Notes for CE 584

Notes for CE 5810

Karl Terzaghi -= Father of Soil Mechanics

Lecture 1

Review the void density relationships for soils and effective stress concepts

Also review the effective stress concept using an example that has pore pressure with static and excess components. Remember that

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

Three equations and nine unknowns

But by summing moments

xy = yx

therefore three equations and six unknowns

Using strain definitions

Six equation and nine unknowns

Finally for homogenous materials

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

and the principal stresses are

The intermediate stress is

Some Mathcad solutions

Mohr’s Circle of stress

Gen solution

Perpendicular Planes

Lecture 4

Discuss geostatic stresses in vertical and horizontal direction

For a situation with no stresses applied at the surface

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

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

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

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) polar coordinates

If the stress function is

by differential calculus and substitution results in

Use a mohr’s circle to develop the x-z coordinate system and

Solution to these equations

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

where  is measure

from the horizontal.

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

Lecture 7

Integrate line load to get the effect of strip load on semi-infinite space. Include numerical integration to show the solutions are obtained in several ways. Do several examples.

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 and

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)

Other loading types are easily used by defining dq in an appropriate manner. For example

(linear decreasing)

(parabolic)

(circle)

Results are given in the

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:

b = distance from center to edge of uniform load and a is he distance of the decreasing load.

Embankment Load

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.

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.

by integration where sum of vertical stress is equal to Q

and

Eq 3.47

by transformation

where

`

and

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

Lecture 12

A point horizontal load can be developed in a similar manner and the equation for

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)

Lecture 13

The point load equation is integrated over a circular loaded area at the surface to produce the following.

Eq 3.61

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.

For uniform load

For Parabolic load

For conical

NonCircular Load

Lecture 14

Loads on a rectangular area are done in the same manner. Integrate a differential area over the boundaries as shown below

A typical solution from the Mathcad program is

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

Where a and b are the dimensions of the footing

Some solutions are

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

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

as 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

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

solve for b/r ratios when I is set to even increments such as .1, .2, .3, etc and the results are as follows

Ib/z

0.1.27

0.2.40

0.3.52

0.4.64

0.5.77

0.6.92

0.71.11

0.81.39

0.91.91

0.10

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 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 POISSONS

LAYER MODULUS RATIO THICKNESS

1 4000000. .150 8.000 IN

2 30000. .400 6.000 IN

3 5000. .450 SEMI-INFINITE

TWO LOAD(S), EACH LOAD AS FOLLOWS

TOTAL LOAD..... 4500.00 LBS

LOAD STRESS.... 75.00 PSI

LOAD RADIUS.... 4.37 IN

LOCATED AT

LOAD X Y

1 .000 .000

2 13.110 .000

RESULTS REQUESTED FOR SYSTEM LOCATION(S)

DEPTH(S)

Z= 8.00

X-Y POINT(S)

X Y

.00 .00

6.56 .00

Z= 8.00 LAYER NO, 1

X Y

.00 .00

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 .1736E-06

UY .0000E+00 .0000E+00

UZ .1451E-01 .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

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

For the point load where the stresses are defined by the equations 3.45, 3.46 and 3.47

The result is

when integrated gives

The point load settlement equation

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

or

Substitute equation 3.61 and 3.62 into the equation and the result when  is 0.5 is

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.

Where I1 is from Table 3.9 and I2 is 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

which end up for  - 0.5 being

resulting in

z1/bIs

000

.5.106.106 s

1.0.293.293 s

2.0.553.553 s

4.0.757.757 s

10..9.9 s

1.01.0 s

Lesson 21

Another approach to the problem is to integrate the point load settlement expression over the area of interest.

As before with  = 0.5

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

when x/b = 0 and z/b = 1, and  = 0.5

The same as obtained from direct integration

Lecture 21

How to calculate average settlement? In general for a circular loaded area

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

For the case above this evaluates to

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

Where A’ and B’ are coefficients from Table 3.9 and 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.

By Eq 8.18 (closed form)

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).

The resulting equations are

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

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

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.

where I3 and I4 are from Table 8.7 and 8.8. At the surface this is

Using superposition as in stress determination the settlement at the center is twice the settlement at the corner.

For non squares the settlement is

where  is from table 8.9. This can be shown by the following example for a rectangular footing with width = 10 and length = 20

This is the same as

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