Engineering and Construction,

Unconfined Compression Test and Design ( Civil Engineering Lab Report )
Objectives:
Almost any type of structure rests either directly or indirectly upon soil. Proper analysis and design of a structure’s foundation based on the condition of the soil on which it sits must be done so that the structure does not collapse or undergo large amounts of settlement. Geotechnical engineering is the sub-field within civil engineering that involves testing soils, analyzing soil behavior, and designing the underlying foundations for structures. In this lab, you will perform one of the simplest but most widely used soils tests and use the results to select the required size for a foundation, much as a civil/geotechnical engineer would do. Your objectives are:

1. Calculate the required size for a building column footing under a given load.
2. Determine the effect of varying water contents on the compressive strength of the soil.

Background:
Soils may be separated into three very broad categories: cohesionless, cohesive, and organic soils. In cohesionless soils (“sandy” soils including gravel, coarse and fine sands, and silt), the soil particles do not stick together. Cohesive soils (“clays”) are more plastic and have very small particles which do tend to stick together, the result of water-particle interaction and attractive forces between particles. Organic soils are spongy, crumbly and compressible and are not desirable for use in supporting structures. Soils encountered in practice are often mixtures of different types of soil and thus exhibit characteristics from each.

Foundations can be either shallow or deep. Shallow foundations (footings and mats) are located just below the lowest part of the structures they support, while deep foundations (piles and drilled shafts) extend far down into the earth. The individual footing (Fig. 1) is the simplest of all foundations, and may be thought of as a simple flat plate or slab, usually square in plan, and acted on by a concentrated load Q (coming from the column pushing down) and a distributed load q (coming from soil pressure pushing up) as shown in Fig. 2. The enlarged size of a footing, compared to the column it supports, gives an increased contact area (area = B2 for a square footing as seen in Fig. 2) between the footing and the underlying soil. The increased area reduces the footing contact pressure on the soil to an allowable amount (allowable bearing capacity qa) and prevents excessive settlement or bearing failure of the foundation. The size of the footing will depend on the column load Q that the footingmust support (see Fig. 2) and the softness of the ground that it will stand on, as given by Eq. 1.

q = Q/B2 = qa EQ (1)
where q = footing contact pressure in psf = lbs per ft2
Q = total vertical load from column in lbs
B2 = area of square footing in ft2
qa = allowable bearing capacity of underlying soil in psf = lbs/ft2

Figure 1 – Footings under construction

Figure 2 – Individual Footing
Soil, like concrete and rock, cannot resist large tension forces. It is capable of resisting compression to some extent, but large compressive forces (such as the load P from a column) cause failure usually in the form of shearing along some internal surface within the soil. Thus, structural strength of soil, as measured by the soil’s bearing capacity is primarily a function of the soil’s shear strength, where shear strength refers to the soil’s ability to resist sliding along internal surfaces within a mass of the soil.

Since the ability of soil to support an imposed load is determined by its shear strength, the shear strength of soil is very important in foundation design. Determination of shear strength is one of the most frequent and important problems in soil mechanics (the underlying theory used in geotechnical engineering). Soil gains its shear strength from two sources – internal friction and cohesion. Internal friction is the main source of shear strength for cohesionless soils (sands), while cohesion predominates for clays.

The unconfined compression test is a quick and simple way to measure the unconfined compressive strength qu and undrained shear strength su of cohesive soils. This information can be used to estimate ultimate bearing capacity and choose a suitable allowable bearing capacity qa for footing design. A cylindrical soil specimen is formed with a length of 2 to 2 ½ times its diameter. The specimen is placed in a compression testing machine and subjected to an axial force, as shown in Fig. 3. The force is applied to produce axial strain at a rate of ½ to 2% per minute, and the resulting stress and strain can be calculated from measurements of force and deflection, using Equations 2, 3 and 4.

e1= ?h/ho EQ (2)

A = A0/(1 – e1) EQ (3)

s1 = P/A EQ (4)

where e1 = strain (dimensionless)
?h = axial deformation = change in length of soil sample in inches
ho = initial length of soil sample in inches
A = cross sectional area of sample in in2
A0 = initial area of sample = pd02/4
d0 = initial diameter of the soil sample in inches
s1 = stress in psi = lbs/in2 (Note: multiply by 122 = 144 to get psf = lb/ft2)
P = applied force in lbs

Figure 3 – Typical Unconfined Compression Test Setup

Note that as the sample is compressed, cross-sectional area A increases a small amount as the initial diameter d0 “bulges out” under load. For an applied load P, the corrected cross sectional area A is calculated from original area A0 if strain e1 is known.

A graph is plotted of stress versus strain (see Fig. 4), and the unconfined compressive strength qu is defined as either the maximum stress at the peak of the stress-strain curve (stiff clays), or the stress when e1 = 0.15 (or 15%) strain, whichever comes first. Once the unconfined compressive strength has been determined, the shear strength su (equal to the cohesion for clays) is equal to one-half the unconfined compressive strength qu, as seen in Fig. 5.
qu = unconfined compressive strength
Figure 4 – Stress-Strain Diagram from Unconfined Compression Test

From Mohr’s Circle geometry, shear strength su = ½ qu

Figure 5 – Mohr’s Circle for an Unconfined Compressive Test

There are many different equations used to calculate ultimate bearing capacity once the unconfined compressive strength and undrained shear strengths are known. As our soil samples consist of clay, we will use the Prandtl equation (Eq. 5), which is one of the simplest models available and is a suitable assumption for foundations built on saturated clay.

qult=(p + 2)c = 5.14su EQ (5)

whereqult = ultimate bearing capacity in psf = lbs/ft2
c = cohesion = undrained shear strength su = ½ qufor clays

To design a foundation, we need the allowable bearing capacityqa which is found by dividing the ultimate bearing capacity from Eq. 5 by a factor of safety FS. Geotechnical engineers use a factor of safety = 3 for foundations built on clay, and putting it all together we end up with the allowable bearing capacity as given by Eq. 6.

qa = qult/FS = 5.14( ½ qu)/3 = 0.86qu EQ (6)

Setting the footing contact pressure q in Eq. 1 equal to the allowable bearing capacity qa (from Eq. 6), the required size of footing B for a clay soil with unconfined compressive strength quandknown column load Q can be calculated, as given in Eq. 7.

B = v(Q/qa) EQ (7)

where v = square root
B = size of square footing in feet
Q = column load in lbs
qa = allowable bearing capacity in psf = lb/ft2 (from Eq. 6)

Lab Procedure:
1. Prepare a test specimen of soil with a known percentage of water (by weight). Choose a water content between 11% and 15% for your test specimen. For example, if you chose 15% water content, then a 150 gram mix would consist of:

150 grams * 0.15 = 22.5 grams of water
150.0 grams – 22.5 grams = 127.5 grams of dry clay.

Weigh out enough soil and water to ensure that you will have more than enough mixed soil to overfill your soil mold.

2. Once appropriate amounts of soil and water have been weighed out and mixed together, fill the mold with the soil-water mixture, compacting the mixed soil in the mold in three equal layers. Compact each layer with 25 blows using the tamping device. Score the top of the first and second layers to ensure a good bond to the succeeding layers.

3. Remove the soil sample from the mold, and weigh and measure dimensions. Record initial length Lo, average diameter do (take measurements of diameter at the top, middle, and bottom of the specimen and average them), and moist mass of the specimen in grams.

4. Perform an unconfined compression test, according to the ASTM (American Society of Testing and Materials) test procedure ASTM D2166: Standard Test Method for Unconfined Compressive Strength of Cohesive Soil. Place the soil sample in the load frame and load it at a strain rate between 0.5-2.0% per minute. Load the sample until e1> 0.15 (i.e. 15%). Take readings of load and deformation frequently enough to fully define the peak of the stress-strain curve during the test. If your load frame is configured with a load cell, your reading will be load P directly, in units of force (note that your calculations will need P inlbs). If your load frame is configured with a proving ring instead of a load cell, your reading will be GP and will be in units of divisions. For proving rings, P is calculated as GP x KP where KP = proving ring constant. Your lab instructor can show you how to read the load frame for deformations and loads, and will help you find the proving ring constant, if necessary. Record all data on the Unconfined Compressive Strength Test Data Sheet.

Calculations and Lab Report:
1. From the measurements of load and deformation taken in lab, calculate strainse1, corrected areas A, and corresponding stresses s1from Eqs. 2, 3, and 4.

2. Make a graph of stress s1 versus strain e1 (plot e1 on the horizontal (“x”) axis and s1 on the vertical (“y”) axis). Use the graph to find the unconfined compressive strength quas either the peak value of s1 or the value of s1 at e1 = 0.15 (15% strain), as shown in Fig. 4. Report this value in lbs/ft2 (psf). As your measurements were made in inches, your graph will give the unconfined compressive strength qu in lbs/in2 (psi). To convert unconfined compressive strength to the more common lbs/ft2 (psf) units used in foundation design, you multiply qu in psi by 144 (1 ft2 = 122 in2 = 144 in2) to get qu in psf. Oncequ is known, the soil’s shear strength su can be found as shown in Fig. 5.

3. Is your sample defined as a soft or stiff clay? Why?

4. Calculate the allowable bearing capacity (in psf) for your soil sample using Eq. 6, and then choose an appropriate size square footing for a column supporting a vertical load Q = 35,000 lbs from Eq. 7.

5. Compare the strength of your soil sample to the other samples tested in your lab section, by plotting the unconfined compressive strength quversus water content (qu on the vertical “y” axis versus % water content on the horizontal “x” axis). Your lab instructor will post results from all lab groups. In your lab report, discuss how % water content influenced the strength of the soil samples.

6. Discuss potential errors that could have occurred during the lab and provide recommendations to improve the accuracy of the lab results.

Unconfined Compression Test ASTM D2166Group Number: 4 Lab session (Time): Fri 12 – 2
10-05-12

Weight of dry soil = 127.5 g
Total weight of soil (after adding water) = 150 g
Water content = 15% %
Initial height h0 = 2.79 inches
Average initial diameter d0 = 1.29 inches
Initial area A0 = 1.31 in2
Mass of wet sample = 122.3 g
Sample wet density = mass of sample/(A0* h0) = 33.46 g/in3
Proving ring constant KP = 0.923 if GP < 705 Divisions
*If > 705 divisions – use equation from chart.
Unconfined compressive strength qu = psi = lbs/in2
Unconfined compressive strength qu = psf = lbs/ft2
Unconfined shear strength su = ½qu psi = lb/in2

Displacement (in)
Strain = e1(in/in)

Load Dial Reading
In Divisions 1 Division =0.0001”
(x0.0001
“)

Load = P (lbs)
Area = A (in2)
Stress = s1
(psi = lbs/in2)
?h = ?h / h0 GP =KP x GP* = A0/(1-e1) = P/A
0.01 0.5
0.02 0.5
0.03 1
0.04 1
0.05 2
0.06 2
0.07 1
0.08 1
0.09 1.5
0.10 2
0.11 2
0.12 2.5
0.13 2.5
0.14 3
0.15 3
0.16 3.5
0.17 4
0.18 4
0.19 4.5
0.20 4.5
0.21 5
0.22 5
0.23 5.5
0.24 5.5

Displacement (in)
Strain = e1(in/in) Load Dial Reading
In Divisions 1 Division =0.0001”
Load = P (lbs)
Area = A (in2)
Stress = s1
(psi = lbs/in2)
?h = ?h / h0 GP =KP x GP* = A0/(1-e1) = P/A
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