zstress2.0 documentation

7/28/2019 ZStress2.0 Documentation

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Virginia Polytechnic Institute

And State University

The Charles E. Via, Jr.

Department of

Civil and Environmental Engineering

CENTER FOR

GEOTECHNICAL PRACTICE AND RESEARCH

ZSTRESS 2.0: A Computer Program

for Calculation of Vertical Stresses

Due to Surface Loads

By

Bingzhi Yang

and

J. Michael Duncan

Center for Geotechnical Practice and Research

200 Patton Hall, Virginia Tech

Blacksburg, Virginia 24061-0105

July 2002


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INTRODUCTION

This computer program is an updated version of a program with the same name writtenin 1987 by Rick Allen and Mike Duncan. The original program, which operated under DOS,

has become outdated and inconvenient by todays standards. This new version, which is

Windows-based, is easier to use. The program (called Zee-Stress) computes vertical stressesdue to surface loads, which are useful in computing settlements. The program calculates

stresses due to point loads and loads distributed over rectangular areas, using the Boussinesq

and the Westergaard equations.

ZSTRESS 2.0 was written in Microsoft Visual Basic 6.0. The program is interactive,

and has facilities for creating, storing and editing data files, and for printing results.

METHOD OF ANALYSIS

ZSTRESS 2.0 calculates changes in vertical stresses due to surface loads using both the

Boussinesq and the Westergaard solutions for point loads and for loads distributed uniformlyover rectangular areas. The equations used within the program are:

Boussinesq Equations (Poulos and Davis, 1974)

Point Load:5

3

2

3

R

zQ

=z

Where:

z = Change in vertical stress (stress)

= magnitude of point load (force)Q

z = depth of stress point (length)

R = 222 zyx ++ (length)x = distance from load to stress point in x-direction (length)y = distance from load to stress point in y-direction (length)

Rectangular Load:

+

+

=

2

2

2

133

1 11tan2 rrr

BLz

zr

BLqz

Where:

z = change in vertical stress beneath cornerof rectangular loaded area (stress)

q = magnitude of surface pressure (stress)

B = width of rectangle (length)L = length of rectangle (length)z = depth of stress point (length)

22

1 zBr += (length)

22

2 zLr += (length)

222

3 zLBr ++= (length)

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Westergaard Equations (Fadum, 1948)

Point Load:5.122 )5.0(22 zr

zQ

+=

z

Where r=22

y+x (length)

The other terms are the same as for the Boussinesq point load equation.

Rectangular Load:

++=

5.0222

1

)5.0(

2tan

2 zLBz

BLqz

Where the terms are the same as for Boussinesq rectangular load equation.

Superposition is used to compute stresses due to multiple loads, and to compute stressesat points that are not beneath the corners of distributed loads.

The Boussinesq equations apply for any value of Poissons ratio. The Westergaard

equations are for Poissons ratio equal to zero; it has become standard practice in geotechnicalengineering to use these forms of the Westergaard equations. The stresses calculated using

these Westergaard equations are smaller than those calculated using the Boussinesq equations.The equations for the Westergaard solution for Poissons ratio larger than zero lead to even

smaller stresses, and are seldom used.

The positions of the loads and the locations of the points at which the stresses are

calculated are specified using an x-y-z coordinate system, as shown in Figure 1. The

coordinates x and y are horizontal, whereas z is vertical and positive downward.

The origin of coordinates can be placed at any convenient location, and the value of x, y

and z for loads and stress points can be positive or negative.

Loads can be applied at any value of z. However, the equations used in calculating the

stresses were derived for loads applied at the surface of an elastic half-space. The program is

therefore best suited for calculating stresses due to loads applied at the surface or near thesurface.

The formulas for stresses due to distributed loads used in ZSTRESS 2.0 are for thevertical change in stress beneath the corner of a uniformly loaded rectangular area. The

program uses superposition techniques to calculate changes in stress at other positions.

Stress points are the points where stresses are calculated. As seen in plan, the stresspoint can be inside, outside, or on the boundary of the loaded area. Stress points can be

located at any depth. If a stress point coincides with the position of a point load, the stress is

theoretically infinite, and no value is given. If a stress point falls within a rectangular loadedarea and is at the same depth as the load, the calculated stress is equal to the applied load (q).

If a stress point falls on the edge of a loaded area (at the same depth), the change in stress is

0.5q. If a stress point falls on the corner (at the same depth), the change in stress is 0.25q.

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z = depth, positive down(vertical)

Vertical change in stresses at any location= combined effect of all loads.

z

(Horizontal)x

B

LUniformly loaded rectangularareas at any locations.Load intensity = q

Point loads at any locations.Load magnitude = Q

(Horizontal)y

(a) Coordinate system

y

x

B

L

(x,y)x>0y>0

Origin of coordinates

B(x,y)

y0

L

B(x,y)

y>0x


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REPRESENTATION OF BUILDING LOADS,EXCAVATION LOADS AND FILL LOADS

With ZSTRESS, three methods can be used to represent the loads of buildings:

1) If each column is supported by an individual spread footing, each footing can be

represented as a rectangular distributed load, with the size of the rectangle equal to

footing size. This is the most accurate procedure, and it requires the most input.

2) Alternatively, each column can be represented as a concentrated point load.

Although the stress at the point of application is infinite, the stress at distancesequal to twice the footing width from the point of application will be essentially

the same as those calculated using procedure 1). This procedure is used sometimes

because less input is required.

3) A third possibility is to treat the entire load of the building as if it is uniformlydistributed over the building area. This is the logical choice if the building is

supported on a continuous mat foundation. It may also be a convenient andsufficiently accurate procedure for buildings founded on spread footings. At

distances comparable to the spacing between footings, this procedure gives

essentially the same results as procedures 1) and 2). Because only one load need bespecified for the entire building, this alternative requires the least input.

When construction of buildings is accompanied by excavation, the excavation results ina negative component of load. If the excavation covers only the area of the building, the

negative changes in stress due to excavation can be calculated by treating the reduction insurface pressure due to excavation as a negative distributed load. If the excavation covers a

very large area around the building, the reduction in stress due to excavation will not decrease

much in magnitude with depth, and no stress distribution calculations need be made todetermine the magnitude of the stress change due to the excavation.

Similarly, when construction of buildings is accompanied by placement of fill, the fillload results in an increase in stress that adds to the stress due to the building loads. The

principles of calculating the stresses due to the fill are the same as for calculating stress

changes due to excavation, except that fills result in positive changes in stresses in the

underlying soil, whereas excavations result in negative changes in stresses.

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USE OF ZSTRESS 2.0

The program can be installed by running the SETUP.EXE file that accompanies this

users manual.

During execution, the program displays the window shown in Figure 2. The user mayinput units for length and force, point loads and rectangular loads, the locations of stress

points, and comments in any order. For rectangular loads, B is the width of the rectangle in

the x direction and L is the length of rectangle in the y direction, as shown in Figure 1. Thereis no limit for the number of point loads, rectangular loads, or stress points.

Whenever the ENTER key is pressed or the mouse clicks at another input cell, theprogram checks the input data. If the user inputs an invalid number, the program pops up a

message Invalid data. When the user clicks the Compute Stresses button, the program

checks all the data and computes the stresses. If invalid data or missing data are found, itdisplays messages in the Error messages section of the window. Otherwise, it computes the

stresses.

The user may change the display format of the computed stresses by clicking on theFormat menu. The default format is floating point with two digits after the decimal point.

The user may save the input data to file at any time. Up to five files can be open at thesame time. The program assumes that data files have the extension DAT (for example,

PROBLEM1.DAT), and it is suggested that this convention be used.

The program provides two procedures to print the results. One is to click the print

command in the File menu. This procedure prints the output without page settings. Another isto click the Save Output as command in the File menu. This procedure saves the output as a

text file, which may be opened using a word processor, and thus can be printed with page

setup. The commands Save data and Save data as in the File menu are for saving data to adata file, which can be opened later by ZStress program.

It should be noted that the program only displays the units the user inputs. It does notconvert units. Consistent units must be used.

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Figure 2 ZSTRESS 2.0 window

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EXAMPLE

An example problem is shown in Figure 3. An eight-foot deep excavation results in a

negative distributed load of q = -1000 psf over the building area, and nine columns apply

positive loads through spread footings. The net load (due to both excavation and the building)is positive.

The excavation load and footing loads can be represented in three ways:

1) Represent the excavation load and the footing loads as distributed loads. The results

are shown in Figure 4. This is the closest representation of the actual conditions.

2) Represent the excavation load as a distributed load and the column loads as point

loads. The results are shown in Figure 5.

3) Represent the excavation load as a distributed load and total building load as a singleuniform distributed load. The results are shown in Figure 6.

The results for these three methods of computing stresses are compared in Table 1. The

data files EXAMPLE1.DAT, EXAMPLE2.DAT, EXAMPLE3.DAT are included in the

program disk.

Table 1 Summary of Results for the Example Problem

Stress Point 1 Stress Point 2 Stress Point 3Method

Boussinesq Westergaard Boussinesq Westergaard Boussinesq Westergaard

1 523 383 369 273 233 176

2 650 482 401 300 243 185

3 423 303 223 171 115 95

Table 1 shows that treating the building load as a single uniform distributed load(Method 3) gives stress changes smaller than the most precise solution (method 1), while

treating column loads as point loads (method 2) gives stress changes larger than the most

precise solution. For stress points beneath the corner or the edge of the building, treatingcolumn loads as point loads gives results with reasonable precision. However, for the stress

point beneath the center of the building, which is directly beneath the largest footing, and

which carries the largest load, treating column loads as point loads gives results considerably

higher than the most precise solution.

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(corner)stress point 3

1ft

y

Excavation 8 ft deep

stress point 2 (wall)stress point 1 (center)

= 125 lb / ft 3

original ground level

29 ft 29 ft1 ft

x

Footing = 6 ft X 6 ftQ = 300,000 lb

Corner column load:

Wall column load:

Q = 550,000 lbFooting = 8 ft X 8 ft

Center column load:

Q = 1,000,000 lbFooting = 11 ft X 11 ft

24 ft

24 ft

1 ft

1 ft

2 ft X 2 ft columns

22 ft

8 ft

2 ft

z

Excavation Load = -(8)(125) = -1000 psf

Corner Footings: q = (300,000)/(66) = 8,300 psf

Wall footings: q = (550,000)/(88) = 8,600 psf

Center footings: q = (1,000,000)/(1111) = 8,260 psf

Equivalent uniform building load:

q = [(4300,000)+(4550,000)+1,000,000]/(6050) = 1467 psf

Figure 3 Example problem

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---------------------------------------------------------------------------

-------------------- Summary of Results and Input --------------------

---------------------------------------------------------------------------

Data File Name: Example1.dat

Date: 7/19/2002

Time: 11:05:12 AM

---------------------------------------------------------------------------

-------------------- Comments --------------------

---------------------------------------------------------------------------

Example problem #1 for the users guide

Treating excavation load and footings as distributed loads.

---------------------------------------------------------------------------

---------------- Stress Point Coordinates and Stresses ----------------

---------------------------------------------------------------------------

x y z Boussinesq Westergaard

No. foot foot foot lbs/square foot

----- --------- --------- --------- --------- ---------

1 30 25 22 523 383

2 30 0 22 369 273

5 0 0 22 233 176

---------------------------------------------------------------------------

-------------------- Point Loads --------------------

---------------------------------------------------------------------------

x y z Magnitude

No. foot foot foot lbs

----- --------- --------- --------- ---------

---------------------------------------------------------------------------

-------------------- Rectangular Pressure Loads --------------------

---------------------------------------------------------------------------

x y z B L Pressure

No. foot foot foot foot foot lbs/square foot

----- --------- --------- --------- --------- --------- ---------

1 0 0 0 60 50 -1000

2 -2 -2 2 6 6 83003 26 -3 2 8 8 8600

4 56 -2 2 6 6 8300

5 -3 21 2 8 8 8600

6 24.5 19.5 2 11 11 8260

7 55 21 2 8 8 8600

8 -2 46 2 6 6 8300

9 26 45 2 8 8 8600

10 56 46 2 6 6 8300

Figure 4 Output for example, treating the excavation load

and the footings as distributed loads

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----------------------------------------------------------------------------------------------- Summary of Results and Input --------------------

---------------------------------------------------------------------------

Data File Name: example2.dat

Date: 7/19/2002

Time: 11:10:20 AM

---------------------------------------------------------------------------

-------------------- Comments --------------------

---------------------------------------------------------------------------

Example problem for user's guide

Treating excavation load as distributed load, and column loads as point loads.

---------------------------------------------------------------------------


---------------------------------------------------------------------------


No. foot foot foot lb/square foot

----- --------- --------- --------- --------- ---------

1 30 25 22 650 482

2 30 0 22 401 300

3 0 0 22 243 185

---------------------------------------------------------------------------

-------------------- Point Loads --------------------

---------------------------------------------------------------------------

x y z Magnitude

No. foot foot foot lb

----- --------- --------- --------- ---------

1 1 1 2 300000

2 30 1 2 550000

3 59 1 2 300000

4 1 25 2 550000

5 30 25 2 1000000

6 59 25 2 550000

7 1 49 2 300000

8 30 49 2 5500009 59 49 2 300000

---------------------------------------------------------------------------


---------------------------------------------------------------------------

x y z B L Pressure

No. foot foot foot foot foot lb/square foot

----- --------- --------- --------- --------- --------- ---------

1 0 0 0 60 50 -1000

Figure 5 Output for example, treating the excavation load

as a distributed load, and the column loads as point loads

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---------------------------------------------------------------------------

-------------------- Summary of Results and Input --------------------

---------------------------------------------------------------------------

Data File Name: example3.dat

Date: 7/19/2002

Time: 11:20:20 AM

---------------------------------------------------------------------------

-------------------- Comments --------------------

---------------------------------------------------------------------------

Example problem for user's guide

Treating excavation load as a single distributed

load, and building load as a single uniform distributed load

---------------------------------------------------------------------------


---------------------------------------------------------------------------


No. foot foot foot lb/square foot

----- --------- --------- --------- --------- ---------

1 30 25 22 423 303

2 30 0 22 223 171

3 0 0 22 115 95

---------------------------------------------------------------------------

-------------------- Point Loads --------------------

---------------------------------------------------------------------------

x y z Magnitude

No. foot foot foot lb

----- --------- --------- --------- ---------

---------------------------------------------------------------------------


---------------------------------------------------------------------------

x y z B L Pressure

No. foot foot foot foot foot lb/square foot

----- --------- --------- --------- --------- --------- ---------1 0 0 0 60 50 -1000

2 0 0 2 60 50 1467

Figure 6 Output for example, treating the excavation load as a distributed load,and the building load as a single uniform distributed load

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REFERENCES

Fadum, R.E. (1948), Influence Values for Estimating Stresses in Elastic Foundations,

Preceedings 2nd

Int. Conf. on Soil Mech. And Found. Eng., Vol. 3, pp. 74-84

Poulos, H. G. and Davis, E. H. (1974), Elastic Solutions for Soil and Rock Mechanics, JohnWiley & Sons, New York.

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