Stresses in a Soil Mass Chapter Eight

1. Stress caused by a point load:

5/222

3

v)z(r

z2π3p

Δσ

or

5/222v])

z

r([(1

1

z 2π

3pΔσ

where, 22 yxr

5/222

3

v)z(r

z2π3p

Δσ

P1=150 t P2=125 t P3=150 t

4 m4 m

3 m

Example 1: Three point loads, shown in figure, calculate the increase in vertical

stress at point B.

Solution:For the effect of p2

z = 3 m and r = 02

5/222

3

v t/m631.6)3(0

32π125*3Δσ

For the effect of p1, z= 3 m and m404yxr 2222

25/222

3

v t/m619.0)3(4

32π150*3Δσ

2 totalv t/m869.7619.0*2631.6Δσ

2. Vertical stress caused by a uniform line load:

22v]1πz[(x/z)

2qΔσ

22v

]1π[(x/z)

2q/z

Δσ

or

592.1]1)2/2[(π2

20*2

]1πz[(x/z)

2qΔσ

22221

v1

Example 2: For the point A, shown in figure below calculate the increase of vertical stress due to the two line loads.Solution:

095.0]1)2/6[(π2

30*2

]1πz[(x/z)

2qΔσ

22222

v2

2v2v1 totalv kN/m 1.6870.0951.592ΔσΔσΔσ

3. Vertical stress caused by a uniform strip load: (finite width and infinite length)

Example 3: q = 200 kN/m2, B = 6m, and z = 2m. Determine the vertical stress increase at ± 9, ± 6, ± 3, and 0.

Solution: a = B/2 = 6/2 = 3m

qΔσv

vΔσxx/az/a

± 9± 30.670.120

± 6± 20.670.240

± 3± 10.670.71142

000.670.93186

4. Vertical stress caused by a triangular strip load:

5. Vertical stress caused by a uniformly loaded circular area:

Example 4: A circular foundation of diameter 10m transmits a uniform contact pressure of 150 kN/m2 . plot the following vertical stress profiles induced by this loading:a) beneath the center, down to 10m, andb) on a horizontal plane 6m below the foundation, between the center and a distance of 12m from the center.

Solution: a) the vertical stress at various depths below the center of the foundation

z, m012345678910

z/R00.20.40.60.811.21.41.61.82

∆qv/∆qs10.970.920.850.750.640.540.460.390.340.28

∆qv150145.5138127.5112.596816958.55142

b) the vertical stress values on a horizontal plane 6m, below the foundation for various offsets from center.

z/R = 6/5 = 1.2

x, m012345681012

x/R00.20.40.60.811.21.622.4

∆qv/∆qs0.540.530.50.450.380.30.220.140.060.03

5

∆qv8179.57567.55745332195.25

6. Vertical stress caused by a uniform rectangular loaded area:

Example 5: The plan of a rectangular foundation shown below transmits a uniform contact pressure of 120 kN/m2. Determine the vertical stress induced by this loading: (a) at a depth of 10m below point A, and (b) at a depth of 5m below B.

Solution: )a (

Rectanglem = B/zn = L/zF(m,n)

110/10 = 15/10 = 0.50.12

210/10 = 120/10 = 20.199

35/10 = 0.520/10 = 20.135

45/10 = 0.55/10 = 0.50.085

2 (A) v kN/m 64.680.085)0.1350.1990.12(120Δσ

v4v3v2v1 (A) v ΔσΔσΔσΔσΔσ

]n)f(m,n)f(m,n)f(m,n)[f(m, q 4321

)b ( ]n)f(m,n)f(m,n)f(m,n)[f(m, qΔσ 4321 (B) v

Rectanglem = B/zn = L/zf(m,n)

119/5 = 3.831/5 = 6.20.2478

219/5 = 3.86/5 = 1.20.216

34/5 = 0.831/5 = 6.20.185

44/5 = 0.86/5 = 1.20.165

2 (B) v kN/m 1.4160.165)0.1850.2160.2478(120Δσ

Example 6: Plan view of a loading shown in the figure below. Find the vertical stress increase at a depth of 10 m below point A.

Solution:

Example 7: The flexible area is uniformly loaded. Given: q = 300 kN/m2. Determine the vertical stress increase at point A.

Solution:

v3v2v1 (A) v ΔσΔσΔσΔσ

for semicircle part (area 1)

21.53

Rz 0

1.50

Rx , and from figure

0.29q

Δσv

5.43300*5.0*29.0Δσv1

for rectangular part (area 1 and 2)

667.23

8zLm , and 5.0

3

5.1zBn , from figure , f(m,n) = 0.136

6.81300*2*136.0Δσ 3 v2,

2 (A) v kN/m 125.181.643.5Δσ

7. Influence (Newmark) chart for vertical stress distribution:

8. Approximate methods for stress distribution (2:1 method)

also called as (linear stress distribution)

Z)Z)(L(B

qBLΔσv

for rectangular or square

2

2

vZ)(D

qDΔσ

for circular

Example 9: A 4.5 m2 foundation exerts a uniform pressure of 200 kN/m2 on a soil,

determine, the due to the load to a depth of 10m below its center using:

vΔσ

1. m,n chart method2. approximate method

3. Newmark chart at depth of 5m.

Solution: (1)

q*n)f(m, *4Δσv zm = B/zn = L/zf(m,n)4* f(m,n)

2.52.25/2.5 = 0.90.90.16250.65130

52.25/5 = 0.450.450.0740.29659.2

7.52.25/7.5 = 0.30.30.03750.1530

102.25/10 =

0.2250.2250.0230.09218.4

(2) 2v kN/m65.82

)5.25.4)(5.25.4(

5.4*5.4*200

Z)Z)(L(B

qBLΔσ

vΔσz2.55.07.510.0

82.6544.87528.12519.263

(3) No. of elements enclosed by plan = 13.9 for one quarter

I = 0.005

2v kN/m 55.6200*0.005*13.9*4q*I *elements of No.Δσ

Problem 1: (H.W.) foundation exerts a uniform pressure of 175 kN/m2 on a soil,

determine, the due to the load to a depth of 8m below points A, O, and P.

vΔσ

The footing shown in the figure below is subjected to a uniform load of 300 kPa. Calculate the vertical stress increase at a depth of

3m below its center and point A.

Problem 2: (H.W.)

A raft foundation of the size given in carries a uniformly distributed load of 300 kN/m2. Estimate the vertical pressure increase at a depth 9 m below the point A.

Determine the stress increase at depth of 5m below point Aif the q= 175 kPa