Eng.

Ramy Adel
01 Phase Relationships 04 Effective stress

Index properties and

02 soil classifications 05 Slope stability

Consolidation and Stability of

03 06
differential settlement retaining structures
• Out lines :
• Foundations types
• Effective stress
• Bearing capacity
• Foundations:
• Types :
• Shallow foundation
• Deep foundation
• Shallow foundation :

• Definition:
• Foundations that transmit structural loads to the near surface soils.

• Types
• Spread footing

• Mat foundation
• Spread footing

• Definition :
• An enlargement at the bottom of a column or bearing wall that spreads the
applied structural loads over a sufficiently large soil area.

• Types :
• Square spread footings

• Circular spread footings

• Strip/continuous footings

• Combined footing
Spread footing types
• Mat foundation :

• Definition :
• A very large spread footing that usually encompasses the entire footprint of the
structure.

• The advantages of the mat foundation over individual spread footings

• Spreads the structure load over a larger area, thus reduces bearing pressure.

• Provides much more structural rigidity and thus reduces the potential for excessive differential
settlements.

• Is easier to water proof.

• Has a greater weight and thus is able to resist greater uplift pressure.
• Stress
• Normal stress
• Force per unit area

• 𝝈=P/A

• UPLIFT STRESS

• it is an upward vertical stress created due to penetration of the water into the
porous material at the dam basement.

• U=𝜸𝒘. 𝒉𝒘
• Effective stress :
• The difference between normal stress and the uplift stress

• 𝜎 ` =𝜎 -U

• Shear stress
• Results from the force acting in a direction parallel to a surface or to a planar cross
section of a body.

• τ = T/A
• T = shearing force
• mode of failure
• Illustrating
• The applied load causes a a slight downward movement
of the footing resulting in a total rapture
of the underlying soil, which make the soil to be pushed up
on both sides of the footing ,

• Due to the shear failure of soil, sudden collapse occurs,

accompanied by tilting of the footing.

• The value of The applied load is considered to be ultimate

Qult which is the maximum load vale applied to this footing
• The shear value that causes the previous failure is found using the following equation

• Equation used : F = c + σN tan φ

• where

• c = cohesion factor φ = angle of internal friction

• in a cohesion less soil (sand) these values is as follows:

c=zero φ=has a value

• In a cohesion soil (clay) these values is as follows:

c= has a value φ= zero

• The value of Qult should be considered in the footing

design by calculating the corresponding
ultimate bearing capacity of the soil
Example :

Solution:
Given : h=20 ft 𝛾=115 pcf 𝜑 =35` 𝜏𝑓 =? ?

Applying H.B equation 𝜏F = c + σN tan φ

σN = 𝛾.h=20*115=2300 psf

The soil is Sandy so the value of C =0

F=0+2300*tan35=1610 psf 𝜏𝑓 = =11.2 psi
• Bearing capacity
• the capacity of soil to support the loads applied to the ground.

• Types

• Ultimate bearing capacity

• the theoretical maximum pressure which can be supported without failure

• Allowable bearing capacity

• the ultimate bearing capacity divided by a factor of safety
• Equation used :

• Where :

• c = cohesion factor

• Nc = bearing capacity factor for cohesion

• 𝜸` = the specific unit weight of the soil

• Nq = bearing capacity factor for depth

• Df = depth of footing below ground surface

• B = width of strip footing

• Nγ = bearing capacity factor for unit weight

Example :
A 2 m wide continuous wall footing is designed to support an axial column load of 600 KN (per
meter of wall length) and a moment of 250 kN·m (per meter of wall length), as shown. The
footing is placed 1 m into a sandy soil with a density of 2000 kg/m3 , a cohesion of 0.5 Pa, and
an angle of internal friction of 30°. most nearly, what is the ultimate bearing capacity per
meter of footing length?(Nc= 37.2, N q = 22.5, and N 𝛾 = 19.7)

Solution :
Given : P= 600KN , M= 250 KN.m 𝜌= 2000 kg/m3
𝜑 = 30` , D= 1 m B= 2 m C=.5 pa
qult=??
Applying the H.B equation
…
Slope stability
 Slope stability
• Definition
• the potential of soil forming slopes to withstand
the undergoing movement.
 Slope stability
• Basic definitions
• Slip or failure zone

• is a thin zone of soil that reaches the critical state resulting in movement of the upper soil mass.

• Slip plane or failure plane

• the surface of sliding.

• Sliding mass

• the mass of soil within the slip plane and the ground surface

• Slope angle

• the angle of inclination of a slope to the horizontal.

 Slope stability
• Causes of Slope Failure.
• Erosion. Rainfall.

• Earthquakes. Geological Features.

• Construction Activities. External Loading

•
Slope stability
In order to secure such slop we need to determine a safety factor that ensures no failure is
going to happen

• F.S=TFF/TMOB

• In order to establish the F.S we need to get the following values

• The resistance that could be found.

In our case the only resistance that we have got
is the soil characteristics as follows :

• The cohesion force along the slip surface

• The soil weight

• The internal friction angel of the soil

• The previous factors are found in the following equation

• TFF = cLS + WM cos αS tan φ TFF = available shearing resistance along slip surface
 Slope stability
• The reason for the slip .which is the weight component at the direction of the slip surface
which can be found as follows :

• TMOB = WM sin αS

• TMOB = mobilized shear force along slip surface

• Note that :

• TFF = represents the shear resistance due to the sliding of the soil

• Tmob= represents the value of the shear force which causes the slip
 Slope stability
Example :
An excavated slope in uniform soil is shown. The specific weight is 14 kN/ m3
, the cohesion is 15 kPa, and the angle of internal friction is 15°. what is most nearly the
available shearing resistance along the assumed failure plane?

Solution :
𝛾=14 KN/m3 , 𝛼𝑠 = 15 𝜑 = 15 C=15 kpa

Applying H.B equation

Sin 30=10/Ls Ls= 10/sin 30=38.64m

TFF = cLS + WM cos αS tan φ 30

Weight= 𝛾.V 10 m
V=X-Area * length for 1 meter strip L=1 m
 Slope stability
Soil slip area:
• Out lines :
• Introduction
• Vertical Stress
• Vertical Stress Without surcharge
• Vertical Stress With surcharge
• Horizontal Stress Profiles and Forces
• Retaining Walls
• Retaining structures
• Definition :
• An engineered structure built to

• support or confine soils in excess of 1.2 meters in height

• retain soil and/or rock accommodated with a changes in grade

• retain lateral forces from soil and/or water are composed of various materials

• the most used Types for the retaining structures area:

• Gravity Piling Cantilever Anchored
Retaining structures examples
Retaining structures types
• Vertical Stress
• Vertical Stress Without surcharge
• Vertical Stress With surcharge
• Pore Water Pressure
• The pressure exerted on its surroundings by water held in pore spaces in rock or soil.

• 𝑢 = 𝛾𝑤. ℎ𝑤

• Effective Vertical Stress

• the difference between the total stress (σ) and the pore pressure (u) in a saturated soil
has been defined as the effective stress (σ').

• σ‘= σ-u σ=

• The pore pressure exists when the soil is under the ground water table

• When the soil is above the water pressure the pore water pressure equals
zero
Vertical Stress Without surcharge
• In this type the most important concerns are:

• the soil types with different specific weights

Which creates a different stress values

• The ground water table (G.W.T) which develops

a pore water pressure that reduces the value
of the total pressure creating an effective pressure

• The total pressure , pore water pressure and

the effective vertical stress usually starts 𝛾1
with a zero value

𝛾2
• Vertical Stress Without surcharge

• Unlike the previous type the value

of the total vertical stress starts
with a surcharge value ,
which is reflected to the value
of the effective vertical stress

𝛾1

𝛾2
• Horizontal Stress Profiles and Forces
• Basic definitions

• Active earth pressure (the wall moves away from the soil )
• When the retaining wall, tends to slide down ,
or moving towards the soil in front of the wall ,
due to load from the retained soil massive then
the pressure from the retained soil
becomes active pressure

• Passive earth pressure (the wall moves towards the soil mass)
• the pressure from the soil in front of the wall, resisting against the wall moving towards it
• Effective Horizontal Force

• To calculate the horizontal force we use a theory called Rankine theory that
have the following characteristics
• neglect the wall friction

• the soil movement is uniform.

• The force acts on the lower

one-third of the stress triangle

• The value of the force whether it`s a passive force or active force equals the area of the stress
triangle multiplied by a specific factor (Rankine coefficient )

• The stress used is the effective stress not the total stress
• Rankine theory relations
• Rankine coefficients
• These coefficients depends on pressure case

• Rankine active earth pressure coefficient (KA)

• KA= 𝑡𝑎𝑛 (45° - φ/2)

• Rankine passive earth pressure coefficient (KP)

• Kp= 𝑡𝑎𝑛 (45° +φ/2)

• at rest earth pressure coefficient (Ko)

• For normal consolidation Ko ≈1– sin φ

• K0 = (1– sin φ) 𝑂𝐶𝑅sin φ 𝑂𝐶𝑅= over consolidation ratio = Pc/Po

• Note that :

• The following equations maybe used to represent the values of these coefficient

• KA= KP=

• .

• .

• Ka and Kp are function only off φ . C has no effect on them.

• Force calculations depends on the side as follows
• The active case
• The total Lateral Earth Active force per unit length of the wall (Pa)

• PA=the Area of the active Earth pressure

diagram multiplied by rankine coeff.

• Note that the water table creates an active pressure

• PA= ½ x KA x x
• Point of application of Pa

• H/3 from the base

Example :
A retaining wall extends 3 m from the top of bedrock to the ground surface. The soil
is cohesionless and has the properties shown. Using Rankine theory, the total active
earth pressure per unit width of retaining wall is most nearly.

Solution :
H=3 m φ=35 =1834 kg/m3
PA=.5*H* *KA
= . = .g.h g=9.81 m/s2

KA= (45° - φ/2)= (45° - 35/2)=.27

𝐻
PA=.5*H* *KA=.5*H* .g.h*KA PA
=.5*3*1834*9.81*.27=21940 N/m
𝜎
• The passive case
• The total Lateral Earth Active force per unit length of the wall (PB)

• PB=Area of the passive Earth pressure diagram

multiplied by rankine coeff.

• PB= ½ x KB x x
• Point of application of Pa

• H/3 from the base

Horizontal Stress Profiles and Forces

𝛾1
𝐻1
1
𝜎`1
𝛾2
𝐻2 2
4
𝜎`1 3
The area of the 1st shape (PA1) 𝜎`2-𝜎`1
PA1=.5*𝜎`1*𝐻1 ∗ 𝐾𝐴 𝜎`1=𝛾1.H1
PA1=.5* 𝛾1. KA. 𝐻1
The area of the 2nd shape (PA2)
PA2=𝜎`1. ℎ2.KA 𝜎`1=𝛾1.H1
The area of the 3 shape (PA3)
rd

PA3=.5*(𝜎`2 − 𝜎`1) ∗ 𝐻2*KA 𝜎`2=𝛾2.H1+𝛾2.H2- 𝛾𝑤.H2

The area of the 4 shape (PA4)
th

Paw=.5* 𝛾w*hw*hw
EXAMPLE:

Solution :
Given : H1=5 ft H2=7 ft Hw=7 =30
R.PA =???

U=hw. 7*62.4=436.8 pcf

KA= (45° - 30/2)=.333
PA1=.5*𝜎`1*𝐻1∗𝐾𝐴
𝜎`1=𝛾1.H1-u U=0 H1
𝜎`1=120*5=600pcf 𝛾
PA1=.5*120*5*5*.333=500 lb PA1
1

PA2= 𝜎`1*𝐻2∗𝐾𝐴 PA2 H2

PA2=120*5*7*.333=1400 lb 𝛾𝑤 2
4 PA3
3
PA3=.5*(𝜎`2-𝜎`1)*𝐻2∗𝐾𝐴
𝜎`2= 𝜎2-u H1=5 ft
𝜎2= 𝛾.H1+𝛾.H2=120*5+120*7=1440 pcf H2=7 ft
U=436.8 pcf Hw=7 𝛾 = 120𝑝𝑐𝑓
𝜎`2=1440-436.8=1003.2 pcf 𝜑=30 𝛾𝑤 = 62.4𝑝𝑐𝑓
KA=.333
PA3=.5* (1003.2-600)*7*.333= 470 pcf
Pw=.5* 𝛾w*hw*hw
=.5*62.4*7*7=1529 pcf ∑ 𝑃𝐴 =500+1400+470+1529=3899 pcf
 Failures modes for the retaining walls

Bearing
Sliding Over turning
capacity
• Bearing capacity
• a result of structure settlement as result the the applied stress is larger than the ultimate
stress

• Equation used :

• ∑ 𝑣 =the sum of the vertical weights and loads

• B = footing width e= eccentricity qult = is calculated previously

• MR=resistance moment Mo= over turning moment

• Sliding
• a result of excessive lateral earth pressures with relation to retaining wall resistance thereby
causing the retaining wall system to move away (slide) from the soil it retains.

• Equation used :

• FR: resisting force

• FD: driving force

• ∑ 𝑣 =the sum of the vertical weights and loads

• d = external friction angle=2/3 f

• Ca=adhesion for concrete on soil only =factor* C C= cohesion

• Failures modes for the retaining walls
• Over turning
• is a result of excessive lateral earth pressures
with relation to retaining wall resistance thereby
causing the retaining wall system heel toe
to topple or rotate (overturn).

• Equation used :

• MR: resistance moment resulting from the weight of

the structure and the passive force

• Mo: the over turning moment resulting

from the lateral forces
Example :
determine the factor of safety for the concrete retaining wall to resist sliding due to lateral earth pressures
exerted on the wall. The wall foundation is on soils with a cohesion of 23.9 kN/m2 . The retaining wall is not
threatened by earthquakes, so omit the dynamic component. The retaining wall characteristics are provided
below. Knowing that the resultant active stress is
R = 398 kN/m
The passive stress could be neglected
Given :
𝛾𝑠𝑜𝑖𝑙 = 21.2 kN/m3
𝛾𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 =23.6 kN/m3 2.1m
angle of Internal Friction, 𝜑 = 36 degrees
𝛿 = (𝜑)2/3 = 24 degrees =𝛼
c = 23.9 kN/m2
wall height, H = 2.1 m
.3 m
wall thickness, h = 0.30 m
footing thickness, t = 0.30 m .46 m .3 m
footing width, B = 2.1 m
distance from the footing edge (toe) to face of wall in front of wall, 0.46 m 2.1 m
Solution :
Applying H.B equation R
2.1m
FR= ∑ 𝒗 + 𝒕𝒂𝒏𝛿+Bca+Pp 24 deg

• Vertical loads

*Total weights *Vertical component of the active resultant force .3 m

The passive side is neglected in both calculations .46 m .3 m
• Total vertical weight are computed for a strip 1m/1ft
2.1 m
which are resulted from

• The weight of soil above heel (W1)

• The weight of concrete for the wall (W2)

• The weight of concrete for the footing base (W4)

W1 = 𝛾 soil(width of soil block above footing)(height of soil block above footing)
= 21.2 kN/m3(1.68 m)(1.83 m) = 65.1 kN/m

W2 = 𝛾 concrete(width of wall)(height of wall above footing)

= 23.6 kN/m3(0.253 m)(1.83 m) = 10.9 kN/m metric

W3 = 𝛾 concrete(width of footing)(height of footing)

= 23.6 kN/m3(2.13 m)(0.30 m) = 15.1 kN/m

Wt = W1 + W2 + W3 = 91.1 kN/m

• Vertical component of the active resultant force

• RV=R sin 𝛿 = (398 kN/m)sin 24 = 162 kN/m

FR= ∑ 𝒗 + 𝒕𝒂𝒏𝛿 +Bca+Pp
=(162+ 91.1)+tan24+2.1*23.9=214 KN/m

FS=Rcos𝛿
= (398 kN/m)cos 24 = 364 kN/m

F.S. = RSL/RH = (214 lb/ft)/(364 kN/m) = 0.6

Example:
Resolve the previous equation to get the F.S for overturning R
2.1m

Knowing that the distance from the R point to the toe is .98m

Solution :
Applying H.B equation : .3 m

.46 m .3 m 1.8 m
Get the moment about point o

∑ 𝑀𝑟 =X1w1+x2w2+x3w3+RV.y=∑ 𝑤𝑖. 𝑥𝑖 + 𝑅𝑉

W1,w2,w3,and w4 is calculated before

X1=(width of footing in front of wall) + (width of wall) + (1/2 of width of soil block above footing)
=.46+.3+1.8*.5=1.66 m
3
x2 = (width of footing in front of wall) + (1/2 of wall width)
= 0.46 m + 0.5(0.3 m) = 0.61 m R
2.1m
x3 = (1/2 width of footing)
= 0.5(2.1m) = 1.05 m

∑ 𝑤𝑖. 𝑥𝑖 = 65.1*1.66+10.9*.61+15.1*1.05=130.57 KN.m

.3 m
RV=R sin 𝛿 = (398 kN/m)sin 24 = 162 kN/m
.46 m .3 m 1.8 m
𝑀𝑟=130.57+162*.98=289.33

Mo=Rcos𝛿 = (398 kN/m)cos 24 = 364 kN/m

F.S=289.33/364=.79

3
Finally , you did it
 The End
Any questions??

CREDITS: This presentation template was

created by Slidesgo, including icons by Flaticon,
and infographics & images by Freepik
Please keep this slide for attribution