Notes for CE 5810
Karl Terzaghi -= Father of Soil Mechanics
Lecture 1
Review the void density relationships for soils and effective
stress concepts
density   
MT M s  M w

VT
VT
unit  weight   
 w  9.81
Ma
 a
VT
kN
lb
 62.4 3
3
m
ft
specific  gravity  G s 
void  ratio  e 
porosity  n 
s  s

w  w
Vvoid
Vsolid
Vvoid
Vtotal
water  content  w 
saturation  S 
Mw
100
M solid
Vw
100
Vvoid
misc  equations
n
1 n
Se  WG s (english )
e
 w Se  w s (metric)
m 
G s  Se
 w (english )
1 e
Also review the effective stress concept using an example that has
pore pressure with static and excess components. Remember that
 '   total  (u static  u excess )
Lecture 2
Develop equilibrium equations, strain equations and
constitutive relationships. Show that the number of equations is
equal to the number of unknowns. Develop plain strain and plain
stress conditions
Equations for elastic solutions





X 0
x
y
z
xy
x

yx
x


y

y
xz


yz
z
Y  0




Z 0
x
y
z
zx
zy
z
Three equations and nine unknowns
But by summing moments
xy = yx
therefore three equations and six unknowns
Using strain definitions
 
u
x
 
v
y
x
y
w
z
v
u


x
y
w
v


y
z
u
w


z
x
 
z

xy

yz

xz
Six equation and nine unknowns
Finally for homogenous materials
1

    
E
E
1

     
E
E
1

     
E
E
2(1   )


E
2(1   )


E
2(1   )


E
 
x
y
z


y
y
z
x


z
z
x
y


xy

yz

xz
x
xy
yz
xz
Six equations and two experimentally determined constants.
Total 15 equations and 15 unknowns make a solvable set with
appropriate boundary conditions.
Lecture 3
Develop mohr’s circle of stress for normal and general
conditions
General equation- positive angle counter clockwisefrom the plane
of the largest normal stress
n 
y x
2

y x
2
  xy sin( 2 )
and
n 
y x
2
sin( 2 )   xy cos( 2 )
if
n  0
then
tan( 2 ) 
 2 xy
y x
and the principal stresses are
1 / 3 
y x
2
 /   ((
y x
2
) 2   2 xy )
The intermediate stress is
 2   (1   3 )
Some Mathcad solutions
Mohr’s Circle of stress
Gen solutionhttp://www.cee.mtu.edu/~balkire/ce584/mohrcircle.mcd
Perpendicular Planes
http://www.cee.mtu.edu/~balkire/ce584/assign2.mcd
Lecture 4
Discuss geostatic stresses in vertical and horizontal direction
For a situation with no stresses applied at the surface
 x  y  xy  yx



x
y
y
x
therfore
 yx
x

 xy
y

 z
  0
z
and if the shear stresses are zero, ie principal stresses then the
vertical stress by integration is equal to z.
If a stress is applied at the surface over an infinite area then the
equilibrium equations reduce to
 xz
0
z
 yz
0
z
 z

z
resulting
 xz  q sin(  v )
 yz  0
 z  z  q sin(  v )
Lecture 5
Horizontal geostatic stresses, anisotropy and hydrostatic stresses.
If there is no stress applied at the surface it is a principal plane and
the equilibrium equations reduce to
 x
0
x
 y
0
y
 z

z
int egrate
 x  f ( z)
 y  f ( z)
 z  z
What is the function of z?
To answer the question it is necessary to consider a specific
loading condition. For example, confined compression
gives strain in the x and y direction equal to zero and from the
constitutive relations
 x   ( y   z )
 y   ( x   z )
therefore
 x   ( ( x   z )   z )

x  z
1
 x  f  z 
Using other stress conditions the same approach can be used, a
different function of will be obtained.
Effect of anisotropy can be obtained as discussed in notes obtained
from Lambe and Whitman
Some mention of hydrostatic stresses should also be considered
because they are of importance in engineering problems
Lecture 6
Introduce polar coordinates and develop the equations for
vertical line load. Required a stress function that satisfies the
compatibility equations (See the handout from Timinshico’s book)
1  1  2

r r r 2  2
 2
  2
r
1 
(
)
 r   r 
r
r 
polar coordinates
If the stress function is

qr

 sin(  )
by differential calculus and substitution results in
2q cos( )
r
  0
r 
 r  0
Use a mohr’s circle to develop the x-z coordinate system and
 z   r cos 2 ( )
 x   r sin 2 ( )
 xz   r sin(  ) cos( )
substitute
r
and
2q
cos 3 ( )
r
2q
x 
cos( ) sin 2 ( )
r
2q
 xz 
cos 2 ( ) sin(  )
r
but
z 
z
cos( ) 
(z 2  r 2 )
and
z 
2qz 3
x 
2qzx 2
 xz 
2qz 2 x

(
1
)
(z  r 2 )2

2
(

1
)
(z  r 2 )2
(
2
1
)
(z  r 2 )2
2
Solution to these equations http://www.cee.mtu.edu/~balkire/ce584/line1.mcd
Introduce horizontal line load equations develop in the same way
and can be combined with the vertical. For example the combined
load equation for an inclined load is
z 
2qi sin(  ) z 3

(
2qi cos( ) xz 2
1
1
)

( 2
)
2
2

(z  r )
(z  r 2 )
where  is measure
from the horizontal. http://www.cee.mtu.edu/~balkire/ce584/pr37.mcd
It can be shown that a line drawn perpendicular to the inclined
load separates the infinite half space into a region of tension and
compression for the vertical stress.
The line separating the tension and compression regions is the
neutral axis.
It is also possible to moments as line load combinations. (A force
couple and distance). Moment/couple
http://www.cee.mtu.edu/~balkire/ce584/couplecal.mcd
Lecture 7
Integrate line load to get the effect of strip load on semiinfinite space. Include numerical integration to show the solutions
are obtained in several ways. Do several examples.
dq  qdA  qdx(1)
d z 
z 
x2

x1
2 z 3 dq

2z 3q

(
(
1
2 z 3 qdx
1
)

( 2
)
2
2 2

(x  z )
(x  z 2 )2
1
)dx
(x  z 2 )2
2
for
x1  0
x2  b
q  1 b
zb 
 tan ( )  2


z
(z  b2 ) 
 z  qI
z 
Where I is an influence factor that can be tabulated. Das uses a
different coordinate system and his equation result are given in
Table3.3. In this table b is one half the strip width instead
of the full width as given above as x2.
Strip Load http://www.cee.mtu.edu/~balkire/ce584/strip.mcd and
http://www.cee.mtu.edu/~balkire/ce584/stripdas.mcd
It is possible to use the same approach to find stress due to a
uniform horizontal loading (Eq. 3.29 In Das, Table 3.6)
Numerical Integration can also be use to obtain results for a given
loading as shown in the example
Lecture 8
Generalize the integration procedure to include various
loading conditions such as inclined loading, parabolic, etc. Do
examples. Show Das’s method and compare the differences.
For inclined loading (linear increasing)
qxdx
b
2 z 3 dq
d z 
 (z 2  b2 )2
dq 
b
2qz 3
x
z 
dx
2

b 0 ( z  b 2 ) 2
z 
qz
z2
(1  2
)
b
(z  b2 )
Other loading types are easily used by defining dq in an
appropriate manner. For example
(linear decreasing)
x
dq  q(1  )dx
b
(parabolic)
dq  q(1 
x2
)dx
b2
(circle)
dq 
q 2
(b  x 2 )1 / 2 dx
b
Results are given in the
http://www.cee.mtu.edu/~balkire/ce584/incload.mcd
http://www.cee.mtu.edu/~balkire/ce584/varstripload.mcd
Lecture 9
Develop the equations for embankment loads and give
several examples.
Embankment loads are combinations of linear increasing, linear
decreasing and uniform strip loads. Each component is calculated
and the result is the sum of the combinations. For example the
increase in vertical stress beneath the center of an embankment
load is:
z 
2qz 3

a
1
2qz 3 
( x  b) 
1
dx

1
dx


2
0 ( x 2  z 2 ) 2

 b  (a  b _  ( x  z 2 ) 2
b
b = distance from center to edge of uniform load and a is he
distance of the decreasing load.
Embankment Load http://www.cee.mtu.edu/~balkire/ce584/embankment.mcd
Lecture 10
Expand on the equations for strip loads and show how they can be
used to develop general equations for practically any type of
loading. http://www.cee.mtu.edu/~balkire/ce584/junk.mcd
Use Fig 3.16 and show how superposition can be used to obtain the
same solutions as done with the integration procedures
Lecture 11
Extend the point load problem to x, y and z space to get the
equations for vertical and horizontal stress.
r 
kQ cos 
R2
R  r2  z2
r
y2  x2
by integration where sum of vertical stress is equal to Q
k
3
2
and
z 
3Qz 3
2R 5
Eq 3.47
by transformation




3Q
1
z 


2z 2  r 2
5/ 2 
(  1)
 z 2

where
` z

Q
I
z2
and




3
1
I
 2

2  r
5/ 2 
(  1)
 z 2

Obviously, it is possible to use several point loads to simulate
footing loads and calculate the increase in stress at a point as the
sum of the stresses contributed by each of the footings.
x and xz can also be developed from the mohr’s circle.
Mathcad solution for I http://www.cee.mtu.edu/~balkire/ce584/pointload.mcd
Lecture 12
A point horizontal load can be developed in a similar manner
and the equation for
Z 
3Qxz 2
2R 5
Eq 3.54
It is possible to extend the point load application by applying the
load over an incremental area and integrating
Point Load(horz)
http://www.cee.mtu.edu/~balkire/ce584/pointloadhorstress.mcd
Lecture 13
The point load equation is integrated over a circular loaded area at
the surface to produce the following.
dQ  qda  qrdrd 
3 z 3 dQ
2R 5
2 rad
3z 3 q
rdrd 
z 
2


2 0 0 (r  z 2 ) 5 / 2
dz 

 z  1 

Eq 3.61

z3
q
2
2 3/ 2 
(b  z ) 
Note: When b is 0 the influence value is 0 and b is infinite then b
is infinite.
A more generalized solution for stresses away form the center can
be obtained from equation 3.63 where the coefficients are a
function of the s/b and z/b ratios and are obtained from table 3.9
and 3.10.
A sample is used to show that the circular loaded area gives
stresses beneath a loaded area that are different then the stress from
an equal point load.
Other non uniform loading over a circular area can be derived in a
similar manner and the results for parabolic load is given by
equation 3.68 and for a uniform increasing (conical) load by
equation3.69.
z
3
3z q

2
2
b
r
5
( r2
0
0
For uniform load
For Parabolic load
z2 )
2
dr d
2
z
b
3z
q



2  b
3
1
2
(b
r ) r
2 2
5
0
( r2
z2 )
(b
r) r
dr d
2
0
For conical
z
3
3z q

2 b
2
b
5
( r2
0
z2 )
dr d
2
0
NonCircular Load http://www.cee.mtu.edu/~balkire/ce584/cirnonunf.mcd
Lecture 14
Loads on a rectangular area are done in the same manner.
Integrate a differential area over the boundaries as shown below
dQ  qdA  qdxdy
and
z 
3qz 3
2
a b
  (x
0 0
2
dxdy
 y 2  z 2 )5 / 2
A typical solution from the Mathcad program is
4.5
1.5
3   3
qz
2 
z( x, y )
1.5
1
2
x
y
2
d xd y
2
2.5
z
1.5
The beauty of the integrated form is it can be used to find the stress
beneath any x,y coordinate by adjusting the limits of integration in
the equation. For example at the center of the loaded area the
integration would be
z 
3qz 3

b a
2 2

b a
 
2 2
1
(x 2  y 2  z 2 )
5
2
dxdy
Where a and b are the dimensions of
the footing
Some solutions are http://www.cee.mtu.edu/~balkire/ce584/rectload.mcd
http://www.cee.mtu.edu/~balkire/ce584/ce584sqf.mcd
http://www.cee.mtu.edu/~balkire/ce584/straineq.mcd
The textbook solution is usually written in terms of dimensionless
parameters m = B/z and n = L/z and are available in Fig 3.28.
Superposition may be required when using tables particularly when
the point of interest is inside or outside the loaded area.
See example for the center of a square loaded area.
Lecture 15
Numerical integration is also possible where the solution is in the
form
z 
3qz 3
2

a
b
0
0
xy
(x  y 2  z 2 )5 / 2
2
x and y are measured along the
axis and must be constant ( it is possible to use non uniform
lengths but the math becomes difficult) .
For an area divided into increments x and y the Trapezoidal
reduces to a reoccurrence equation of the form
x y
1 f ( x, y)(corners )  2 f ( x, y)(edges)  4 f ( x, y)(int erior ) as
z  k
2 2
demonstrated in the example It is also possible to use Simpson’s
rule to increase accuracy. This reduces to a formula of the form
with reoccurrence coefficients of 1 at corners, 4 at side points and
16 at the middle point for a single area of x and y
z 
3qz 3 x y
1 f ( x, y, z)(corners )  4 f ( x, y, z)(edge)  16 f ( x, y, z)(center)
2 3 3
It should be noted that for more extensive areas the reoccurrence
coefficients change. (See the class handout)
Lecture 16
Another procedure used to find stress beneath a rectangular loaded
area is the use of a Newmark Chart. The approach is to use the
uniform circular load equation in the form
 z  qI
where




1

I  1  2
3
b


2
 ( z 2  1) 
solve for b/r ratios when I is set to even increments such as .1, .2,
.3, etc and the results are as follows
I
b/z
0.1
.27
0.2
.40
0.3
.52
0.4
0.5
0.6
0.7
0.8
0.9
0.10
.64
.77
.92
1.11
1.39
1.91

Select a convent scale (z = 1in) and draw the circles with radius b
as determined from the ratios. The results is a series of concentric
circles. Divide the circles into some number of sectors (20 degrees
for example). Then each segment (intersected areas from the
sector line and concentric circles ) loaded produces the effect of
1/number of sector X 10 from the concentric circle. This number
is called the influence factor of the chart.
Lecture 17
SAP 90 – Use of this structural software to simulate loads of
interest to the geotechnical engineer.
SAP90
To use SAP90 requires a data file and a couple of commands from the DOS window on a
computer in Rm. 211 or 213. Actually, it is pretty easy and the solutions can provide a
lot of interesting facts from studying different load configuration and boundary
conditions. Before you start the computer exercise it is best to read the information
provided in class about preparing the data to run SAP90. I will provide a sample data set
for a simple problem (e-mail). The beauty of it is the same data can be used over and
over with minor changes to solve a variety of problems.
To Use SAP90
1. In your H: drive make a directory to place your SAP data and results.
H: mkdir name of directory <enter>
2. In the directory just made, move the data file provided by BDA.
3. In your SAP directory enter the following:
H:\your directory\ce426 <enter.> Read the screen for print out info
H:\your directory\SAP90 <enter>
The SAP 90 screen will come up
<enter>
A prompt will ask for the name of the data file. Type: prob1as <enter>
The program will execute and you can observe the results scroll past on the screen. The
results will be put into files with different extensions. You can use any standard
technique to read the contents of the files or print them out.
4. To use the graphical editor provided by SAP90 type
SAPLOT <enter>
and follow the prompts
GOOD LUCK
Lecture 18
There are several tables in the book that involve various layered
systems
Line load - soft layer over rigid layer – stress is higher in the
upper layer. As the layer get thicker the stress approaches the
infinite solution. See Fig 3.4
Circular loaded area – two layers with different modulus
values – the stiffer layer at the surface tend to take up the stress
resulting in lower stresses in the lower layer. See Fig 3.30
Circular loaded area over three layers with different modulus
ratio values - The solution is obtained by interpolation using the
tables in the appendix. Solution is only at the interface of the layers
For a more generalized solution use the program elsym5.
This can be used to find stresses at any number of points (x,y) in at
most a five layer system. See handout for details on preparing the
input file and reading the output file.
Use the software elsym5 http://www.cee.mtu.edu/~balkire/ce584/elsym5.exe or in the
directory R:/classes/ce584 This gives very good results for up to
five layers.
Typical Results
Output
ELASTIC SYSTEM - ELSYM5 GR LAB 5
ELASTIC
MODULUS
4000000.
30000.
5000.
LAYER
1
2
3
POISSONS
RATIO
.150
.400
.450
THICKNESS
8.000 IN
6.000 IN
SEMI-INFINITE
TWO LOAD(S), EACH LOAD AS FOLLOWS
TOTAL LOAD.....
LOAD STRESS....
LOAD RADIUS....
LOAD
1
2
4500.00 LBS
75.00 PSI
4.37 IN
LOCATED AT
X
Y
.000
.000
13.110
.000
RESULTS REQUESTED FOR SYSTEM LOCATION(S)
DEPTH(S)
Z=
8.00
X-Y POINT(S)
X
Y
.00
6.56
Z=
.00
.00
8.00 LAYER NO,
X
.00
Y
.00
1
6.56
.00
NORMAL STRESSES
SXX
.1301E+03 .1155E+03
SYY
.1565E+03 .1562E+03
SZZ
-.1585E+01 -.1424E+01
SHEAR STRESSES
SXY
.0000E+00 .0000E+00
SXZ
.3108E+00 -.5270E-03
SYZ
.0000E+00 .0000E+00
PRINCIPAL STRESSES
PS 1
.1565E+03 .1562E+03
PS 2
.1301E+03 .1155E+03
PS 3 -.1586E+01 -.1424E+01
PRINCIPAL SHEAR STRESSES
PSS 1 .7903E+02 .7882E+02
PSS 2 .1319E+02 .2035E+02
PSS 3 .6584E+02 .5846E+02
DISPLACEMENTS
UX
-.1717E-03
UY
.0000E+00
UZ
.1451E-01
.1736E-06
.0000E+00
.1416E-01
NORMAL STRAINS
EXX
.2672E-04 .2307E-04
EYY
.3430E-04 .3477E-04
EZZ
-.1114E-04 -.1055E-04
SHEAR STRAINS
EXY
.0000E+00 .0000E+00
EXZ
.1787E-06 -.3030E-09
EYZ
.0000E+00 .0000E+00
PRINCIPAL STRAINS
PE 1
.3430E-04 .3477E-04
PE 2
.2672E-04 .2307E-04
PE 3 -.1114E-04 -.1055E-04
PRINCIPAL SHEAR STRAINS
PSE 1 .4544E-04 .4532E-04
PSE 2 .7581E-05 .1170E-04
PSE 3 .3786E-04 .3362E-04
Lecture 19
Settlement is soils usually considered to be made up of
primary, secondary and elastic components. The elastic
component has its base in the theory of elasticity and can be
developed using equations based on equilibrium, strain and
constitutive relationships. Each loading condition has a different
set of equations and the results are different. In comparison to the
stress equations the settlements equations are more involved and
the integration is more difficult.
In general settlement is the integration of strain over some depth
s    z dz
where the strain is defined in terms of stress and the material
properties Poisson’s ratio and the modulus of elasticity.
For example for a point load the equation in cylinderical
coordinates is
z 
1
( z   ( r   )
E
For the point load where the stresses are
defined by the equations 3.45, 3.46 and 3.47
The result is
s    z dz  
1
( z   ( r   )) dz
E
but
r
 tan 
z
and
dz 
rd
sin 2 

s
Q(1   ) 2
cos( )(3 cos( )  2 )d
2Er 0
when integrated gives
Q(1   2 )
s
Er
The point load settlement equation
http://www.cee.mtu.edu/~balkire/ce584/setpointload.mcd
http://www.cee.mtu.edu/~balkire/ce584/straineq.mcd
Lecture 20
Deflection due to other loading conditions follow in a similar
manner. For a circular loaded area the equation is simplified due to
axial symmetry and reduces to

1
s    z  2 r dz
E0




1

 z  q 1  2
31
b


2
 ( z 2  1) 




q
2(1   )
1

   r  (1  2 )  2

1
3
2
2
b
b

( 2  1) 2 ( 2  1) 2 

z
z

or
Substitute equation 3.61 and 3.62 into the equation and the result
when  is 0.5 is
s
1.5qb
E
at the center of the loaded area. Settlement at points away from
the center can be determined by more complicated integration or
by using Eq 8.18 in the book.
(1   )  z

s  qb
 I 1  (1   ) I 2  Where I1 is from Table 3.9 and I2 is
E
b

from Table 8.6 Another possibility is to use a Newmark chart for
settlement. They are formed in the same manner as for stress and
are described earlier and shown below
The effect of layers can also be accounted for by changing
the limits on the integration

z
1 1
1
s
(  21 )dz 
( z  2 2 )dz

E1 0
E 2 z1
which end up for  - 0.5 being

1.5qb 
1
1 
s
2
1
E 
z1
2
(

1
)

b2







resulting in
z1/b
I
s
0
.5
0
.106
1.0
.293
2.0
.553
4.0
.757
10.
.9

1.0
0
.106 s
.293 s
.553 s
.757 s
.9 s
1.0 s
Lesson 21
Another approach to the problem is to integrate the point load settlement expression over the area of
interest.
Q (1   2 )
rE
dQ (1   2 )
ds 
rE
dQ  qrdrd 
s
2 b
s
q (1   2 )
E
s
2q (1   )b
E
  rdrd
0 0
2
As before with  = 0.5
s
1.5qb
E
The main point here is that settlement can be obtained at various points including the center,
edge and at any depth by using one or more of the techniques above or by using the generalized
equation 8.18 and the Tables 3.9 and 8.6
For example
Let x/b = 1, z/b = 0 then I1 = 0 and I2 =1.27
s
qb(1  .5)
1.27qb
.95qb
(0  (1  .5)1.27) 
(1  .5 2 )  .
 .635s center
E
E
E
when x/b = 0 and z/b = 1, and  = 0.5
s
qb(1  .5)  z
 1.06qb
.2429  (1  .5)(.8284) 

E
E
b

The same as obtained from direct integration

1.5qb 
1
s
1
2
1
E 
z1
2
(

1
)

b2



 1.5qb 
1  1   1.06qb

1

E 
E

2
2




Lecture 21
How to calculate average settlement? In general for a circular loaded area
s avg  .843s center
at the surface. This can be approximated by finding the average of I2 value from Table
8.6 for x/b ratios up to 1. For example 2.0+1.97+1.91+1.8+1.62+1.27 = 10.37 /6 =1.76/2 = .88 the
settlement at the center. This can be developed in a more exact way by numerical integration
I 2 avg 
 (I
2(i )
 I 2(i 1) ) Ai
A
i
For the case above this evaluates to
I 2 avg  1.68
and
1.68qb(1   2 )
E
1.26qb
s avg 
E
or
s
s avg  .84 s center
Lecture 21
Another approach to settlement is to evaluate the strain at the average point in a layer
and multiply that value by the depth of the layer (basically the method of Schmertmann). In
this case the strain definition under a circular loaded area is
z 
q(1   )
(1  2 ) A '  B ' Where A’ and B’ are coefficients from Table 3.9 and
E


Table 3.10. A’ and B’ are evaluated at z/b based on the depth to the center of the layer of
interest . The s/b value is 0.
10
5
10
Rigid Base
q(1   )
(1  2 ) A '  B '
E
q1.5
5.4q
0  .358 
s  10
E
E
s    z dz  H avg  H


By Eq 8.18 (closed form)
q(1   )b  z

I 1  (1   ) I 2 

E
b

q(1.5)10
0  .5(2)  1(.293),5(.828)  4.4q
s
E
E
s
The difference is not too great and would get smaller if the layer thickness is larger or
the depth to the average point is large.
This approach can be used to account for non homogenous soils by finding the strain at
the center of each layer with a different modulus value and calculating the settlement for each
layer and then summing the settlement for each layer to get the total. (See the worked example )
Another way of getting the settlement at any location and with any profile is to use the
program Elsym5. This program was developed for finding stresses and strains beneath vehicle
loadings. The general description was given above and can be used to find settlement below a circular
footing (watch the Units). http://www.cee.mtu.edu/~balkire/ce584/elsym5.exe
10
10
E=10000 psf
10
E=20000 psf
20
E=50000 psf
The resulting equations are
q1.5
0  .35777  10 q1.5 0  .25602  20 q1.5 0  .09487
10000
20000
50000
s  0.537  0.192  0.057  0.786
s  H avg1  10
Remember the s/b and z/b values are determined at the center of each of the layers.
Lecture 22
In a similar manner the point load expression can be integrated to find the settlement
under rectangular areas
s
q(1   2 )
E  
1
(x  y )
2
2
1
2
dxdy
It is possible to approximate this integral using finite difference recognizing that at the origin
the function approaches infinity and leads to errors. In difference form the equation is
s
q(1   2 ) x y
 f (0,0)  2 f (5,0)  f (10,0)  2 f (0,5)  4 f (5,5)  .....
E
2 2
Assuming the origin is at the corner of a square area 10X10 units in dimension and an
increment of 5 units. The results are not too good at the corner but as the origin gets further
away from the loaded area the approximation gets better.
With the origin of the approximate at (0,0) the settlement is 70% of the closed form solution.
When the origin is 5 units away from the corner the approximate solution is 130% of the closed
form solution.
The closed form integration produces an elliptical integral and is evaluated with the aid of
tables. Selected results are given below.
s e (corner ) 
qB
1  2 

(1   2 )  I 3  (
)I 4 
2E
1


where I3 and I4 are from Table 8.7 and 8.8. At the surface this is
se 
qB
(1   2 ) I 3
2E
Using superposition as in stress determination the settlement at the center is twice the settlement
at the corner.
s e (center ) 
qB
(1   2 ) I 3
E
For non squares the settlement is
s e (center) 
qB
(1   2 )
E
where  is from table 8.9. This can be shown by the following example for a rectangular footing
with width = 10 and length = 20
4qB
(1   2 ) I 3
2E
4q5
s e (center ) 
(1   2 )1.532
2E
q
s e (center )  (1   2 )15.32
E
but
s e (center ) 
B  10
therefore
s e (center ) 
qB
(1   2 )1.532
E
This is the same as
s e (center) 
qB
(1   2 )
E
where  is 1.532 for a L/B ratio of 2.
The same superposition can be used to find the settlement at other locations. For example at
position A on the center of the edge of a square footing the value is
2q B
(1   2 )1.532
2E 2
qB

(1   2 )0.766
E
s edge 
s edge
The average settlement for a rectangular footing can be obtained by numerical procedures as in
calculating the average stress. In this case the influence factor that is associated with each
location is multiplied by the node coefficient and the sum is divided by the total number of node
coefficients to get the answer. For example for a square footing with the following coefficients
.561
.766
1.12
2
The result is 1x(4x.56) + 4(2x .766) + 1x(4x1.122) = 12.848 divided by the sum of
4x1 + 4x2 +1x4 = 16 with the result the average influence value is .803 which is 0.72 times the
center value. Better results can be obtained using Simpson’s rule or by using more subdivisions.
The same result can be obtained from
s avg 
qB
(1   2 ) '
E
where ’ is in Table 8.9
Like wise the settlement for a rigid footing is assumed to be 7% less then average settlement
s rigid 
qB
(1   2 ) r
E
where r is also from Table 8.9 In general
 '  .85
and
 r  .93 '  .79
For a square footing
Lecture 23
Settlement in finite layers can be done for rectangular footings as was done with
circular footings.
s e  s ez    s eZto
For example for a layer with z = B the result is
s e (center ) 
qB
qB
(1   2 )1.122  .84  .282
(1   2 )
E
E
For z = 10B the result is
s e (center ) 
qB
qB
(1   2 )1.122  .126  .996
(1   2 )
E
E
As the z/B ratio gets large the value approaches 1.122 for the infinite layer case (The influence
factors are from Table 8.7).
It is possible to approximate settlement using the increment layer technique as with the circular
footings.
se   avgi H i
where the average strain should be obtained for the actual footing type.
http://www.cee.mtu.edu/~balkire/ce584/strainrectftg.mcd In many cases this may not be available and
the use of the circular footing equation (Eq. 8.17) may be used as an approximation.
Embedment of footings general causes a reduction in the settlement. There are equations and
charts in Poulos and Davis’s book and they can be consulted for specific cases. A general
solution is to multiply the settlement by some factor as suggested in the book Eq 8.33
s e (avg)  1 0
qB
E
with the factor 0(Fig 8.13 a) being used to modify the computed value for embedment. Note
the correction for shallow footings (Df =B) is not great and amounts to .97 or less. the correction
for depth to rigid base is 1 and for large values of H/B is the same as obtained by using the
general equations
s e (avg)  .84 x.75 x1.122
qB
qB
 .7
E
E
As the value of L/B approaches five the difference between the curve value and a computed
value using the incremental approaches increases. for more detail on how the curve values were
determined see the article by Christian and Carrier.
http://www.cee.mtu.edu/~balkire/ce584/settrectftg.mcd
Lecture 24
Schmertmann’s technique
This technique is quite straight forward and is easy to justify for the circular footing. In this
case the strain at a depth is calculated using Eq. 8.17 and the settlement is determined by
determining th4e area under the strain versus depth curve. If it is assumed that the maximum
strain of 0.5 occurs at b/Z = 1 or B/Z = .5 and is 0 at b/Z = 4 or B/Z = 2 then the equation for a
homogenous soil is
s e (center ) 
qB
2
.5qB
(.5 x ) 
E
2
E
or
s e (center ) 
qB
.84qB
(.75)(1.122) 
E
E
The difference between Schmertmann and theoretical is substantial and is modified by using
corrections to the Schmertmann equation. For depth of foundation
C1  1  0.5(
q
)
q'q
Where q’ is the stress applied at the footing level and q is the effective stress associated with the
soil above the footing level. The other correction is for creep effects and is
t
C 2  1  0.2 log( )
.1
Where t is time in years after construction. Together the equation for settlement is
se (center)  C1C 2 (q'q)
Iz
z
Ez
In this form it is possible to account for different modulus values
I avgn 
 I avgi
se (center)  C1C 2 (q'q) 
 ... 

En 
 Ei
The modulus values are difficult to determine and Schmertmann used the Dutch cone
penetration test to obtain E. In metric units E = 3.5qc (L/B.10) and qc = 2,5qc (L/B,10) where
the qc is the penetration value in kN/m2.
Influnce factors for Schmertmann's Method
z/b
0
0.3
0.5
1
2
4
6
8
Table 3.9 Table 3.10
A'
B'
u=.5
1
0
0
0.71265 0.26362 0.39543
0.55279 0.35777 0.536655
0.29289 0.35355 0.530325
0.10557 0.17889 0.268335
0.02986 0.05707 0.085605
0.01361 0.02666 0.03999
0.0072 0.01526 0.02289
Circle Loaded Area
b =radius
Strain
u=.3
u=0
.6-4b cur
0.52
1
0
0.713284 0.97627
0.36
0.752552 0.91056
0.6
0.611918 0.64644
0.515
0.287453 0.28446
0.343
0.089718 0.08693
0
0.041735 0.04027
0.023582 0.02246
Influnce values
Circle Load
Influnce Value
1
u = 0.3
u = 0.5
u=0
4b-.6
0.5
0
0
2
4
6
Z/b Ratio
8
10
Rectangular Footing calculations
Z/B
0
0.25
0.5
1
2
4
6
8
10
Table 8.7 Table 3.10
L/B=1
L/B=10
L/B=5 L/B=1
1.122
2.544 2.105
1.095
2.525 2.085 0.108
1.025
2.473 2.032
0.28
0.838
2.322 1.878 0.374
0.552
2.026 1.571 0.286
0.306
1.621 1.147 0.123
0.208
1.355 0.884 0.049
0.158
1.16
0.71 0.025
0.126
1.009 0.589 0.016
Strain Avg
Area = Strain *del H
L/B=10 L/B=5 L/B =1 L/B=10 L/B=5 2B - 5 Cur
4B-.5 Cur
X val
0.1
0.2
0
0.076 0.08 0.027 0.019 0.02
0.5 0.35 0.5
0.208 0.212 0.07 0.052 0.053 0.25
0.5
1
0.302 0.308 0.187 0.151 0.154
0 0.25
2
0.296 0.307 0.286 0.296 0.307
0
4
0.2025 0.212 0.246 0.405 0.424
0.133 0.132 0.098 0.266 0.263
0.0975 0.087 0.05 0.195 0.174
0.0755 0.061 0.032 0.151 0.121
0.57 0.923 0.958
Influnce value rectangular foot
0.6
Strain
0.5
L/B = 1
L/B = 10
L/B =5
2B-.5
4B-.5
0.4
0.3
0.2
0.1
0
0
5
10
15
Z/B
The calculations above show that the influence values for the curve given by Schmertmann
under estimate the values obtained from the influence calculations. For example the square
footing result is
Theoretical
s e (center ) 
qB
qB
(1   2 )1.122  .84
E
E
Schmertmann
s e (center ) 
qB
1
qB
(.3x.5  x1.5 x.5)  .53
E
2
E
Tabulated above
s e (center)  .57
qB
E
For L/B = 10 and  = .5 the results are
Closed form
s e (center ) 
qB
qB
(1   2 )2.544  1.90
E
E
Schmertmann
s e (center ) 
qb
1
qB
(.35 x1  3x.5)  1.1
E
2
E
Tabulated above
s e (center)  .923
qB
E
It can be seen that the closed form values are greater then the Schmertmann values and
for large L/B ratio gets quite large. This is expected as the Schmertman technique does not take
into account the strain that occur in the soil below the distance of 4B or greater.
The spread sheet above is located in http://www.cee.mtu.edu/~balkire/ce584/schmert.xls
Lecture 25
In general it can be shown that a general state of strain can be reduced to two
components associated with the change in volume (spherical)and the change in shape
(deviatoric). For certain loading conditions the constitutive equations reduce to relatively
simple expressions that can be easily evaluated. For example volumetric compression when all
stresses are equal and the shear stresses are 0 have
 vol   x   y   z
1
 z   (2 z )   z   (2 z )   z   (2 z )
E
3(1  2 )
(1  2 )

 z  3 z 
z
E
E
 vol 
 vol
With the result that the volumetric settlement is three time the settlement in the vertical
direction. A similar development can be done for confined compression where the strain in the
x and y direction is 0. In this case the volumetric and vertical compression are equal and the
vertical strain is
 vol   z 
(1   )(1  2 )
z
(1   ) E
and vertical and volumetric settlement are the same. Note that the vertical stress under a
circular load of radius b is


z3
 z  q 1  2
2 3/ 2 
 (b  z ) 
and settlement is
se    z dz 
(1   )(1  2 )
 z dz
(1   ) E 
Which is
se  2
(1   )(1  2 )
qb
(1   ) E
for a circular loaded area at the surface of a infinite half space with radius b.
Lecture 26
A more general relationship for tri axial loading can be obtained in a similar manner
where
x  y
and
x  y
with the result that
 vol 
z
E
(1   ) 
2 x
(1  2 )
(1  2 ) 
( z  2 x )
E
E
by defining x and z in terms of x, z,  and E and solving the two equations it is possible to
determine E and . Experimental measurements can then be used to solve for these parameters.
From this discussion it can be seen that volumetric strain can be defined in terms of elastic
parameters and using conventional one dimensional consolidation tests it is possible to define the
compression index in terms of these parameters. (See Lambe and Whitman for more detail)
Lecture 27
As explained above it is possible to describe volumetric strain in terms of elastic
parameters and it should be possible to evaluate vertical strain in terms of vertical stress and
some parameters. The problem is in determining the parameters and various tests are used to
define the relationships. In the one dimensional test (constrained compression), the relationship
was defined in terms of E and . These terms are not usually defined in the one dimensional
test, but it is possible to relate the the values obtained in the usual test to these values. for
example, the compression index obtained in the (1-D test) is given as
e
Cc 
log
 zf
 zi
but
 z   vol 
z 
e
1  eo
 zf (1  2 )(1   )
Cc

log

z  z
1  eo
 zi
(1   ) E
D
therefore
Cc 
(1  eo ) zavg
.435 D
Thus, one dimensional settlement is a subset of settlement in general and elastic approaches are
valid if the elastic parameters are determined from suitable tests. However it is observed that
the elastic parameters are non linear and other more suitable techniques are used to evaluate
settlement. These will be explained below.
Lecture 28
If it is valid that the elastic parameters are linear in a certain range of stress application
then the resulting settlement ijn the range can be evalfuated in a atraight forward manner using
the relationship
s e    zii H i
where
 ziavg  K i iavg
In this expression the coefficient K is assume to be constant in the interval H and is determined
from one of the coefficients used to describe the relationship between stress and change in strain
such as mv, av, Cc and/or D. In using this approach the only problem is to determine where the
average stress is located in the layer. The usual procedure is to assume the center of the layer
but there are errors involved in using this assumption as can be shown with the following
calculation.
trueaverage  
 ( z )dz
z top  z bot
and
error  
 ( z )dz
z top  z bot

 ( z top )   ( z bot )
z top  z bot
where


z3
 ( z )  1  2
2 3/ 2 
 (b  z ) 
then
error 
z top
q
 z bot

 

z 3 top
z 3 bot
 1  2

1




2
2
zb o t
  (b  z top ) 3 / 2   (b 2  z bot ) 3 / 2  


z3

 1  (b 2  z 2 ) 3 / 2 dz  
2
zto p 





The result above is for a circular loaded area with radius of b. Other shapes could be
considered it (z) can be defined. The solution to the equations given above is
http://www.cee.mtu.edu/~balkire/ce584/avgstress.mcd
Lecture 29
The use of the equation for finding the average stress in a layer can
be applied to the following conditions:
B = 10, q = 1000, Ztop=10 and Zbot=20 with the following results
avg=.438q and error = 0%
trap=.465q and error = ((.465-.438)/.438)*100 = 6.2%
simpson=.437q and error = ((.437-.438)/.438)*100 = .2%
centerline=.423q and error = ((.424-.438)/.438)*100 = 3.4%
As can be seen the amount of error is small and for Simpson’s rule is
average = top+4*centerline+bot This may not always be the came ratios as the
actual values will be a function of location with respect to depth and the
thickness of the layer.
From the discussion above the following method is proposed for evaluating
the average stress or strain for a soil layer.
Best is the actual average obtained from the integration of the function
followed by Simpson’s approx., Centerline approx. and finally the
trapezoidal rule approximation.
Lecture 30
Summary of the main points covered in the course including
techniques to determine the effects of loading on the development of
stress and strain in the elastic half space. Particular emphasis is on the
use of integration techniques to obtain the actual solutions. Various
approximations and software packages were used in cases where the
true solutions were not readily obtained by closed form techniques.
Other areas that could be discussed include modulus
determination, more SAP work and may be some thing associate
with the elliptical integrals.
Settlement Square http://www.cee.mtu.edu/~balkire/ce584/settsqftg.mcd
Strain Circle http://www.cee.mtu.edu/~balkire/ce584/straincalc.mcd
Mathcad Solutions and Assignments
Solutions
Beam on Elastic Found
http://www.cee.mtu.edu/~balkire/ce584/bending.mcd
Regression Eq http://www.cee.mtu.edu/~balkire/ce584/regress.mcd
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