Shallow Foundations - Eccentricity - Problem 1

Consider the figure on the left. The following are given:

y = 120 pcf

phi = 30 degrees

Foundation soil = sand

Square Footing, B = 3.0 ft.

Df = 3.0 ft.

a. Calculate the ultimate bearing capacity. Applied moment, M = 50 kips-ft, applied load, P = 100 kips

b. Calculate the ultimate bearing capacity. Applied moment, M = 20 kips-ft, applied load, P = 100 kips

c. Calculate the ultimate bearing capacity. Applied moment, M = 100 kips-ft, applied load, P = 100 kips

I created these problems in a hurry. Please let me know if there are any errors. I'll post the solution tomorrow.

Bearing Capacity Equations - Which to use?

One of the readers of this blog recently posted a very good question that led me to dig deeper. The question is, "The 6 minute solutions uses a different equation for circular foundations than the DAS book. Can you reconcile these two equations?

DAS: 1.3cNc + qNq + 0.3yBNy
6 min: (1.2)cNc + qNq + 0.5yBNy(0.7)

Is it something that will allow me to get the correct answer on the test or are they different enough that the solutions could be similar and I would have to decide between to very close sol'ns?"

What do we do?

For PE Exam, lets just stick to what CERM recommends. The problem with this is the fact that shape factors also depend on the type of soil (See Table 4-1, 4-3, 4-5a of Bowles, 5th Ed.) apart from the geometry of the foundation. So Table 36.4 and 36.5 in CERM (11th Ed.) is not clear in that way and it is my opinion that those tables are provided for an approximate solution. So although I recommend using CERM for the PE exam, I would also recommend you to look for other methods (field or theoretical) for calculating bearing capacity in your work.

Lets look at the shape factors proposed by several scientists:

Terzaghi

For Strip Round Square

Sc 1.0 1.3 1.3

Sy 1.0 0.6 0.8

Meyerhof

Sc = 1 + 0.2(Kp)(B/L) for any phi

Sq = Sy = 1 + 0.1(Kp)(B/L) for phi > 10 degrees

Sq = Sy = 1 for phi = 0

Vesic

Sq = 1 + (B/L)tan(phi)

Sc = 1 + (Nq/Nc)(B/L)

Sy = 1 - 0.4(B/L)

If you use all these equations and evaluate bearing capacity, you'll notice a small difference in the answer.

Now, as far as the equations on Das and Six Minute Solutions are concerned:

DAS: 1.3cNc + qNq + 0.3yBNy
6 min: (1.2)cNc + qNq + 0.5yBNy(0.7) = 1.2cNc +qNq + 0.35yBNy

Now, lets take an example: y = 120 pcf, B = 3.0 feet, phi = 30 degrees, Df = 3 feet; c = 500 psf

DAS: 1.3cNc + qNq + 0.3yBNy = 1.3(500 psf)(37.2) + (120 pcf)(3 ft.)(22.5) + (0.3)(120 pcf)(3 ft.)(19.7) = 34407.6 psf
6 min: (1.2)cNc + qNq + 0.35yBNy = (1.2)(500 psf)(37.2) + (120 pcf)(3 ft.)(22.5) + (0.35)(120 pcf)(3 ft.)(19.7) = 32902.2 psf

For practical purposes, both the answers are the same because we are going to apply a very high factor of safety anyway. If I were you, I'd choose the more conservative one.

It is my understanding that the Terzaghi's equation gives the most conservative answer.

Over the period of years, several scientists and engineers have suggested various factors, i.e. shape, depth, inclination, and the bearing capacity factors. There is no significant advantage of one method over another. If you want to dig deeper, I would suggest Foundation Analysis and Design by J.E. Bowles.

Shallow Foundations - Eccentricity

Foundations can often be subjected to eccentric loading due to moment(s) applied to it. The figure illustrates an eccentrically loaded footing. The pressure at the bottom of the foundation, therefore, can be evaluated as

qmax = P/A + Me/I = P/A(1+6e/B)
qmin = P/A - Me/I = P/A(1-6e/B)

where e = M/P = eccentricity; P = applied load; M = applied moment; I = moment of inertia; A = Area of the footing (B x L)

One of the implications of the above equations is the fact that qmin = 0 when e = B/6
Thus, for e > B/6, qmin < 0 which shows that there would be a development of tension in such a case. Now, we all know soil cannot take tension, therefore, the foundation would separate from the underlying soil in such a case.

We'll start with a problem tomorrow.

Shallow Foundations - Solution 1 (e, f, g, h)

e. Net Bearing Capacity if GWL = 3 feet deep

Here, the GWL is below the footing, therefore
Tult = yDfNq + 0.5By'Ny
= (120 pcf)(3.0 feet)(22.5) + (0.5)(3.0 feet)(120 pcf - 62.4 pcf)(19.7)
= 9802.08 psf
Tnet = Tult - yDf = 9802.08 - (120 pcf)(3.0 feet) = 9442.08 psf or 9.4 ksf

f. Net Bearing Capacity if GWL = 4 feet deep. For a water table located so that 0 =< d <= B, we use an equivalent unit weight i.e., yeff = [yd + y'(B-d)]/B in the part of the equation that represents the failure zone below the footing, where d is the depth of GWT measured from the base of the footing. In this case, d = 1.0'

Tult = yDfNq + (0.5)B(yeff)Ny
Tult = (120 pcf)(3.0')(22.5) + (0.5)[(120 pcf)(1.0') + (120 - 62.4)(1)](19.7)
= 8100 psf + 2316.72 psf
= 10416.72 psf or 10.4 ksf
Tnet = Tult - yDf = 10416.72 - (120 pcf)(3.0') = 10056.72 psf or 10.1 ksf

I am sure you'd be able solve part g and h without much help. If you still have doubts, I'll be more than happy to help you.

Also, I'd always draw a figure (if possible) before solving a problem. It makes things a bit clearer.

Shallow Foundations - Solution 1 (Part a, b, c, d)

Solution:
a. Ultimate bearing capacity with deep GWL
Tult = cNc + yDfNq + 0.5ByNy
c = 0 (sand)
For phi = 30 deg; Nq = 22.5, Ny = 19.7 ...(Table 36.2 CERM 11th ed.)
Tult = 0 + (120 pcf)(3 feet)(22.5) + 0.5(3 feet)(120 pcf)(19.7)
Tult = 8100 psf + 3546 psf
Tult = 11646 psf or 11.6 ksf

b. Net bearing capacity with deep GWL
Tnet = Tult - yDf = 11646 psf - (120 pcf)(3 feet) = 11286 psf or 11.3 ksf

c. Allowable bearing capacity with deep GWL
Tall = Tnet/(factor of safety) = 11286 psf/3.0 = 3762 psf or 3.8 ksf

d. Net bearing capacity with GWL = 2 feet deep below the ground surface
Here, we use the concept of effective unit weight to the submerged part.
y' = effective unit weight = y - yw where yw = unit weight of water = 62.4 pcf
Therefore, the second part of the equation becomes, y(2 feet)Nq + y'(1 feet)Nq
And the last part of the equation becomes 0.5(B)(y')Ny

Tult = (120 pcf)(2 feet)(22.5) + (120 pcf - 62.4 pcf)(1 feet)(22.5) + (0.5)(3
feet)(120 pcf - 62.4 pcf)(19.7)
Tult = 5400 psf + 1296 psf + 1702.08 psf
Tult = 8398.08 psf or 8.4 ksf
Tnet = Tult - overburden = 8398.08 psf - [(120 pcf)(3 feet) + (62.4 pcf)(1 feet)]
Tnet = 7975.68 psf or 8 ksf

Note how the presence of water has reduced the bearing capacity.

I'll post the solutions to rest of the problem tomorrow. I'll also add another problem on shallow foundation tomorrow with increased complexity, i.e. soil layers, eccentric loads etc. Also, please let me know if you find any errors on my solution, I'll try to rectify it.

Shallow Foundations - Problem 1

1. A continuous footing in sand has a width of 3 feet and the depth to the base of foundation is 3 feet. Assume angle of internal friction, phi = 30 degrees; unit weight = 120 pcf; factor of safety = 3.0; ground water level = deep; assume general shear failure and use Terzaghi's bearing capacity factors.

Determine

a. Ultimate bearing capacity, Tult
b. Net bearing capacity, Tnet
c. Allowable bearing capacity, Tall
d. Net bearing capacity if GWL = 2 feet deep
e. Net bearing capacity if GWL = 3 feet deep
f. Net bearing capacity if GWL = 4 feet deep
g. Net bearing capacity if GWL = 6 feet deep
h. Net bearing capacity if GWL = 7 feet deep

Let me know if I have missed any information in this question and/or you have any questions/concerns.

Shallow Foundations

I've decided to jump to shallow foundations. I'll come back to phase relationships and compaction later.

What do we mean by shallow foundation?
In simple terms, a foundation with depth to width ratio less than or equal to one typically, i.e. Df/B <= 1 ; however, it may be more.(Bowles, 5th Ed.)

Sources of uncertainties in foundation design:
1. Load estimation
2. Variability of soil conditions at the site
3. Evaluation of engineering properties of soil and rock at the site
4. uncertainties in modeling

The bearing capacity of shallow foundation (strip footing) can be evaluated with the use of Terzaghi's equation:

Ultimate Bearing Capacity, T(ult) = cNc + qNq + 0.5ByNy

or in words, T(ult) = contribution for cohesion + contribution from overburden (and surcharge) + contribution from soils (failure zone) below the footing.

c = cohesion, psf
y = unit weight, pcf
B = width of the footing, ft
q = yDf where Df is the depth of the footing below the ground level.
Nc, Nq, Ny = bearing capacity factors

The coefficients of each term in the equation changes with the shape of foundation, e.g. for square footing,

T(ult) = 1.3cNc + qNq + 0.4ByNy

Sand: There is negligible cohesion in sands, so we can remove the cohesion term from the equation to calculate the bearing capacity of a shallow foundation in sand.
T(ult) = qNq + 0.5ByNy (assuming strip footing)

Saturated clay: For phi = 0, we'll have Ny = 0. Therefore,
T(ult) = cNc + qNq (assuming strip footing)
one of the implications of the above equation is:

for angle of internal friction, phi = 0, Nc = 5.7, Nq = 1.0 (refer to Table 36.2 in CERM 11th edition)
T(ult) = 5.7c + yDf
Net bearing capacity, T(net) = T(ult) - yDf ...apply overburden correction to get net bearing capacity.
Therefore,
T(net) = 5.7c
Allowable Bearing Capacity, T(all) = T(net)/Factor of safety, assume FS = 2.5, we get
T(all) = 5.7c/2.5 = 2.3c ~ 2c (for practical purposes)

Hence, we see that the allowable bearing capacity of shallow foundation on saturated clay is approximately twice the cohesive strength of the soil.

Also, since c = qu/2 ... qu = unconfined compressive strength,
T(all) = qu ...for practical purposes

I hope you enjoyed the start. We'll solve a problem tomorrow...

Solution 1

From the figure (notice red lines) provided above,

Coefficient of uniformity, Cu = D60/D10 = 0.2/0.025 = 8

Coefficient of gradation (also known as coefficient of curvature), Cz = (D30)^2/(D60 x D10) = (0.12)^2/(0.2 x 0.025) = 2.88

What do these values mean?

Cu > 10 : Well graded

Cu < 4 : Uniform

Cu = 1 : All grains same size

1 <= Cz <= 3 and Cu > 4 : Well Graded Gravel

1 <= Cz <= 3 and Cu > 6 : Well Graded Sand

Also, Gravel sizes are usually > 4.75 mm

0.075 mm <= Sand <= 4.75 mm

silt and clay < 0.075 mm

By looking at our solution chart above, we don't see any gravel, and we have less than 35% fines. Also, Cu > 6 and 1 <= Cz <= 3. Moreover, since the given sample has more than 12% fines, we can, therefore, classify the soil as Silty Sand (SM).

Problem 1: Soil Particle Size Distribution

Soil particle size distribution helps us in classifying the soil. The soil can be classified based on several methods. Two of the most used are the AASHTO (American Association of State Highway Officials) and USCS (Unified Soil Classification System). Pavement engineers usually use AASHTO whereas other classical geotechnical engineers use USCS method.

Problem: Using the figure provided above, classify the soil with the help of USCS method. Evaluate

1. The uniformity coefficient, Cu

2. Coefficient of gradation, Cz

Books

Let me start with suggesting a few books that I've used over the years. If you are someone just looking for passing the PE, the following books should be enough for you:

1. Civil Engineering Reference Manual, latest edition

2. Fundamentals of geotechnical engineering by Braja M. Das

3. Principles of Foundation Engineering by Braja M. Das

4. Six Minute Solutions

However, if you are someone who likes to delve deeper into the thick of the things, you should use the following books:

1. Foundation Analysis and Design by J. E. Bowles

2. Soil Mechanics by Terzaghi, Peck, and Mesri

3. Introduction to Geotechnical Engineering by Holtz and Kovacs

Day 1

This blog is dedicated to people planning to take PE exam with geotechnical engineering in the afternoon. During my preparation, I found that some of the available references were very inadequate in the area of soil mechanics. I welcome questions and problems in geotechnical engineering from anyone who is preparing for the PE exam (you can ask even if you are not preparing for the PE exam). I'll try to post one or two problems every day and their solutions a day after, that'll give everybody sometime to work out their solutions. Alright guys, the problems start tomorrow!