12-ELAST

Elastic Theory for Use in Soils -- GEOTECHNICAL ENGINEERING-1997 -- Prof. G.P. Raymond© 85

Figure 1. Defined stressdirections in cylindrical

coordinates for a verticalpoint load on the surface of

a semi-infinite solid.

ELASTIC THEORY FOR USE IN SOILS

SYMBOLS

Notation Dimensional Analysis

a slope length of loaded area Lb half width of loaded rectangular area LB full width of loaded rectangular area LI influence factor or coefficient -K coefficient -L full length of loaded rectangular area Lm B/z -n L/z -N number of areas covered on influence chart -q intensity of loading per unit area M L-1 T-2

Q load M L T-2

r radius on horizontal plane LR distance from surface origin Lx distance in x direction from origin Ly distance in y direction from origin Lz distance in z direction from origin Lσ normal stress M L-1 T-2

τ shearing stress M L-1 T-2

ν Poisson's ratio -θ,θo,θ1,θ2 angle Angle

Subscripts where not defined above

r in the radial directionx in the x directiony in the y directionz in the z directionv in the vertical directiont in the tangential directionl major principal value3 minor principal valuerz in the radial direction on a plane perpendicular to z directionxz in the radial direction on a plane perpendicular to z direction

1. INTRODUCTION In many problems, particularly those related to settlements, it is

necessary to determine the stresses or increases in stresses in a soil mass.It has generally been found that if the factor of safety of the soil mass withrespect to ultimate bearing capacity failure exceeds a value of three (andin many cases less) the increases in stress in the soil are approximatelyequal to those computed on the assumption that the soil is perfectly linearelastic. One of the main reasons for this similarity is that according tolinear elastic theory, constant ratios always exist between stresses andstrains. For the theory to be applicable, the real requirement is not that thematerial necessarily be elastic, but that there must be constant ratiosbetween stresses and the corresponding strains for the applied loading.Therefore, in non-elastic soil masses the elastic theory may be applied toany case in which stresses and strains may reasonably be assumed toadhere to constant ratios. Some types of loading on soils cause strainswhich are everywhere approximately proportional to the stresses; underother loading conditions such as those in which failure in shear isimminent, the strains are anything but proportional to the stresses. Clearlyelastic theory would be applicable to the former case only.

Extensive use has been made ofelastic theory applied to the case of anhomogeneous isotropic elastic halfspace assuming that the elasticparameters are uniform throughout thehalf-space. In nature the rigidity ofsoils generally increase with depth asa consequence of the increasingeffective overburden pressure.Gibson (1967) has shown that forincompressible soils (ν = 0.5) theelastic half space stress changes,caused by a surface vertical load, areidentical irrespective of whether theelastic modulus is uniformthroughout, or increases linearly withdepth from a surface value of zero. Itis therefore reasonable to assume that,in relatively homogeneous isotropicsoils where the pseudo-elastic modulus increases approximately linearwith depth but is not necessarily zero at the surface, the stress increases,due to a vertical surface load, will approximate those given by theassumption of uniformity of elastic parameters throughout. Note that theassumption of stress similarity does not imply strain similarity, indeed thedeformations obtained in Gibson's comparison are greatly different.

2. BOUSSINESQ EQUATIONSThe equations expressing the stress components caused by a

perpendicular, point, surface force, at points within an elastic, isotropic,homogeneous mass which extends infinitely in all directions from a levelsurface, are attributed to Boussinesq (1885). Other point load solutionsare given in the Appendix. The equations given for the Boussinesqproblem, based on the coordinates shown in Figure 1 are

σz '3 Q z 3

2 π (r 2 % z 2)5/2'

3 Q2 π z 2

(cos5 θ) (1)

σr 'Q

2 π3 r 2 z

(r 2 % z 2)5/2&

1 & 2 ν

r 2 % z 2 % z (r 2 % z 2)

'Q

2 π z 23 sin2θ cos3θ &

(1 &2ν) cos2θ1 % cosθ

(2)

σt ' &Q (1&2ν)

2 πz

(r 2%z 2)3/2&

1

r 2%z 2% z (r 2%z 2)

' &Q (1&2ν)

2πz 2cos3θ &

cos2θ(1 % cosθ)

(3)

τrz '3 Q r z 2

2 π (r 2 % z 2)5/2'

3 Q sinθ cos4θ2 π z 2 (4)

where σz is the vertical increment of stressσr is the radial increment of stressσt is the tangential increment of stressτrz is the shear increment of stressQ is the applied increment of surface loadz is the depth of the required stressr is the radial distance of the required stress from the point ofapplication, and thus:

θ ' tan&1 rz (5)

Examination of equation (1) shows that the increase in vertical stress


Figure 2. Strip loading on surface of semi-infinite solid.

Figure 3. Contours of equal vertical stress below (a)uniform strip loading; (b) circular loading; (c) square

loading.

is independent of the Poisson's ratio ν. Unfortunately the lateral increasesin stress require a knowledge of Poisson's ratio. Where the lateral increasein stress are required they are commonly, but not always, obtained usingν = 0.5. That is the soil is assumed incompressible. This is a reasonableassumption for dense granular soils and for the initial loading of saturatedclays. Even so it is not uncommon to use the same stresses, whererequired, during and after consolidation due to the difficulty of estimatingan effective stress Poisson's ratio.

3. STRESSES BELOW UNIFORMLY LOADED AREASA load Q applied to a given surface area A can be divided into an

infinite number of discrete point loads (Q dA/A). If the foundation soilis assumed to behave similar to an elastic solid then the stresses producedby the total load are equal to the sum of the stresses produced by the pointloads. Thus the resultant state of stress for the applied load can be foundby integration of a number of point loads.

(a) Line Load: Integration of the Boussinesq equations for the stresses atan arbitrary point within an elastic solid result, for the stresses within theplane perpendicular to the line load, in the equations

σz '2 QR z

3

π (x 2 % z 2)2'

2 QR

π zcos4θ (6)

σx '2QR z x

2

π(x 2 % z 2)2'

2 QR

π zcos2θ sin2θ (7)

τxy '2QR z

2 x

π(x 2 % z 2)2'

2 QR

π zcos3θ sinθ (8)

where QR = the load per unit lengthθ ' tan&1 x/z (9)

x = the horizontal distance of the point from the vertical plane belowthe load

It should be noted that the equations are independent of Poisson's ratio.The stress parallel to the line load would, of course, be dependent onPoisson's ratioσy ' ν (σz % σx) (10)

(b) Strip Loading: Similarly for the strip loading with a uniform loadingper unit area as shown in Figure 2.

σz 'qπ

sinθ cosθ % θθ2θ1 (11)

σx 'qπ

&sinθ cosθ % θθ2θ1 (12)

τxz 'qπ

sin2θθ2θ1 (13)

where

θ1 ' tan&1 x & bz (14)

θ2 ' tan&1 x % bz (15)

b = half width of the loaded stripx = horizontal distance from a vertical plane along the centre line of theloaded strip

For this case the principal stresses may be obtained as

σ1 'qπ

(θo % sinθo) (16)

σ2 'qπ

(θo & sinθo) (17)

whereθo ' θ2 & θ1 (18)

Hence for every point on a circle through the edges of the loaded strip andthe given point the principal stresses have the same intensity and thedirection of those principal stresses passes through the intersectionbetween the circle and the plane of symmetry of the loaded strip.

(c) Circular and Rectangular Loaded Areas: The integration of theBousinesq Equation for circular and rectangular areas has been solved byLove (1929). The solutions are given in the Appendix. Unfortunately thesolutions involve comprehensive equations which do not lend themselvesto easy calculation.They can be easily solved on a digital computer and typical solutions forthe vertical stress, shown in graphical form, are given in Figure 3.Examination of these figures indicates that the curves are in the form ofbulbs. These bulbs are commonly called "bulbs of pressure".


4. INFLUENCE CHARTSIn order to calculate the stresses at a point below irregular shaped

uniformly loaded area Newmark (1942) developed what are known asinfluence charts. These charts are similar to that shown in Figure 4.Figure 4 may be used for the calculation of the vertical stress as follows:

(1) the uniformed loaded area is first drawn to scale such that thedistance OQ equals the depth at which the stress is required,

(2) the drawn to scale area is then located so that the centre of the chartis directly above the location of the point on the drawn scaled areaof the required vertical stress,

(3) the vertical stress is then calculated from the product of threequantities

(a) the chart influence values (I) times(b) the number of influence blocks (N) covered by the area times(c) the value of the uniform surface pressure (q) below the loaded

area.i.e. ∆σv ' I.N.q (19)

Other charts have been prepared for vertical stresses and variousshaped areas. Typical of these charts for a rectangular loaded area isFigure 5 and for an embankment loadings is Figure 6.

5. ASSUMED CONTACT PRESSURE FOR THE CALCULATIONOF FOUNDATION STRESSES

The term contact pressure indicates the normal stress at the surface ofcontact between a footing and the supporting earth. On an elastic semiinfinite solid with uniform parameters a uniformly loaded area producesa bowl-shape settlement. On the other hand, if the elastic modulusincreases linearly with depth from zero at the surface and Poisson's ratioequals a half, the settlement would be uniform (Gibson 1967). Such aresponse of uniform settlement or settlement proportional to the contactpressure is known as a "Winkler" foundation and may be modelled as anelastic spring foundation.

The distribution of contact pressure on a rigid footing may be expectedto depend very much on the uniformity of the elastic constants within adepth of one or two footing widths below the contact surface and therelative magnitude of the load increase in relation to the overburdenstresses. Large footings on heavily overconsolidated clays and smallmodel footings on dense sands tend to behave as footings on semi-infiniteelastic solids with constant elastic parameters. Thus the contact pressuresbelow rigid footings are greater at the footing perimeters than at theircentres. Large footings on sands perform much closer to thoserepresented by a "Winkler" foundation. Foundations on soft claysgenerally have low factors of safety and the soft clays are subject tononlinear behaviour. Linear elasticity is generally inappropriate.

Fortunately as the point in the foundation soil where the requiredstresses are to be estimated moves further from the contact area the contactstress distribution has less and less effect on the estimated stresses. Thusfor the calculation of foundation soil stresses it is normal practice toassume a uniform distribution of contact stress. This assumption,however, is not necessarily valid when calculating the foundation designmoments as discussed elsewhere.

6. REFERENCES

Boussinesq, J. (1885) "Applications Des Potentiels a l'etude de l'equilibreet du Mouvement Des Solids Elastiques", Gather-Villars.

Fadum, R.E. (1948) "Influence Values for Estimating Stresses in ElasticFoundations", Proceedings of the Second International Conference on SoilMechanics and Foundation Engineering, Rotterdam, Volume 3, pp. 77-84.

(TA710.I6t).

Gibson, R.E. (1967) "Some Results Concerning Displacements andStresses in a Non-homogeneous Elastic Half-Space", Geotechnique,Volume 17, No. 1, pp. 58-67. (TA1.G3).

Love, A.E.H. (1929) "The Stress Produced in a Semi Infinite Solid byPressure on Part of the Boundary", Philosophical Transactions of theRoyal Society of London, Series A, Volume 228, pp. 377-420. (Q1.L84).

Newmark, N.M. (1942) "Influence Charts for Computation of Stresses inElastic Foundations", University of Illinois, Engineering ExperimentalStation, Bulletin Series no. 338, p. 28.

Osterberg, J.O. (1957) "Influence Values for Vertical Stresses in a SemiInfinite Mass Due to an Embankment Loading", Proceedings of the FourthInternational Conference on Soil Mechanics and Foundation Engineering,London, Volume 1, pp. 393-394. (TA710.I6t).


Figure 4. Influence chart for vertical stress under any shaped uniform surface load. (e.g. Newmark, 1942)


Figure 5. Vertical stress under corner of a rectangular area carrying a uniform pressure (e.g. Fadum, 1948).


Figure 6. Vertical stress below centre line of half an embankment (e.g. Osterberg, 1957).


APPENDIX

KELVIN SOLUTION:- Point load within an infinite elastic mass.Solutions given are for a vertical point load Q acting downwards in the

z-direction, and at the origin of the axis (z-axis positive downwards)within an infinite elastic mass. These are as follows (symbols as forBoussinesq solution):

σz 'Q

8π(1 & ν)(1 &2ν) z

(z 2 % r 2)3/2%

3z 3

(z 2 % r 2)5/2(20)

σr ' &Q

8π (1 & ν)3r 2 z

(z 2 % r 2)5/2&

(1 & 2ν) z(z 2 % r 2)3/2 (21)

σt 'Q

8π(1 & ν)(1 & 2ν) z(z 2 % r 2)3/2

(22)

τrz ' &Q

8π(1 & ν)(1 & 2ν) r(z 2 % r 2)5/2

%3r z 2

(z 2%r 2)5/2 (23)

CERRUTI SOLUTION:- Point load on surface in x-direction.Solutions given are for a point load Q, acting at the origin and in the x-

direction, on and parallel to the surface of a semi-infinite (i.e, half space)elastic mass. These are as follows (symbols and axis as for Boussinesqsolution):

σx 'Q x

2πR 3

3x 2

R 2&

(1 & 2ν)(R % z)2

R 2 & y 2 &2R y 2

R % z(24)

σy 'Q x

2πR 3

3 y 2

R 2&

(1 & 2ν)(R % z)2

3R 2 & x 2 &2R x 2

R % z(25)

σz '3Q x z 2

2 π R 5 (26)

τxy 'Q y

2πR 3

3x 3

R 2%

(1 & 2ν)(R % z)2

& R 2 % x 2 %2R x 2

R % z(27)

τyz '3Q x y z

2π R 5 (28)

τzx '3Q x 2 z2π R 5 (29)

where R ' x 2 % y 2 % z 2 (30)

MINDLIN PROBLEM 1 SOLUTION:- Vertical point load 'c' belowsurface.

Solutions given are for a vertical point load Q acting downwards adistance 'c' below the surface of a semi-infinite (i.e, half space) elasticmass. The origin of the axis is a distance 'c' above the surface and directlyabove the point load. These are as follows (symbols and axis as forBoussinesq solution; note different origin):

σr '& Q

8π(1&ν)[

3z13

R15&

2(1 % ν) z1R1

3

%4(1 & ν) (1 & 2ν)

R(R % z )%

2(7ν & 5)z % 12(1 & ν) cR 3

%3(3 & 4ν)z 3 & 6(7 & 2ν)c z 2 % 24c 2 z

R 5

%30c z 3 (z & c )

R 7]

(31)

σt '& Q

8π(1 & ν)[

(1 & 2ν) z1R1

3

&4(1 & ν) (1 & 2ν)

R (R % z )%

(1 & 2ν) (3 & 4ν) & 6(1 & 2ν)cR 3

%6(1&2ν)c z 2 & 6c 2 z

R 5]

(32)

σz 'Q

8π(1 & ν)[

3 z13

R15

%(1 & 2ν) z1

R13

&(1 & 2ν) (z & 2c)

R 3

%3(3 & 4ν)z 3 & 12(2 & ν) c z 2 % 18c 2 z

R 5

%30c z 3 (z & c )

R 7]

(33)

τrz 'Q r

8π(1 & ν)[

3 z13

R15

%1 & 2νR1

3&

1 & 2νR 3

%3(3 & 4ν)z 2 & 6(3 & 2ν)c z % 6c 2

R 5

%30c z 3 (z & c )

R 7]

(34)

z1 ' z & 2c (35)

R1 ' r 2 % z 21 (36)

R ' x 2 % y 2 % z 2 (37)

MINDLIN PROBLEM 2 SOLUTION:- Horizontal point load 'c'below surface.

Solutions given are for a horizontal point load Q acting in the x-direction a distance 'c' below the surface of a semi-infinite (i.e, half space)elastic mass. The origin of the axis is a distance 'c' above the surface anddirectly above the point load. These are as follows (symbols and axis asfor the Boussinesq solution; note different origin):

σx '& Q x

8π(1 & ν)[ &

3 x 2

R15

&(1 & 2ν)R1

3%

1 & 2νR 3

%3x 2 & 6(3 & 2ν)c z % 18 c 2

R 5

%30c x 2 (z & c )

R 7]

(38)

σy '& Q x

8π(1 & ν)[ &

3 y 2

R15

%1 & 2νR1

3%

1 & 2νR 3

%3y 2 & 6(1 & 2ν)c z % 6 c 2

R 5

%30c y 2 (z & c )

R 7]

(39)

σz '& Q x

8π(1 & ν)[ &

3 z 21

R15

%1 & 2νR1

3%

1 & 2νR 3

%3z 2 & 6(1 & 2ν)c z % 6 c 2

R 5

%30c z 2 (z & c )

R 7]

(40)


τxy '& Q x

8π(1 & ν)[ &

3 x2

R15

&1 & 2νR1

3%

1 & 2νR 3

%3x 2 & 6c (z & c)

R 5

%30c x 2 (z & c )

R 7]

(41)

τzy '& Q x y8π(1 & ν)

[ &3 z1R1

5

%3z % 6(1 & 2ν)c

R 5%

30c z (z & c )R 7

] (42)

τzx '& Q

8π(1 & ν)[ &

3 x 2 z1R1

5&

(1 & 2ν)z1R 3

1

%(1 & 2ν)z1

R 3%

3x 2 zR 5

%6(1 & 2ν)c x 2 & 6c z (z & c)

R 5

%30c x 2 z (z & c )

R 7]

(43)

wherez1 ' z & 2c (44)

R1 ' r 2 % z 21 (45)

R ' x 2 % y 2 % z 2 (46)

RECTANGULAR LOAD: Uniform load on footing 2a by 2b in size.Footing has 2a-dimension in x-direction and 2b-dimension in y-

direction. Origin is at centre of surface loaded area on a semi-infiniteelastic solid. Z-axis is downwards

σx 'p

2π2νVz & (1 & 2ν)Xxx & z Vxx (47)

σy 'p

2π2ν Vz & (1 & 2ν)Xyy & z Vyy (48)

σz 'p

2πVz & z Vzz (49)

τyz ' &p

2πz Vyz (50)

τzx ' &p

2πz Vzx (51)

τxy ' &p

2π(1 & 2ν) Xxy % z Vxy (52)

Xxx ' tan&1 b & ya & x

% tan&1 b % ya & x

& tan&1 z (b & y)A (a & x)

& tan&1 z (b % y)D (a & x)

% tan&1 b & ya % x

% tan&1 b % ya % x

& tan&1 z (b & y)B (a % x)

& tan&1 z (b % y)C (a % x)

(53)

Xyy ' tan&1 a & xb & y

% tan&1 a % xb & y

& tan&1 z (a & x)A (b & y)

& tan&1 z (a % x)B (b & y)

% tan&1 a & xb % y

% tan&1 a % xb % y

& tan&1 z (a & x)D (b % y)

& tan&1 z (a % x)C (b % y)

(54)

Xxy ' loge(z%A) (z%C)(z%B) (z%D) (55)

Vxx ' &a & x

(a & x)2 % z 2

b & yA

%b % yD

&a % x

(a % x)2 % z 2

b & yB

%b % yC

(56)

Vyy ' &b & y

(b & y)2 % z 2

a & xA

%a % xB

&b % y

(b % y)2 % z 2

a & xD

%a % xC

(57)

Vzz ' & Vxx % Vyy (58)

Vxy 'z

(a & x)2 % z 2

b & yA

%b % yD

&z

(a % x)2 % z 2

b & yB

%b % yC

(59)

Vxy 'z

(b & y)2 % z 2

a & xA

%a % xB

&z

(a % x)2 % z 2

b & yB

%b % yC

(60)

Vzx '1A&

1B%

1C

&1D (61)

Vz ' &2π

% cos&1 (a & x) (b & y)

(a & x)2 % z 2 (b & y)2 % z 2

% cos&1 (a & x) (b % y)

(a & x)2 % z 2 (b % y)2 % z 2

% cos&1 (a % x) (b & y)

(a % x)2 % z 2 (b & y)2 % z 2

% cos&1 (a % x) (b % y)

(a % x)2 % z 2 (b % y)2 % z 2

(62)

A ' (x & a)2 % (y & b)2 % z 2 (63)

B ' (x % a)2 % (y & b)2 % z 2 (64)

C ' (x % a)2 % (y % b)2 % z 2 (65)

D ' (x & a)2 % (y % b)2 % z 2 (66)cos-1 between 0 and π, and tan-1 between -π/2 and +π/2.

CIRCULAR LOAD: Acting on area of radius aFooting has radius a in r-direction. Origin is at centre on loaded area

that is on the surface of a semi-infinite elastic solid. z-axis downwards.

σr 'p

2π2ν Vz & (1 & 2ν) Xrr & z Vrr (67)

σt 'p

2π2ν Vz & (1 & 2ν) Xrr

&p

2π(1 & 2ν)

rXr %

zrVr

(68)

σz 'p

2πVz & z Vzz (69)

τrz ' &z

2πVrz (70)

τ&t r ' τrz ' 0 (71)where

Xr 'πa 2

r%z Br

Ε &z(2a 2 % 2r 2 % z 2)

B rΚ

&a 2 & r 2

rΚ Ε (k,θ) & (Κ & Ε) Γ(k,θ)

(72)


Xrr ' &π a 2

r 2&z Br 2

%z(2a 2 % z 2)B r 2

Κ

%a 2 % r 2

r 2Κ Ε(k,θ)&(Κ & Ε) Γ(k,θ)

(73)

Vz ' & 2 Κ Ε(k,θ) & (Κ & Ε) Γ(k,θ) & zB

Κ (74)

Vr ' &Br

1 %A 2

B 2Κ & 2Ε (75)

Vrr '(r 2 % a 2 % z 2) (r 2 & a 2 & z 2)

B 2 A r 2Ε

&2B

Κ %Br 2

1 %A 2

B 2Κ & Ε

(76)

Vrz 'zB r

1 %B 2

A 2Ε & 2Κ (77)

Vzz '2B

Κ

&Br 2

%(r 2 % a 2 % z 2) (r 2 & a 2 & z 2)

a 2 B rΕ

(78)

A ' (a & r )2 % z 2 (79)

A ' (a % r )2 % z 2 (80)

Κ ' m

π2

0

1

1 & 1 &A 2

B 2sin2θ

dθ

Complete elliptic integral of 1st. kind :&

modulus 1 &AB

2

(81)

Ε ' m

π2

0

1 & 1 &A 2

B 2sin2θ dθ

Complete elliptic integral of 2nd. kind :&

modulus 1 &AB

2

(82)

Γ(k,θ) ' mθ

0

1

1& AB

2sin2θ

dθ

Elliptic integral of 1st. kind :&

modulus AB

(83)

Ε(k,θ) ' mθ

0

1 &AB

2sin2θ dθ

Elliptic integral of 2nd. kind :&

modulus AB

(84)

when r < a π > θ > π2and

θ ' tan&1 za & r

(85)

when r > a π2

>θ >0 and

θ ' tan&1 zr & a

(86)

On the centre line when r=0 then:Vr60 ' 2π ( a 2 % z 2 & z) (87)

Vrr r60'Vrr r60

' &12Vzz r60

(88)

Xrr r60'Xrr r60

' &12Vz r60

(89)

The elliptic integrals may be evaluated from the following where k iswritten as the modulus and must have a value not greater than 1:

Κ 'π2

1 %12

22k 2 %

12 32

22 42k 4

%π2

12 32 52

22 42 62k 6 etc.

(90)

Ε 'π2

1 &12

22k 2 &

12 32

22 42k 4

%π2

&12 32 52

22 42 62k 6 etc.

(91)

Γ (k,θ) π θ2

Κ

& sinθ cosθ 12A2 k

2 %1 32 4

A4 k4

& sinθ cosθ 1 3 52 4 6

A6 k6 etc.

(92)

Ε(k,θ) ' π θ2

Ε

% sinθ cosθ 12A2 k

2 %1

2 4A4 k

4

% sinθ cosθ 1 32 4 6

A6 k6 etc.

(93)

where

A2 '12 (94)

A4 '3

2 4%

14

sin2θ (95)

A6 '3 5

2 4 6%

54 6

sin2θ %16

sin4θ (96)

A8 '3 5 7

2 4 6 8%

5 74 6 8

sin2θ

%7

6 8sin4θ %

18

sin6θ(97)


EXAMPLE 1A building 20 m x 20 m results in a uniform surface contact

pressure of 150 kPa. Using the Newmark Influence Chart obtain thevertical pressure depth of 10 m below (a) the centre of the building (b)a corner of the building. Check your result from the Fadum Chart.Estimate the additional pressure at both locations of a tower 5 m x 5m placed at the centre of the building imposing 300 kPa uniformadditional pressure._________________________________________________________ Increase in stress below centre of building from influence chart

= 4 x No. of sq. in one quarter x q x I

= 4 x 177 x 150 x 0.001 = 106 kPa

Increase in stress below corner of building from influence chart

= No of sq. x q x I

= 232 x 150 x 0.001 = 35 kPa

Increase in stress below centre from Fadum chart

For one quarter of building L = 10 m; B = 10 m; z = 10 m

n = L/z = 10/10 = 1

m = B/z = 10/10 = 1

k from chart = 0.177

σ = 4 x k x q = 4 x 0.177 x 150 = 106 kPa

Increase in stress below corner from Fadum chart

h = 20 m; B = 20 m; z = 10 m

n = L/z = 20/10 = 2

m = B/z = 20/10 = 2

k from chart = 0.233

σ = k x q = 0.233 x 150 = 35 kPa

Increase in stress below centre due to tower

= 4 x 27 x 300 x 0.001 = 32 kPa

Increase in stress below corner due to tower

= 7 x 300 x 0.001 = 2 kPa


EXAMPLE 2The bearing capacity failure of strip load on a rigid plastic soil is (π

+ 2)cu. Compare this with the load at first yield for an elastic plasticmaterial (φu = 0)._________________________________________________________

σ1 = q(ψ + sinψ)/π

σ3 = q(ψ - sinψ)/π

σ1 - σ3 = 2q sinψ/π

d (σ1 & σ3)dψ

' 0 for max ' cos ψ

ˆ ψ ' 90E for max σ1 & σ3

(σ1 & σ3)max '2 qπ

For a φu material cu '(σ1 & σ3)max

2at failure

ˆ q (internal failure) = π cu

Compare with Bearing Capacity q = (π + 2)cu

ˆ q(U.B.C.)q(Elastic)

'π % 2π

' 1.64

EXAMPLE 3Calculate the vertical stresses at 30 m depth below the ground

surface supporting an embankment of height 10 m constructed of soilweighing 20 kN/m3, shoulder slopes of 2H:1V, a top width 20 m, andbase width of 60 m for the following positions:(a) vertically below centre line.(b) vertically below shoulder.(c) vertically below toe.(d) vertically below 10 m beyond toe.

q = ∆σmax due to fill = 10 x 20 = 200 kPa.Chart embankment dimensions are:

a = 20 m; b = varies; z = 30 m.(a) VERTICAL STRESS ON CENTRE LINE

(1) a/z = 0.666; b/z = 0.333; I1 = 0.325.(2) a/z = 0.666; b/z = 0.333; I2 = 0.325.σz = (I1 + I2)q = (0.325 + 0.325) 200 = 130 kPa.

(b) VERTICAL STRESS BELOW SHOULDER(1) a/z = 0.666; b/z = 0.0; I1 = 0.180.(2) a/z = 0.666; b/z = 0.666; I2 = 0.400.σz = (I1 + I2)q = (0.180 + 0.400) 200 = 116 kPa.

(a) VERTICAL STRESS BELOW TOE(1) a/z = 0.666; b/z = 1.333; I1 = 0.470.(2) a/z = 0.666; b/z = 0.0; I2 = 0.180.σz = (I1 + I2)q = (0.470 - 0.180) 200 = 58 kPa.

(b) VERTICAL STRESS BELOW SHOULDER(1) a/z = 0.666; b/z = 1.666; I1 = 0.480.(2) a/z = 0.666; b/z = 0.333; I2 = 0.325.σz = (I1 + I2)q = (0.480 - 0.325) 200 = 31 kPa.


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