Shear Strength
Eng. Ahmed Yahiya Barakat Mohamed
Shear Failure Theory
https://www.youtube.com/watch?v=HkjVWkyLfig
https://www.youtube.com/watch?v=MS4H_u0ARpo
https://www.youtube.com/watch?v=6tSnA9I6uL4
https://www.youtube.com/watch?v=Kc0i6BQguV0
Mohr-Coulomb Failure Criterion

 f  c   tan 

friction angle
cohesion
f
c


f is the maximum shear stress the soil can take without failure, under
normal stress of .
12
Mohr Circles & Failure Envelope

Y
X
X
Y

Soil elements at
different locations
X
~ failure
Y
~ stable
Mohr Circles & Failure Envelope
The soil element does not fail if
the Mohr circle is contained
within the envelope
GL

c
Y
c
c
Initially, Mohr circle is a point

c+
Mohr Circles & Failure Envelope
As loading progresses, Mohr
circle becomes larger…
GL

c
Y
c
c
.. and finally failure occurs
when Mohr circle touches the
envelope
Effective Stress
 Archimedes’ principle
“The loss of weight of a body
submerged in water is equal to the
weight of water displaced, and that
a floating body displaces its own
weight of water.”
Effective Stress
Stresses in Soils under Static Hydraulic Forces.
The stress at any point A is
Where σ is the total stress
This total stress σ can be visualized to consist of TWO components : (1) the pressure
upon the soil skeletal component, called the effective stress σ‘ and, (2) the pressure
from the water component in the voids continuum, called the pore water pressure u,
also known as the neutral stress.
Effective Stress
OVERBURDEN STRESSES
Pore water pressure
Effective stresses
Total stresses
VERTICAL STRESSES
EXAMPLE
Determine the total stress,
pore water pressure, and
effective vertical stress at
A, B, and D.
At D
sD = 6x16.5+13x19.25
sD = 349.25 kN/m2
uD = 13x9.81 =127.53 kN/m2
s’D = 6x16.5+13x(19.2 – 9.81)
s’D = 6x16.5+13x(19.2 – 9.81) = 221.72 kN/m2
VERTICAL STRESSES
Vertical And Horizontal Stresses
PRESSURE FROM APPLIED DISTRIBUTED LOADS:
Az=(x +z)(y+ z)
Ϭz= P/ Az
Vertical And Horizontal Stresses
Example
A 1.5 kN point load is applied to the soil surface by the tire of a parked motorcycle. Using
Boussinesq's equation , most nearly, what is the increase in vertical pressure at a point 1.0 m below
the surface and 0.22 m radial from the tire?
(A) 470 Pa
(B) 540 Pa
(C) 640 Pa
(D) 710 Pa
The answer is (C).
Example
Solution
Selection of Type of Foundations
•
•
•
•
•
Approximate Loads of Structures:
Residential and Housing, 1.0 up to 1.2 t/m2 per floor
Commercial and Office, 1.2 up to 1.5 t/m2 per floor
Schools and Hospitals, 1.5 to 2.0 t/m2 per floor
These loads are multiplied by the number of floors, then divided by the
foundation area to determine the soil stresses, then the type of foundation.
Three options will be available:
• 1- Stress on soil < qall soil, R.C foundation area ≤ 2/3 foundation area  Isolated
footings.
• 2- Stress on soil < qall soil, R.C foundation area > 2/3 foundation area  Raft
foundation.
• 3- Stress on soil > qall soil  Pile foundation.
The failure zone can be divided to three zones:
Zone I:
The soil remains in the elastic state. It acts as if it was a part of the sinking footing and
penetrates the soil like a wedge.
Zone II:
It is the zone of radial shear, straight lines radiate from the outer edge of the footing.
The lines of the other set are logarithmic spirals for (c-) or circular arc for cohesive
soil.
Zone III
The passive
Bearing capacity equation (Egyptian Code) for Homogeneous
Soil under vertical and Centric Loads:
𝑞𝑢𝑙𝑡.𝑔𝑟𝑜𝑠𝑠 = 𝐶 𝑁𝑐 𝑐 + 1 𝐷𝑓 𝑁𝑞 𝑞 + 2 𝐵 𝑁 
Where:
qult.gross = ultimate gross bearing capacity of soil
Df = depth of foundation
B = width of foundation
1 = unit weight of soil above F.L
2 = unit weight of soil below F.L
𝑐 , 𝑞 ,  = shape factors (from tables)
𝑁𝑐 , 𝑁𝑞 , 𝑁 = Bearing capacity factors (from table related to ˚ )
q ult.net = q ult.gross – overburedn pressure
q ult.net = q ult.gross – ∑1Df
qallowable = qall. =
𝑞𝑢𝑙𝑡
F.O.S
𝐹.𝑂.𝑆
= 3---5
Note:
(F.O.S) against bearing capacity failure =
𝑞𝑢𝑙𝑡.𝑛𝑒𝑡
𝜎𝑛𝑒𝑡
≥3
The effect of ground water table
1- Ground water table between ground surface and foundation
level
𝑞𝑢𝑙𝑡.𝑔𝑟𝑜𝑠𝑠 = 𝐶 𝑁𝑐 𝑐 + 1 𝐷𝑓 𝑁𝑞 𝑞 + 2 𝐵 𝑁 
𝑞𝑢𝑙𝑡.𝑔𝑟𝑜𝑠𝑠
= 𝐶 𝑁𝑐 𝑐 + 𝑏 𝐷1 + 𝑠𝑢𝑏 𝐷2 . 𝑁𝑞 𝑞 + 𝑠𝑢𝑏 𝐵 𝑁 
2. Ground water table at foundation level
𝑞𝑢𝑙𝑡.𝑔𝑟𝑜𝑠𝑠 = 𝐶 𝑁𝑐 𝑐 + 1 𝐷𝑓 𝑁𝑞 𝑞 + 2 𝐵 𝑁 
𝑞𝑢𝑙𝑡.𝑔𝑟𝑜𝑠𝑠 = 𝐶 𝑁𝑐 𝑐 + 𝑏 𝐷𝑓 𝑁𝑞 𝑞 + 𝑠𝑢𝑏 𝐵 𝑁 
1=== bulk
2=== sub.
3. Ground water table below foundation level
Case adw ≥ B
𝑞𝑢𝑙𝑡.𝑔𝑟𝑜𝑠𝑠 = 𝐶 𝑁𝑐 𝑐 + 1 𝐷𝑓 𝑁𝑞 𝑞 + 2 𝐵 𝑁 
Case adw< B
1 = b
2 mod. = [2sub + dw (2 above water - 2 sub)]
Example
Required:
a. qall.net
a. Is the footing safe?
a. F.O.S against shear failure
b. Find the allowable load if the allowed
settlement of soil = 5.5 cm (immediate and
Consolidation Settlement), knowing that Cc =
0.1, and the soil is normally consolidated, Es =
15000 KN/m2 and initial void ratio = 0.55, and
thickness of the given layer is 4m.
qall.net
For square footing
𝑐 =𝑞 = 1.3
 = 0.7
f = 30˚ 𝑁𝑐 = 30 , 𝑁𝑞 = 18 , 𝑁 = 10
𝑞𝑢𝑙𝑡.𝑔𝑟𝑜𝑠𝑠 = 𝐶 𝑁𝑐 𝑐 + 1 𝐷𝑓 𝑁𝑞 𝑞 + 2 𝐵 𝑁 
qult.gross = 1.3 x 6 x 30+1.3 x1.8 x1.5 x18 + 0.7 x 0.95 x 2 x10=
310.48 t/m2
qult.net= qult.gross- overburden pressure
qult.net= 310.48 – (1.8 x 1.5) = 307.78 t/m2
qult net
qall.net= .
𝐹.𝑂.𝑆
=
307.78
3
= 102.59 𝑡/𝑚2
Is the footing safe?
𝜎𝑛𝑒𝑡 =
F.O.S =
𝑃
𝑎𝑟𝑒𝑎
𝑞𝑢𝑙𝑡.𝑛𝑒𝑡
𝜎𝑛𝑒𝑡
=
200
2𝑥2
=
307.78
50
= 50 𝑡/𝑚2
= 6.15 > 3 𝑠𝑎𝑓𝑒
F.O.S against shear failure
F.O.S =
𝑞𝑢𝑙𝑡.𝑛𝑒𝑡
𝜎𝑛𝑒𝑡
=
307.78
50
= 6.15
Allowed settlement = 5.5 cm, the allowable
load is?
1st Immediate settlement.
𝑃𝑎𝑙𝑙
𝑃𝑎𝑙𝑙
∆𝜎𝑡 =
=
2×2
4
𝑃𝑎𝑙𝑙
𝑃𝑎𝑙𝑙
∆𝜎𝑚 =
=
(2 + 2) × (2 + 2)
16
𝑃𝑎𝑙𝑙
𝑃𝑎𝑙𝑙
∆𝜎𝑚 =
=
(2 + 4) × (2 + 4)
36
1 𝑃𝑎𝑙𝑙 4𝑃𝑎𝑙𝑙 𝑃𝑎𝑙𝑙
𝑃𝑎𝑙𝑙
∆𝜎𝑎𝑣 =
+
+
=
6 4
16
36
3456
−4
= 2.89 × 10 𝑃𝑎𝑙𝑙
𝟏
𝑺𝒆 =
× 𝟐. 𝟖𝟗 × 𝟏𝟎−𝟒 × 𝑷𝒂𝒍𝒍 × 𝟒
𝟏𝟓𝟎𝟎𝟎
2nd Primary Consolidation
𝜎𝑜 = 1.5 × 18 + 2 × 9.5 = 46 𝐾𝑁 𝑚2
𝑃𝑎𝑙𝑙
46 + 16
4
𝑆𝑃 =
0.1𝑙𝑜𝑔
1 + 0.55
46
𝑆𝑡 = 𝑆𝑒 + 𝑆𝑃 = 0.055 𝑚
𝑃𝑎𝑙𝑙
46
+
1
4
16 = 0.055 𝑚
−4
𝑆𝑡 =
× 2.89 × 10 × 𝑃𝑎𝑙𝑙 × 4 +
0.1𝑙𝑜𝑔
15000
1 + 0.55
46
𝑃𝑎𝑙𝑙 = 455 KN, 45.5 ton