Stresses in a Soil Mass

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Stresses in a Soil Mass. I will assume you have had strengths and covered mohrs circles, so I will not cover When soil is subjected to a vertical load, such as a foundation, stresses are created in the soil The stresses spread laterally with depth

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Stresses in a Soil Mass

• I will assume you have had strengths and covered mohrs circles, so I will not cover
• When soil is subjected to a vertical load, such as a foundation, stresses are created in the soil
• The stresses spread laterally with depth
• As you go deeper, the stress decreases but affects a larger plan area
• For example – a 5’ x 5’ footing at z=0 exerts a stress of 2000 psf over the 5’x5’ area
• At z=5, the stress is may only be 1000psf, but over an area of say 10’x10’
• (numbers only an example, actual calculated stresses will be different)

Stresses in a Soil Mass

• In simpler terms:
• Load P over area A results in stress q
• With depth
• P stays the same
• A increases
• q decreases

Stresses in a Soil Mass

Boussinesq Case

Boussinesq developed an equation to model stress distribution with depth for a homogeneous, isotropic material (modulus of elasticity and poissons ratio constant in all directions)

For point load Q, stress at A at depth z and lateral distance r from Q is:

Δσv = Change in vertical stress

Δσv= (Q/z2) * 3 /( 2π[1+(r/z)2]5/2)

Max at r = 0

Stresses in a Soil Mass

• A point load is never encountered in practice
• The Boussinesq equation for a point load must be converted to a load over an area:
• Isolated footing
• Perimeter footing
• Slab
• How would this be done?

Stresses in a Soil Mass

• This has already been done and put into tabular form for different shape foundations
• Table 9-7 on 246 shows the results for rectangular footings
• Using the following:
• m = B/z (B is always the short dimension)
• n = L/z (L is always the long dimension)
• Plot n and m to obtain I3
• I3 * the footing pressure is the stress beneath the corner of footing

Stresses in a Soil Mass

• Table 9-8 on 249 is the same but under the center of a rectangular footing
• Using the following:
• m1 = L/B (B is always the short dimension)
• n1 = z/b (L is always the long dimension)
• b = B/2
• Plot n1 and m1 to obtain I4
• I4 * the footing pressure is the stress beneath the center of footing

Stresses in a Soil Mass

• Example:
• A load of 125,000 lbs is placed on a footing
• The footing is 5’ x 5’
• Find the stress below the center of the footing at z = 5’
• q = Q/A = 125,000/25 = 5000psf
• m1 = L/B = 5/5 = 1
• b = B/2 = 5/2 = 2.5
• n1 = z/b = 5/2.5 = 2
• Table 9.8 says I4 = 0.336
• Therefore Δσv = 0.336 * 5000 = 1680psf

Stresses in a Soil Mass

• If you need the stress under a rectangular footing, but you need it at a point other than the corner or center, you would use superposition
• By creating a series of rectangular shapes in the area needed, you can sum the calculated stresses to obtain the total stress at that point

Stresses in a Soil Mass

• By breaking the larger rectangle into 4 smaller ones, you can find the stress under point A’, which is not under the center or corner of the larger area, but under the corner of each of the smaller areas.

1

2

A’

B

4

3

L

Stresses in a Soil Mass

• Find the increase in vertical stress due to the loaded area at a point 5’ below B

2’

B

q = 2000 psf

3

4

Stresses in a Soil Mass

• By using 2 areas you can find the stress 5’ below B

2

B

q = 2000 psf

3

6

• Solve by finding Δσv for the 6’x3’ area and subtracting off the Δσv for the 2’x3’ area.

Stresses in a Soil Mass

• Solve Δσv for the 6’x3’ area 1st

B

q = 2000 psf

3

6

• m = B/z = 3/5 = 0.6
• n = L/z = 6/5 = 1.2
• Table 9.7 says I3 = 0.1431

Stresses in a Soil Mass

• Solve Δσv for the 2’x3’ area.

2

B

q = 2000 psf

3

• m = B/z = 2/5 = 0.4
• n = L/z = 3/5 = 0.6
• Table 9.7 says I3 = 0.0801

Stresses in a Soil Mass

• Δσv = 2000 (.1431 - .0801) = 126psf

2’

B

q = 2000 psf

3

4

Stresses in a Soil Mass

• To use superposition, all shapes MUST share the point in question under one of their corners

Stresses in a Soil Mass

• Solve for next class

10’

q = 5000psf

4’

A

9’

4’

• Find stress at A at z = 5’

2 : 1 Method

• This method approximates stresses due to a foundation by assuming the load spreads at a rate of 2V to 1H

For a square footing (B x B): Δσv= Q/(B+z)2

For a rectangular footing (B x W): Δσv= Q/(B+z)(W+z)

Stresses in a soil mass

• Plotting change in vertical stress vs. depth

Δσv

z

Stresses in a soil mass

• Assume there is a footing at z = 0

Δσv

• Also assume bearing pressure is 2000 psf
• The Δσv at z = 0 is 2000 psf
• Using either Boussinesq or the 2:1 method, plot stress vs. depth
• Now, assume a layer of fill placed on an entire site

z

Stresses in a soil mass

• Conceptually, it is like placing a 100’x100’ footing on a 100’x100’ site.
• Looking at the plot, what does this mean?
• What is the stress at
• z = 5’
• z = 10’
• z = 20’
• etc.

Δσv

z