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Published byDamion Hosmer Modified about 1 year ago

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1 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)

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2 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

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3 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: Δσ v = Change in vertical stress Δσ v = (Q/z 2 ) * 3 /( 2π[1+(r/z) 2 ] 5/2 ) Max at r = 0

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4 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?

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5 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 I 3 I 3 * the footing pressure is the stress beneath the corner of footing

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6 Stresses in a Soil Mass Table 9-8 on 249 is the same but under the center of a rectangular footing Using the following: m 1 = L/B (B is always the short dimension) n 1 = z/b (L is always the long dimension) b = B/2 Plot n 1 and m 1 to obtain I 4 I 4 * the footing pressure is the stress beneath the center of footing

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7 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 m 1 = L/B = 5/5 = 1 b = B/2 = 5/2 = 2.5 n 1 = z/b = 5/2.5 = 2 Table 9.8 says I 4 = 0.336 Therefore Δσv = 0.336 * 5000 = 1680psf

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8 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

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9 Stresses in a Soil Mass B L A’ 12 4 3 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.

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10 Stresses in a Soil Mass 3 4 B 2’ q = 2000 psf Find the increase in vertical stress due to the loaded area at a point 5’ below B

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11 Stresses in a Soil Mass 3 6 B q = 2000 psf By using 2 areas you can find the stress 5’ below B 2 Solve by finding Δσv for the 6’x3’ area and subtracting off the Δσv for the 2’x3’ area.

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12 Stresses in a Soil Mass 3 6 B q = 2000 psf Solve Δσv for the 6’x3’ area 1st m = B/z = 3/5 = 0.6 n = L/z = 6/5 = 1.2 Table 9.7 says I 3 = 0.1431

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13 Stresses in a Soil Mass 3 B 2 Solve Δσv for the 2’x3’ area. q = 2000 psf m = B/z = 2/5 = 0.4 n = L/z = 3/5 = 0.6 Table 9.7 says I 3 = 0.0801

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14 Stresses in a Soil Mass Δσv = 2000 (.1431 -.0801) = 126psf 3 4 B 2’ q = 2000 psf

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15 Stresses in a Soil Mass To use superposition, all shapes MUST share the point in question under one of their corners

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16 Stresses in a Soil Mass Solve for next class 4’ 10’ 9’ 4’ q = 5000psf A Find stress at A at z = 5’

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17 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): Δσ v = Q/(B+z) 2 For a rectangular footing (B x W): Δσ v = Q/(B+z)(W+z)

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18 Stresses in a soil mass Plotting change in vertical stress vs. depth z Δσ v

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19 Stresses in a soil mass Assume there is a footing at z = 0 z Δσ v Also assume bearing pressure is 2000 psf The Δσ v 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

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20 Stresses in a soil mass z Δσ v 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.

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