stress of soil
DESCRIPTION
CHAPTER 3. Stress of soil. §3 Stress of soil. § 3.1 General § 3.2 Self-weight stress § 3.3 Stress at the bottom of foundation § 3.4 Additional stress in foundation. §3 Stress of soil. §3.1 General. Weight of soil (effective stress) Surface loads Fill large area Point loads: - PowerPoint PPT PresentationTRANSCRIPT
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Stress of soil
CHAPTER 3
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§3 Stress of soil
§3.1 General§3.2 Self-weight stress §3.3 Stress at the bottom of foundation §3.4 Additional stress in foundation
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§3.1 General
§3 Stress of soil
Weight of soil (effective stress)
Surface loads Fill large area Point loads:
Hydro pole, light stand, column, etc Lines loads
Rack or rail loading, strip foundation Rectangular area
Raft or rectangular footing Circular area
Tank Earth embankment
Road, railway, fill, ice, etc.
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§3.2 Self-weight stress
§3 Stress of soil
natural ground
cz
cx
cy 11
z
zcz
z
σcz
σcz= z
Due to soil’s self weight
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i
n
iinncz hhhh
1
2211
natural ground
h1
h2
h33
2
1
ground water level
1 h1
1 h1 + 2h2
1 h1 + 2h2 + 3h3
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czcycx K 0
natural ground
z
cz
cx
cy
zcz
lateral earth pressure coefficient
at restAssumption:no shear stresses on the vertical planes bounding the column in horizontal direction.
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Example 1: How to calculate the stress in soils?
Natural ground
Ground water level
Top roof of the aquitard
Clay
sand
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57.0kPa
80.1kPa
103.1kPa150.1kPa
194.1kPa
i
n
iinncz hhhh
1
2211
Solution:
Natural ground
Ground water level
Top roof of the aquitard
Clay
sand
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F
§3.3 Stress at the bottom of foundation §3 Stress of soil
Filled soil
Base pressureGeotextile polymer
Smaller load
Extreme load
Smaller load
Extreme load
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A
GFp
l
e
bl
GF
p
p 61
min
max
W
M
A
GF
p
p
min
max
Outdoor ground
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l
e
bl
GF
p
p 61
min
maxDiscussion:
pmax
pmin
e<l/6
pmax
pmin=0
e=l/6
e>l/6
pmax
pmin<0
pmax
pmin=0
redistribution of the base pressure
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(1) Point load
M(x,y,z)
Po
y
x
z
xy
z
r
R M(x,y,0)
325
3
cos2
3
2
3
R
P
R
Pzz
2z
PKz
Note:z is independent of
elastic modulus Poisson’s ratio.
§3.4 Additional stress in foundation§3 Stress of soil
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Stresses due to point load Rules of the stress distribution
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Pa
z
Pb
a b
superposition principle
several point loads:
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(2) Strip area carrying uniform pressure
pKcz
blm /bzn /
),( nmfKc
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pKKKK ccccz ⅣⅢⅡⅠ
z
M
o
IV
IIIII
I
oI
IIIII
IV
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pKK ccz ⅡⅠ II
I
oo
I
IIIo
IVo
pKKKK ccccz ⅣⅢⅡⅠ
II
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I
o o
III
IIIV
pKKKK ccccz ⅣⅢⅡⅠ
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(3) Strip area carrying linearly increasing pressure
pK tz 11
ttz pK 22
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Strifoundation Load :F=400kN/m , M=20kN•m,
? Additional stress on the middle of base? Plot the distribution of the stress
2m
F
M
0 = 18.5kN/m3
0.1m
1.5m
Example 2
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Solution
1.Base pressure
l
e
bl
GF
p
p 61
min
max
F=400kN/m
0 = 18.5kN/m3
M=20kN •m
0.1m
2m
1.5m
319.7kPa140.3kPa
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2.Additonal pressure
1.5m
292.0kPa112.6kPa
dp
p
p
p0
min
max
min0
max0
0.1mF=400kN/m
M=20kN •m
2m
0 = 18.5kN/m31.5m
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3.Additioanl pressure at the middle point
2m
F=400kN/m
M=20kN •m
0.1m
1.5m0 = 18.5kN/m3
179.4kPa
112.6kPa
292.0kPa112.6kPa
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2m
F=400kN/m
M=20kN •m
0.1m
1.5m0 = 18.5kN/m3
202.2kPa193.7kPa
165.7kPa
111.2kPa
80.9kPa
62.3kPa
1m1m
2m
2m
2m
result