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.