ch-06

Introduction

Under normal conditions, square and rectangular footings such as those described inChapters 3 and 4 are economical for supporting columns and walls. However, undercertain circumstances, it may be desirable to construct a footing that supports a lineof two or more columns. These footings are referred to as combined footings. Whenmore than one line of columns is supported by a concrete slab, it is called a mat foundation. Combined footings can be classified generally under the following categories:

8. Rectangular combined footingb. Trapezoidal combined footingc. Strap footing

Mat foundations are generally used with soil that has a low bearing capacity. A briefoverview of the principles of combined footings is given in Section 6.2, followed bya more detailed discussion on mat foundations.

Combined Footings

Rectangular Combined Footing

In several instances, the load to be carried by a column and the soil bearing capacityare such that the standard spread footing design will require extension of the column foundation beyond the property line. In such a case, two or more columns canbe supported on a single rectangular foundation, as shown in Figure 6.1. If the netallowable soil pressure is known, the size of the foundation (B x L) can be determined in the following manner:

8. Determine the area of the foundation

A = Ql + Q2qnet(all)

where Ql' Q2 = column loadsqnet(all) = net allowable soil bearing capacity

(6.1)

255

256 Chapter 6 Mat Foundations

~--x----<~

Section

B . qnet(all)/unit length

Figure 6. 1 Rectangular combined footing

d. Once the length L is determined, the value of L 1 can be obtained as follows:

(6.3)

(6.2)

(6.4)

L = 2(Lz + X)

where L = length of the foundation

c. For a uniform distribution of soil pressure under the foundation, the resultant of the column loads should pass through the centroid of the foundation.Thus,

Note that the magnitude of L z will be known and depends on the location of theproperty line.

e. The width of the foundation is then

b. Determine the location of the resultant of the column loads. From Figure 6.1,

Q2L3X=---Ql + Q2

L

1• •Property- B Planline 1

B=AL

(6.5)

Trapezoidal Combined Footing

Trapezoidal combined footing (see Figure 6.2) is sometimes used as an isolatedspread foundation of columns carrying large loads where space is tight. The size of

6.2 Combined Footings 257

B1 . qnet(all)/unit length

~---X----+i

~----L3--+------t'1>'f

B2 • qnet(all)/unit length

Section

Propertyline

I-+--------L-------.....

Figure 6.2 Trapezoidal combined footing

the foundation that will uniformly distribute pressure on the soil can be obtained inthe following manner:

a. If the net allowable soil pressure is known, determine the area of the foundation:

A = Ql + Q2qnet(all)

From Figure 6.2,

(6.6)

b. Determine the location of the resultant for the column loads:

X =_Q_2L_3_

Ql + Q2

(6.7)

(c)

Section

Strap

[Jr------:..-Strap ---i~ Plan

Wall

c. From the property of a trapezoid,

_(B1 + 2Bz)LX+Lz - -B1 + Bz 3

With known values of A, L, Ar, and L z, solve Eqs. (6.6) and (6.7) to obtain B1 andBz. Note that, for a trapezoid,

Section Section

Strap Strap

~ ~ [£J EEUStrap Plan Strap Plan

(a) (b)

L L-<X+L <3 z 2

Figure 6.3 Cantilever footing-use of strap beam

Cantilever Footing

Cantilever footing construction uses a strap beam to connect an eccentrically loadedcolumn foundation to the foundation of an interior column. (See Figure 6.3). Cantilever footings may be used in place of trapezoidal or rectangular combined footings when the allowable soil bearing capacity is high and the distances between thecolumns are large.

Common Types of Mat Foundations

The mat foundation, which is sometimes referred to as a raft foundation, is a combined footing that may cover the entire area under a structure supporting severalcolumns and walls. Mat foundations are sometimes preferred for soils that have low


6.3 Common Types of Mat Foundations 259

load-bearing capacities, but that will have to support high column or wall loads. Under some conditions, spread footings would have to cover more than half the building area, and mat foundations might be more economical. Several types of matfoundations are used currently. Some of the common ones are shown schematicallyin Figure 6.4 and include the following:

1. Flat plate (Figure 6.4a). The mat is of uniform thickness.2. Flat plate thickened under columns (Figure 6.4b).3. Beams and slab (Figure 6.4c). The beams run both ways, and the columns are lo

cated at the intersection of the beams.4. Flat plates with pedestals (Figure 6.4d).5. Slab with basement walls as a part of the mat (Figure 6.4e). The walls act as stiff

eners for the mat.

Mats may be supported by piles, which help reduce the settlement of a structure built over highly compressible soil. Where the water table is high, mats are often placed over piles to control buoyancy. Figure 6.5 shows the difference betweenthe depth Df and the width B of isolated foundations and mat foundations.

• • • •... • • • • ..., -------------------- -

, Plan--

• • • •

Section

.. .. .., ..Section

'.' '.' '.' ....~--~ ~--~ ~--~

I I I I I 1

' I ' I ' I

... ... ._--.. _--.._-- ....._I Plan I ~---:_-~-_~--:---~-- _' Plan

' , ' I ' ,

• • • •---1 ---, ---,I I I I I II I I I I I' I ' I , ,

• • • •

.-, ,

• '.' '.' '.'... ' ,I --------------------

'.' '.' '.' '.''.' '.' '.' '.'

Section

••••(a) (b) (c)

HE &roOD

I I

Section

IiJ IiJ IiJ IiJ

... IiJ IiJ IiJ IiJ,----------------------

IiJ IiJ IiJ IiJ

liJ liJ IiJ Iil

(d)

...-' Plan

,- - - - - .. ,- - - - - ... - - - - - ..I " 'I I

I '. " II II '. II II " I~ JL ~L J.. :-----~:-----~:-----~ ..

I : ~~-----:~-----~ _I PlanI " 'I IL JL JL J,- - - - - .. ,- - - - - .. 1- - - - - ..

I, II III " I" I, I

I " " IL JL JL J

(e)

Figure 6.4 Common types of mat foundation

(6.8)

(3.21)

(3.14)

Figure 6.5 Comparison of isolated foundationand mat foundation (B == width, Df == depth)

Bearing Capacity of Mat Foundations

TIle gross ultimate bearing capacity of a mat foundation can be determined by thesame equation used for shallow foundations (see Section 3.7), or

(Chapter 3 gives the proper values of the bearing capacity factors, as well as theshape depth, and load inclination factors.) The term B in Eq. (3.21) is the smallest dimension of the mat. The net ultimate capacity of a mat foundation is

where eu :=: undrained cohesion

A suitable factor of safety should be used to calculate the net allowable bearing capacity. For rafts on clay, the factor of safety should not be less than 3 underdead load or maximum live load. However, under the most extreme conditions, thefactor of safety should be at least 1.75 to 2. For rafts constructed over sand, a factorof safety of 3 should normally be used. Under most working conditions, the factor ofsafety against bearing capacity failure of rafts on sand is very large.

For saturated clays with 4J == 0 and a vertical loading condition, Eq. (3.21) gives

(Note: Nc == 5.14, Nq == 1, and Ny == 0.)

From Eqs. (3.25) and (3.28), for <p == 0,

F = 1 + B (Nq) == 1 + (B)(_l_) = 1 + O.195B

cs L Nc L· 5.14 L

and


6.4 Bearing Capacity of Mat Foundations 261

Substitution of the preceding shape and depth factors into Eg. (6.8) yields

(6.9)

Hence, the net ultimate bearing capacity is

For FS == 3, the net allowable soil bearing capacity becomes

qu(net) (0.195B)( Dt )qnet(all) = ----Ps = 1.713cu 1 + L 1 + O.4B

(6.10)

(6.11 )

The net allowable bearing capacity for mats constructed over granular soil deposits can be adequately determined from the standard penetration resistance numbers. From Eg. (5.79), for shallow foundations,

2 _ (3.28B + 1)2 (Se)qnet(all)(kN/m ) - 11.98(Nl)6o 3.28B Fd 25

where (N1)60 == corrected standard penetration resistanceB == width (m)Fd == 1 + 0.33 (Dt / B) ::s 1.33Se == settlement, in mm

When the width B is large, the preceding equation can be approximated (assuming that 3.28B + 1 ~ 3.28B) as

(6.12)

In English units, Eq. (6.12) may be expressed as

(6.13)

(6.14)

(6.15)

(6.16)

Unit weight = 'Y

Qq = - - yDfA

IFigure 6.6 Definition of net pressure on soil caused by a mat foundation

The net pressure applied on a foundation (see Figure 6.6) may be expressed as

and

Note that Eq. (6.13) could have been derived from Eqs. (5.80) and (5.82).Note that the original equations (5.79) and (5.82) were for a settlement of

25 mm (1 in.), with a differential settlement of about 19 mm (0.75 in.). However,the width of the raft foundations are larger than those of the isolated spread footings. As shown in Table 5.3, the depth of significant stress increase in the soil belowa foundation depends on the width of the foundation. Hence, for a raft foundation, the depth of the zone of influence is likely to be much larger than that of aspread footing. Thus, the loose soil pockets under a raft may be more evenly distributed, resulting in a smaller differential settlement. Accordingly, the customaryassumption is that, for a maximum raft settlement of 50 mm (2 in.), the differentialsettlement would be 19 mm (0.75 in.). Using this logic and conservatively assuming that Fd = 1, we can respectively approximate Eqs. (6.12) and (6.13) as

where Q = dead weight of the structure and the live loadA = area of the raft

In all cases, q should be less than or equal to qnet(all)'


6.5 Differential Settlement of Mats 263

•

(6.18)

Differential Settlement of Mats

In 1988, the American Concrete Institute Committee 336 suggested a method forcalculating the differential settlement of mat foundations. According to this method,the rigidity factor K r is calculated as

E'IK ::::: _b (6.17)

r E B3s

where E'::::: modulus of elasticity of the material used in the structureEs ::::: modulus of elasticity of the soilB == widtll of foundationIb == moment of inertia of the structure per unit length at right angles to B

The term E'Ib can be expressed as

E'h = E'(IF + 2: Ib'+ 2:~~)


where E'Ib = flexural rigidity of the superstructure and foundation perunit length at right angles to B

'i.E'Ib= flexural rigidity of the framed members at right angles to B'i. (E'ah3/12) = flexural rigidity of the shear walls

a = shear wall thicknessh = shear wall height

E'IF = flexibility of the foundation

Based on the value of K" the ratio (8) of the differential settlement to the total settlement can be estimated in the following manner:

1. If K r > 0.5, it can be treated as a rigid mat, and 8 = O.2. If K r = 0.5, then 8 = O.l.3. If K r = 0, then 0 = 0.35 for square mats (B/ L = 1) and 0 = 0.5 for long

foundations (B/L = 0).

Field Settlement Observations for Mat Foundations

Several field settlement observations for mat foundations are currently availablein the literature. In this section, we compare the observed settlements for somemat foundations constructed over granular soil deposits with those obtained fromEqs. (6.12) and (6.13).

Meyerhof (1965) compiled the observed maximum settlements for mat foundations constructed on sand and gravel, as listed in Table 6.1. In Eq. (6.12), if thedepth factor, 1 + 0.33 (Df / B), is assumed to be approximately unity, then

qnet(all)

Se(mm) = 25 11.98(N1

)60 (6.19)

From the values of qnet(all) and (N1)60 given in Columns 6 and 5, respectively, ofTable 6.1, the magnitudes of Se were calculated and are given in Column 8.

Column 9 of Table 6.1 gives the ratios of calculated to measured values of SeThese ratios vary from about 0.83 to 3.5. Thus, calculating the net allowable bearingcapacity with the use of Eq. (6.12) or (6.13) will yield safe and conservative values.

Stuart and Graham (1975) reported the case history of the 13-story AshbyInstitute building of Queens University, Belfast, Ireland, the construction ofwhich began in August 1960. The building was supported by a mat foundation54.9 m (180 ft) X 19.8 m (65 ft). Figure 6.7a shows a schematic cross section ofthe building. The nature of the subsoil, along with the field standard penetrationresistance values at the south end of the building, is shown in Figure 6.7b. Thebase of the mat was constructed about 6.1 m (20 ft) below the ground surface.

The average corrected standard penetration number (N1 )60 between the bottom of the mat and a depth of about B/2 below the mat is about 17. The engineersestimate that qnet(all) was about 161 kN/m2(3360 Ib/ft2

). From Eq. (6.12),

S ( ) __ 25qnet(all)mm 0~

e 11.98(Nd 6{1 + 0.33(~) ]

NC')CJ1

Table 6.1 Settlement of Mat Foundations on Sand and Gravel (based on Meyerhof, 1965)

Observed Calculatedqnet{all) maximum maximum

Case B Average kN/m2 settlement, settlement, calculated Se

No. Structure Reference m (ft) (N')60 (kiP/ft2) 5e mm (in.) Se mm (in.) observed Se

(1) (2)' (3) (4) (5) (6) (7) (8) (9)

1 T. Edison Rios and Silva (1948) 18.29 229.8 15.24 31.97Sao Paulo, Brazil (60) 15 (4.8) (0.6) (1.26) 2.1

2 Banco do Brazil Rios and Silva (1948); 22.86 239.4 27.94 27.75Sao Paulo, Brazil Vargas (1961) (75) 18 (5.0) (1.1) (1.09) 0.99

3 Iparanga Vargas (1948) 9.14 304.4 35.56 70.58Sao Paulo, Brazil (30) 9 (6.4) (1.4) (2.78) 1.99

4 C.B.I., Esplanda Vargas (1961) 14.63 383.0 27.94 36.33Sao Paulo, Brazil (48) 22 (8.0) (1.1) (1.43) 1.3

5 Riscala Vargas (1948) 3.96 229.8 12.7 23.98Sao Paulo, Brazil (13) 20 (4.8) (0.5) (0.94) 1.89

6 Thyssen Schultze (1962) 22.55 239.4 24.13 19.98Dusseldorf, Germany (74) 25 (5) (0.95) (0.79) 0.83

7 Ministry Schultze (1962) 15.85 220.2 20.32 22.98Dusseldorf, Germany (52) 20 (4.6) (0.8) (0.9) 1.13

8 Chimney Schultze (1962) 20.42 172.4 10.16 35.98Cologne, Germany (67) 10 (3.6) (0.4) (1.42) 3.54

,I

Borehole 4-South end

2.3 m (7.5 it)

Made ground

Medium red-brown silty sand

Loose brown silty fine sand

Firm brown silty clay with sandFirm brown sandy silt with sand layersMedium dense brown finesand with small stone pieces

Medium dense brown fine sand

Medium-compact sandy gravel

Medium dense gravel with fine tomedium sand

Figure 6.7 Ashby Institute Building of Queens University as reported by Stuart and Graham (1975): (a) cross section of building; (b) subsoil conditions at south end

N60---------

(b)

(a)

Compensated Foundation

Substituting the appropriate values into the preceding equation yields

(25)(161)Se = = 17.9 mm (0.7 cm)

(11.98)(17)[1 + 0.33(1~~) ]

The construction of the building was completed in February 1964. Figure 6.8shows the mean settlement of the mat at the south end from 1960 to 1972. In the latter year (eight years after completion of the building), the mean settlement wasabout 14 mm (0.55 in.). Thus, the estimated settlement of 17.9 mm (0.7 in.) wasabout 30% higher than that actually observed.

Figure 6.6 and Eq. (6.16) indicate that the net pressure increase in the soil under amat foundation can be reduced by increasing the depth Df of the mat. This approachis generally referred to as the compensated foundation design and is extremely use-


6.7 Compensated Foundation 267

Year

o 1960 1961~196219631964 1965 1966 1967 196X 19691970 1971 1972

0.2 5~

.5E0.4 10...E~ 0.6 15vU1

0.8 20

I.0 '-----'-_....I..-----'"------'-_....I..-----'_---'-_....I..----l_---...L.._...J...1_....J1L....----J 25

Figure 6.8 Mean settlement at the south end of the mat foundation, as reported by Stuartand Graham (1975)

". "'1 I.' .....

I I Figure 6.9 Compensated foundation

ful when structures are to be built on very soft clays. In this design, a deeper basement is made below the higher portion of the superstructure, so that the net pressureincrease in soil at any depth is relatively uniform. (See Figure 6.9.) From Eq. (6.16)and Figure 6.6, the net average applied pressure on soil is

Qq = - - yDfA

For no increase in the net pressure on soil below a mat foundation, q should bezero. Thus,

QDf =-Ay

(6.21)

This relation for Df is usually referred to as the depth of a fully compensatedfoundation.

The factor of safety against bearing capacity failure for partially compensatedfoundations (i.e., Df < Q/Ay) may be given as

FS = qnet(u) = qnet(u) ( )q Q 6.22

A - yDf


(6.23)

For saturated clays, the factor of safety against bearing capacity failure can thus beobtained by substituting Eg. (6.10) into Eg. (6.22):

6.7 Compensated Foundation 269

-~2-in"""·.T.-:-.--.,....-~ 30 m x 40 m.. ..,..'-'2m .

Sand·'y~15:7kN/m3

T Groundwater table

,--- -

Sand'Ysat = 19.1 kN/m3

- ~ _ Normally consolidated clay -~ - 6 m - -- -= 'Ysat = 18.6 kN/m3 ::.--=

=-- .. ~~---==:=-~~---==:=----=-~. Cc = 0.28; eo = 0.9~

··Sand

Consolidas~ttlelnellt under a


t1u:n == 89.3 kN/m2

z 21n - - -~-= 1.41 - (B/2) - (30/2)

Structural Design of Mat Foundations

The structural design of mat foundations can be carried out by two conventionalmethods: the conventional rigid method and the approximate flexible method.Finite-difference and finite-element methods can also be used, but this sectioncovers 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 bystep with reference to Figure 6.11:

6.8 Structural Design of Mat Foundations 271

1. Figure 6.11a shows mat dimensions of L x B and column loads of Qil Q2'

Q3' .... Calculate the total column load as

(6.24)

2. Determine the pressure on the soil, q, below the mat at points A, B, C, D, ... ,by using the equation

(6.25)

where A = BLIx = (1/12)BL3 = moment of inertia about the x-axisIy = (1/12) LB3 = moment of inertia about the y-axis

M x = moment of the column loads about the x-axis = Qey

My = moment of the column loads about the y-axis = Qex

y'

_....L.... ~X'

Y

Bl~Bl---t--Bl----l-Bl--lDI Bel T

Q9 : QIO II QIl : Q12I I BI I I 1

I I II II I I

----lr--------l---------1-----i 'rex-l I: I Te :I I II II I ey I

---+--.---~j • --t---Qs I Q6 i Q7 : Q8

I I II I II I II I II I II I I

----~--------i---------4-----I I II I II I II I II I II I II Q2 I Q3 II I I

1

L ]

II+-..-----B·-----.....I(a)

Figure 6.11 Conventional rigid mat foundation design

-----


I F

B1 • q.v (modified)

unit length

II-."-----B------+-I.. I(b)

L"

(6.27)

(6.28)

(6.29)

L'

bo = 2(L' + L")

L'

Edgeof mat

(c)

Be = x' --

x 2

Ley = y' -"2

L"bo = L' + L"

QIyi + Q2YZ + Q3Y3 + ...y' = -----=-'----=-------=--=-------=-=---=-----

Q

Edge ofmat

L"

bo = 2L' + L"

and

The load eccentricities, ex and ey, in the x and y directions can be determined by using (x', y') coordinates:

Qlx i + Q2XZ + Q3X3 + ...x' = (6.26)

Q

and

Similarly,

3. Compare the values of the soil pressures determined in Step 2 with the net allowable soil pressure to determine whether q ~ qall(net).

4. Divide the mat into several strips in the x and y directions. (See Figure 6.11). Letthe width of any strip be B1•

Edge ofmat

Figure 6.11 (continued)

(6.30)


5. Draw the shear, V, and the moment, M, diagrams for each individual strip (inthe x and y directions). For example, the average soil pressure of the bottomstrip in the x direction of Figure 6.11a is

q/ + qpqav = 2

where q/ and qF = soil pressures at points 1 and F, as determined from Step 2.

The total soil reaction is equal to qavBjB. Now obtain the total column loadon the strip as Q, + Q2 + Q3 + Q4' The sum of the column loads on the stripwill not equal qavBIB, because the shear between the adjacent strips has notbeen taken into account. For this reason, the soil reaction and the column loadsneed to be adjusted, or

Now, the modified average soil reaction becomes

(average load)

qav(modified) = qav B Bqav 1

and the column load modification factor is

(6.31)

(6.32)

(6.33)average load

F = ------=-----Ql + Q2 + Q3 + Q4

SO the modified column loads are FQl> FQ2' FQ3, and FQ4' This modifiedloading on the strip under consideration is shown in Figure 6.11b. The shear andthe moment diagram for this strip can now be drawn, and the procedure is repeated in the x and y directions for all strips.

6. Determine the effective depth d of the mat by checking for diagonal tensionshear near various columns. According to ACI Code 318-95 (Section 11.12.2.1c,American Concrete Institute, 1995), for the critical section,

(6.34)

where U = factored column load (MN), or (column load) X (load factor)4> = reduction factor = 0.85f~ = compressive strength of concrete at 28 days (MN/m2

)

The units of bo and d in Eq. (6.34) are in meters. In English units, Eq. (6.34)may be expressed as

(6.35)

where U is in lb, bo and d are in in., and f~ is in Ib/in2

The expression for bo in terms of d, which depends on the location of thecolumn with respect to the plan of the mat, can be obtained from Figure 6.11c.


7. From the moment diagrams of all strips in one direction (x or y), obtain the maximum positive and negative moments per unit width (i.e., M' = Mj B1).

8. Determine the areas of steel per unit width for positive and negative reinforcement in the x and y directions. We have

M" ~ (M') (load factor) ~ </>AJy( d - ~)

and

(6.36)

(6.37)

where As = area of steel per unit widthfy = yield stress of reinforcement in tension

M u = factored moment<p = 0.9 = reduction factor

Examples 6.5 and 6.6 illustrate the use of the conventional rigid method of matfoundation design.

Approximate Flexible Method

In the conventional rigid method of design, the mat is assumed to be infinitelyrigid. Also, the soil pressure is distributed in a straight line, and the centroid of thesoil pressure is coincident with the line of action of the resultant column loads.(See Figure 6.12a.) In the approximate flexible method of design, the soil is assumed to be equivalent to an infinite number of elastic springs, as shown in Figure 6.12b. This assumption is sometimes referred to as the Winkler foundation. Theelastic constant of these assumed springs is referred to as the coefficient of subgrade reaction, k.

To understand the fundamental concepts behind flexible foundation design,consider a beam of width B j having infinite length, as shown in Figure 6.12c. Thebeam is subjected to a single concentrated load Q. From the fundamentals of mechanics of materials,

(6.38)

where M = moment at any sectionEF = modulus of elasticity of foundation materialIF = moment of inertia of the cross section of the beam = az)B j h3 (see

Figure 6.12c).

However,

dMdx = shear force = V

d


~Q

... ,', ... ',

..... ','. ...

Resultant ofsoil pressure

: ".:.., " :.

A

Sectionat A - A A

Point load

z (c)

-----..x

Figure 6.12 (a) Principles of design by conventional rigid method; (b) principles of approximate flexible method; (c) derivation of Eq. (6.42) for beams on elastic foundation

and

dV '1'dx = q = SOl reactIon

Hence,

d 2Mdx2 = q

Combining Eqs. (6.38) and (6.39) yields

d4zEFIFdx4 = q

--._-----------

(6.39)

(6.40)


However, the soil reaction is

q = -zk'

where z = deflectionk' = kB]k = coefficient of subgrade reaction (kN/m3 or Ib/in3)

So

Solving Eq. (6.41) yields

z = e-ax(A' cos{3x + A" sin{3x)

where A' and A" are constants and

(6.41 )

(6.42)

(6.43){3 - 4[B;k\/4E;i;

The unit of the term {3, as defined by the preceding equation, is (length) -1.

This parameter is very important in determining whether a mat foundation shouldbe designed by the conventional rigid method or the approximate flexible method.According to the American Concrete Institute Committee 336 (1988), mats shouldbe designed by the conventional rigid method if the spacing of columns in a strip isless than 1.75/{3. If the spacing of columns is larger than 1.75/{3, the approximateflexible method may be used.

To perform the analysis for the structural design of a flexible mat, one mustknow the principles involved in evaluating the coefficient of subgrade reaction, k.Before proceeding with the discussion of the approximate flexible design method,let us discuss this coefficient in more detail.

If a foundation of width B (see Figure 6.13) is subjected to a load per unit area ofq, it will undergo a settlement d. The coefficient of subgrade modulus can be defined as

(6.44)

The unit of k is kN/m3 (or Ib/in3). The value of the coefficient of subgrade reaction

is not a constant for a given soil, but rather depends on several factors, such as thelength L and width B of the foundation and also the depth of embedment of thefoundation. A comprehensive study by Terzaghi (1955) of the parameters affectingthe coefficient of subgrade reaction indicated that the value of the coefficient decreases with the width of the foundation. In the field, load tests can be carried out bymeans of square plates measuring 0.3 m x 0.3 m (1 ft x 1 ft), and values of k can


\\'4-..-----B------~\

q

!.'\f~:;,\~_____________ ~ j<>y}.) Figure 6.13 Definition of coefficient of subgrade reaction, k

be calculated. The value of k can be related to large foundations measuring B x Bin the following ways:

Foundations on Sandy Soils For foundations on sandy soils,

(6.45)

where ko.3 and k = coefficients of subgrade reaction of foundations measuringO.3m X O.3mandB(m) X B(m),respectively(unitiskN/m3)

In English units, Eq. (6.45) may be expressed as

(6.46)

where k1 and k = coefficients of subgrade reaction of foundations measuring1 ft X 1 ft and B (ft) X B (ft), respectively (unit is Ib/in3

)

Foundations on Clays For foundations on clays,

(6.47)

The definition of k in Eq. (6.47) is the same as in Eq. (6.45).In English units,

(6.48)


The definitions of k and k] are the same as in Eq. (6.46).For rectangular foundations having dimensions of B X L (for similar soil and q),

""•II

(6.49)

~B4 E

k' = Bk = 0.65 12 _s- s 2Eplp 1 - J.ts

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 Committee336 (1988), is described step by step. The use of the design procedure, which is basedprimarily on the theory of plates, allows the effects (i.e., moment, shear, and deflec-

where k = coefficient of subgrade modulus of the rectangular foundation(L X B)

k(sxs) = coefficient of subgrade modulus of a square foundation having dimension of B X B

Equation (6.49) indicates that the value of k for a very long foundation with a widthB is approximately 0.67k(sxs)'

The modulus of elasticity of granular soils increases with depth. Because thesettlement of a foundation depends on the modulus of elasticity, the value of k increases with the depth of the foundation.

Table 6.2 provides typical ranges of values for the coefficient of subgrade reaction, k 1, for sandy and clayey soils.

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

(6.50)

(6.51)

or

1EB' Ek = 0.65 12 _s- s 2

Eplp B(l - J.ts)

where Es = modulus of elasticity of soilB = foundation width

Ep = modulus of elasticity of foundation materialI p = moment of inertia of the cross section of the foundationJ.ts = Poisson's ratio of soil

For most practical purposes, Eq. (6.50) can be approximated as

Esk = -----'=----B(l - J.tD


Table 6.2 Typical Subgrade Reaction Values, k j

k,

Soil type MN/m3 Ib/in.3

Dry or moist sand:Loose 8-25 30-90Medium 25-125 90-450Dense 125-375 450-1350

Saturated sand:Loose 10-15 35-55Medium 35-40 125-145Dense 130-150 475-550

Clay:Stiff 10-25 40-90Very stiff 25-50 90-185Hard >50 >185

tion) of a concentrated column load in the area surrounding it to be evaluated. Ifthe zones of influence of two or more columns overlap, superposition can be employed to obtain the net moment, shear, and deflection at any point. The method isas follows:

1. Assume a thickness h for the mat, according to Step 6 of the conventional rigidmethod. (Note: h is the total thickness of the mat.)

2. Determine the flexural ridigity R of the mat as given by the formula

(6.52)

where EF = modulus of elasticity of foundation materialfJ-F = Poisson's ratio of foundation material

3. Determine the radius of effective stiffness-that is,

L'= if (6.53)

where k = 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 (see Figure 6.14a). The formulas to use are

....;;1.....-- _

(1 - 7.)AzjL'

(6.54)


6

5

III

II. \

{~/\Az /

// \h/ r--.... A4

/ V'~// Y ~I __ 1-_-

~----\ -....:..t'- __

-I--

0.40.30.20.1o

4

2

1

o-0.4 -0.3 -0.2 -0.1

rL' 3

a

(a)

y

F----'-----.x

(b)

Figure 6.14 Approximate flexible method of mat design

and

. Qf (1 - JLp) A 2lM, ~ langentlalmomenl ~ -4('FAl + {, J (6.55)

where r = radial distance from the column loadQ = column load

AI, A z = functions of r/L'

The variations of Al and A zwith r/L' are shown in Figure 6.14b. (For details seeHetenyi, 1946.)

In the Cartesian coordinate system (see Figure 6.14a),

(6.56)

and

(6.57)

(6.58)

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

QV = 4L,A3

The variation of A 3 with r/L' is shown in Figure 6.14b.6. If the edge of the mat is located in the zone of influence of a column, determine

the moment and shear along the edge. (Assume that the mat is continuous.)

Moment and shear opposite in sign to those determined are applied at theedges to satisfy the known conditions.

-I.J

6.8 Structural Design ofMat Foundations 281

7. The deflection at any point is given by

(6.59)

The variation of A 4 is presented in Figure 6.14.

Example 6.5

Th~:plan of ama.tX.<?undatio~with column loads is shown in Figure 6.15. UseEq:(6.25) to c late the soil pres~ures at points A, B, c: D, E, F, G, H, 1,1, K, L, M,

The s' at is76 ftX96 ft, are 24 in. X 24 in. in section,

2 £t 24 ft 24 £t 24 £t 2ft

-1 1..- ~I"- -1..- ~1 ~

A B C D E F C ~I I I 3 £t• I • • I • .....

DL = 100 kip DL = 180 kip DL = 190 kip DL= 110 kipLL = 60 kip LL'" 120 kip LL = 120 kip I:,L = 70 kip

I II II I 30 £tI II II II II I• I • • I • - e-

DL = 180 kip DL= 360 kip DL= 400 kip DL= 200 kipLL = 120 kip LL = 200 kip r.IU:.250kiP

LL = 120 kipI II ex I 30 ftI II x IeyI

• TI

I II I

• I • • I • - f-

DL"" 190 kip DL=400 kip DL = 440 kip DL= 200 kipLL = 130 kip LL = 240 kip LL= 300 kip LL = 120 kip

I I

t y' I II I 30 £t

I I I

I I II I

IDL = 120 kip DL = 180 kip I DL = 180 kip DL= 120 kipt:.= 70 kip LL = 120 kip I LL = 120 kip LL= 70 kip x I

-+---------+---.---t-- ....3 £tI

N M L K J I H TNote: DL = dead load

LL = live load


a~g~~!1(net) = 1.5 kip/fe. Verify that the soil pressures are les$ than the net allow..;ab\~l?earing capacity.

SolutionFrom Figure 6.15,

C('jJl1.i:mmdead load (DL) = 100 + 180 + 190 + 110 + 180 + 360 + 400 + 200

+ 190 + 400 + 440 + 200 + 120 +180 + 180 + 120

= 3550 kip -

+ 120+ 70 + 120

240 +300 + 120


or

y' =44.273 ft

and

ey = 44.273 - ~ = - 0.727 ft

The moments caused by eccentricity are

Mx == Qey == (8761) (0.727) == 6369 kip-ft

________________ --J


24 + d (in.)

J

d2

II

36 + ~ (in.)

24 in.

T

I

12 in. I

24 in.

,------------------,IIIIII


we

</Jl!c = </J(4)Vj;bod = (0.85)(4) (V3000) (96 + 2d)d

so

d2

24 in. 24 + d (in.)

d24 in. 2

24 + d (in.)

(0.85) (4) (v3000) (96 + 2d)d1000 . .~ 487

(96 + 2d)d ~ 2615.1

d:::::: 19.4 in.

L _

.~


Part c: Reinforcement Requirements,Figure 6.18 gives tHe design of strip BCDKLM and shows the load diagram, in which

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

,,Q2 == (1.4) (360) + (1.7) (200) == 844 kip

"Q3 == (1.4) (400) + (1.7) (240) :::; 968 kip

Q4 == (1.4) (180) + (1.7) (120) == 456 kip


(a) Load diagram

Q, =456 kIp Qz =844 kip QJ = 968 kip Q4 =456 kIp

E~30_f(-t_30_ft _1_30_ft_5

~~ ~EC"'l 0. . 0.~~ r-._r--- 0l~

0lr-:~01

~.II

(d) Location of main reinforcement

Top bars Top bars Top barsHI-...-----~I~k-...-----~Hk-...----~I_1Bottom Bottom Bottom Bottombars bars bars bars

452.5 kip358.5 kip

462.2 kip

2281.1 kip- ft

306.6 kip

(b) Shear diagram

(c) Moment diagram

388.3 kip

2447.8 kip-ft

366.4 kip

Problems 289

I~ b ~ I r-0.85 J:-.J

I T a/I T(31 = a

1 C I

hd Neutral axis

t---..... T

(:1) (b)

RectaJngttlar section in bending; (a) section, (b) assumed stress U1i>U1"t.j.'j'Vll

6.1 Determine the net ultimate bearing capacity of a mat foundation measuring45 ft X 30 ft on a saturated clay with ell = 1950 Ib/ft2

, 4> = 0, and DJ = 6.5 ft.Use Eq. (6.10).

6.2 Repeat Problem 6.1, but now let Cll = 120 kN/m2, 4> = 0, B = 8 m, L = 18 m,

and DJ = 3 m.6.3 What will be the net allowable bearing capacity of a mat foundation with di

mensions of 15 m X 10 m constructed over a sand deposit? Let Df = 2 m, allowable settlement = 30 mm, and corrected average penetration number(N1)60 = 10. Use Eq. (6.12).

6.4 Repeat Problem 6.3 for an allowable settlement of 50 mm.6.5 Consider a mat foundation with dimensions of 20 m X 13 m. The dead and

live load on the mat is 42 MN. The mat is to be placed on a clay withCll = 40 kN/m2

• The unit weight of the clay is 17.5 kN/m3. Find the depth DJof the mat for a fully compensated foundation.

6.6 What will be the depth DJ of the mat considered in Problem 6.5 for FS = 3against bearing capacity failure?


Size of mat = B X L

6.7 Consider the mat foundation shown in Figure P6.7. LetQ = 25 MN, Df = 1.5 m, Xl = 2 m, X2 = 3 m, and X3 = 4 m. The clay is normally consolidated. Estimate the consolidation settlement under the centerof the mat.

6.8 Estimate the consolidation settlement under the corner of the mat foundationdescribed in Problem 6.7.

6.9 Redo Problem 6.7, assuming that the preconsolidation pressure of the clay is120 kN/m2 and the swelling index is about 1/4Cc-

6.10 For the mat shown in Figure P6.1O, Ql = Q3 = 40 tons, Q4 = Qs = Q6 =60 tons, Q2 = Q9 = 45 tons, and Q7 = Qs = 50 tons. All columns are 20 in.X 20 in. in cross section. Use the procedure outlined in Section 6.8 todetermine the pressure on the soil at points A, B, C, D, E, F, G, and H.

6.11 The plan of a mat foundation with column loads is shown in Figure P6.11. Calculate the soil pressure at points A, B, C, DJ E, and F. (Note: All column sections are planned to be 0.5 m X 0.5 m.)

6.12 Divide the mat shown in Figure P6.11 into three strips, such as AGHF(Bl = 4.25 m), GllH (Bl = 8 m), and lCDl (Bl = 4.25 m). Use the resultsof Problem 6.11, and determine the reinforcement requirements in the y direction. Here, f~ = 20.7 MN/m2

, fy = 413.7 MN/m2, and the load factor is 1.7.

6.13 From the plate load test on a plate of dimensions 1 ft x 1 ft in the field, thecoefficient of subgrade reaction of a sandy soil is determined to be 80 Ib/in3•

What will be the value of the coefficient of subgrade reaction on the same soilfor a foundation with dimensions of 30 ft x 30 ft?

6.14 The subgrade reaction of a sandy soil obtained from the plate load test on aplate of dimensions 1 m x 0.7 m is 18 kN/m3

• What will be the value of k onthe same soil for a foundation measuring 5 m X 3.5 m?

i

. .. .

.~ Gt9uridw~~er table

FigureP6.7

y'

ABC----- -----

T24 ft

H • 0

Q4 Q5 Q6

24 ft

Q7 Q8 Q9 1x'

G F Er---16 ft-r---16 ft ~I Figure P6. 10

Figure P6. 11

7m--.. x

10.25m

T7m

450kN

y

•I

500kN

•1500 kN

400kN

1500 kN

I400 kN I 500 kN

I I

: II I

: II I

: II I

1500 kN : .1500 kNI I

4.25 I ----.LJ 8m-"·"",,,I"I---~ i m... m

L --j _IIIIII

: 1200 kNIIIIIIII

i 350 kNI

y'

I0.25m:----r---..... x'

F

-I I- 8 m-_·I~"--8m·-__I l-0.25m 0.25 m

""'r- _


fences

American Concrete Institute (1995). ACI Standard Building Code Requirements for Reinforced Concrete, ACI 318-95, Farmington Hills, ML

American Concrete Institute Committee 336 (1988). "Suggested Design Procedures for Combined Footings and Mats," Journal of the American Concrete Institute, Vol. 63, No. 10,pp.1041-1077.

Hetenyi, M. (1946). Beams ofElastic Foundations, University of Michigan Press, Ann Arbor, MI.Meyerhof, G. G. (1965). "Shallow Foundations," Journal of the Soil Mechanics and Founda

tions Division, American Society of Civil Engineers, Vol. 91, No. SM2, pp. 21-31.Rios, L., and Silva, F. P. (1948). "Foundations in Downtown Sao Paulo (Brazil)," Proceedings,

Second International Conference on Soil Mechanics and Foundation Engineering, Rotterdam, Vol. 4, p. 69.

Schultze, E. (1962). "Probleme bei der Auswertung von Setzungsmessungen," Proceedings,Baugrundtagung, Essen, Germany, p. 343.

Stuart,1. G., and Graham,1. (1975). "Settlement Performance of a Raft Foundation on Sand,"in Settlement ofStructures, Halsted Press, New York, pp. 62-67.

Terzaghi, K. (1955). "Evaluation of the Coefficient of Subgrade Reactions," Geotechnique, Institute of Engineers, London, Vol. 5, No.4, pp.197-226.

Vargas, M. (1948). "Building Settlement Observations in Sao Paulo," Proceedings Second International Conference on Soil Mechanics and Foundation Engineering, Rotterdam,Vol. 4, p. 13.

Vargas, M. (1961). "Foundations of Tall Buildings on Sand in Sao Paulo (Brazil)," Proceedings, Fifth International Conference on Soil Mechanics and Foundation Engineering,Paris, Vol. 1, p. 841.

Vesic, A. S. (1961). "Bending of Beams Resting on Isotropic Solid," Journal of the EngineeringMechanics Division, American Society of Civil Engineers, Vol. 87, No. EM2, pp. 35-53.

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