CE366 Foundation Engineering 1 Middle East Technical University

2018-2019 Spring Department of Civil Engineering

Recitation 2
Field Load Test-Settlement

P1) PLATE LOAD TEST (FIELD LOAD TEST)

a) Footing on Clay (load test on clay)

The results of a plate load test in stiff clay are shown in the Figure 3. The size of the test plate is
0.305 m x 0.305 m. Determine the size of a square column footing that should carry a load of 2500
kN. (FS = 2.0; maximum permissible settlement is 40 mm)

Load (kN/m2)
0 500 1000
0

5
Settlement (mm)

10

15

20

25

Figure 3. Plate load test in stiff clay

Solution:
Load (kN/m2)
0 500 1000 Cohesive Soil
0 (qult)ftg = (qult)test
1
5 k B
Sftg = Stest x
b
Settlement (mm)

10 Where;

15 Sftg and Stest : settlements of footing and test plate which

are loaded with the same pressure, respectively.
20
B (or BF such as in Lecture Notes): width of footing
25 b (or BP such as in Lecture Notes): width of test plate

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 CE366 Foundation Engineering 1 Middle East Technical University
2018-2019 Spring Department of Civil Engineering

(qult)test= 700 kN/m2 = (qult)ftg

700 2500
q all = = 350kPa ⇒ = 350 ⇒ B = 2.7 m
FS = 2.0 BxB

350 x 5 2.7
S ftg = x = 31mm < 40mm
500 0.305

Stest under 350 kPa loading

Settlement c alculation, how ever, i s no t v ery reliable; b ecause i t can not r epresent c onsolidation
settlement. G enerally, bearing cap acity cr iteria governs, not t he s ettlement, i n t he de sign o f
foundations resting on clays.

q
Coefficient of s ubgrade reaction: k = (kN / m 3) ( k i s t he s lope of q vs S g raph, f or m ore
S
information about “k” see Lecture Notes p.112-113 or Ordemir p. 28)

500
k test = = 100000 kN / m 3 = 100 MN / m 3
0.005

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 CE366 Foundation Engineering 1 Middle East Technical University
2018-2019 Spring Department of Civil Engineering

b) Load test on sand

The results of a plate load test in a sandy soil are shown in the Figure 4. Size of the test plate:
0.305 m x 0.305 m. Determine the size of square footing that should carry a load of 2500 kN with
a maximum permissible settlement of 40 mm.

load (kN/m2)
0 200 400 600 800
0

10
settlement (mm)

20

30

40

50

Figure 4. Plate load test in sandy soil

Solution:

load (kN/m2)
0 200 400 600 800 Cohesionless Soils
0
B
(qult)ftg= (qult)testx
10 b
settlement (mm)

2
20  B(b + 0.3) 
Sftg=Stest  
 b( B + 0.3) 
30
(For typical plate dimension, b=0.3 m, this becomes
40 equation 2.13 page 52 in Lecture Notes)

50
B and b are in meters!

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 CE366 Foundation Engineering 1 Middle East Technical University
2018-2019 Spring Department of Civil Engineering

Q (kN) Assumed B (m) q=Q/B2 (kPa) Stest (mm) Sftg (mm)

2500 4 156 5.2 (20/600x156) 17.7<40
3 278 9.3 30.2<40
2.5 400 13.3 41.7>40
2.6 370 12.3 38.9
2.55 384 12.8 40.4

Square column footing of 2.55 x 2.55 m dimensions will be appropriate.

-OR-
2
2500 20  B(0.305 + 0.3) 
= 40 = 2 x × ⇒ B = 2.56m
600  0.305( B + 0.3 
S ftg
B

Stest

Thus,

B 2.55
qftg = qtestx = 700 x = 5852kN / m 2
b 0.305

Thus, factor of safety against bearing capacity (F.S.) is,

5852
F .S . = = 15.2 ( settlement governs design)
384.5

Thus, coefficient of subgrade reaction (k) is,

600
k= = 30 MN / m 3
0.020

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 CE366 Foundation Engineering 1 Middle East Technical University
2018-2019 Spring Department of Civil Engineering

P2) IMMEDIATE SETTLEMENT

A foundation 4 m × 2 m, carrying a net uniform pressure of 200 kN/m2, is located at a depth of 1.5

m in a layer of clay 5 m thick for which the value of Eu is 45 MN/m2. The layer is underlain by a

second layer, 10 m thick, for which the value of Eu is 80 M N/m2. A hard stratum lies below the

second layer. Ground water table is at the depth of foundation. Determine the average immediate

settlement under the foundation.

Hint: Since soil is SATURATED CLAY, νs =0.5. So the following equation can be used:

q⋅B
S i = µ 0 ⋅ µ1 ⋅
Eu

Solution:
q = 200 kN/m2

1.5 m

2m
3.5 m
Eu = 45 MN/m2

Eu = 80 MN/m2
10 m

Hard stratum

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2018-2019 Spring Department of Civil Engineering

B is the smaller dimension !

D
B We obtain,
H µ0 from D / B
µ1 from H / B and L / B

Hard stratum

D / B = 1.5 / 2 = 0.75  µ0 = 0.95 (Figure 3.3, p.62 Lecture Notes)

(1) Consider the upper layer with Eu = 45 MPa.

D H / B = 3.5 / 2 = 1.75 µ1 = 0.65

Eu = 45 MPa L/B=4/2=2
H = 3.5 m
q⋅B (200) ⋅ 2
S i 1 = µ 0 ⋅ µ1 ⋅ = (0.95) ⋅ (0.65) ⋅ = 5.49mm
Eu 45
Hard stratum

(2) Consider the two layers combined with Eu = 80 MPa.

D H / B = (3.5 + 10) / 2 = 6.75 µ1 = 0.9

L/B=4/2=2
Eu = 80 MPa
q⋅B (200) ⋅ 2
H = 13.5 m S i 2 = µ 0 ⋅ µ1 ⋅ = (0.95) ⋅ (0.9) ⋅ = 4.28mm
Eu 80

Hard stratum

(3) Consider the upper layer with Eu = 80 MPa.

D H / B = 3.5 / 2 = 1.75 µ1 = 0.65

Eu = 80 MPa L/B=4/2=2
H = 3.5 m q⋅B (200) ⋅ 2
S i 3 = µ 0 ⋅ µ1 ⋅ = (0.95) ⋅ (0.65) ⋅ = 3.08mm
Eu 80
Hard stratum

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 CE366 Foundation Engineering 1 Middle East Technical University
2018-2019 Spring Department of Civil Engineering

Using the principle of superposition, the settlement of the foundation is given by;

Si = Si 1 + Si 2 - Si 3

Si = 5.49 + 4.28 – 3.08

Si = 6.69 mm

P3) SCHMERTMAN

A soil profile consists of deep, loose to medium dense sand (γdry = 16 kN/m3 , γsat = 18 kN/m3). The
ground water level is at 4 m depth. A 3.5 m x 3.5 m square footing rests at 3 m depth. The total
(gross) load acting a t t he f oundation l evel ( footing w eight + c olumn l oad + w eight of s oil or
footing) is 2000 kN. Estimate the elastic settlement of the footing 6 years after the construction
using influence factor method (Schmertman, 1978).
End resistance values obtained from static cone penetration tests are;
Depth (m) qc (kN/m2)
0.00 – 2.00 8000
2.00 - 4.75 10000
4.75 - 6.50 8000
6.50 – 12.00 12000
12.00 – 15.00 10000

Note that;
for square footing; z (depth)(from foundation level) Iz (strain factors)
0 0.1
B/2 0.5
2B 0.0
Where B is width of footing

• Es = 2.0 qc

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 CE366 Foundation Engineering 1 Middle East Technical University
2018-2019 Spring Department of Civil Engineering

Solution:
Iz
Si = C1 C2 qnet Σ ∆z
E
qnet = net foundation pressure

σ 'o
C1 = 1 − 0.5 correction factor for footing depth
qnet

σ'o = effective overburden pressure at foundation level

t
C 2 = 1 + 0.2 log correction factor for creep
0.1
t = time at which the settlement is required (in years)

qgross = 2000 kN
4m 3m γdry = 16 kN/m3

deep loose to medium dense sand

γsat = 18 kN/m3

2000
q net = − 3 x16 = 115.26 kPa
3.5 x3.5

gross pressure. initial effective overburden pressure

σo’ = 3x16 = 48 kPa

48
C1 = 1 − 0.5 = 0.792
115.26
6
C 2 = 1 + 0.2 log = 1.356
0.1

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2018-2019 Spring Department of Civil Engineering

ground surface qc (kN/m2)

0.0 m
8000

0.1 0.2 0.3 0.4 0.5

Iz
10000
4 Layer 1 ∆z1 1 0.5B = 1.75m

4.75 2
8000 Layer 2 ∆z2
6 3
6.5
Layer 3 ∆z3 4

8 5 enter from mid-height

of each layer
6

12000 Layer 4 ∆z4

10 7

2B=2x3.5 = 7m

Width of foundation,
12
B = 3.5 m

10000
14

15
Es = 2.0 qc

Layer No Depth(m) ∆z(m) qc(kPa) Es(kPa) Iz (Iz /Es) ∆z

1 3.00-4.75 1.75 10.000 20.000 0.3 2.65x10-5
2 4.75-6.50 1.75 8.000 16.000 0.416 4.55x10-5
3 6.50-8.25 1.75 12.000 24.000 0.249 1.82x10-5
4 8.25-10.00 1.75 12.000 24.000 0.083 0.605x10-5
Σ = 9.625x10−5
Si = (0.792) (1.356) (115.26) (9.625x10-5)
= 0.01191 m Si = 11.91 mm

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 CE366 Foundation Engineering 1 Middle East Technical University
2018-2019 Spring Department of Civil Engineering

P4) CONSOLIDATION SETTLEMENT

Ignore th e imme diate s ettlement, a nd c alculate to tal c onsolidation s ettlement o f s oil p rofile
composed of t wo di fferent t ypes of clay, i .e. C lay 1 and C lay 2 due t o 1 50 kP a net foundation
loading. Take unit weight of water as 10 kN /m3 and assume that Skempton-Bjerrum Correction
Factor i s 𝜇𝜇 = 0.7 for bot h c lay layers. N ote t hat σ’c (or s ometimes s hown a s σ’p) is th e
preconsolidation pressure.

10 m x 10 m

qnet = 150 kPa

CLAY 1 2m z

γdry = 19 kN/m3
γsat= 20 kN/m3
CLAY 1
6m Cr = 0.05; Cc = 0.15
eo = 0.80; σ’c = 80 kPa

γsat = 20 kN/m3
Cr = 0.03; Cc = 0.10
CLAY 2 6m eo = 0.60; σ’c = 200 kPa

INCOMPRESSIBLE

Solution:

Settlement w ill ta ke p lace d ue to lo ading ( qnet = 150 kP a) a pplied at a d epth of 2 m . T hus, all
(consolidation) settlement calculations will be performed for clayey soil beneath the foundation (z
> 2 m).

Reminder: General equation of 1D consolidation settlement (one dimensional vertical

consolidation) for an overconsolidated clay is;

Cr σ′c Cc σ′v,f
Sc,1D = H log � ′ � + H log � ′ �
1 + eo σv,o 1 + eo σc

Note that, all calculations are done for the mid-depth of the compressible layers under the loading.

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 CE366 Foundation Engineering 1 Middle East Technical University
2018-2019 Spring Department of Civil Engineering

Consolidation settlement in Clay 1:

Initial effective overburden stress, σ’v,o = (2*19) + (3*(20-10)) = 68 kPa

Stress increment due to foundation loading, Δσ= [150*(10*10)] / [(10+3)*(10+3)] = 88.8 kPa

Final stress, σ’v,f = 68 + 88.8 = 156.8 kPa

This is an overconsolidated clay (overconsolidation ratio OCR = σ’c / σ’v,o = 80 / 68 > 1.0) ; and the
final st ress, σ’v,f is greater t han σ’c ( σ’v,f > σ’c ) t herefore w e s hould u se bot h C r and C c in
consolidation settlement calculation (see figure below).
e (void ratio) Recompression curve: Cr = ∆e / ∆logσ’

2 Compression curve (virgin

line): Cc = ∆e / ∆logσ’

σ’c = 80 kPa log σ’

σ’v,0 = 68 kPa σ’v,f = 156.8 kPa

0.05 80 0.15 156.8

𝑆𝑆𝑐𝑐,1𝐷𝐷 = � (6) log � �� + � (6) log � �� = 0.158𝑚𝑚 = 15.8 𝑐𝑐𝑐𝑐
1 + 0.80 68 1 + 0.80 80

1 2

Consolidation settlement in Clay 2:

Initial effective overburden stress, σ’v,0 = (2*19) + (6*(20-10)) + (3*(20-10)) = 128 kPa

Stress increment due to foundation loading, Δσ= [150*(10*10)] / [(10+9)*(10+9)] = 41.6 kPa
Final stress, σ’v,f = 128 + 41.6 = 169.6 kPa

This is an overconsolidated clay (overconsolidation ratio OCR = σ’c / σ’v,o = 200 / 128 > 1.0) ; and
the final stress, σ’v,f is less than σ’c ( σ’v,f < σ’c ) therefore we should use only C r in consolidation
settlement calculation (see figure below).

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 CE366 Foundation Engineering 1 Middle East Technical University
2018-2019 Spring Department of Civil Engineering

[Note: If a soil would be a normally consolidated clay (OCR = σ’c / σ’v,o = 1.0), we would use only
Cc in consolidation settlement calculation.]

e (void ratio) Recompression curve

Compression curve (virgin

Line)

log σ’
σ’c = 200 kPa
σ’v,0 = 128 kPa σ’v,f = 169.6 kPa

0.03 169.6
𝑆𝑆𝑐𝑐,1−𝐷𝐷 = � (6) log � �� = 0.014 𝑚𝑚 = 1.4 𝑐𝑐𝑐𝑐
1 + 0.60 128

Total Consolidation Settlement (1D):

𝑆𝑆𝑐𝑐,1𝐷𝐷 = 15.9 + 1.4 = 17.3𝑐𝑐𝑐𝑐

Corrected Settlement for 3D Consolidation (Skempton-Bjerrum Factor):

𝑆𝑆𝑐𝑐,3𝐷𝐷 = 𝑆𝑆𝑐𝑐,1𝐷𝐷 ∗ 𝜇𝜇 = 17.3 ∗ 0.7 = 𝟏𝟏𝟏𝟏. 𝟏𝟏 𝒄𝒄𝒄𝒄

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