Module 5 - NPTELnptel.ac.in/courses/105104137/module5/lecture19.pdf · NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Module 5 (Lecture 19) MAT FOUNDATIONS Topics 1.1 STRUCTURAL DESIGN

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 𝐿𝐿 × 𝐵𝐵 and columns loads of 𝑄𝑄1,𝑄𝑄2,𝑄𝑄3, … Calculate the total column load as 𝑄𝑄 = 𝑄𝑄1 + 𝑄𝑄2 + 𝑄𝑄3 … [5.24]


2. Determine the pressure on the soil, q, below the mat at points 𝐴𝐴,𝐵𝐵,𝐶𝐶,𝐷𝐷, …, by using the equation 𝑞𝑞 = 𝑄𝑄

𝐴𝐴± 𝑀𝑀𝑦𝑦𝑥𝑥

𝐼𝐼𝑦𝑦± 𝑀𝑀𝑥𝑥𝑦𝑦

𝐼𝐼𝑥𝑥 [5.25]

Where

𝐴𝐴 = 𝐵𝐵𝐿𝐿

𝐼𝐼𝑥𝑥 = (1/12)𝐵𝐵𝐿𝐿3 = moment of inertia about the 𝑥𝑥 axis

𝐼𝐼𝑦𝑦 = (1/12)𝐿𝐿𝐵𝐵3 = moment of inertia about the 𝑦𝑦 axis

𝑀𝑀𝑥𝑥 = moment of the column loads about the 𝑥𝑥 axis = 𝑄𝑄𝑒𝑒𝑦𝑦

𝑀𝑀𝑦𝑦 = moment of the column loads about the 𝑦𝑦 axis = 𝑄𝑄𝑒𝑒𝑥𝑥

The load eccentricities, 𝑒𝑒𝑥𝑥 and 𝑒𝑒𝑦𝑦 , in the 𝑥𝑥 and 𝑦𝑦 directions can be determined by using (𝑥𝑥′ , 𝑦𝑦′) coordinates:

𝑥𝑥′ = 𝑄𝑄1𝑥𝑥′1+𝑄𝑄2𝑥𝑥′2+𝑄𝑄3𝑥𝑥′3+⋯𝑄𝑄

[5.26]

And

𝑒𝑒𝑥𝑥 = 𝑥𝑥′ − 𝐵𝐵2 [5.27]

Similarly

𝑦𝑦′ = 𝑄𝑄1𝑦𝑦′1+𝑄𝑄2𝑦𝑦′2+𝑄𝑄3𝑦𝑦′3+⋯𝑄𝑄

[5.28]

And

𝑒𝑒𝑦𝑦 = 𝑦𝑦′ − 𝐿𝐿2 [5.29]

3. Compare the values of the soil pressures determined in step 2 with the net allowable soil pressure to determine whether 𝑞𝑞 ≥ 𝑞𝑞all (net ).

4. Divide the mat into several strips in x and y directions (see figure 5.8a). Let the width of any strip be 𝐵𝐵1.

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 𝑞𝑞𝑎𝑎𝑎𝑎 ≈

𝑞𝑞𝐼𝐼+𝑞𝑞𝐹𝐹2

[5.30]


Where 𝑞𝑞𝐼𝐼and 𝑞𝑞𝐹𝐹 = soil pressures at poins 𝐼𝐼 and 𝐹𝐹 as determined from step 2.

The total soil reaction is equal to 𝑞𝑞𝑎𝑎𝑎𝑎𝐵𝐵1𝐵𝐵. Now obtain the total column load on the strip as 𝑄𝑄1 + 𝑄𝑄2 + 𝑄𝑄3 + 𝑄𝑄4. The sum of the column loads on the strip will not equal 𝑞𝑞𝑎𝑎𝑎𝑎𝐵𝐵1𝐵𝐵 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 = 𝑞𝑞𝑎𝑎𝑎𝑎𝐵𝐵1𝐵𝐵+(𝑄𝑄1+𝑄𝑄2+𝑄𝑄3+𝑄𝑄4)2

[5.31]

Now, the modified average soil reaction becomes

𝑞𝑞𝑎𝑎𝑎𝑎(modified ) = 𝑞𝑞𝑎𝑎𝑎𝑎 �average load

𝑞𝑞𝑎𝑎𝑎𝑎𝐵𝐵1𝐵𝐵� [5.32]

And the column load modification factor is

𝐹𝐹 = average load𝑄𝑄1+𝑄𝑄2+𝑄𝑄3+𝑄𝑄4

[5.33]

So, the modified column loads are 𝐹𝐹𝑄𝑄1,𝐹𝐹𝑄𝑄2,𝐹𝐹𝑄𝑄3, and 𝐹𝐹𝑄𝑄4. 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, 𝑈𝑈 = 𝑏𝑏𝑜𝑜𝑑𝑑[𝜙𝜙(0.34)�𝑓𝑓′𝑐𝑐] [5.34] Where 𝑈𝑈 = factored column load (MN), or (column load) × (load factor) 𝜙𝜙 = reduction factor = 0.85 𝑓𝑓′𝑐𝑐 = compressive strength of concrete at 28 days (MN/m2)

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

𝑈𝑈 = 𝑏𝑏𝑜𝑜𝑑𝑑 (4𝜙𝜙�𝑓𝑓′𝑐𝑐 ) [5.35]


Where

𝑈𝑈 is in lb,𝑏𝑏𝑜𝑜 and 𝑑𝑑 are in in. , and 𝑓𝑓′𝑐𝑐 is in lb/in2

The expression for 𝑏𝑏𝑜𝑜 in terms of 𝑑𝑑, 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, 𝑀𝑀′ = 𝑀𝑀/𝐵𝐵1).

8. Determine the areas of steep per unit width for positive and negative reinforcement in the x and y directions. 𝑀𝑀𝑢𝑢 = (𝑀𝑀′)(load factor) = 𝜙𝜙𝐴𝐴𝑠𝑠𝑓𝑓𝑦𝑦 �𝑑𝑑 −

𝑎𝑎2� [5.36]

And 𝑎𝑎 = 𝐴𝐴𝑠𝑠𝑓𝑓𝑦𝑦

0.85𝑓𝑓′𝑐𝑐𝑏𝑏 [5.37]

Where 𝐴𝐴𝑠𝑠 = area of steel per unit width 𝑓𝑓𝑦𝑦 = yield stress of reinforcement in tension 𝑀𝑀𝑢𝑢 = factored moment 𝜙𝜙 = 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 𝐵𝐵1 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,

𝑀𝑀 = 𝐸𝐸𝐹𝐹𝐼𝐼𝐹𝐹𝑑𝑑2𝑧𝑧𝑑𝑑𝑥𝑥 2 [5.38]

Where

𝑀𝑀 = moment at any section

𝐸𝐸𝐹𝐹 = modulus of elasticity of foundation material

𝐼𝐼𝐹𝐹 = moment of inertia of the cross section of the beam = � 112�𝐵𝐵1ℎ3 (see figure 5.9c)

However

𝑑𝑑𝑀𝑀𝑑𝑑𝑥𝑥

= shear force = 𝑉𝑉


And

𝑑𝑑𝑉𝑉𝑑𝑑𝑥𝑥

= 𝑞𝑞 = soil reaction

Hence

𝑑𝑑2𝑀𝑀𝑑𝑑𝑥𝑥 2 = 𝑞𝑞 [5.39]

Combining equations (38 and 39) yields

𝐸𝐸𝐹𝐹𝐼𝐼𝐹𝐹𝑑𝑑4𝑧𝑧𝑑𝑑𝑥𝑥 4 = 𝑞𝑞 [5.40]

However, the soil reaction is

𝑞𝑞 = −𝑧𝑧𝑧𝑧′

Where

𝑧𝑧 = deflection

𝑧𝑧′ = 𝑧𝑧𝐵𝐵1

𝑧𝑧 = coefficient of subgrade reaction (kN/m3 or lb/in3)

So

𝐸𝐸𝐹𝐹𝐼𝐼𝐹𝐹𝑑𝑑4𝑧𝑧𝑑𝑑𝑥𝑥 4 = −𝑧𝑧𝑧𝑧𝐵𝐵1 [5.41]

Solution of equation (41) yields

𝑧𝑧 = 𝑒𝑒−𝛼𝛼𝑥𝑥 (𝐴𝐴′ cos𝛽𝛽𝑥𝑥 + 𝐴𝐴" sin 𝛽𝛽𝑥𝑥) [5.42]

Where 𝐴𝐴′and 𝐴𝐴" are constants and

𝛽𝛽 = � 𝐵𝐵1𝑧𝑧4𝐸𝐸𝐹𝐹𝐼𝐼𝐹𝐹

4 [5.43]

The unit of the term 𝛽𝛽 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/𝛽𝛽. If the spacing of columns is larger than 1.75/𝛽𝛽, 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, Δ. The coefficient of subgrade modulus, k, can be defined as

𝑧𝑧 = 𝑞𝑞Δ [5.44]

Figure 5.10 Definition of coefficient of subgrade reaction, 𝑧𝑧

The unit of 𝑧𝑧 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, 𝐿𝐿, and width, 𝐵𝐵, 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 𝐵𝐵 × 𝐵𝐵 in the following ways.

Foundations on Sandy Soils

𝑧𝑧 = 𝑧𝑧0.3 �𝐵𝐵+0.3

2𝐵𝐵�

2 [5.45]

Where

𝑧𝑧0.3 and 𝑧𝑧 = coefficients of subgrade reaction of foundation measuring 0.3m ×0.3m and 𝐵𝐵(m) × 𝐵𝐵(m), respectively (units is kN/m3

In English units, equation (45) may be expressed as

𝑧𝑧 = 𝑧𝑧1 �𝐵𝐵+12𝐵𝐵�

2 [5.46]


Where

𝑧𝑧1 and 𝑧𝑧 = coefficient of subgrade reaction of foundation measuring 1 ft ×a ft and 𝐵𝐵 (ft) × 𝐵𝐵, respectively (units is lb/in3)

Foundations on Clays

𝑧𝑧(kN/m3) = 𝑧𝑧0.3(kN/m3) �0.3 (m)𝐵𝐵 (m)

� [5.47]

The definition of k in equation (47) is the same as in equation (45).

In English units,

𝑧𝑧(lb/in3) = 𝑧𝑧1(lb/in3) �1 (ft)𝐵𝐵 (ft)

� [5.48]

The definitions of 𝑧𝑧 and 𝑧𝑧1 are the same as in equation (46).

For rectangular foundations having dimensions of 𝐵𝐵 × 𝐿𝐿 (for similar soil and q),

𝑧𝑧 =𝑧𝑧(𝐵𝐵×𝐵𝐵)�1+0.5𝐵𝐵𝐿𝐿�

1.5 [5.49]

Where

𝑧𝑧 = coefficient of subgrade modulus of the rectangular foundation (𝐿𝐿 × 𝐵𝐵)

𝑧𝑧(𝐵𝐵×𝐵𝐵) =coefficient of subgrade modulus of a square foundation having dimension of 𝐵𝐵 × 𝐵𝐵

Equation (49) indicates that the value of k of a very long foundation with a width B is approximately 0.67𝑧𝑧(𝐵𝐵×𝐵𝐵).

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 𝑧𝑧1 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(> 50𝑀𝑀𝑀𝑀/m3

Scott (1981) proposed that for sandy soils the value of 𝑧𝑧0.3 can be obtained from standard penetration resistance at any given depth, or

𝑧𝑧0.3(MN/m3) = 18𝑀𝑀cor [5.50]

Where

𝑀𝑀𝑐𝑐𝑜𝑜𝑐𝑐 = 𝑐𝑐𝑜𝑜𝑐𝑐𝑐𝑐𝑒𝑒𝑐𝑐𝑐𝑐𝑒𝑒𝑑𝑑 standard penetration resistance

In English units,

𝑧𝑧1(U. S. ton/ft3) = 6Ncor [5.51]

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

𝑧𝑧′𝐵𝐵𝑧𝑧 = 0.65 �𝐸𝐸𝑠𝑠𝐵𝐵4

𝐸𝐸𝐹𝐹𝐼𝐼𝐹𝐹

12 𝐸𝐸𝑠𝑠1−𝜇𝜇𝑠𝑠2

[5.52]

Where

𝐸𝐸𝑠𝑠 = modulus of elasticity of soil

𝐵𝐵 = foundation width

𝐸𝐸𝐹𝐹 = modulus of elasticity of foundation material

𝐼𝐼𝐹𝐹 = moment of inertia of the cross section of the foundation

𝜇𝜇𝑠𝑠 = Poisson′sratio of soil

For most practical purposes, equation (52) can be approximated as


𝑧𝑧 = 𝐸𝐸𝑠𝑠𝐵𝐵(1−𝜇𝜇𝑠𝑠2)

[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: 𝑅𝑅 = 𝐸𝐸𝐹𝐹ℎ3

12(1−𝜇𝜇𝑠𝑠2) [5.54]

Where 𝐸𝐸𝐹𝐹 = modulus of elasticity of foundation material 𝜇𝜇𝐹𝐹 = Poisson′s ratio of foundation material

3. Determine the radius of effective stiffness:

𝐿𝐿′ = �𝑅𝑅𝑧𝑧

4 [5.55]

Where 𝑧𝑧 = 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):


𝑀𝑀𝑐𝑐 = radial moment = −𝑄𝑄4�𝐴𝐴1 −

(1−𝜇𝜇𝐹𝐹)𝐴𝐴2𝑐𝑐𝐿𝐿′

� [5.56]

𝑀𝑀𝑐𝑐 = tangential moment = −𝑄𝑄4�𝜇𝜇𝐹𝐹𝐴𝐴1 + (1−𝜇𝜇𝐹𝐹)𝐴𝐴2

𝑐𝑐𝐿𝐿′

� [5.57]

Figure 5.11 Approximate flexible method of mat design

Where 𝑐𝑐 = radial distance from the column load 𝑄𝑄 = column load 𝐴𝐴1,𝐴𝐴2 = functions of 𝑐𝑐/𝐿𝐿′ The variations of 𝐴𝐴1 and 𝐴𝐴2 with 𝑐𝑐/𝐿𝐿′ are shown in figure 5.11b (for details, see Hetenyi, 1946). In the Cartesian coordinates system (figure 5.11a), 𝑀𝑀𝑥𝑥 = 𝑀𝑀𝑐𝑐 sin2𝛼𝛼 + 𝑀𝑀𝑐𝑐 cos2𝛼𝛼 [5.58] 𝑀𝑀𝑦𝑦 = 𝑀𝑀𝑐𝑐 cos2𝛼𝛼 + 𝑀𝑀𝑐𝑐 sin2𝛼𝛼 [5.59]

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


𝑉𝑉 = 𝑄𝑄4𝐿𝐿′𝐴𝐴3 [5.60]

The variation of 𝐴𝐴3 with 𝑐𝑐/𝐿𝐿′ 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 (𝛿𝛿) at any point is given by 𝛿𝛿 = 𝑄𝑄𝐿𝐿′ 2

4𝑅𝑅𝐴𝐴4 [5.61]

The variation of 𝐴𝐴4 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 𝐴𝐴,𝐵𝐵,𝐶𝐶,𝐷𝐷,𝐸𝐸,𝐹𝐹,𝐺𝐺,𝐻𝐻, 𝐼𝐼, 𝐽𝐽,𝐾𝐾, 𝐿𝐿,𝑀𝑀, and 𝑀𝑀. The size of the mat is 76 ft × 96 ft, all columns are 24 in.× 24 in. in section, and 𝑞𝑞all (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 (𝐷𝐷𝐿𝐿) = 100 + 180 + 190 + 110 + 180 + 360 + 400 + 200 +190 + 400 + 440 + 200 + 120 + 180 + 180 + 120 = 3550 kip

Column live load (𝐿𝐿𝐿𝐿) = 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, 𝑈𝑈 = (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

𝐼𝐼𝑥𝑥 = 112

(76)(96)3 = 5603 × 103 ft4

𝐼𝐼𝑦𝑦 = 112

(96)(76)3 = 3512 × 103 ft4

And

∑𝑀𝑀𝑦𝑦 = 0

So

5780𝑥𝑥′ = (24)(300 + 560 + 640 + 300) + (48)(310 + 650 + 740 + 300) +(72)(180 + 320 + 320 + 190)

𝑥𝑥′ = 36.664 ft

And

𝑒𝑒𝑥𝑥 = 36.664 − 36.0 = 0.664 ft

Similarly,

∑𝑀𝑀𝑥𝑥′ = 0

So

5780𝑦𝑦′ = (30)(320 + 640 + 740 + 320) + (60)(300 + 560 + 650 + 320) +(90)(160 + 300 + 310 + 180)

𝑦𝑦′ = 44.273 ft

And

𝑒𝑒𝑦𝑦 = 44.273 − 902 = −0.727 ft

The moments caused by eccentricity are

𝑀𝑀𝑥𝑥 = 𝑄𝑄𝑒𝑒𝑦𝑦 = (8761)(0.727) = 6369 kip − ft

𝑀𝑀𝑦𝑦 = 𝑄𝑄𝑒𝑒𝑥𝑥 = (8761)(0.664) = 5817 kip − ft


From equation (25)

𝑞𝑞 = 𝑄𝑄𝐴𝐴

± 𝑀𝑀𝑦𝑦𝑥𝑥𝐼𝐼𝑦𝑦

± 𝑀𝑀𝑥𝑥𝑦𝑦𝐼𝐼𝑥𝑥

= 8761(76)(96)

± (5817)(𝑥𝑥)3512×103 ± (6369)(𝑦𝑦)

5603×103

Or

𝑞𝑞 = 1.20 ± 0.0017𝑥𝑥 ± 0.0011𝑦𝑦 (kip/ft2)

Now the following table can be prepared.

Point 𝑄𝑄𝐴𝐴

(kip/ft2)

𝑋𝑋 (ft) ±0.0017𝑋𝑋 (ft) 𝑦𝑦 (ft) ±0.0011𝑦𝑦 (ft) 𝑞𝑞(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 𝑞𝑞all (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, 𝐴𝐴𝐵𝐵𝑀𝑀𝑀𝑀,𝐵𝐵𝐶𝐶𝐷𝐷𝐾𝐾𝐿𝐿𝑀𝑀,𝐷𝐷𝐸𝐸𝐹𝐹𝐼𝐼𝐽𝐽𝐾𝐾, and 𝐹𝐹𝐺𝐺𝐻𝐻𝐼𝐼)

and determine the average soil reaction at the ends of each strips. c. Determine the reinforcement requirements in the y direction for 𝑓𝑓′𝑐𝑐 =

3000 lb/in2 and 𝑓𝑓𝑦𝑦 = 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

𝑈𝑈 = 1.4(𝐷𝐷𝐿𝐿) + 1.7(𝐿𝐿𝐿𝐿) = (1.4)(190) + (1.7)(130) = 487 kip

𝑏𝑏𝑜𝑜 = 2(36 + 𝑑𝑑/2) + (24 + 𝑑𝑑) = 96 + 2𝑑𝑑(in).

From ACI 318-95

𝜙𝜙𝑉𝑉𝑐𝑐 ≥ 𝑉𝑉𝑢𝑢

Where

𝑉𝑉𝑐𝑐 = nominal shear strength of concrete


𝑉𝑉𝑢𝑢 = factored shear strength

𝜙𝜙𝑉𝑉𝑐𝑐 = 𝜙𝜙(4)�𝑓𝑓′ 𝑐𝑐𝑏𝑏𝑜𝑜𝑑𝑑 = (0.85)(4)(�3000)(96 + 2𝑑𝑑)𝑑𝑑

So

(0.85)(4)(�3000)(96+2𝑑𝑑)𝑑𝑑1000

≥ 487

(96 + 2𝑑𝑑)𝑑𝑑 ≥ 2615.1

𝑑𝑑 ≈ 19.4 in.

For the critical internal column shown in figure 5.14,

Figure 5.14 Critical internal column

𝑏𝑏𝑜𝑜 = 4(24 + 𝑑𝑑) = 96 + 4𝑑𝑑(in. )

𝑈𝑈 = (1.4)(440) + (1.7)(300) = 1126 kip

And

(0.85)(4)(�3000)(96+4𝑑𝑑)𝑑𝑑1000

≥ 1126

(96 + 4𝑑𝑑)𝑑𝑑 ≥ 6046.4

𝑑𝑑 ≈ 28.7 in.


Use 𝑑𝑑 = 29 in.

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

ℎ = 29 + 3 + 1 = 33 in.

Part b: Average Soil Reaction

Refer to figure 5.12. For strip ABMN (width = 14 ft)

𝑞𝑞1 = 𝑞𝑞(at 𝐴𝐴)+𝑞𝑞(at 𝐵𝐵)

2= 1.082+1.106

2= 1.094 kip/ft2

𝑞𝑞2 = 𝑞𝑞(at 𝑀𝑀 )+𝑞𝑞(at 𝑀𝑀)

2= 1.212+1.188

2= 1.20 kip/ft2

For strip BCDKLM (width = 24 ft)

𝑞𝑞1 = 1.106+1.127+1.1473

= 1.127 kip/ft2

𝑞𝑞2 = 1.253+1.233+1.2123

= 1.233 kip/ft2

For strip DEFIJK (width = 24 ft)

𝑞𝑞1 = 1.147+1.167+1.1883

= 1.167 kip/ft2

𝑞𝑞2 = 1.294+1.273+1.2533

= 1.273 kip/ft2

For strip FGHI (width = 14 ft)

𝑞𝑞1 = 1.188+1.2122

= 1.20 kip/ft2

𝑞𝑞2 = 1.318+1.294 2

= 1.306 kip/ft2

Check for Σ 𝐹𝐹𝑎𝑎 = 0:

Soil reaction for strip 𝐴𝐴𝐵𝐵𝑀𝑀𝑀𝑀 = 12(1.094 + 1.20)(14)(96) = 1541.6 kip

Soil reaction for strip 𝐵𝐵𝐶𝐶𝐷𝐷𝐾𝐾𝐿𝐿𝑀𝑀 = 12(1.127 + 1.233)(24)(96) = 2718.7 kip

Soil reaction for strip 𝐷𝐷𝐸𝐸𝐹𝐹𝐼𝐼𝐽𝐽𝐾𝐾 = 12(1.167 + 1.273)(24)(96) = 2810.9 kip

Soil reaction for strip 𝐹𝐹𝐺𝐺𝐻𝐻𝐽𝐽 = 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

𝑄𝑄1 = (1.4)(180) + (1.7)(120) = 456 kip

𝑄𝑄2 = (1.4)(360) + (1.7)(200) = 844 kip

𝑄𝑄3 = (1.4)(400) + (1.7)(240) = 968 kip

𝑄𝑄4 = (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,𝐶𝐶 = 0.85𝑓𝑓′ 𝑐𝑐𝑎𝑎𝑏𝑏

∑Tensile force,𝑇𝑇 = 𝐴𝐴𝑠𝑠𝑓𝑓𝑦𝑦

𝐶𝐶 = 𝑇𝑇

Note that for this case 𝑏𝑏 = 1 ft = 12 in.

(0.85)(3)(12)𝑎𝑎 = 𝐴𝐴𝑠𝑠(60)

𝐴𝐴𝑠𝑠 = 0.51𝑎𝑎

From equation (36),

𝑀𝑀𝑢𝑢 = 𝜙𝜙𝐴𝐴𝑠𝑠𝑓𝑓𝑦𝑦 �𝑑𝑑 −𝑎𝑎2�

(95.05)(12) = (0.9)(0.51𝑎𝑎)(60) �29 − 𝑎𝑎2�

𝑎𝑎 = 1.47 in.


Thus

𝐴𝐴𝑠𝑠 = (0.51)(1.47) = 0.75 in2

Minimum reinforcement, 𝑠𝑠min (𝐴𝐴𝐶𝐶𝐼𝐼 318 − 95) = 200/𝑓𝑓𝑦𝑦 = 200/60,000 =0.00333

Minimum 𝐴𝐴𝑠𝑠 = (0.00333)(12)(29) = 1.16 in2/ft. Hence use minimum reinforcement with 𝐴𝐴𝑠𝑠 = 1.16 in2/ft.

Use no. 9 bars at 10 in. center-to-center (𝑨𝑨𝒔𝒔 = 𝟏𝟏.𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐/𝐟𝐟𝐟𝐟) 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, 𝐴𝐴𝑠𝑠 ≤ 𝐴𝐴𝑠𝑠(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),

𝑧𝑧 = 𝑧𝑧1 �𝐵𝐵+12𝐵𝐵�

2

Where

𝑧𝑧1 = 80 lb/in2

𝐵𝐵 = 30 ft

So

𝑧𝑧 = 80 � 30+1(2)(30)

�2

= 21.36 in3


Part b

From equation (49),

𝑧𝑧 =𝑧𝑧(𝐵𝐵×𝐵𝐵)�1+0.5𝐵𝐵𝐿𝐿�

1.5

𝑧𝑧(30 ft×30 ft) = 21.36 lb/in3

So

𝑧𝑧 =(21.36)(1+0.530

451.5

= 19 lb/in3

PROBLEMS

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

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

Depth (m) Field value of 𝑀𝑀𝐹𝐹

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, 𝐷𝐷𝑓𝑓 = 1.5 m, and allowable settlement mm. assume that the unit weight of soil 𝛾𝛾 = 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, 𝑐𝑐𝑢𝑢 = 30 kN/m2, 𝛾𝛾clay = 18.5 kN/m3.


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

bearing capacity failure?

4. Repeat problem 4 part b for 𝑐𝑐𝑢𝑢 = 20 kN/m2. 5. A mat foundation is shown in figure P-1. The design considerations are 𝐿𝐿 =

12 m,𝐵𝐵 = 10 m,𝐷𝐷𝑓𝑓 = 2.2 m,𝑄𝑄 = 30 MN, 𝑥𝑥1 = 2 m, 𝑥𝑥2 = 2 m, 𝑥𝑥3 = 5.2 m, and preconsolidation pressure 𝑝𝑝𝑐𝑐 = 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,

𝑄𝑄1,𝑄𝑄3 = 40 tons,𝑄𝑄4,𝑄𝑄5,𝑄𝑄6 = 60 tons,𝑄𝑄2,𝑄𝑄9 = 45 tons, and 𝑄𝑄7,𝑄𝑄8 = 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

Module 5 - NPTELnptel.ac.in/courses/105104137/module5/lecture19.pdf · NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Module 5 (Lecture 19) MAT FOUNDATIONS Topics 1.1 STRUCTURAL DESIGN

Documents