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2.10 Problems

Example 2.1: Design of Shallow Foundation in Saturated Clay

Design a square footing to support a column load of 667 kN. The base of the footing will be

located 1 m below the ground level and the soil is saturated clay having the following properties:

Cu = 53.0 kPa , = 0 , γ = 17.3 kN/m3

mv = 1.04 x 10-4m2/kN , Cv = 1.29 x 10-2cm2/min

i. Determine the safe bearing capacity of the clay and hence obtain a suitable sized

footing.

ii. If the clay extends 5 m below the base of the footing, calculate the oedometer

settlement assuming that the load is distributed at 2 to 1 and that mv is constant over

the stratum. (see note below.)

iii. When is the oedometer settlement 75% complete? Assume that the initial pore

pressure is uniform over the stratum and that the clay is underlain by an

incompressible pervious layer.

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Fig Ex

2.1a

a) Skempton’s Nc values depend upon . Hence the footing size will have to be determined by

trial and error.

Try B = 2.4 m

= 0.42

Nc = 7.0 (from Graph Fig Ex 2.1a)

Net

Net = 123 kPa

The area of footing required = = 5.42m2

A 2.4m square footing is satisfactory in bearing.

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b) The solution given below is intended to illustrate the basic approach to settlement

calculation. There are however several alternative methods in common use. Divide layer into

5-1 m strips. The corresponding oedometer settlement calculations are given below.

Fig E 2.1b

Layer

number

Mean Depth

‘m’

Area Mid Section

‘m2’

due to Load kPa

1 1.5 2.9 79.3 0.0083

2 2.5 3.9 43.9 0.0046

3 3.5 4.9 27.8 0.0029

4 4.5 5.9 19.2 0.0020

5 5.5 6.9 14.0 0.0015

667 kN

1 m

6 m

2.4 m

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Settlement = 1.93 cm (0.76 ins.)

The oedometer settlement is 75% complete when the time factor T ≈ 0.48. The drainage path, H,

cannot be defined precisely in the above problem. The minimum time for 75% consolidation

would be given by assuming H = 2.5 m.

T = 0.42

0.48 =

t=

=

Example 2.2: Check of Transcona Grain Elevator Failure (field data approx. only)

Problem:For the soil and building details given in Fig Ex. 2.2 determine Factor of Safety against

bearing capacity failure.

i. Using Skempton’s formula.

ii. Using circular arc method.

Dead load = 20,000 tons

Live load = 26,000 tons

Total = 46,000 tons

Qgross applied = = 3.06 tsf

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Fig Ex. 2.2

Method (i)

(Using average value of Cu = 0.4 tsf)

(qgross ult = 2.2 + 0.62 = 2.82 tsf)

Net applied building load = 3.6 – soil removed.

F.S.

qgross

30'

20'

77' x 195'

R =

95'

app

rox.

O

7'12'

= 103 pcf

Cu= 0.3tsf

A

C D

E

B

Cu= 0.5tsf

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and failure would (did) occur.

[i.e., Maximum allowable building load should have been limited to (for F.S. = 3)]

0.6

0.6

Method (ii)

Assuming the centre of rotation of the probable failure arc to be located at 0 with R = 95 ft.

approx.,

Then AC = 25 ; CD = 127 ; DE = 40

(for strip footing loading) (216 kPa)

For the grain elevator, 77 195 , use shape factor

Therefore

This value is to be compared with using Skempton’s formula. The difference

is mainly due to the greater distance of the failure arc located in the soft stratum.

Example 2.3:

Given: A square footing as shown in the Fig Ex. 2.3a is to carry the loading indicated. The water

table will be temporarily lowered during construction. The average S.P.T. N value is 25

as measured in the field. of the sand is 18.0 kN/m3 and is 11.0 kN/m3.

Determine: The size of a square footing to ensure that the settlement does not exceed 25 mm.

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Fig Ex. 2.3a

Fig Ex. 2.3b: Chart for correction of N-values in sand for influence of overburden pressure

Solution: Neglect the difference between the unit weights of soil and concrete and compute the

original value of effective vertical stresses at a depth of 0.5B below the base of the

footing.

Assume B = 2m.

Then,

D.L. = 356 kN

L.L. = 712 kN

1.80 m1.52 m

B

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Also,

25 mm =

=

Area of footing required =

=

or B = 2.16m

(If the original assumption regarding the value of B was not close another iteration

would be required)

Example 2.4: Immediate Settlement Calculation of Foundation on Saturated Clay

Given: The continuous wall footing, shown in the Fig Ex.2.4a, is 17m long and is founded on a

deposit of ‘stiff’ saturated clay as shown.

Determine: The ‘immediate’ settlement of the footing.

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Fig Ex.2.4a

255 kN/m

0.3 m

1 m

17 m

B= 1.7 m

0.3 m

s = 20 kN/m3

Eu= 48000 kN/m2

concrete = 23.6 kN/m3

= 0.5

Clay:

D

H

B

= Average settlement

01 (qB

= 0.5L = length

q

1.0

0.9

0.8

5 10 15 20

0

D/B

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Fig Ex.2.4b: Values of and for settlement calculations

Solution: The net pressure applied by the footing is the pressure applied by the footing in

excess of the pressure that existed at the depth of the footing before the footing was

constructed.

i.e.

kN/m (Usually the difference between the unit weights of soils and

concrete is neglected)

Now

= 1.3 and = 1.0 (from Fig Ex.2.4b)

Then,

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Example 2.5:

Given: (i) the continuous wall footing, shown in the Fig Ex.2.5, is 17m long and is founded on a

deposit of ‘stiff’ saturated clay as shown.

(ii) the soil is sand with a value of = 36°

(iii) the depth to the water table = 2m

(iii) kN/m3 ; kN/m

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Determine: Factor of Safety against bearing capacity failure.

Fig Ex.2.5a

N.T.S

255 kN/m

0.3 m

1 m

17 m

B= 1.7 m

0.3 m

2 m

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Fig Ex.2.5b

Solution: Using the following equation,

And from the Fig Ex.2.5a, ;

Where, = 10.0 +

= 14.7 kN/m3

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Also,

F.S. =