Lateral Earth Pressure

My inability to be able to steal some time out of my schedule leaves me no option but to provide you with some links that might help you. I'll, however, post my take on lateral earth pressure in near future which may be too late for the candidates taking their exam day after tomorrow but may help those who are taking their exams next year or later. I really apologize for the inconvenience and wish all the examinees the very best!

The following link provides a pretty good (and brief) lesson on lateral earth pressure.
http://www.pdhonline.org/courses/c155/c155content.pdf

For any questions, feel free to ask me.

A question by a visitor

Anonymous said...

I'm confused in calculating the effective vertical stress in a sandy gravel soil layer below the graound water level and it is between a sand layer on the top and a clay layer on the bottom.
What unit weight i have to consider the sand unit waight or I need the porosity(n)of the sandy gravel to get the unit waight?

Thanks

geotechie said...

suppose we have a situation like the following:

_____________________________________________

Sand wet unit weight = Ys,w
layer thickness = d1
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
water table @ depth dw, unit weight of water = Yw
Sand saturated unit weight = Ys,sat
______________________________________________

Sandy Gravel
saturated unit weight = Ysg
layer thickness = d2
______________________________________________

Clay
saturated unit weight = Yc
layer thickness = d3

______________________________________________

Effective Stress = Total Stress - Pore Water Pressure

Total Stress at the bottom of clay layer = (Ys,w)(dw) + (Ys,sat)(d1-dw)+ (Ysg)(d2) + (Yc)(d3)

Pore water pressure = (Yw)(d1-dw) + (Yw)(d2) + (Yw)(d3)

Therefore, Effective Stress =
(Ys,w)(dw) + (Ys,sat)(d1-dw)+ (Ysg)(d2) + (Yc)(d3) - (Yw)(d1-dw) - (Yw)(d2) - (Yw)(d3)

= (Ys,w)(dw) + (Ys,sat-Yw)(d1-Yw) + (Ysg-Yw)(d2) + (Yc-Yw)(d3)

Let me know if it helps (or if it does not).

Sorry guys...

I am extremely sorry guys...its been a very busy several weeks...I'll try to post something on retaining walls before the PE exam.

Shear Strength - Problem #3

A Consolidated Undrained (CU) triaxial test is conducted on silt. The effective cell pressure is 4.32 ksf. The deviator stress is 3.45 ksf. The total pore pressure for this test is 2.88 ksf. Draw Mohr's circle of the initial and final condition. Also, show circles that represent total and effective stresses.

Shear Strength-Triaxial Test

Advantages:
1. Suitable for cohesive soils as well
2. Samples can be saturated
3. Principal planes do not rotate
4. Failure will seek the weakest plane

Three permissible drainage cases:
UU or Unconsolidated Undrained
CD or Consolidated Drained
CU or Consolidated Undrained

Think of practical situations where these tests would be appropriate.

Shear Strength - Solution #2

The problem can also be solved using equations and I left it to you for exercise. I recommend using Mohr circle, it is faster and makes it easy to visualize stresses. Please let me know if you have questions.

Shear Strength - Problem #2

A direct shear test is run on a dense silty sand with a normal stress = 12 ksf. Take coefficient of earth pressure at rest, Ko = 0.5. At failure, the normal stress = 12 ksf and the shear stress is 10 ksf. Draw the Mohr circles for the initial and at failure and determine:

a) the principal stresses at failure 

b) the orientation of the failure plane

c) the orientation of the major and minor principal planes at failure

d) the orientation of the plane that the maximum shear stress acts upon.

Important Laboratory Tests to Obtain Shear Strength of Soils

Three most widely used laboratory tests are:
1. Direct Shear
2. Triaxial
     a. Unconsolidated Undrained (UU)
     b. Consolidated Drained (CD)
     c. Consolidated Undrained (i.e., rapid drawdown of a reservoir)
3. Unconfined Compression

Details of these tests can be read from any decent textbooks available in the market. I would personally recommend Holtz and Kovacs!

Shear Strength - III: Solution #1

Again, I am very sorry for the delay (have been flooded with work lately)!

I have not been able to provide with a solution circle because I am facing some problem with uploading that particular file. In any case, any decent book on soil mechanics will resolve your issues concerning solving Mohr's Circle problems.

Do let me know if there is any error in my solution or if you need more explanation.

Shear Strength - III: Problem #1

The following problem has been taken directly from my notes (in my graduate school):

For the soil element shown, determine the maximum and minimum principal stresses, the maximum shear stress and its corresponding normal stress, and the normal and shear stresses on plane AB. Use the following methods,
a.) by using equations
b.) by using Mohr's circles
c.) by using the Origin of Planes

I'll post the solution tomorrow. If anyone is interested in reading further, I would recommend, "An Introduction to Geotechnical Engineering" by Holtz and Kovacs.

Shear Strength - II

Consider a mass of soil acted upon by several forces:

 

Figure 1

Assumptions:
1. Forces act in 2-dimensional plane
2. Resolve forces into components on a small element, say at point O above, and on a plane passing through O at an angle alpha to the horizontal.
3. Sign convention: Use the following

 

Fig. 2: Sign convention

Resolution of forces of first figure into components on a small element at point O:

Fig. 3: Resolution of forces in Fig. 1 into components on a small element at point O.

Length CA = 1; Depth into the screen = 1; Length AB = cos(alpha); CB = sin(alpha)

4. Equilibrium and development of Mohr's circle:
   I've provided this because I believe it is essential to understand the practicality of an issue from theoretical point of view as well.

Please do let me know if you I have missed anything in my attempt to dissect shear strength. You'll observe that the discussions and problems in shear strength will become interesting as we move along.

Shear Strength

I think it would be prudent to understand the concept of shear strength before starting lateral earth pressures and consolidation.

Here are a few excerpts from my graduate course (I attribute a lot of my current abilities as a geotechnical engineer to my graduate advisor) on shear strength:

What is "failure" and "strength"?

We can answer the question is several steps:

1. If the load or stress in a foundation or an earth slope is increased until the deformations become unacceptably large, we say the soil has "failed."

2. In case 1, we are referring to the "strength" of the soil. This is the maximum or ultimate stress the soil can sustain.

3. In geotechnical engineering, we are generally concerned about shear strength, since most of our problems in foundations and earthwork engineering see failure resulting from excessive applied shear forces.

Most of the relationships that are used for characterization of strength and stress-deformation properties of soils are empirical. The Mohr-Coulomb theory is by far the most widely used:

shear resistance (on failure plane at failure)  = cohesion + normal stress (on failure plane at failure) x tan(phi)
phi = angle of internal friction

In reality the shearing resistance of the soil depends on many factors and a complete equation might be of the form:

shearing resistance = f(e,phi,c,effective normal stress, c',strain, strain rate,T,S,H)  where
e = void ratio
phi = angle of internal friction
c' = effective cohesion
c = cohesion
T = temperature
S = soil structure
H = stress history

There are a variety of types of friction angles, cohesion, etc:
-Total stress
-Effective stress
-Drained
-Undrained
-Peak Strength
-Residual Strength

I have inserted a typical stress vs strain plot (to the best of my drawing ability) for you to have an idea.

The values of cohesion and friction angle applicable in practice depend on such factors as
- whether the problem is one of loading or unloading
- whether short or long term stability is of interest (drainage).

A variety of test types are employed:
- Direct shear
- Triaxial compression
- Triaxial extension
- Simple shear
- Vane shear

I hope this helps form the basis of some of the further discussions and problem solving sessions we'll have from tomorrow.

Very Busy last week

Sorry guys, I couldn't post any questions last week due to an extremely busy week. I am hoping to continue with my discussion on shallow foundations and introduce retaining walls next week. Till then, sayonara and have a good weekend.

Shallow Foundations - Eccentricity - Solution 1 (c)

(c) e = M/P = 100 ft-kips/100 kips = 1.0

Therefore, e > B/6  i.e., 3 ft./6 = 0.5
qmin = P/A(1 - 6e/B) = P/A(1 - 6x1/3) < 0
qmax = P/A(1 + 6e/B) = 3P/A = (3 x 100 kips)/(3.0 ft. x 3.0 ft.)
or, qmax = 33.33 kips/ft^2

Ultimate Bearing Capacity, Tult


Here, we use the concept of effective width, Beff


Beff is the effective width where the pressure on the foundation is acting (see figure)


In this case, Beff = B - 2e = 3.0 - 2(1) = 1 ft
Leff = L - 2e = 3 - 2(0) = 3.0 ft. ...because its just one-way eccentricity in this problem

Tult = yDfNq + (0.85)(0.5)(Beff)yNy         ......Square Footing/Sand

Tult = (120 pcf)(3.0 ft.)(22.5) + (0.85)(0.5)(1.0 ft.)(120 pcf)(19.7)


Please note that although the effective footing is now a rectangle with Beff/L = 0.33, I've still not changed the shape factor multiplier. I've not done so because a B/L ratio of 0.33 will yield a higher shape factor multiplier and hence, this is more conservative.


Tult = 9104.7 psf

Let's go a step further and evaluate the factor of safety against bearing capacity failure,
Total ultimate load the foundation can sustain is Qult = Tult x Effective Area = Tult x Beff x B = 27314.1 lbs or 27.3 kips.

The factor of safety against bearing capacity failure is, FS = Qult/P = 27.3 kips/100 kips = 0.273


And this foundation, my friends, is doomed until something is done about it. 


A few things that we can do to improve the factor of safety are discussed below:


1. Increase the width of the foundation: Let's see how this affects the FS - 


Earlier our foundation was 3' x 3' , now make it 5' x 5'
so, B/6 = 5/6 = 0.834' which is still less than eccentricity, e.   


Tult = (120 pcf)(3.0 ft.)(22.5) + (0.85)(0.5)(3.33 ft.)(120 pcf)(19.7)


Tult = 11445.65 psf or 11.4 ksf


Qult = Tult x Effective Area = 11445.65 psf x 5 ft. x 3.33 ft. = 190570 psf or 190.6 ksf


FS = 190.6/100 = 1.9 which is a great improvement from 0.27, however, still less than what I would like to have for shallow foundations.


2. Increase the depth of the foundation to 5 feet (keep the dimensions of the foundation same): Let's see how this affects the FS - 

Tult = (120 pcf)(5.0 ft.)(22.5) + (0.85)(0.5)(1 ft.)(120 pcf)(19.7)

Tult = 14504.7 psf or 14.5 ksf

Qult = Tult x Effective Area = 14504.7 psf x 3 ft. x 1 ft. = 43514.1 psf or 43.5 ksf

FS = 43.5/100 = 0.435 which is an increase from 0.27, however, still too less.


Notice that the factor of safety is more sensitive to the increase in the width as opposed to the depth of the foundation.

3. A combination of the above, i.e. increase the footing width and lower the foundation. I'll leave this for you to try yourself.

I hope the above exercises were helpful in understanding how load eccentricity affects shallow foundations and ways to improve factors of safety.

Shallow Foundations - Eccentricity - Solution 1 (b)

(b) e = M/P = 20 kips-ft/100 kips = 0.2 ft.
Therefore, e < B/6  i.e., 3 ft./6 = 0.5
qmin = P/A(1 - 6e/B) = P/A(1 - 6x0.2/3) = 6.67 kips/ft^2
qmax = P/A(1 + 6e/B) =  (100 kips)(1.4)/(3.0 ft. x 3.0 ft.) or, qmax = 15.55 kips/ft^2

Ultimate Bearing Capacity, Tult
Tult = yDfNq + (0.85)(0.5)(B)yNy         ......Square Footing/Sand

Tult = (120 pcf)(3.0 ft.)(22.5) + (0.85)(0.5)(3.0 ft.)(120 pcf)(19.7)
Tult = 11114.1 psf

Let's go a step further and evaluate the factor of safety against bearing capacity failure,
Total ultimate load the foundation can sustain is Qult = Tult x Area = Tult x B x B = 100026.9 lbs or 100 kips.

The factor of safety against bearing capacity failure is, FS = Qult/P = 100 kips/100 kips = 1.0


Thus, we can see that the factor of safety is same as that of scenario a earlier. However, since we have a different pressure distribution between a and b, we know where we would want more reinforcement.

Shallow Foundations - Eccentricity - Solution 1 (a)

a. eccentricity, e = M/P = 50 kips-ft/100 kips = 0.5 ft.
    Therefore, e = B/6  i.e., 3 ft./6 = 0.5
     qmin = P/A(1 - 6e/B) = P/A(1 - 6x1/6) = 0
     qmax = P/A(1 + 6e/B) = 2P/A = (2 x 100 kips)/(3.0 ft. x 3.0 ft.)
     or, qmax = 22.2 kips/ft^2

   Ultimate Bearing Capacity, Tult
   Tult = yDfNq + (0.85)(0.5)(B)yNy               ......Square Footing/Sand

   Tult = (120 pcf)(3.0 ft.)(22.5) + (0.85)(0.5)(3.0 ft.)(120 pcf)(19.7)
   Tult = 11114.1 psf

Let's go a step further and evaluate the factor of safety against bearing capacity failure,
Total ultimate load the foundation can sustain is Qult = Tult x Area = Tult x B x B = 100026.9 lbs or 100    kips.

The factor of safety against bearing capacity failure is, FS = Qult/P = 100 kips/100 kips = 1.0

Do you think this Factor of Safety is good enough in such a case?

It is my opinion that this factor of safety is very inadequate for such a foundation.

Your comments and insights are welcome.
I'll post solution for rest of the problem tomorrow.

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!