StressesStresses in a Soil Massin a Soil Mass

updated April 15, 2008

Haryo Dwito Armono, M.Eng, Ph.D

S il th t t f d ti bj t d tSoil that support foundation subjected to net stresses increases

Net stresses increases depend onNet stresses increases depend on Load per unit area to which the foundation is subjectedsubjectedDepth below foundation at which the stress estimation is desiredestimation is desired

This topic discusses the principles ofThis topic discusses the principles of estimation of vertical stress increase in

il d t i t f l disoil due to various types of loading.

Boussinesq MethodBoussinesq MethodSimplicity

J. Boussinesq, 1885

p yBased on Linear theory of elasticity

Assumptions:The system (loads & soil) in a state of equilibriumAll loads have been applied gradually and no kinetic energy pp g y gypassed onThe system is conservative and independent of time *)The soil is weiightless, continuous, homogeneous, isotropic, g , , g , p ,and linearly elastic *)The material constants are known from experiments and independent of time *)

*) cannot be satisfied by the real system

Point LoadPoint Load

Boussinesq (1883)Boussinesq (1883)

P load - kN

x

yr

32 2

5

2 2 2 2 2

3

2z

P zp r x y

LπΔ = = +

Δpzy

y

x

L

z

2 2 2 2 2

.z

L x y z r z

p Q I

= + + = +

Δ =

Δpx

zA

z

2

5/ 2

3 1

PQ

z

I

=

=Δpy

z( )

5/ 22

2

2 / 1

/

r z

P kN

p kN m

π ⎡ ⎤+⎣ ⎦=

Δ = /zp kN m

kPa

Δ ==

Line LoadLine LoadInfinite Length

( )3

22 2

2qzp

x zπΔ =

+

load/unit length - kN/m

x

.p Q IΔ =

Δp

x

z

2

qQ

z

I

=

Δp

A

z

x

( )22

/ 1I

x zπ=

⎡ ⎤+⎣ ⎦z

2

/

/

q kN m

p kN m

kPa

=

Δ == kPa

Strip LoadStrip Loadload/unit area - kN/m2

Finite width, Infinite Length

1 1

2 2 2

tan tan( / 2) ( / 2)

( / 4)

z z

x B x Bqp

B B

− −⎧ ⎫⎡ ⎤ ⎡ ⎤−⎪ ⎪⎢ ⎥ ⎢ ⎥− +⎣ ⎦ ⎣ ⎦⎪ ⎪

Δ = ⎨ ⎬⎡ ⎤

load/unit area kN/m

x

B

2 2 2

2 2 2 2 2 2

( / 4)

( / 4)

pBz x z B

x z B B z

π ⎨ ⎬⎡ ⎤− −⎣ ⎦⎪ ⎪−⎪ ⎪⎡ ⎤+ − +⎣ ⎦⎩ ⎭Δp z

δβ.p Q I

Q q

Δ =

=

Ax

δβ

[ ]sin cos( 2 )

and in radians

Q q

Iβ β β δ

πβ δ

+ +=

z

β−δ

2

2

/ 0

/2 2

q kN m

p kN m

β

β ππ πδ

= < <

Δ = − < <

β

2 2kPa=

Circular LoadCircular LoadUnder circular center

( )3/ 22

11

/ 1p q

R z

⎡ ⎤⎢ ⎥Δ = −⎢ ⎥⎡ ⎤+⎢ ⎥⎣ ⎦⎣ ⎦

R

( )

.p Q I

⎢ ⎥⎣ ⎦⎣ ⎦

Δ =

load/unit area - kN/m2

x

3/ 22

11

Q q

I

=

= −⎡ ⎤

R

( )3/ 22

2

/ 1

/

R z

q kN m

⎡ ⎤+⎣ ⎦

=Δp

z

2

/

/

q kN m

p kN m

kPa

=

Δ ==

ΔpA

Circular LoadCircular LoadAny location

I =

1.5

load / unit area = kPaoq =

r = 2x = 8 ; x/r = 4z = 7 ; z/r = 3.5;

Figure 8.22

Rectangular LoadRectangular Loady

Under rectangular cornerFinite width, Finite Length

2 2 2 2

2 2 2 2 2 2

2 21

2 1 2

1 1

4 2 1tan

mn m n m n

m n m n m nqp

mn m nπ−

⎡ ⎤⎛ ⎞+ + + +⎢ ⎥⎜ ⎟+ + + + +⎝ ⎠⎢ ⎥

Δ = ⎢ ⎥⎛ ⎞+ +⎢ ⎥+ ⎜ ⎟⎢ ⎥⎜ ⎟

L

2 2 2 2tan

1

given in a chart in Figure. 3.25

interchangeable

m n m n

B Lm n

z zp Q I m n

+ ⎜ ⎟⎢ ⎥⎜ ⎟+ − +⎢ ⎥⎝ ⎠⎣ ⎦

= =

Δ

B

x

2 2 2 2

. , interchangeable

2 1

p Q I m n

Q q

mn m n m n

Δ =

=

+ + + + 2⎡ ⎤⎛ ⎞⎢ ⎥

load/unit area - kN/m2

x2 2 2 2

2 1

11

4

mn m n m n

m n m nI

π

+ + + ++ + +

=2 2

2 21

2 2 2 2

2

1

2 1tan

1

m n

mn m n

m n m n−

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟+ +⎝ ⎠⎢ ⎥⎢ ⎥⎛ ⎞+ +⎢ ⎥+ ⎜ ⎟⎢ ⎥⎜ ⎟+ − +⎢ ⎥⎝ ⎠⎣ ⎦

x

L

2

2

/

/

q N m

p N m

⎢ ⎥⎝ ⎠⎣ ⎦

=

Δ =

z

p

Pa=ΔpA

Rectangular Load load / unit area = kPaoq =

Rectangular Load

z = 5x = 1.5 :x/z = 0.3y = 2 :y/z = 0.4

I = 0.46

Figure 8.21 Figure 3.25

Triangular LoadTriangular LoadFinite width, Finite Length

L

y

B Bm n

z z= =

L load / unit area = kPaoq =

given in a chart in Figure. 8.24

.

z z

p Q IΔ =

B

x

2

Q q=load/unit area - kN/m2

x

A

2

2

/

/

q N m

p N m

Pa

=

Δ ==

z

L

B

ΔpA

B

B

Figure 8.24

Trapezoidal LoadTrapezoidal LoadFinite width, Infinite Length

qo = load of embankment/areakN/m2kN/m2

a b.p Q IΔ =

x

h

( )

. .

1

oQ q g h

a b bI

ρ

β δ δ

= =

⎡ ⎤+⎛ ⎞= + −⎜ ⎟⎢ ⎥

Δp z

δβ

( )

2/

Ia a

kN kP

β δ δπ

= +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

A

2

2

/

/

oq kN m kPa

p kN m

kPa

= =

Δ ==

TrapezoidTrapezoid

l d / it kP load / unit area = kPaoq =

Figure 8.23

SuperpositionSuperpositionFor elastic material superposition is valid

x

q/unit length - kN/m q/unit length - kN/m

x

+

q/unit length - kN/m

x=

x1

Az

Az

x2

Az

x1 x2

A B A B A B A B

=C D C D

+C D

+C D

+

=3/4

= + -

= -

= -

+

-

+

1/2= + 1/2

= + 1/2

-=

- +- +

A +=

- +B

BA C

C

-

Newmark’s Influence ChartNewmark s Influence ChartAdopt a scale such that AB orAdopt a scale such that AB or OQ is equal to depth (z)

Plot th eplan of loaded area on the chart

BA BAon the chart

Place the plan in such a way that point P is located directly above the center of the chart

Count the number of blocks N

Calculate ΔpCalculate Δpz

. .

I i fl lzp I q NΔ =

I = influence value

q = pressureon the loaded area (kPa)

n = number of blocks

Assignment 1Assignment 1A vertical structural load F is to be applied above point O at the surface of a mass of elastic soil as shown on Figure 1elastic soil, as shown on Figure 1.Due to the presence of a soft layer at 2m depth, an estimate of the vertical stress increase Δpz is needed at points A and B indicated in the Figures.Compute Δpz at A and B for the following cases:

1. F is applied as a point load at point O (Figure 2a)2. F is uniformly distributed along a line of length 8m (Figure 2b)3 F is uniformly distributed over an area of length 8m and width 0 5m (Figure 2c)3. F is uniformly distributed over an area of length 8m and width 0.5m (Figure 2c)

a. Treat the area as a strip footingb. Treat the area as a finite rectangular flexible foundation

4. F is uniformly distributed over a circular area centered at O so that the contact pressure is the same as in case 3 (Figure 2d)is the same as in case 3 (Figure 2d)

5. F is uniformly distributed over a square area centered at O so that the contact pressure is the same as in case 3 and 4 (Figure 2e)

6. F is uniformly distributed over a trapezoid load ofT t th l b k t (Fi 2f)a. Treat the area as a long embankment (Figure 2f)

b. Treat the area as a finite trapezoidal embankment (Figure 2g)

Present your calculation for all cases (8 of them) and summarize your results in a Table y ( ) yfor points A and

F = 100 kN

8m

F = 100 kN

8mF = 100 kNF = 100 kNF = 100 kNF = 100 kN1 2a 2b

x

Ox

Ox

Ox

O

F 100 kN

x

O

F 100 kN

x

O

y 2my 2my 2my 2my 2my 2m

z

BA 2m

z

BA 2m

z

BA 2m

z

BA 2m

z

BA 2m

z

BA 2m

F = 100 kNF = 100 kN F = 100 kNF = 100 kN

F = 100 kNF = 100 kN2c 2d 2e

x

Ox

Ox

Ox

Ox

O

8m0.5m

x

O

8m0.5m

x

y 2m

x

y 2m

x

y 2m

x

y 2m

x

y 2m

x

y 2m

z

BA 2m

z

BA 2m

z

BA 2m

z

BA 2m

z

BA 2m

z

BA 2m

Assignment 2Assignment 2A highway embankment as shown in Figure 3a and 3b. Assume the average

d it f th t i l i th b k t i 2 0 M / 3density of the material in the embankment is 2.0 Mg/m3

Due to the presence of a soft layer at 2m depth, an estimate of the vertical stress increase Δpz is needed at points A and B indicated in the Figures 3.

Compute Δp at A and B for the following cases:Compute Δpz at A and B for the following cases:

Load is uniformly distributed over an area of length 8m and width 8m (Figure 3a)a. Treat the embankment as a very long embankmenty gb. Treat the embankment as a finite rectangular and triangular flexible embankment

ρ = 2 Mg/m3ρ = 2 Mg/m3

x

8m

2

1m

8m

x

8m

22

1m

8m

ρ = 2 Mg/m3

2m

1m

8m

21

21ρ = 2 Mg/m3

2m

1m

8m

21

21

yL

2m 1y

L

2m

L

2m 11

z

BA

2m

z

BA

2m

z

BA 2m

z

BA 2m

z

BA 2m

BibliographyBibliographyHoltz, R.D and Kovacs, W.D., “An Introduction to GeotechnicalHoltz, R.D and Kovacs, W.D., An Introduction to Geotechnical

Engineering”

Das, B.M., “Soil Mechanics”

Das, B.M., “Advanced Soil Mechanics”

Problem Examples

Holtz and Kovacs

Examples 8 18Examples 8.18The 3 x 4 m rectangular footing is loaded uniformly by 117 kPa

Required :a. Find the vertical stress under the corner of the footing at a depth of 2mb. Find the vertical stress under the center of the footing at a depth of 2m

Solution:a. x = 3m, y=4m, z= 2m; therefore

m=x/z = 3/2 = 1.5, n = y/z = 4/2=2y

From figure 8.21, I = 0.223Δpz = qo. I = 117 x 0.223 = 26 kPa

a. To compute the stress under the center, it is necessary to divide the 3 x 4 rectangular footing into 4 sections of 1.5 x 2m in size. Find the stress under one corner and multiply this value by 4 to take into account the four quadrants of the uniformly loaded area.

x = 1 5m y = 2m z = 2m; thenx = 1.5m, y = 2m, z = 2m; thenm= x/z = 1.5/2 = 0.75, n= y/z = 2/2 = 1

The corresponding value of I from Fig 8.21 is 0.159Δpz = qo. I = 4 x 117 x 0.159 = 74 kPapz qo 0 59 a

Example 8 19Example 8.19A 5 x 10 m area uniformly loaded with 100 kPa

Required:

a. Find the stress at a depth of 5m under point A in Figure 8.19

b. Find the stress at point A if the right half of the 5 x 10m area were loaded with an p gadditional 100 kPa

Solution:

a. Refer to Figure 8.19 and the numbered points as shown Add the rectangles in the followingas shown. Add the rectangles in the following manner (+ for loaded areas and – for unloaded areas)+A123 – A164 – A573 +A584 result in the

1

Loaded area (8627)

A

5

5

loaded rectangle we want 8627. Find four separate influence values from Figure 8.21 for each rectangle at a depth of 5m then add and subtract the computed5

689

5

4

5m, then add and subtract the computed stresses.Note that it is necessary to add rectangle A584 because it was subtracted twice as part

3

10

5

27

5

10

of rectangles A164 and A573Figure 8.19

105

The computation are shown in the following tablep g

+A123 -A164 -A573 +A584

x 15 15 10 5

10 5 5 5 5

A 5 1

Loaded area (96210)

11

y 10 5 5 5

z 5 5 5 5

m=x/z 3 3 2 1

5

5

48 69

Loaded area (96210)

8

n=y/z 2 1 1 1

I 0.238 0.209 0.206 0.180

Δpz 23.8 -20.9 -20.6 +18.0

5

21073

Total Δpz = 23.8 – 20.9 – 20.6 + 18 = 0.3 kPa

b. When rectangle 78910 is loaded with 100 kPa and rectangle 96210 is loaded with 200

5 10

b. When rectangle 78910 is loaded with 100 kPa and rectangle 96210 is loaded with 200 kPa, repeat part (a) above to obtain the stress under point A at 5m depth for the entire rectangle 8627 loaded with 100 kPa.Next, a second set of four rectangles would have to be calculated just as for part (a) b b t l t l 96210 ld b l d d ith +100 kP th th ld babove but only rectangle 96210 would be loaded with +100 kPa; the others would be

– 100 kPa.The total Δpz equals 0.3 kPa from part (a) plus (+A123 – A164 – A11103 +A1194) = 23.8 – 21.0 – 23.2 + 20.6 or 0.3 + 0.2 = 0.5 kPa

Example 8 20Example 8.20A circular tank 3.91 m in diameter is uniformly loaded with 117 kPaRequiredRequireda. Compute the stress under the center of the tank at a depth of 2m below

the tankb. Compute the stress under the edge of the tank, also a depth of 2m

Solutiona. Refer to Figure 8.22 :

z = 2m r = 3 91 / 2 = 1 95m x = 0; thenz = 2m, r = 3.91 / 2 = 1.95m, x = 0; thenz/r = 2/1.95 = 1.02 x/r = 0/1.95 = 0Find I = 0.63. Δpz = qo. I = 117 x 0.63 = 74 kPa(This compares exactly with Δpz = 74 kPa at the center for a 3 x 4m ( p y pzrectangular area in Example 8.18, In both cases, the area is 12m2)

b. Again refer to Figure 8.22, For the edge of the circular loaded area:2 1 95 1 95z = 2m, r = 1.95m, x = r = 1.95m

z/r = 2/1.95 = 1.02 x/r = 1.0Find I = 0.33 Δpz = qo. I = 117 x 0.33 = 39 kPa(This compares with with Δp = 26 kPa at the corner for a 3 x 4m(This compares with with Δpz = 26 kPa at the corner for a 3 x 4m uniformly loaded rectangular area in Example 8.18, In both cases, the area of the loaded area is the same : 12m2)

Example 8 21Example 8.21A highway embankment, as shown below. Assume the average density of the material in the

Solution :

First, calculate the applied surface stress qo and the

embankment is 2.0 Mg/m3. Compute the vertical stress under the centerline at depth of 3 and 6m

5m5m

dimensions of the embankment in terms of a and b

From Figure 8.23

b = 5m a = 2 x 3m = 6m

Next calxulate the vertical stress for z = 3m3m

3m2

12

1

Next calxulate the vertical stress for z = 3m

a/z = 6/3 = 2 b/z = 5/3 = 1.67

From Figure 8.23, I = 0.49

3 22.0 9.81 3 59Mg/m m/s m = kPaoq ghρ= = × ×

B

A3m

g

For one half of the embankment, or 58 kPa for the entire embankment.

Th s at this shallo depth Δp is almost the same as the

59 0.488 29 kPa kPAz op q IΔ = = × =

Thus at this shallow depth, Δpz is almost the same as the contact stress.

Finally, calculate the vertical stress for z = 6m

a/z = 6/6 = 1 b/z = 5/6 = 0.83

From Figure 8.23, I = 0.4459 0.44 2 52 kPa kPAz op q IΔ = = × × =

Example 8 22Example 8.22

A uniform load of 250 kPa isA uniform load of 250 kPa is applied to the loaded area shown in Figure Compute the stress at depth ofCompute the stress at depth of 80m below the ground surface due to the loaded area under point O

BAO

40

h z

= 8

0

Uniform Load 250 kPa

1020D

epth

O

604020Q

Draw the loaded area such that the length of the line OQ is scaledthe length of the line OQ is scaled to 80m. For example, the distance AB in Figure above is 1.5 times the distance OQ,

BA

0

OQ = 80m and AB = 120mNext, place point O, th epoint where the stress is required, over

Dep

th z

= 8

0

O

the center of the influence chartThe number of blocks (and partial blocks) are counted under the loaded area In this case abouth

BA

loaded area. In this case, abouth 160 blocks are found. The vertical stress at 80m is then indicated by

p q IΔ = × no of blocksept

h z

= 8

0

To compute the stress at other depths the process is repeated by

250 0.001 160 40z op q IΔ = ×

= × ×

no of blocks

kPa blocks= kPa

De

O

depths, the process is repeated by making aother drawings for the different depths, changing the scale each time to correspond to the distance OQ on the influence chart

BibliographyBibliographyHoltz, R.D and Kovacs, W.D., “An Introduction to GeotechnicalHoltz, R.D and Kovacs, W.D., An Introduction to Geotechnical

Engineering”

Das, B.M., “Soil Mechanics”

Das, B.M., “Advanced Soil Mechanics”