CON4341
Soil Mechanics and Geology
(Effective Stress)
Roger CHAN (rogerchchan@vtc.edu.hk)
MHKIE MICE MIEAust MIMMM MASCE CEng NECReg UKRoGEP MSc BEng
Question from Student
What is gw (unit weight of water)?
gw = rw x g (g is acceleration due to gravity = 9.81 m/s²)
If you recall physics at secondary school level…
Newton’s Second Law F = ma (F is force (N); m is mass 質量 (kg) and a is acceleration (m/s²)
And if you recall that the unit for weight 重量 is N…
For unit weight, so a unit volume is considered. Unit volume means 1m³
Mass of water under unit volume = 1000kg/m³ (rw) x 1m³ = 1000kg.
F = ma -> F = gw, m = rw, a = g
Question from Student
How to get this equation without the aid of excel?
Method 1: Directly read from the graph. As long as you can read the number
close enough to 43.8% (say 43.5% to 44.1%) then, you probably get the correct
answer
Method 2: Use the math you learnt in secondary school. y = mx + c
Cone Penetration Plot
25
y = 39.468x + 2.7227
20
A
Penetration
B
15
10
5
0
0.00%
10.00%
20.00%
30.00%
Moisture Content
40.00%
50.00%
60.00%
Question from Student

As the required point lies between point A and B, so you may simply take
point A and B to work out the linear equation.

A (0.5223, 23.3) and B (0.4297, 19.6), therefore the slope gradient will be
(23.3-19.6) / (0.5223 – 0.4297) = 39.9 -> this is “m” in y = mx + c

To find out c, you may simply substitute point A or B into the equation.
Say substitute point A, 23.3 = 39.9 (0.5223) + c, c = 2.4.

The linear equation is y = 39.9x + 2.4

To find the liquid limit, substitute 20 into y then you will get x = 44.1%

The equation is slightly different to the equation given by the excel, but no
worry such minor difference is acceptable. Also, in civil engineering world,
there is no fixed answer for every calculation. Different designer to handle
the same question could get different answers because they may use
different approach and assumption.
But you all applied the same
fundamental concept. So the difference is still acceptable.
Stress on Soil

Normal Stress (s)
i.e. perpendicular to the section (Fy)
s = Fy / A

Shear Stress
i.e. parallel to the section (Fx)
t = Fx / A
F
Fx
Fy
A
Total Stress (s)
rd or gd
H1
G.W.L
rsat or gsat
H2
A

Stress due to all forces on the soil sample

Unit for Stress : kPa, kN/m², N/mm², MPa….

Unit for Unit Weight : kN/m³

So Stress (kN/m²) = Unit Weight (kN/m³) x Height (m)

Total Vertical Stress at A (s) = s 1 + s 2
= gd x H1 + gsat x H2
Total Stress
Total Head
Pore Water Pressure (u)
rd or gd
H1
G.W.L
rw or gw
H2
A

rsat or gsat
Pore Water Pressure at A (u) = gw x H2
Effective Stress (s’)

It is the normal stress to which soil particles are subjected

Total Stress (s) = Effective Stress (s’) + Pore Water Pressure (u)
(carried by soil grain) (carried by water)

It is critical for : Shear Strength determination
Bearing Capacity calculation
Slope Stability evaluation
Rate of Settlement calculation
Illustration Example

Calculate total stress (sv) and effective stress (sv’) at depth 0m, 6m and 14m.

The given unit weight is bulk unit weight
Illustration Example (1)

At point 1,
Total stress = 0 kN/m², Effective stress = 0 kN/m², pore water pressure = 0 kN/m²

At point 2,
Total stress = unit weight x height = 20 x 6 = 120 kN/m²
Pore water pressure = 9.81 x 6 = 58.86 kN/m²
Effective stress = Total stress – pore water pressure = 61.14kN/m²

At point 3,
Total stress = 120 + 8 x 17 = 256 kN/m²
Pore water pressure = 9.81 x 14 = 137.32 kN/m²
Effective stress = 256 – 137.32 = 118.68 kN/m²
Hydrostatic Condition

Hydrostatic means there is no flow

Water level at all standpipes are the same

Hydrostatic Pressure increases proportionally with depth
At Point A,
gw
h1
s = gw h1; u = gw h1; s’= 0
At Point B,
z
s = gw h1 + gsat z; u = gw (h1+z); s’= (gsat – gw)z
At Point C,
h2
A
gsat
B
Pressure head
s = gw h1 + gsat h2; u = gw (h1+h2); s’= (gsat – gw)h2
C
Pressure head
Valve closed
Upward Seepage
dh
h1
h2
gw
A
gsat
B
C
Valve closed
h1
h2
gw
A
gsat
B
C
Valve open
At Point A, s = gw h1; u = gw h1; s’= 0
At Point C, s = gw h1 + gsat h2; u = gw (h1+h2+dh); s’= (gsat – gw)h2 – gwdh
Effective stress is reduced because of seepage (gwdh)
Upward Seepage
dh’
h1
h2
gw
gsat
A
z
B
C
Valve open
At Point B,
dh
s = gw h1 + gsat z;
For steady 1D flow, the
u = gw (h1+z+dh’);
hydraulic gradient is constant
dh/h2 = dh’/z
dh’ = (dh/h2)z
s’= (gsat – gw)z – gw(dh/h2)z
= gsub z – gw i z
gwdh’ or gw i z is termed as seepage pressure
Upward Seepage
h
A
a
B
d
C
Let’s set datum at C
At Point A,
Elevation Head = a + d
Pressure Head = 0
Total Head = a + d
At Point B,
Elevation Head = d
Pressure Head = a
Total Head = a + d
At Point C,
Elevation Head = 0m (because at datum)
Pressure Head = h+a+d
Total Head = h+a+d
Head Difference between B and C = h
Therefore, energy at C is higher than B.
So water flow from C to B
Upward Seepage
Stress at A-A,
s = 0; s’ = 0 (No soil -> 0) u = 0; (at Phreatic level -> 0)
Stress at B-B,
u = gw x a; s’ = 0; s = s’ + u = gw x a
h
A
A
a
B
d
gw
B
gsat
C
C
Total Stress at Point C-C,
s = gsat x d + gw x a
Imagine summation
of overburden
Pore water Pressure at C-C
u = gw (h + a + d)
Effective Stress at Point C-C
s’ = s - u
= gsat x d + gw x a - gw (h + a + d)
= (gsat – gw) x d - gw x h
= gsub x d – gw x h
Upward Seepage
Constant water level
Sheetpile
length
Formation
level
Photo Reference:
https://www.reddit.com/r/nextfuckinglevel/comments/eu69cq/the_power
_of_sheet_piles/?utm_source=share&utm_medium=web2x&context=3
Illustration Example
Consider the upward water flow through a layer of soil in a
tank. For the soil, void ratio (e) = 0.5; specific gravity = 2.65
1.5m
0.5m
2.5m
gw
gsat
Calculate the total stress, pore water pressure and effective
stress at the mid-point of the tank.
Valve open

How to find gsat? (recall lecture 2)
gsat = (total weight of soil + total weight of water) / total volume
= (Gsgw + wGsgw) / (1 + e)
= (Gsgw + egw) / (1 + e) (Sr = 1 because fully saturated)
= (2.65 x 9.81 + 0.5 x 9.81) / (1 + 0.5)
= 20.6kN/cu.m

Unit weight of water (gw)
= Density of Water x acceleration due to gravity
= 1000 x 9.81
= 9.81kN/cu.m
Illustration Example
At the mid point,
1.5m
s = 0.5 gw + z gsat (z = 1)
= 0.5 x 9.81 + 20.6
= 25.51kPa
u = [(0.5 + 1) + i(z)]gw
(i is hydraulic gradient h/L)
0.5m
gw
z
2.5m
gsat
=20.6kPa
s’= s – u
= 4.91kPa
Valve open
h/L (z)
1.5/2.5
Critical Hydraulic Gradient

For upward seepage case, if hydraulic gradient increases gradually…

what scenario will make the hydraulic gradient increase???
E.g. deeper excavation

When the effective stress becomes ZERO (the soil cannot take any load),
maximum hydraulic gradient occurs. It is so called Critical Hydraulic
Gradient.

Recall, s’ = gsub z – gw i z or gsub x d – gw x h, substitute 0 into s’
i = gsub / gw or h/d = gsub / gw
we use ic to represent when effective stress = 0 as critical hydraulic gradient
Critical Hydraulic Gradient

Recall the Phase Diagram
Mass (M)
Mw = wMs
Volume (V)
VOID
Vv = e (e = Vv/Vs)
(Ms = Gsrw)
(= eSrrw)
Ms = Gsrw
SOLID
GRAIN
Vs = 1 (for unit volume)
Critical Hydraulic Gradient

We know ic = gsub / gw (gsub = gsat – gw)

gsat = [(e + Gs) gw / (1 + e)]

ic = [(e + Gs) gw / (1 + e)] – gw / gw
= (Gs – 1) / (1+e)
Critical hydraulic gradient is used to determine the maximum depth can be
excavated the soil will not fail.
Past Paper
Past Paper
Depth (m)
Total (kN/m²)
Pore Water Pressure
(kN/m²)
Effective Stress (kN/m²)
0
0
0
0
6
20 x 6 =120
9.81 x 6 = 58.86
61.14
14
120 + 8 x 17 = 256
58.86 + 8 x 9.81 = 137.34
118.66
0
Total stress
6
14
120
58.86
61.14
256
118.66
137.34
Past Paper
Past Paper
Depth (m)
Total Stress (kPa)
PWP (kPa)
Effective Stress (kPa)
3
1.85 x 9.81 x 3 = 54.4
0
54.4
6
54.4 + 2.1 x 9.81 x 3 =
116.2
3 x 9.81 = 29.4
86.8
16
116.2 + 1.92 x 9.81 x 10
= 304.6
(10 + 6 + 2)x 9.81 =
176.6
128.0
Pressure head is 18m,
Water pressure = 18 x gw
Home Exercise 1
The underground soil profile of a site was revealed as follows:
Determine the vertical effective stress at 0m, 3m, 5m, 16m below ground
Home Exercise 1

At 0m,
s, s’, u = 0

At 3m,
s = 19 x 3 = 57kPa; u = 0; s’ = 57kPa

At 5m,
s = 57 + 21 x 2 = 99kPa; u = 2 x 9.81 = 19.62kPa; s’ = 79.38kPa

At 16m,
s = 99 + 22 x 11 = 341kPa; u = 13 x 9.81 = 127.53kPa; s’ = 213.47kPa
Home Exercise 2
The underground soil profile of a site was revealed as follows:
Determine the vertical effective stress at 0m, 3m, 5m, 12m, 16m below ground
dh
L
Home Exercise 2

There is an upward seepage. So in calculation, you need to consider hydrostatic pressure and
seepage pressure.

Hydraulic gradient (i) has to be calculated = h/L

Identify hydraulic head (level difference between two water level)

Capillary water means negative pore water pressure

Recall s’ = gsub z – gw i z
i = h/L = 5 / 13 = 0.385
At 0m, s = 0; u = -(3) x 9.81 = -29.43kPa; s’ = 29.43kPa (because capillary rise)
At 3m, s = 19 x 3 = 57kPa; u = 0kPa; s’ = 57kPa
At 5m, s = 57 + 2 x 21 = 99kPa; u = [2 + (0.385 x 2)]x 9.81 = 27.17kPa; s’ = 71.83kPa
At 12m, s = 99 + 7 x 22 =253kPa; u = 14 x 9.81 = 137.34kPa; s’ = 115.66kPa
At 16m, s = 253 + 4 x 20 = 333kPa; u = (13+5) x 9.81 = 176.58kPa; s’ = 156.42kPa
Home Exercise 3
The underground soil profile and ground water conditions at a site are as shown
in the following figure.
Home Exercise 3
(a) Recall Total Head = Elevation Head + Pressure Head
Total Head = 12m at A, B and C
(b) v = ki
i = h/L = 3/10 = 0.3
k = 2.5 x 10-8
v = 2.5 x 10-8 x 1 x 0.3 = 75 x 10-8 m/s
(c) s = 17 x 3 + 18 x 10 = 231kPa
(d) u = (10 +3) x 9.81 = 127.5kPa
Refer to Ch 6.6
Flow velocity (m/s)
(e) s ’ = s - u
= 103.5kPa
(f) i = h’/7.5
= 2.25m
Alternatively, q = kAi
Assume unit area,
q/A = 75 x 10-8 m/s