Notes for CE 5810

Karl Terzaghi -= Father of Soil Mechanics

Lecture 1Review 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 conditionsEquations 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 3Develop mohr’s circle of stress for normal and general

conditionsGeneral 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 solutionhttp://www.cee.mtu.edu/~balkire/ce584/mohrcircle.mcd

Perpendicular Planes http://www.cee.mtu.edu/~balkire/ce584/assign2.mcd

Lecture 4Discuss 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 5Horizontal 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 compressiongives 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 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

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 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 insteadof 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)

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

(parabolic)

(circle)

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:

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 11Extend 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 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

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.

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

z3 z3. q.

2 p. b.

0

2 p.

0

b

rb2 r2( )

1

2 r.

r2 z2( )5

2

d d.

For conical

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

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 http://www.cee.mtu.edu/~balkire/ce584/rectload.mcdhttp://www.cee.mtu.edu/~balkire/ce584/ce584sqf.mcdhttp://www.cee.mtu.edu/~balkire/ce584/straineq.mcd

z3 z3. q.

2 p. b.

0

2 p.

0

br

b r( ) r

r2 z2( )5

2

d d.

z x y,( )

1.5

4.5y

1.5

1.5x3

2 p.q. z3. 1

x2 y2 z2 2.5. d d

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 15Numerical 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 formwith 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 16Another 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

I b/z0.1 .270.2 .400.3 .520.4 .640.5 .770.6 .920.7 1.110.8 1.390.9 1.910.10

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 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 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

http://www.cee.mtu.edu/~balkire/ce584/setpointload.mcdhttp://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

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/b I s0 0 0.5 .106 .106 s

1.0 .293 .293 s

2.0 .553 .553 s

4.0 .757 .757 s

10. .9 .9 s

1.0 1.0 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). http://www.cee.mtu.edu/~balkire/ce584/elsym5.exe

10 105Rigid Base

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

10

101020 E=10000 psfE=20000 psfE=50000 psf

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

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

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

where ’ is in Table 8.9

Like wise the settlement for a rigid footing is assumed to be 7% less then average settlement

where r is also from Table 8.9 In general

.561 .7661.122

For a square footing

Lecture 23

Settlement in finite layers can be done for rectangular footings as was done with circular footings.

For example for a layer with z = B the result is

For z = 10B the result is

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.

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

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

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

The difference between Schmertmann and theoretical is substantial and is modified by using corrections to

the Schmertmann equation. For depth of foundation

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

Where t is time in years after construction. Together the equation for settlement is

In this form it is possible to account for different modulus values

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.

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

Schmertmann

Tabulated above

For L/B = 10 and = .5 the results are

Closed form

Schmertmann

Tabulated above

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

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

and vertical and volumetric settlement are the same. Note that the vertical stress under a circular load of

radius b is

and settlement is

Which is

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

with the result that

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

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

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.

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 isaverage = 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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