Structural Design And Analysis Of Mat Foundation.pdf 5g3513

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]

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 2. Determine the pressure on the soil, q, below the mat at points 𝐴𝐴, 𝐵𝐵, 𝐶𝐶, 𝐷𝐷, …, by using the equation 𝑄𝑄

Where

𝑞𝑞 = 𝐴𝐴 ±

𝑀𝑀𝑦𝑦 𝑥𝑥 𝐼𝐼𝑦𝑦

±

𝑀𝑀𝑥𝑥 𝑦𝑦 𝐼𝐼𝑥𝑥

[5.25]

𝐴𝐴 = 𝐵𝐵𝐵𝐵

𝐼𝐼𝑥𝑥 = (1/12)𝐵𝐵𝐵𝐵3 = moment of inertia about the 𝑥𝑥 axis

𝐼𝐼𝑦𝑦 = (1/12)𝐿𝐿𝐿𝐿 3 = 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 ′ =

𝑄𝑄1 𝑥𝑥′ 1 +𝑄𝑄2 𝑥𝑥′ 2 +𝑄𝑄3 𝑥𝑥′ 3 +⋯

[5.26]

𝑄𝑄

And 𝐵𝐵

𝑒𝑒𝑥𝑥 = 𝑥𝑥 ′ − 2

[5.27]

Similarly 𝑦𝑦 ′ =

𝑄𝑄1 𝑦𝑦′ 1 +𝑄𝑄2 𝑦𝑦′ 2 +𝑄𝑄3 𝑦𝑦′ 3 +⋯

[5.28]

𝑄𝑄

And

𝐿𝐿

𝑒𝑒𝑦𝑦 = 𝑦𝑦 ′ − 2

[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 ≈

𝑞𝑞 𝐼𝐼 +𝑞𝑞 𝐹𝐹 2

[5.30]

NPTEL - ADVANCED FOUNDATION ENGINEERING-1

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

The total soil reaction is equal to 𝑞𝑞𝑎𝑎𝑎𝑎 𝐵𝐵1 𝐵𝐵. 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 𝑞𝑞𝑎𝑎𝑎𝑎 𝐵𝐵1 𝐵𝐵 because the shear between the adjacent strips has not been taken into . For this reason, the soil reaction and the column loads need to be adjusted, or Average load =

𝑞𝑞 𝑎𝑎𝑎𝑎 𝐵𝐵1 𝐵𝐵+(𝑄𝑄1 +𝑄𝑄2 +𝑄𝑄3 +𝑄𝑄4 )

[5.31]

2

Now, the modified average soil reaction becomes average load

𝑞𝑞𝑎𝑎𝑎𝑎 (modified ) = 𝑞𝑞𝑎𝑎𝑎𝑎 �

𝑞𝑞 𝑎𝑎𝑎𝑎 𝐵𝐵1 𝐵𝐵

�

[5.32]

And the column load modification factor is average load

𝐹𝐹 = 𝑄𝑄

[5.33]

1 +𝑄𝑄2 +𝑄𝑄3 +𝑄𝑄4

So, the modified column loads are 𝐹𝐹𝐹𝐹1 , 𝐹𝐹𝐹𝐹2 , 𝐹𝐹𝐹𝐹3 , and 𝐹𝐹𝐹𝐹4 . 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 = 𝑏𝑏𝑜𝑜 𝑑𝑑[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 = 𝑏𝑏𝑜𝑜 𝑑𝑑 (4𝜙𝜙�𝑓𝑓′𝑐𝑐 )

[5.35]

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Where 𝑈𝑈 is in lb, 𝑏𝑏𝑜𝑜 and 𝑑𝑑 are in in. , and 𝑓𝑓′𝑐𝑐 is in lb/in2

The expression for 𝑏𝑏𝑜𝑜 in 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. 𝑎𝑎

𝑀𝑀𝑢𝑢 = (𝑀𝑀′ )(load factor) = 𝜙𝜙𝐴𝐴𝑠𝑠 𝑓𝑓𝑦𝑦 �𝑑𝑑 − 2 � And

𝐴𝐴𝑠𝑠 𝑓𝑓𝑦𝑦

𝑎𝑎 = 0.85𝑓𝑓′ Where

𝑐𝑐 𝑏𝑏

[5.36]

[5.37]

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.

NPTEL - ADVANCED FOUNDATION ENGINEERING-1

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, 𝑑𝑑 2 𝑧𝑧

𝑀𝑀 = 𝐸𝐸𝐹𝐹 𝐼𝐼𝐹𝐹 𝑑𝑑𝑑𝑑 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 = �121 �𝐵𝐵1 ℎ3 (see figure 5.9c) However 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

= shear force = 𝑉𝑉

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 And 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

= 𝑞𝑞 = soil reaction

Hence 𝑑𝑑 2 𝑀𝑀 𝑑𝑑𝑑𝑑 2

= 𝑞𝑞

[5.39]

Combining equations (38 and 39) yields 𝑑𝑑 4 𝑧𝑧

𝐸𝐸𝐹𝐹 𝐼𝐼𝐹𝐹 𝑑𝑑𝑑𝑑 4 = 𝑞𝑞

[5.40]

However, the soil reaction is 𝑞𝑞 = −𝑧𝑧𝑧𝑧′ Where

𝑧𝑧 = deflection 𝑘𝑘 ′ = 𝑘𝑘𝐵𝐵1

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

𝑑𝑑 4 𝑧𝑧

𝐸𝐸𝐹𝐹 𝐼𝐼𝐹𝐹 𝑑𝑑𝑑𝑑 4 = −𝑧𝑧𝑧𝑧𝐵𝐵1

[5.41]

𝑧𝑧 = 𝑒𝑒 −𝛼𝛼𝛼𝛼 (𝐴𝐴′ cos 𝛽𝛽𝛽𝛽 + 𝐴𝐴" sin⁡𝛽𝛽𝛽𝛽)

[5.42]

Solution of equation (41) yields

Where 𝐴𝐴′ and 𝐴𝐴" are constants and 4

𝐵𝐵 𝑘𝑘

𝛽𝛽 = �4𝐸𝐸1 𝐼𝐼

𝐹𝐹 𝐹𝐹

[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.

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 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 𝑞𝑞

𝑘𝑘 = Δ

[5.44]

Figure 5.10 Definition of coefficient of subgrade reaction, 𝑘𝑘

The unit of 𝑘𝑘 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 𝐵𝐵+0.3 2

𝑘𝑘 = 𝑘𝑘0.3 � Where

2𝐵𝐵

�

[5.45]

𝑘𝑘0.3 and 𝑘𝑘 = 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 𝐵𝐵+1 2

𝑘𝑘 = 𝑘𝑘1 � 2𝐵𝐵 �

[5.46]

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Where 𝑘𝑘1 and 𝑘𝑘 = coefficient of subgrade reaction of foundation measuring 1 ft × a ft and 𝐵𝐵 (ft) × 𝐵𝐵, respectively (units is lb/in3 ) Foundations on Clays

0.3 (m)

𝑘𝑘(kN/m3 ) = 𝑘𝑘0.3 (kN/m3 ) � 𝐵𝐵 (m) �

[5.47]

The definition of k in equation (47) is the same as in equation (45). In English units, 1 (ft)

𝑘𝑘(lb/in3 ) = 𝑘𝑘1 (lb/in3 ) �𝐵𝐵 (ft)�

[5.48]

The definitions of 𝑘𝑘 and 𝑘𝑘1 are the same as in equation (46).

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

𝐵𝐵 𝐿𝐿

𝑘𝑘 (𝐵𝐵 ×𝐵𝐵 ) �1+0.5 �

Where

1.5

[5.49]

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

𝑘𝑘(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 0.67𝑘𝑘(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 𝑘𝑘1 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 )

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 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 𝑘𝑘0.3 can be obtained from standard penetration resistance at any given depth, or 𝑘𝑘0.3 (MN/m3 ) = 18𝑁𝑁cor

[5.50]

Where 𝑁𝑁𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 standard penetration resistance In English units,

𝑘𝑘1 (U. S. ton/ft 3 ) = 6Ncor

[5.51]

For long beams, Vesic (1961) proposed an equation for estimating subgrade reaction: 12

𝐸𝐸 𝐵𝐵 4

𝑘𝑘 ′ 𝐵𝐵𝐵𝐵 = 0.65 �𝐸𝐸𝑠𝑠 𝐼𝐼 Where

𝐹𝐹 𝐹𝐹

𝐸𝐸𝑠𝑠

1−𝜇𝜇 𝑠𝑠2

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

[5.52]

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 𝐸𝐸

𝑠𝑠 𝑘𝑘 = 𝐵𝐵(1−𝜇𝜇 2)

[5.53]

𝑠𝑠

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 deg 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: 𝐸𝐸 ℎ 3

𝐹𝐹 𝑅𝑅 = 12(1−𝜇𝜇 2)

Where

𝑠𝑠

[5.54]

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

3. Determine the radius of effective stiffness: 4

𝑅𝑅

𝐿𝐿′ = �𝑘𝑘

[5.55]

Where 𝑘𝑘 = 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):

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 𝑄𝑄

𝑀𝑀𝑟𝑟 = radial moment = − 4 �𝐴𝐴1 − 𝑄𝑄

(1−𝜇𝜇 𝐹𝐹 )𝐴𝐴2 𝑟𝑟 𝐿𝐿′

𝑀𝑀𝑡𝑡 = tangential moment = − 4 �𝜇𝜇𝐹𝐹 𝐴𝐴1 +

�

[5.56]

(1−𝜇𝜇 𝐹𝐹 )𝐴𝐴2 𝑟𝑟 𝐿𝐿′

�

[5.57]

Figure 5.11 Approximate flexible method of mat design Where 𝑟𝑟 = radial distance from the column load 𝑄𝑄 = column load 𝐴𝐴1 , 𝐴𝐴2 = functions of 𝑟𝑟/𝐿𝐿′

The variations of 𝐴𝐴1 and 𝐴𝐴2 with 𝑟𝑟/𝐿𝐿′ are shown in figure 5.11b (for details, see Hetenyi, 1946). In the Cartesian coordinates system (figure 5.11a), 𝑀𝑀𝑥𝑥 = 𝑀𝑀𝑡𝑡 sin2 𝛼𝛼 + 𝑀𝑀𝑟𝑟 cos2 𝛼𝛼

𝑀𝑀𝑦𝑦 = 𝑀𝑀𝑡𝑡 cos2 𝛼𝛼 + 𝑀𝑀𝑟𝑟 sin2 𝛼𝛼

[5.58] [5.59]

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

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 𝑄𝑄

𝑉𝑉 = 4𝐿𝐿′ 𝐴𝐴3

[5.60]

The variation of 𝐴𝐴3 with 𝑟𝑟/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 =

𝑄𝑄𝑄𝑄′ 2 4𝑅𝑅

𝐴𝐴4

[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 𝑁𝑁. 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/ft 2 . that the soil pressures are less than the net allowable bearing capacity.

NPTEL - ADVANCED FOUNDATION ENGINEERING-1

Figure 5.12 Plan of a mat foundation Solution From figure 5.12, Column dead load (𝐷𝐷𝐷𝐷) = 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

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 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 = 121 (76)(96)3 = 5603 × 103 ft 4

𝐼𝐼𝑦𝑦 = 121 (96)(76)3 = 3512 × 103 ft 4 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 = 𝑄𝑄𝑄𝑄𝑦𝑦 = (8761)(0.727) = 6369 kip − ft 𝑀𝑀𝑦𝑦 = 𝑄𝑄𝑄𝑄𝑥𝑥 = (8761)(0.664) = 5817 kip − ft

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 From equation (25) 𝑄𝑄

𝑞𝑞 = 𝐴𝐴 ± 8761

𝑀𝑀𝑦𝑦 𝑥𝑥 𝐼𝐼𝑦𝑦

±

𝑀𝑀𝑥𝑥 𝑦𝑦 𝐼𝐼𝑥𝑥

(5817 )(𝑥𝑥)

(6369)(𝑦𝑦)

= (76)(96) ± 3512 ×10 3 ± 5603 ×10 3 Or

𝑞𝑞 = 1.20 ± 0.0017𝑥𝑥 ± 0.0011𝑦𝑦 (kip/ft 2 ) Now the following table can be prepared. 𝑄𝑄

𝑋𝑋 (ft)

±0.0017𝑋𝑋 (ft)

𝑦𝑦 (ft)

±0.0011𝑦𝑦 (ft)

A

(kip/ ft 2 ) 1.2

-38

-0.065

48

-0.053

𝑞𝑞(kip/ ft 2 )

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

Point

𝐴𝐴

1.082

The soil pressures at all points are less than the given value of 𝑞𝑞all (net ) = 1.5 kip/ft 2 .

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 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, 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴, 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵, 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷, and 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹) 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(𝐷𝐷𝐷𝐷) + 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

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 𝑉𝑉𝑢𝑢 = factored shear strength

𝜙𝜙𝑉𝑉𝑐𝑐 = 𝜙𝜙(4)�𝑓𝑓 ′ 𝑐𝑐 𝑏𝑏𝑜𝑜 𝑑𝑑 = (0.85)(4)(�3000)(96 + 2𝑑𝑑)𝑑𝑑 So

(0.85)(4)(�3000 )(96+2𝑑𝑑)𝑑𝑑 1000

(96 + 2𝑑𝑑)𝑑𝑑 ≥ 2615.1

≥ 487

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)𝑑𝑑 1000

(96 + 4𝑑𝑑)𝑑𝑑 ≥ 6046.4 𝑑𝑑 ≈ 28.7 in.

≥ 1126

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 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 =

𝑞𝑞2 =

𝑞𝑞 (at 𝐴𝐴 ) +𝑞𝑞 (at 𝐵𝐵 ) 2

𝑞𝑞 (at 𝑀𝑀 ) +𝑞𝑞 (at 𝑁𝑁 ) 2

=

=

1.082+1.106 2

1.212+1.188 2

= 1.094 kip/ft 2 = 1.20 kip/ft 2

For strip BCDKLM (width = 24 ft) 𝑞𝑞1 =

1.106+1.127+1.147 3

𝑞𝑞2 =

1.253+1.233+1.212

𝑞𝑞1 =

1.147+1.167+1.188

3

= 1.127 kip/ft 2

= 1.233 kip/ft 2

For strip DEFIJK (width = 24 ft) 3

𝑞𝑞2 =

1.294+1.273+1.253

𝑞𝑞1 =

1.188+1.212

3

= 1.167 kip/ft 2 = 1.273 kip/ft 2

For strip FGHI (width = 14 ft) 𝑞𝑞2 =

2

1.318+1.294 2

= 1.20 kip/ft 2

= 1.306 kip/ft 2

Check for Σ 𝐹𝐹𝑣𝑣 = 0:

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

Soil reaction for strip 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 = 12(1.127 + 1.233)(24)(96) = 2718.7 kip Soil reaction for strip 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 12(1.167 + 1.273)(24)(96) = 2810.9 kip Soil reaction for strip 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 = 12(1.20 + 1.306)(14)(96) = 1684.0 kip ∑ 8755.2 kip ≈ ∑ Column load = 8761 kip − OK

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 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.

NPTEL - ADVANCED FOUNDATION ENGINEERING-1

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 ′ 𝑐𝑐 𝑎𝑎𝑎𝑎 ∑ Tensile force, 𝑇𝑇 = 𝐴𝐴𝑠𝑠 𝑓𝑓𝑦𝑦 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), 𝑎𝑎

𝑀𝑀𝑢𝑢 = 𝜙𝜙𝐴𝐴𝑠𝑠 𝑓𝑓𝑦𝑦 �𝑑𝑑 − 2 �

𝑎𝑎

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

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Thus 𝐴𝐴𝑠𝑠 = (0.51)(1.47) = 0.75 in2

 Minimum reinforcement, 𝑠𝑠min (𝐴𝐴𝐴𝐴𝐴𝐴 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), 𝐵𝐵+1 2

𝑘𝑘 = 𝑘𝑘1 � 2𝐵𝐵 � Where

𝑘𝑘1 = 80 lb/in2 𝐵𝐵 = 30 ft So

30+1 2

𝑘𝑘 = 80 �(2)(30)� = 21.36 in3

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Part b From equation (49), 𝑘𝑘 =

𝐵𝐵 𝐿𝐿

𝑘𝑘 (𝐵𝐵 ×𝐵𝐵 ) �1+0.5 � 1.5

𝑘𝑘(30 ft×30 ft) = 21.36 lb/in3 So

𝑘𝑘 =

(21.36)(1+0.530 45 1.5

PROBLEMS

= 19 lb/in3

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/ft 2 , 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) 1.5

Field value of 𝑁𝑁𝐹𝐹 9 12

3.0 11 4.5 7 6.0 13 7.5 11 9.0 13 10.5 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 .

NPTEL - ADVANCED FOUNDATION ENGINEERING-1 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.

NPTEL - ADVANCED FOUNDATION ENGINEERING-1

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.

NPTEL - ADVANCED FOUNDATION ENGINEERING-1

Figure P-3