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SHEAR STRENGTH OF SOIL:
Shear Strength of soil is its ability to resist shear stresses.
Coulumb Equation: S = C + Ơ tan Ø
Where:
S = shear strength of soil
C = cohesiveness of soil
Ơ = effective normal stress
Ø = angle of internal friction
A. Direct Shear Test:
a) Normal Stress: P P = applied load
Ơ = P/A A = area
A = _Πd
2
_ Ơ = normal stress
4 A
b) Shearing Stress:
τ = F/A
F
A
F
F = shear force
τ = Shearing Stress
A = area
Relation of Normal Stress and Shear Stress:
a) For Normally consolidated clay, C = 0 (non-cohesive)
Shearing stress
τ
τ
Ø
Normal Stress, Ơ
tan Ø = τ
τ = Ơ tanØ
b) For over consolidated clay:
Shearing stress
τ h
Ø
C
Normal Stress, Ơ
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τ = c + h
tan Ø = h/Ơ
τ = C + Ơ tanØ
C = cohesive clay
DIRECT SHEAR TEST:
The soil specimen is placed in a metal shear box. The soli specimens may be square or
circular in cross sections with sizes usually 50mm x 50mm or 100mm x 100mm x 25mm high.
The metal shear box is split into halves. The normal stress is applied from the top of the shear
box, then a shear force is applied by moving on half of the box relative to the other to cause failure
in the soil specimen.
Normal force
Shear Force
Direct shear test is performed to determine interface friction angle.
TRI-AXIAL TEST:
a) For normally consolidated clay (non-cohesive soil); C = 0
Effective stress failure envelope from drainage tests in sand and normally consolidated
clay.
Ơ
3
= chamber confining stress, cell stress, lateral stress, or minor stress
Δd = deviator stress
Ơ
1
= major stress, vertical stress
Ø = angle of internal friction
s
h
e
a B
r
r
stress
O Ø Ɵ A r
Ơ
3
Δd
Ơ
1
Ơ
1
= Ơ
3
+ Δd; Δd = Ơ
1
- Ơ
3
r = (Ơ
1
Ơ
3
)/2; OA = Ơ
3
+ r
OA = Ơ
3
+ (Ơ
1
Ơ
3
)/2
OA = 2 Ơ
3
+ Ơ
1
Ơ
3
: OA = Ơ
1
+ Ơ
3
2 2
Sin Ø = ____r____ =
1
Ơ
3
)/2 = Ơ
1
Ơ
3
OA (Ơ
1
+ Ơ
3
)/2 Ơ
1
+ Ơ
3
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Ơ
1
Failure plane
Ơ
3
Ɵ
Ø = angle of internal friction
Failure Plane
s
h
e
a B
r
r
stress
A Ø Ɵ x r
C
x = 2Ɵ
180
o
-
Inclination of the plane of failure caused by shear: ∆ABC
Ø + 90 + 180 - 2Ɵ = 180
2Ɵ = 90 + Ø
Ɵ = 45
o
+ Ø/2
Ɵ = angle that the failure plane makes with the major principal stress
b) Over consolidated clay (Cohesive Soil)
Ơ
1
= Ơ
3
tan
2
(45
o
+ Ø/2) + 2C tan (45
o
+ Ø/2)
B r = Ơ
1
Ơ
3
)/2
r
A ϕ c r
x Ơ
3
C
y
Ơ
1
y = Ơ
3
+ (Ơ
1
Ơ
3
)/2
y = (Ơ
1
+ Ơ
3
)/2 tan ϕ = c/x x =c/tanϕ x = c cot ϕ
triangle ABC: sin ϕ = BC/AC
=
1
Ơ
3
)/2 = __
1
Ơ
3
)/2__ = ___Ơ
1
Ơ
3
___
x + y x + (Ơ
1
+ Ơ
3
)/2 2x + Ơ
1
+ Ơ
3
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where: cot ϕ = cos ϕ/sin ϕ
sin ϕ = _____Ơ
1
Ơ
3
_____ or Ơ
1
Ơ
3
= sin ϕ [2c cot ϕ + Ơ
1
+ Ơ
3
]
2c cotϕ + Ơ
1
+ Ơ
3
Ơ
1
Ơ
3
= 2c sin ϕ [cos ϕ/sin ϕ + Ơ
1
sin ϕ + Ơ
3
sin ϕ]
Ơ
1
Ơ
3
= 2c cos ϕ + Ơ
1
sin ϕ + Ơ
3
sin ϕ -----------------simplify
Ơ
1
- Ơ
1
sin ϕ = 2c cos ϕ + Ơ
3
+ Ơ
3
sin ϕ
Ơ
1
- Ơ
1
sin ϕ = 2c cos ϕ + Ơ
3
[1+ sin ϕ]
Ơ
1
= Ơ
3
[1+ sin ϕ] + _2c cos ϕ_
(1 - sin ϕ) 1 - sin ϕ
where: 1+ sin ϕ = tan
2
(45 + ϕ/2); __cos ϕ_ = tan (45 + ϕ/2)
1 - sin ϕ 1 - sin ϕ
Therefore: Ơ
1
= Ơ
3
tan
2
(45 + ϕ/2) + 2c tan (45 + ϕ/2)
Example:
1. The size of sand specimen in a direct test was 50mm x 50mm x 30mm. It is known
that the sand, tan Ø = 0.75/e, and the specific gravity of solids is 2.65. During the test,
a normal stress of 140kPa was applied. Failure occurred at a shear stress of 105kPa.
a) What is the void ratio?
b) Compute the dry unit weight of sand.
c) What is the weight of sand specimen?
Given: tan Ø = 0.75/e Normal stress (Ơ) = 140kPa
specific gravity of solids is 2.65 Shear stress (τ) = 105kPa
Solution:
a) tan Ø = Normal stress (Ơ) = _105_ = 0.75
Shear stress (τ) 140
0.75 = 0.75; e = 1.0
e
b) γ = Gs γ
w
= _2.65 (9.81)_ = 12.998 or 13kN/m
3
1 + e 1 + 1
c) W = wV = 13kN/m
3
[(0.05) (0.05) (0.03)]m
3
= 0.975N
Exercises:
1. The following are the results of direct shear tests performed on two identical samples of
the soil. In test one, the sample shears at a stress of 70kPa when the compressive normal
stress is 95kPa. In test two, the sample shears at a stress of 105kPa when the normal stress
is 150kPa.
a) Determine the value of the apparent cohesion.
b) Determine the angle of internal friction for the sand.
c) Determine the shear stress at a depth of 4.20m if the unit weight of soil is
15.75kN/m
3
.
2. A consolidated drained axial test was conducted on a cohesion less soil that has a friction
angle of 30
o
and deviator stress at failure of 400kPa.
a) Find the angle that the failure plane makes with the major principal plane.
b) Find the confining pressure.
c) Find the shear stress at the point on the failure plane.

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SHEAR STRENGTH OF SOIL: Shear Strength of soil is its ability to resist shear stresses. Coulumb Equation: S = C + Ơ tan Ø Where: S = shear strength of soil C = cohesiveness of soil Ơ = effective normal stress Ø = angle of internal friction A. Direct Shear Test: a) Normal Stress: Ơ = P/A A = _Πd2_ 4 P P = applied load A = area Ơ = normal stress A b) Shearing Stress: τ = F/A F A F F = shear force τ = Shearing Stress A = area Relation of Normal Stress and Shear Stress: a) For Normally consolidated clay, C = 0 (non-cohesive) Shearing stress τ τ Ø Normal Stress, Ơ tan Ø = τ/Ơ τ = Ơ tanØ b) For over consolidated clay: Shearing stress τ h Ø C Normal Stress, Ơ τ=c+h tan Ø = h/Ơ τ = C + Ơ tanØ C = cohesive clay DIRECT SHEAR TEST: The soil specimen is placed in a metal shear box. The soli specimens may be square or circular in cross sections with sizes usually 50mm x 50mm or 100mm x 100mm x 25mm high. The metal shear box is split into halves. The normal stress is applied from the top of the shear box, then a shear force is applied by moving on half of the box relative to the other to cause failure in the soil specimen. Normal force Shear Force Direct shear test is performed to determine interface friction angle. TRI-AXIAL TEST: a) For normally consolidated clay (non-cohesive soil); C = 0 Effective stress failure envelope from drainage tests in sand and normally consolidated clay. Ơ3 = chamber confining stress, cell stress, lateral stress, or minor stress Δd = deviator stress Ơ1 = major stress, vertical stress Ø = angle of internal friction s h e a r B r stress O Ø Ɵ Ơ3 A r Δd Ơ1 Ơ1 = Ơ3 + Δd; r = (Ơ1 – Ơ3)/2; Δd = Ơ1 - Ơ3 OA = Ơ3 + r OA = Ơ3 + (Ơ1 – Ơ3)/2 OA = 2 Ơ3 + Ơ1 – Ơ3 : OA = Ơ1 + Ơ3 2 2 Sin Ø = ____r____ = (Ơ1 – Ơ3)/2 = Ơ1 – Ơ3 OA (Ơ1 + Ơ3 )/2 Ơ1 + Ơ3 Ơ1 Failure plane Ơ3 Ɵ Ø = angle of internal friction Failure Plane s h e a r B r stress A Ø Ɵ x r C x = 2Ɵ 180 - 2Ɵ o Inclination of the plane of failure caused by shear: ∆ABC Ø + 90 + 180 - 2Ɵ = 180 2Ɵ = 90 + Ø Ɵ = 45o + Ø/2 Ɵ = angle that the failure plane makes with the major principal stress b) Over consolidated clay (Cohesive Soil) Ơ1 = Ơ3 tan2 (45o + Ø/2) + 2C tan (45o + Ø/2) r = Ơ1 – Ơ3)/2 B r A ϕ x c r Ơ3 C y Ơ1 y = Ơ3 + (Ơ1 – Ơ3)/2 y = (Ơ1 + Ơ3)/2 tan ϕ = c/x x =c/tanϕ x = c cot ϕ triangle ABC: sin ϕ = BC/AC = (Ơ1 – Ơ3)/2 = __(Ơ1 – Ơ3)/2__ = ___Ơ1 – Ơ3___ x+y x + (Ơ1 + Ơ3)/2 2x + Ơ1 + Ơ3 where: cot ϕ = cos ϕ/sin ϕ sin ϕ = _____Ơ1 – Ơ3_____ or Ơ1 – Ơ3 = sin ϕ [2c cot ϕ + Ơ1 + Ơ3] 2c cotϕ + Ơ1 + Ơ3 Ơ1 – Ơ3 = 2c sin ϕ [cos ϕ/sin ϕ + Ơ1 sin ϕ + Ơ3 sin ϕ] Ơ1 – Ơ3 = 2c cos ϕ + Ơ1 sin ϕ + Ơ3 sin ϕ -----------------simplify Ơ1 - Ơ1 sin ϕ = 2c cos ϕ + Ơ3 + Ơ3 sin ϕ Ơ1 - Ơ1 sin ϕ = 2c cos ϕ + Ơ3 [1+ sin ϕ] Ơ1 = Ơ3 [1+ sin ϕ] + _2c cos ϕ_ (1 - sin ϕ) 1 - sin ϕ where: 1+ sin ϕ = tan2 (45 + ϕ/2); 1 - sin ϕ __cos ϕ_ = tan (45 + ϕ/2) 1 - sin ϕ Therefore: Ơ1 = Ơ3 tan2 (45 + ϕ/2) + 2c tan (45 + ϕ/2) Example: 1. The size of sand specimen in a direct test was 50mm x 50mm x 30mm. It is known that the sand, tan Ø = 0.75/e, and the specific gravity of solids is 2.65. During the test, a normal stress of 140kPa was applied. Failure occurred at a shear stress of 105kPa. a) What is the void ratio? b) Compute the dry unit weight of sand. c) What is the weight of sand specimen? Given: tan Ø = 0.75/e specific gravity of solids is 2.65 Normal stress (Ơ) = 140kPa Shear stress (τ) = 105kPa Solution: a) tan Ø = Normal stress (Ơ) = _105_ = 0.75 Shear stress (τ) 140 0.75 = 0.75; e e = 1.0 b) γ = Gs γw = _2.65 (9.81)_ = 12.998 or 13kN/m3 1+e 1+1 c) W = wV = 13kN/m3[(0.05) (0.05) (0.03)]m3 = 0.975N Exercises: 1. The following are the results of direct shear tests performed on two identical samples of the soil. In test one, the sample shears at a stress of 70kPa when the compressive normal stress is 95kPa. In test two, the sample shears at a stress of 105kPa when the normal stress is 150kPa. a) Determine the value of the apparent cohesion. b) Determine the angle of internal friction for the sand. c) Determine the shear stress at a depth of 4.20m if the unit weight of soil is 15.75kN/m3. 2. A consolidated drained axial test was conducted on a cohesion less soil that has a friction angle of 30o and deviator stress at failure of 400kPa. a) Find the angle that the failure plane makes with the major principal plane. b) Find the confining pressure. c) Find the shear stress at the point on the failure plane. Name: Description: ...
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