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
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