed that for sandy soils the value of 0.3 can be obtained from standard penetration resistance at any given depth, or

0.3(MN/m3) = 18cor [5.50]

Where

= standard penetration resistance In English units,

1(U. S. ton/ft3) = 6Ncor [5.51]

For long beams, Vesic (1961) proposed an equation for estimating subgrade reaction:

= 0.65 4

12 12 [5.52] Where

= modulus of elasticity of soil = foundation width = modulus of elasticity of foundation material = moment of inertia of the cross section of the foundation = Poissonsratio of soil For most practical purposes, equation (52) can be approximated as


= (12) [5.53]

The coefficient of subgrade reaction is also very useful parameter in the design of rigid highway and airfield pavements. The pavements with a concrete wearing surface are generally referred to as a rigid pavement, and the pavement with an asphaltic wearing surface is called a flexible pavement. For surface load acting on a rigid pavement, the maximum tensile stress occurs at the base of the slab. For estimating the magnitude of the maximum horizontal tensile stress developed at the base of the rigid pavement, elastic solutions involving slabs on Winkler foundations are extremely useful. Some of the early work in this area was done by Westergaard (1926, 1939, and 1947).

Now that we have discussed the coefficient of subgrade reaction, we will proceed with the discussion of the approximate flexible method of designing mat foundations. This method, as proposed by the American Concrete Institute Committee 336 (1988), is described step by step. The design procedure is based primarily on the theory of plates. Its use allows the effects (that is, moment, shear, and deflection) of a concentrated column load in the area surrounding it to be evaluated. If the zones of influence of two or more columns overlap, superposition can be used to obtain the net moment, shear, and deflection at any point.

1. Assume a thickness, h, for the mat, according to step 6 as outlined for the conventional rigid method. (Note: h is the total thickness of the mat).

2. Determine the flexural ridigity R of the mat: = 312(12) [5.54] Where = modulus of elasticity of foundation material = Poissons ratio of foundation material

3. Determine the radius of effective stiffness:

=

4 [5.55] Where = coefficient of subgrade reaction The zone of influence of any column load will be on the order of 3 to 4 L.

4. Determine the moment (in polar coordinates at a point) caused by a column load (figure 5.11a):


= radial moment = 4 1 (1)2

[5.56]

= tangential moment = 4 1 + (1)2

[5.57]

Figure 5.11 Approximate flexible method of mat design

Where = radial distance from the column load = column load 1,2 = functions of / The variations of 1 and 2 with / are shown in figure 5.11b (for details, see Hetenyi, 1946). In the Cartesian coordinates system (figure 5.11a), = sin2 + cos2 [5.58] = cos2 + sin2 [5.59]

5. For the unit width of the mat, determine the shear force, V, caused by a column load:


= 4 3 [5.60] The variation of 3 with / is shown in figure 5.11b.

6. If the edge of the mat is located in the zone of influence of a column, determine the moment and shear along the wedge (assume that the mat is continuous). Moment and shear opposite in sign to those determined are applied at the edges to satisfy the known conditions.

7. Deflection () at any point is given by = 24 4 [5.61] The variation of 4 is given in figure 5.11.

Example 5

The plan of a mat foundation with column loads is shown in figure 5.12. Use equation (25) to calculate the soil pressures at points ,,,,,,,, , ,, ,, and . The size of the mat is 76 ft 96 ft, all columns are 24 in. 24 in. in section, and all (net ) =1.5 kip/ft2. Verify that the soil pressures are less than the net allowable bearing capacity.


Figure 5.12 Plan of a mat foundation

Solution

From figure 5.12, Column dead load () = 100 + 180 + 190 + 110 + 180 + 360 + 400 + 200 +190 + 400 + 440 + 200 + 120 + 180 + 180 + 120 = 3550 kip Column live load () = 60 + 120 + 120 + 70 + 120 + 200 + 250 + 120 + 130 +240 + 300 + 120 + 70 + 120 + 120 + 70 + 2230 kip So Service load = 3550 + 2230 = 5780 kip


According to ACI 318-95, factored load, = (1.4)(Dead load) + (1.7)(Live load). So Factored load = (1.4)(3550) + (1.7)(2230) = 8761 kip The moments of inertia of the foundation are

= 112(76)(96)3 = 5603 103 ft4 = 112(96)(76)3 = 3512 103 ft4 And

= 0 So 5780 = (24)(300 + 560 + 640 + 300) + (48)(310 + 650 + 740 + 300) +(72)(180 + 320 + 320 + 190) = 36.664 ft And

= 36.664 36.0 = 0.664 ft Similarly,

= 0 So 5780 = (30)(320 + 640 + 740 + 320) + (60)(300 + 560 + 650 + 320) +(90)(160 + 300 + 310 + 180) = 44.273 ft And

= 44.273 902 = 0.727 ft The moments caused by eccentricity are

= = (8761)(0.727) = 6369 kip ft = = (8761)(0.664) = 5817 kip ft


From equation (25)

=

= 8761(76)(96) (5817)()3512103 (6369)()5603103 Or

= 1.20 0.0017 0.0011 (kip/ft2) Now the following table can be prepared.

Point

(kip/ft2) (ft) 0.0017 (ft) (ft) 0.0011 (ft) (kip/ft2) A 1.2 -38 -0.065 48 -0.053 1.082

B 1.2 -24 -0.041 48 -0.053 1.106

C 1.2 -12 -0.020 48 -0.053 1.127

D 1.2 0 0.0 48 -0.053 1.147

E 1.2 12 0.020 48 -0.053 1.167

F 1.2 24 0.041 48 -0.053 1.188

G 1.2 38 0.065 48 -0.053 1.212

H 1.2 38 0.065 -48 0.053 1.318

I 1.2 24 0.041 -48 0.053 1.294

J 1.2 12 0.020 -48 0.053 1.273

K 1.2 0 0.0 -48 0.053 1.253

L 1.2 -12 -0.020 -48 0.053 1.233

M 1.2 -24 -0.041 -48 0.053 1.212

N 1.2 -38 -0.065 -48 0.053 1.188

The soil pressures at all points are less than the given value of all (net ) = 1.5 kip/ft2.


Example 6

Use the results of example 5 and the conventional rigid method.

a. Determine the thickness of the slab. b. Divide the mat into four strips (that is, ,,, and )

and determine the average soil reaction at the ends of each strips. c. Determine the reinforcement requirements in the y direction for =3000 lb/in2 and = 60,000 lb/in2.

Solution

Part a: Determination of Mat Thickness

For the critical perimeter column as shown in figure 5.13 *(ACI 318-95),

Figure 5.13 Critical perimeter column

= 1.4() + 1.7() = (1.4)(190) + (1.7)(130) = 487 kip = 2(36 + /2) + (24 + ) = 96 + 2(in). From ACI 318-95

Where

= nominal shear strength of concrete


= factored shear strength = (4) = (0.85)(4)(3000)(96 + 2) So

(0.85)(4)(3000)(96+2)1000 487 (96 + 2) 2615.1 19.4 in. For the critical internal column shown in figure 5.14,

Figure 5.14 Critical internal column

= 4(24 + ) = 96 + 4(in. ) = (1.4)(440) + (1.7)(300) = 1126 kip And

(0.85)(4)(3000)(96+4)1000 1126 (96 + 4) 6046.4 28.7 in.


Use = 29 in. With a minimum cover of 3 in. over the steel reinforcement and 1-in. diameter steel bars, the total slab thickness is

= 29 + 3 + 1 = 33 in. Part b: Average Soil Reaction

Refer to figure 5.12. For strip ABMN (width = 14 ft) 1 = (at )+(at )2 = 1.082+1.1062 = 1.094 kip/ft2 2 = (at )+(at )2 = 1.212+1.1882 = 1.20 kip/ft2 For strip BCDKLM (width = 24 ft) 1 = 1.106+1.127+1.1473 = 1.127 kip/ft2 2 = 1.253+1.233+1.2123 = 1.233 kip/ft2 For strip DEFIJK (width = 24 ft) 1 = 1.147+1.167+1.1883 = 1.167 kip/ft2 2 = 1.294+1.273+1.2533 = 1.273 kip/ft2 For strip FGHI (width = 14 ft) 1 = 1.188+1.2122 = 1.20 kip/ft2 2 = 1.318+1.294 2 = 1.306 kip/ft2 Check for = 0: Soil reaction for strip = 12(1.094 + 1.20)(14)(96) = 1541.6 kip Soil reaction for strip = 12(1.127 + 1.233)(24)(96) = 2718.7 kip Soil reaction for strip = 12(1.167 + 1.273)(24)(96) = 2810.9 kip Soil reaction for strip = 12(1.20 + 1.306)(14)(96) = 1684.0 kip 8755.2 kip Column load = 8761 kip OK


Part c: Reinforcement Requirements

Refer to figure 5.15 for the design of strip BCDKLM. Figure 5.15 shows the load diagram, in which

1 = (1.4)(180) + (1.7)(120) = 456 kip 2 = (1.4)(360) + (1.7)(200) = 844 kip 3 = (1.4)(400) + (1.7)(240) = 968 kip 4 = (1.4)(180) + (1.7)(120) = 456 kip The shear and moment diagrams are shown in figure 5.15b and c, respectively. From figure 5.15c, the maximum positive moment at the bottom of the foundation =2281.1/24 = 95.05 kip ft/ft.


Figure 5.15

Figure 5.16 Rectangular section in bending; (a) section, (b) assumed stress distribution across the section

For the design concepts of a rectangular section in bending refer to figure 5.16.

Compressive force, = 0.85 Tensile force, = = Note that for this case = 1 ft = 12 in. (0.85)(3)(12) = (60) = 0.51 From equation (36),

= 2 (95.05)(12) = (0.9)(0.51)(60) 29 2 = 1.47 in.


Thus

= (0.51)(1.47) = 0.75 in2 Minimum reinforcement, min ( 318 95) = 200/ = 200/60,000 =0.00333 Minimum = (0.00333)(12)(29) = 1.16 in2/ft. Hence use minimum

reinforcement with = 1.16 in2/ft. Use no. 9 bars at 10 in. center-to-center ( = . /) at the bottom of

the foundation.

From figure 5.15c, the maximum negative moment = 2447.8 kip ft/24 = 102 kip ft/ft. by observation, (min ). Use no. 9 bars at 10 in. center-to-center at the top of the foundation.

Example 7

From the plate load test (plate dimension1 ft 1 ft) in the field, the coefficient of subgrade reaction of a sandy soil was determined to be 80 lb/in3. (a) What will be the value of the coefficient of subgrade reaction on the same soil for a foundation with dimensions of 30 ft 30 ft? (b) if the full-sized foundation has dimension of 45 ft 30 ft, what will be the value of the coefficient of subgrade reaction? Solution

Part a

From equation (46),

= 1 +12 2 Where

1 = 80 lb/in2 = 30 ft So

= 80 30+1(2)(30)2 = 21.36 in3


Part b

From equation (49),

= ()1+0.51.5 (30 ft30 ft) = 21.36 lb/in3 So

= (21.36)(1+0.530451.5 = 19 lb/in3 PROBLEMS

1. Determine the net ultimate bearing capacity of mat foundation with the following characteristics: a. = 120 kN/m2, = 0, = 8 m, = 18 m, = 3 m b. = 2500 lb/ft2, = 0, = 20 ft, = 30 ft, = 6.2 ft

2. Following are the results of a standard penetration test in the field (sandy soil):

Depth (m) Field value of

1.5 9

3.0 12

4.5 11

6.0 7

7.5 13

9.0 11

10.5 13

Estimate the net allowable bearing capacity of a mat foundation 6.5 m 5 m in plan. Here, = 1.5 m, and allowable settlement mm. assume that the unit weight of soil = 16.5 kN/m3.

3. A mat foundation on a saturated clay soil has dimensions of 20 m 20 m. Given dead and live load = 48 MN, = 30 kN/m2, clay = 18.5 kN/m3.


a. Find the depth, of the mat for a fully compensated foundation. b. What will be the depth of the mat () for a factor of safety of 2 against

bearing capacity failure?

4. Repeat problem 4 part b for = 20 kN/m2. 5. A mat foundation is shown in figure P-1. The design considerations are =12 m, = 10 m, = 2.2 m, = 30 MN, 1 = 2 m, 2 = 2 m, 3 = 5.2 m, and

preconsolidation pressure = 105 kN/m2. Calculate the consolidation settlement under the center of the mat.

Figure P-1 6. Refer to figure P-2. For the mat,

1,3 = 40 tons,4,5,6 = 60 tons,2,9 = 45 tons, and 7,8 = 50 tons. all columns are 20 in. 20 in. in cross section. Use the procedure outlined in section 7 to determine the pressure on the soil at A, B, C, D, E, F, G, and H.


Figure P-2

7. The plan of a mat foundation with column loads is shown in figure P-3. Calculate the soil pressure at points A, B, C, D, E, and F. note: all columns are 0.5 m 0.5 m in plan.


Figure P-3

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Lecture 19

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