Solution Example 1

Example 1
It is proposed to place a 2 m × 2 m square footing at a depth of 1.5 m in a North Norfolk glacial

clay soil. The 0.5 m thick footing is to support a 0.4 m × 0.4 m centrally located square column
which will carry a vertical characteristic permanent load of 600 kN and a vertical characteristic
transient load of 350 kN. Take the unit weight of reinforced concrete, ɤconcrete as 25 kN/m3. The
soil has the following properties:
Undrained shear strength, cu = 158 kPa
Effective cohesion, c′ = 0 kPa
Angle of shearing resistance, ϕ′ = 28°
Unit weight of soil, ɤ = 20 kN/m3
mv = 0.104 m2/MN
CV = 2.285 m2/Yr
E = 30 MPa
The ground water table is coincident with the base of the foundation. Design foundation
according to Eurocode 7 DA1.
Solution:
Schematic of the foundation:
Verification of ultimate limit state (ULS)

Step 1: Choose design scenario.
EC7 DA1 C1
Step -2: Partial design factors according to EC7 DA1 C1
ɤG = 1.35, ɤQ = 1.5, ɤCu = 1.0, ɤcʹ = 1.0, ɤϕʹ = 1.0, ɤR = 1.0
Step -3: Calculate design load
Fdesign = ɤG x VG + ɤQ x VQ + ɤG x (self-load of footing + soil above footing)
Self-load of footing = (2 x 2 x 0.5 + 0.4 x 0.4 x 1) x 25 = 54 kN
1
Soil/backfill above footing = (2 x 2 x 1 – 0.4 x 0.4 x 1) x 20 = 76.8 kN
Fdesign = 1.35 x 600 + 1.5 x 350 + 1.35 x (54 + 76.8) = 1512 kN
Step -4: Calculate design material properties
cu_design = cu / ɤCu = 158 / 1.0 = 158 kPa
cʹ_design = cʹ / ɤcʹ = 0 / 1.0 = 0 kPa
ϕʹ_design = ϕʹ / ɤϕʹ = 28 / 1 = 28o
ɤ_design = ɤ / ɤɤ = 20 / 1 = 20 kN/m3
Step -5: Calculate the design bearing resistance of soil in undrained condition (short-
term).
R/A = (π + 2) Cu bc Sc ic + q
bc = 1.0 for the level ground surface
Sc = 1.2 for the square footing
ic = 1.0 for load is vertical (at right angle) to the foundation
q = unit weight of soil x depth of foundation = ɤ x Df
R/A = (3.14156 + 2) x 158 x 1 x 1.2 x 1 + 20 x 1.5 = 1004.84 kN/m2
R = 1058 x A = 1004.84 x 2 x 2 = 4019 kN
Rdesign = R / ɤR = 4019/ 1.0 = 4019 kN
Step -6: Calculate over design factor (ODF) and degree of utilization (DOU) in undrained
condition (short-term).
ODF = Rdesign / Fdesign = 4019 / 1512 = 2.66
DOU = (1/ ODF) X 100 = (1 / 2.66) = 38 %
Step -7: Calculate the bearing resistance of soil in drained condition (long-term).
R/A = cʹNcbcScic + ɤDfNqbqSqiq + 0.5ɤʹBNɤbɤSɤiɤ
cʹ = 0 kPa so first term of the above equation will be equal to 0.
Nq = eπtanϕʹtan2(45 + ϕʹ/2) = e3.14tan28tan2(45 + 28/2) = 14.72
bq = 1.0, iq = 1.0
Sq = 1 + sinϕʹ = 1 + sin28 = 1.469
Nɤ = 2(Nq – 1) tanϕʹ = 2(14.72 – 1) tan28 = 14.59
bɤ = 1.0, iɤ = 1.0
Sɤ = 0.7
2
R/A = 0 + 20x1.5x14.72x1x1.469x1 + 0.5x(20-9.81) x2x14.59x1x0.7x1 = 753 kN/m2
R = 753 x 2 x 2 = 3011 kN
Rdesign = R / ɤR = 3011 / 1.0 = 3011 kN
Step -8: Calculate over design factor (ODF) and degree of utilization (DOU) in drained
condition (long-term).
DOU = (1/ ODF) x 100 = (1 / 1.98) = 50 %
…………………………………………………………………………………………………..
Verification of ultimate limit state (ULS)
Step -1: Choose design scenario
EC7 DA1 C2
Step -2: Partial design factors according to EC7 DA1 C2
ɤG = 1.0, ɤQ = 1.3, ɤCu = 1.4, ɤcʹ = 1.25, ɤϕʹ = 1.25, ɤR = 1.0
Step -3: Calculate design load
Fdesign = ɤG x VG + ɤQ x VQ + ɤG x (self-load of footing + soil above footing)
Self-load of footing = (2 x 2 x 0.5 + 0.4 x 0.4 x 1) x 25 = 54 kN
Soil above footing = (2 x 2 x 1 – 0.4 x 0.4 x 1) x 20 = 76.8 kN
Fdesign = 1.0 x 600 + 1.3 x 350 + 1.0 x (54 + 76.8) = 1186 kN
Step -4: Calculate design material properties
cu_design = cu / ɤCu = 158 / 1.4 = 113 kPa
cʹ_design = cʹ / ɤcʹ = 0 / 1.25 = 0 kPa
ϕʹ_design = ϕʹ / ɤϕʹ = 28 /1.25 = 22.4o
ɤ_design = ɤ / ɤɤ = 20 / 1 = 20 kN/m3
Step -5: Calculate the design bearing resistance of soil in undrained condition (short-
term).
R/A = (π + 2) Cu bc Sc ic + q
bc = 1.0 for the level ground surface
Sc = 1.0 for the square footing
ic = 1.0 for load is vertical (at right angle) to the foundation
q = unit weight of soil x depth of foundation = ɤ x Df
R/A = (3.14156 + 2) x 113 x 1 x 1.2 x 1 + 20 x 1.5 = 727.2 kN/m2
3
R = 1058 x A = 727.2 x 2 x 2 = 2909 kN
Rdesign = R / ɤR = 2909/ 1.0 = 2909 kN
Step -6: Calculate over design factor (ODF) and degree of utilization (DOU) in undrained
condition (short-term).
DOU = (1/ ODF) X 100 = (1 / 2.45) = 41 %
Step -7: Calculate the bearing resistance of soil in drained condition (long-term).
cʹ = 0 kPa so first term of the above equation will be equal to 0.
Nq = eπtanϕʹtan2(45 + ϕʹ/2) = e3.14tan22.4tan2(45 + 22.4/2) = 8.14
bq = 1.0, iq = 1.0
Sq = 1 + sinϕʹ = 1 + sin22 = 1.37
Nɤ = 2(Nq – 1) tanϕʹ = 2(7.78 – 1) tan22.4 = 5.89
bɤ = 1.0, iɤ = 1.0
Sɤ = 0.7
R/A = 0 + 20 x 1.5 x 8.14 x 1 x 1.37 x 1 + 0.5 x (20-9.81) x 2 x 5.89 x 1 x 0.7 x 1 = 400.1 kN/m2
R = 400.1 x 2 x 2 = 1603 kN
Rdesign = R / ɤR = 1603 / 1.0 = 1603 kN
Step -8: Calculate over design factor (ODF) and degree of utilization (DOU) in drained
condition (long-term).
DOU = (1/ ODF) X 100 = (1/1.35) = 74 %
…………………………………………………………………………………………………..
4
Verification of Serviceability Limit State (SLS)
ST = S0 + S1 + S2
S2 ≈ 0 mm for inorganic soils
Immediate Settlement:
𝑞𝐵(1 − 𝜇 2 )𝐼𝑓
𝑆0 =
𝐸
(600 + 350 + 54 + 76.8)

𝑞= = 270.2 𝑘𝑁/𝑚2
(2 × 2)
B=2m
μ = 0.49
If = 0.82 (from Table see the slides)
E = 30000 kN/m2
270.2 × 2 × (1 − 0.492 ) × 0.82
𝑆0 = = 11.22 𝑚𝑚
30000
Consolidation Settlement:
𝑆1 = ∑ 𝑚𝑣 . 𝜎 . ℎ
𝑧
𝞂0 = contact stress = (Service load/area) = (600+350+54+76.8)/ (2 x 2) = 270.2 kN/m2
𝞂1 = stress at 1m depth = Fservice / (B+Z) (L+Z) = (600+350+54+76.8) /(2+1) (2+1) = 120 kN/m2
𝞂2 = stress at 2 m depth = Fservice / (B+Z) (L+Z) = (600+350+54+76.8) / (2+2) (2+2) = 68 kN/m2
≈ 10% 𝞂0
So, influence zone (h) = 4 m, 𝞂mid = 68 kN/m2
(𝞂z)avg. = 1/6(𝞂0 + 4 𝞂mid + 𝞂4)
(𝞂z)avg. = 1/6(270 + 4 × 68 + 30) = 95 kN/m2
𝑆1 = ∑ 𝑚𝑣 . 𝜎 . ℎ
𝑧
𝑆1 = ∑ 0.000104 × 95 × 4000 = 39.65 𝑚𝑚
ST = S0 + S1
ST = 11.22 + 39.65 = 50.87 mm ≈ 50 mm (close to Permissible limit)
So, our design is safe and acceptable.
5
Final drawing:
7 bars of 16 mm diameter both in x and y directions, 40 mm cover.

Solution Example 1

Uploaded by

Copyright:

Available Formats

Solution Example 1

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Solution Example 1

Uploaded by

Copyright:

Available Formats

Example 1

It is proposed to place a 2 m × 2 m square footing at a depth of 1.5 m in a North Norfolk glacial

Verification of ultimate limit state (ULS)

(600 + 350 + 54 + 76.8)

𝑆1 = ∑ 0.000104 × 95 × 4000 = 39.65 𝑚𝑚

You might also like