5 Total & Effective Stress

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DUET CE-3301 GEOTECHNICAL ENGINEERING -I NOTE NO. – 05 PRINCIPLES OF TOTAL AND EFFECTIVE STRESSES

PREPARED BY: Dr. Md. Mokhlesur Rahman Professor Civil Engineering Department Dhaka University of Engineering & Technology (DUET), Gazipur Prepared by: Dr. Mokhlesur Rahman, Professor, Department of Civil Engineering.

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EFFECTIVE STRESS Effective Stress (E.S.), Neutral Stress (N.S.) and Total Stress (T.S.): –

At any horizontal depth Z in a soil profile, the total downward pressure is due to the weight of soil above the section (Fig.-1), is called total stress of soil.

Fig. -1 –

Resistance to this pressure is provided, partly by the soil grains, and if the section is below the water table, partly by the upward pressure of the water.

–

Total stress (Fig.-1) at depth Z per unit area, σ = Z1γ + Z2γsat.

–

This is resisted by the inter-granular pressure σ′, which is referred to as the effective stress and by the upward water pressure u, which is referred to as the neutral stress or pore water pressure and equals Z2γw. i.e. Total downward load per unit area = Inter-granular pressure + Upward water pressure. => Total stress = Effective stress + Neutral stress.

–

This relationship between total, effective and neutral stress is of great importance in soil mechanics. PRINCIPLE OF EFFECTIVE STRESS

–

The statement of the amount of stress by the soil skeleton is called the effective stress, the actual stress or granular stress of soil itself is called the effective stress.

Prepared by: Dr. Mokhlesur Rahman, Professor, Department of Civil Engineering.

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Fig.-2 –

Let us consider in Fig.-2, the equilibrium across a horizontal wavy surface passing through the points of contact of the soil grains.

Let, σs =

Average contract stress between the grains.

As =

Area of grain contact.

uw =

Average pore water pressure.

ua =

Average pore air pressure.

Aw =

Area of water surfaces.

Aa =

Area of air surfaces.

σ

=

Total stress of soil.

A

= Total area of soil. Here, A = As + Aw + Aa or, Aa = A - As - Aw

P

=

We have,

Total load on soil.

P  A

Or, P = σ A

For equilibrium, Or,

Or,


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Typically,

≤ 1 % and the term

Writing, σ′ =

We obtain,

Or,

=(

may be ignored.

= Effective stress.

=

+

+

)+

Or,

where, ζ (zeta) =

For a fully saturated soil, ζ =

=

= 1.

So, Or, Where, u = Pore water pressure = Neutral pressure. For dry soil, ζ Or, σ′ =

But

<<< less ≈ 0. So, σ′ = σ

Computation of effective stress: The computation of effective stress requires the separate determination of the total stress (σ) and of the pore pressure (u) the following typical cases may arrive in practical field conditions. a)

Water table is at ground surface.

b)

Water table is above the ground surface.

c)

Water table is below the ground surface.


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d)

The soil mass may be saturated due to capillary rise.

e)

The soil layer height has as tension pressure.

a)

Water table is at ground surface with out seepage (Fig.-3):

Fig.-3 We have, Effective stress, σ′ = Total stress =

= Z γsat.

Neutral pressure or pore water pressure

u

Or, u = Z

Where, γsat = Saturated unit weight of soil. γw = Unit weight of water. Z

= Height or depth of soil from surface.

hp = Depth of water. Here, hp = Z. So σ′ = Z γsat – Z γw = Z (γsat-γw) = Z γ′. Where, γ′ = = b)

Submerged unit weight of the soil. Effective unit weight of the soil.

Water table above the ground surface without seepage (Fig.-4):


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Fig.-4 Total stress, σ = Z1 γw + Z γsat. Neutral stress, u = γwhp = γw (Z1+Z) Effective stress, σ′ = σ-u = Z1γw + Zγsat - γw (Z+Z1) = Z1γw+Zγsat - γwZ - γwZ1 = Z (γsat - γw) = Zγ′. So, effective stress is unchanged for two case (a) & (b).

c)

Water table below the ground surface without seepage (Fig.-5):


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Fig.-5 Total stress,

σ = Z1γ + (Z - Z1) γsat = Z1γ + Zγsat - Z1γsat.

Neutral stress,

u = hpγw = (Z - Z1) γw = Zγw - Z1γw.

Effective stress, σ′ = σ – u = Z1γ + Zγsat - Z1γsat - Zγw + Z1γw. = Z1γ + Z (γsat - γw) - Z1 (γsat - γw). = Z1γ + (Z - Z1) γ′ Where, γ′ = Unit weight of the soil. d)

The soil layer may be saturated by capillary rise (Fig.-6):

hc = Height of capillary rise Negative pressure occurred at capillary rise of water = - hcγw.


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Fig.-6 Total stress,σ = Zγsat. Pore water pressure, u = - γwZ1 + Zγw. Effective stress, σ′ = σ – u = Zγsat - (- γwZ1 + Zγw) = Zγsat + γwZ1 - Zγw = Z (γsat - γw) + γwZ1 = Z γ′+Z1γw e)

The soil layer height has artesian pressure (Fig-7):

Fig.-7 Effective Stresses in saturated soil with seepage: -

a.

The effective stress at any point in a soil mass will change when water seeps through it. It will increase or decrease depending on the direction of seepage. a.

Upward seepage

b.

Downward seepage

Upward seepage


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Fig.-8 Fig.-8 Shows a layer of granular soil in a tank where upward seepage of water is caused by opening the value located below the tank. The rate of water supply is kept constant. The loss of head due to upward seepage between the levels of A and B is h keeping in mind that the total stress at any point in the soil mass is due solely to the wet of soil and water above it the effective stress calculations at A and B are as follows: At A: Total stress σA = H1γw Pore water pressure = uA = H1γw Effective stress, σ'A = σA - uA = 0 At B: Total stress σB = H1γw + H2γsat Pore water pressure = uB = (H1 + H2 + h).γw Effective stress, σ'B = H2 (γsat - γw) - hγw At C: Total stress σC = H1γw + zγsat Pore water pressure = uC = (H1 + z +

).γw

Effective stress, σ'C = σC – uC = z (γsat - γw) = zγ′ –

γw

zγw


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= zγ′ – izγw

[ Where,

]

If σ'C = 0 = zγ′ – iczγw or, ic = b.

Downward seepage (Fig.-9)

Fig.-9 At constant level of water in the soil tank is maintained by adjusting the supply from the top and the outflow at the bottom. The hydraulic gradient due to the downward seepage,

. The total stress

pore pressure of water and effective stress at any point C are, Total stress σC = H1γw + zγsat Pore water pressure = uC = (H + z Effective stress,

).γw

σ'C = (H1γw + zγsat) – (H + z -

).γw

= zγ′ + izγw [Where,

]

Critical Hydraulic gradient The critical hydraulic gradient is the hydraulic gradient at which the soil becomes unstable i.e when the inter-granular pressure i.e effective stress becomes zero. Prepared by: Dr. Mokhlesur Rahman, Professor, Department of Civil Engineering.

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Fig.-10 Consider a sample of soil, length d, with water flowing upwards owing to head h as shown in the above Fig.-10. The hydraulic gradient = Total stress at base of sample = dγsat + aγw = σ Neutral stress = (h + a + d) γw = u Effective stress, σ' = σ– U = dγsat + aγw - (h + a + d) γw = d (γsat - γw) - hγw σ' = dr' - hγw dr' is the submerged weight of soil and must greater than hγ w for there to be any effective pressure hγw known as the seepage force. If the head h is increased until γ'd = hγw, then effective stress G'= 0 and the soil will become unstable. In h



this condition the hydraulic gradient d   and is known as the critical w hydraulic gradient, ic. Also  Gs  6         sat   w  1  6  w w Gs  1 G 1 ic     or , ic  s w w w 1 6 1 6

Quick sand condition: A soil under critical hydraulic gradient will be unstable and is said to be in a “quick” condition. By this definition any granular soil may be “quick sand” but Prepared by: Dr. Mokhlesur Rahman, Professor, Department of Civil Engineering.

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soils with high permeability i.e. gravels and coarse sands require large quantities of water to maintain a critical hydraulic gradient. Quick sand condition is therefore usually confined to fine grained sands. At a quick condition σ' = 0 = σ – u or, σ = u Thus when the pore pressure equals the total pressure on plane a quick condition exists and the pore pressure can only equal the total pressure when Δh > 0, which id a flow condition. Liquefaction: When a fine or medium saturated loose sand deposit is subjected to a sudden shock the mass will temporarily liquefy. This phenomenon is termed as liquefaction. In the situation just described four criteria were given: a particular sand, loose state, saturation and a sudden shock The shock temporarily increases the pore pressure. The total stress is not large when the soil is loose also the structure is some what unstable. The grain size is such that the pore pressure can “float” the grains. The result is a temporary liquefaction a sand mass until pore drainage occurs. During this time lag the very viscous sand water mixture has little shear strength to support any structures on its and if not confined may flow laterally. This phenomenon has been observed to occur in several fairly recent earthquakes. It also sometimes occurs during pile driving i.e. when the pile has great penetration for several of the hammer blows. Liquefaction can be readily observed in the laboratory by building a “quick sand” tank. Example – 1: A layer of saturated clay 4m thick is overlain by sand 5m deep, the water table being 3m below the surface. The saturated unit weights of the clay and sand are 19 kN/m3 and 20 kN/m3 respectively. Above the water table the unit weight of the sand is 17 kN/m3. Plot the values of total vertical stress and effective vertical stress against depth. If sand to a height of 1m above the water table is saturated with capillary water, how are the above stresses affected? Soln: Dry γ sand = 17 kN/m3 Prepared by: Dr. Mokhlesur Rahman, Professor, Department of Civil Engineering.

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Effective γ sand = γsat - γw = 20 - 9.8 = 10.2 kN/m3 Effective γ clay = γsat - γw = 19 - 9.8 = 9.2 kN/m3 At 5m depth, σ = 3×17 + 2×10.2 = 71.4 kN/m3 At 9m depth, σ = 3×17 + 2×10.2 + 4×9.2 = 108.2 kN/m3 Depth σv (kN/m2) 3 3 ×17 5 (3×17) + (2×20) 9

= 51.0 = 91.0

(3×17) + (2×20) + (4×19) = 167.0

u (kN/m2) = 0 2×9.8 = 19.6 6×9.8 = 58.8

σ′v = σv -u 51.0 71.4 108.2

For water table is saturated with capillary water, total vertical stress and effective vertical stress are changed as follows: The only effect of the 1m capillary rise, therefore is to increase the total unit weight of the sand between 2m and 3m the depth from 17 kN/m 3 to 20 KN/m3 an increase of 3 kN/m3 = (20 -17) kN/m3. Both total and effective vertical stresses below 2m depth are therefore increase by the constant amount 3 × 1= 3 kN/m2 pore water pressure being unchanged.

Example – 2: A 1.20m layer of soil is subjected to a seepage head of 1.80m.What depth of coarse sand would be required above the soil to provide a factor of safety 2.0 against piping if Gs = 2.70 and e = 0.035 for both soil and sand respectively? Soiln: Let, Coarse sand depth = x


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Now, Seepage head 1.80m of water. So, upward pressure = 1.80 × 1000 = 1800 Kg/m2 Saturated density of soil, γsat = = 2642.51 Kg/m3

Or, γsat = ( Here, Gs = 2.7, e = 0.035

Effective pressure, σ′ = (γsat - γw) Z = (2642.51 – 1000) Z = 1642.51Z Kg/m2 To provide a factor of safety 2 against piping, so the effective pressure at the bottom of soil must be twice of the upward seepage pressure. So,

= 2.00

Or, 1642.51 Z = 1800 × 2 Or, Z =

= 2.1918 m ≈ 2.20 m

So, Depth of Coarse Sand, x = (Z – 1.2) m (2.20 – 1.20) m = 1.0 m (Ans.)

=

Example – 3: A granular soil deposit is shown in figure. Plot the variation of total stress, pore water pressure and effective stress against depth. For the granular soil given are e = 0.50 and Gs = 2.65, γw = 9.81 KN/m3.


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Soln: Unit weight calculation: Between levels a to b,

= 17.331 kN/m3

=

Between levels b to c,

= = 18.966 kN/m3

Between levels c to d,

= = 20.601 kN/m3

Stress calculation (all answers in kN/m2) Level a, σ = 0, u = 0, σ′ = σ – u = 0. Level b, σ = 2 x 17.331 = 34.662, u = 0(just above b), σ′ = 34.662. σ = 34.662, u (just below b) = -0.5 x 9.81 x 1 = -4.905, σ′ = 34.662 – (4.905) = 39.567. Level c, σ = 34.662 + 1 x 18.966 = 53.658, u = 0, σ′ = 53.658. Level d, σ = 53.658 + 2 x 20.601 = 94.860, u = 2 x 9.81 = 19.62, σ′ = 94.8619.62 = 75.24.


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Exercise 1.

A 5m thick layer of saturated clay is overlain by sand layer of 4m deep. The water table is 3m below the ground surface. The saturated unit weight of clay and sand are 22KN/m3 and 20 KN/m3 respectively. The unit wt. of sand above the water table is 17 KN/m3. Find the total σ′ effective vertical pressure at the top, σ′ middle of the clay layer. What would be the effective pressure at the middle of the clay layer of water table rises up to the ground surface? Plot all the values.

Ans.Top: σ = 71 KN/m2, σ′ = 61.2 KN/m2, Mid: σ = 126 KN/m2, σ′ = 91.7 KN/m2, W.T at G.L: σ = 135 KN/m2, σ′ = 71.3 KN/m2. 2.

A 5m depth of sand overlies a 6m layer of clay, the water table being at the surface; the permeability of the clay is very low. The saturated unit wt. of sand is 19 KN/m3 and that of the clay 20 KN/m3. A 4m depth of fill material of unit weight 20 KN/m3 is placed on the surface over an extensive area. Determine the effective vertical stress at the centre of the clay (a) immediately after the fill has been placed, assuming this to take place rapidly. (b) many years after the fill has been placed.

Ans.Page at 92, R.F Craig’s soil mechanics. 3rd edition. QUESTIONS: 1)

Prove that hydraulic gradient is the ratio of effective unit weight of soil to unit weight of water at critical flow condition. (marks 5)

2)

Prove that effective stresses are unchanged for any depth of soil layer if water table is at or above the ground surface.(marks 5)

3)

Prove that seepage velocity is greater than discharge velocity.(marks 5)

4)

Prove that effective stress is greater for down-ward seepage than that of for upward seepage. (marks 5)


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5)

Prove that the families of curves defined by the Laplace seepage equation always intersect at right angles. (marks 5)


5 Total & Effective Stress

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