Session 4 - Design (examples) of spread foundations.ppt

Design (examples) of spread foundations

1. Session 4
Design examples of spread
foundations

2. Design of a pad foundation
on clay

3. P
k
= 270 kN Q
k
= 70 kN
h
1
= 0,5 m
h
2
= 0,5 m
Ground properties: 
k
= 18kN/m³,

w ater
=10kN/m³,

k
’= 8kN/m³,

k
’= 20°, c
k
’= 5kPa,
c
u,k
= 30kPa
B x B

k
= 25 kN/m³’
k
= 15 kN/m³

4. •Undrained conditions:
1)what are the characteristic and design bearing resistance
for a pad
1.7 m x 1.7 m (DA1)
2.0 m x 2.0 m (DA2 & DA3)?
R
k
/A’= (+2)c
u
b
c
s
c
i
c
+ q = 203 kPa
With c
u,d
= 30/1.4 = 21.4 kPa (DA1-2 and DA3):
R
d/A’ = 150 kPa, thus
R
d
= 433 kN for DA1-2 and R
d
= 600 kN for DA3
With 
R
= 1.0, R
d
/A’=203 kPa, thus
R
d = 586 kN for DA1-1
And with 
R= 1.4 or R
d/A’= 145 kPa, thus
R
d
= 580 kN for DA2

5. •Undrained conditions:
2) what are the characteristic and design value of actions
resistance for a pad 1.7 m x 1.7 m (DA1) and 2.0 m x 2.0 m
(DA2 & DA3)?
V
k= P
k + Q
k + G
pad,k (62)= 402 kN for DA1
V
k= P
k + Q
k + G
pad,k (86)= 426 kN for DA2& DA3
V
d
= 1.35 (P
k
+ G
pad,k
) + 1.5 Q
k
for DA1-1, DA2 and DA3, thus
V
d
= 553 kN for DA1-1 and V
d
= 585 kN for DA2 and DA3
V
d
= 1.0 (P
k
+ G
pad,k
) + 1.3 Q
k
for DA1-2, thus
V
d
= 423 kN pour DA1-2

6. •Undrained conditions:
3) Conclusions for ULS in p&t design situations? We check if V
d
≤ R
d :
DA1-1 & DA1-2 : 553 kN ≤ 586 kN & 423 kN ≤ 433 kN  O.K.
DA 2 : 585 kN ≤ 580 kN  quasi-O.K.
DA 3 : 585 kN ≤ 600 kN O.K.
4) What is the equivalent OFS : we define OFS as R
k/V
k
DA1-1 & DA2-2 : 587/ 402 = 1.46
DA2 and DA
 3 : 812/426 = 1.91
4) Do you have any comment?
yes; some OFS may seem vey low, but one has to check as well
:
SLS : either settlement calculation or ‘simplified’ method
And ULS accidental? Seismic?

7. •Drained conditions:
1)what are the characteristic and design bearing resistance for a
pad 1.85mx1.85m (DA1), 1.95mx1.95m (DA2) and
2.15mx2.15m (DA3)?
R
k/A’= q´N
q;ks
q;k + 0.5´
kB´N
;ks
 + c´
kN
c;ks
c thus
R
k
/A’=193kPa for DA1, 194kPa for DA2 and 196kPa for DA3
With 
R
= 1.0, R
d
/A’=193 kPa, thus
R
d = 660 kN for DA1-1
And with 
R= 1.4 or R
d/A’= 194/1.4= 139 kPa, thus
R
d
= 529 kN for DA2
With c’
d
= 5/1.25 = 4 kPa and ’
d
= Atan(tan20°/1.25)=
16.23°(DA1-2 and DA3):
R
d/A’ = 120 kPa for DA1-2 and 121.5 kPa for DA3, thus
R
d
= 410 kN for DA1-2 and R
d
= 562 kN for DA3

8. •Drained conditions:
2) what are the characteristic and design value of actions
resistance for a pad 1.85mx1.85m (DA1), 1.95mx1.95m
(DA2) and 2.15mx2.15m (DA3)? Note that G’
k
/A’ = 11.5
kPa
V
k= P
k + Q
k + G’
pad,k (39)= 379 kN for DA1
V
k= P
k + Q
k + G
pad,k (44)= 384 kN for DA2
V
k
= P
k
+ Q
k
+ G’
pad,k
(53)= 393 kN for DA3
V
d = 1.35 (P
k + G’
pad,k )+ 1.5 Q
k for DA1-1, DA2 and DA3, thus
V
d = 522 kN for DA1-1, 530 kN for DA2, and 541 kN for DA3
V
d
= 1.0 (P
k
+ G’
pad,k
) + 1.3 Q
k
for DA1-2, thus
V
d = 400 kN pour DA1-2

9. •Drained conditions:
3) Conclusions for ULS in p&t design situations? We check if
V
d ≤ R
d :
DA1-1 & DA1-2 : 522 kN ≤ 660 kN & 400 kN ≤ 410 kN  O.K.
DA 2 : 530 kN ≤ 529 kN  quasi-O.K.
DA 3 : 541 kN ≤ 562 kN O.K.
4) What is the equivalent OFS : we define OFS as R
k
/V
k
DA1-1 & DA2-2 : 660/379 = 1.74
DA2 : 738/384 = 1.92
DA3 : 906/393 = 2.30
4) Do you have any comment?
yes; some OFS may seem vey low, but one has to check as
well :
SLS : either settlement calculation or ‘simplified’ method
And ULS accidental? Seismic?

10. ULS Structural design
•Bearing pressures due to the column load :

d
= V
d
/B² = (
G;unfav
P
k
+ 
Q
Q
k
)/B²
and M
d
= [
d
(B/2)²]/2 = [V
d
/B²(B/2)²]/2 = V
d
/8, in kNm/m

The required reinforcement is determined using EN 1992.

11. Design of spread foundations
with
inclined and/or eccentric loads

12. Examples
-Spread foundation for a tower
in: Frank et al. (2004) Designers’ Guide to EN 1997-1,
Thomas Telford, London
-Strip footing under eccentric and inclined loading
in : Frank R., Schuppener B., Vogt N., Weißenbach A.
(2007). "Design approaches of Eurocode 7 for the
verification of ultimate limit states in geotechnical design
in France and Germany", Revue Européenne de Génie
Civil, vol. 11, n° 5, mai 2007, p. 621-641

13. Ground properties: ’= 20 kN/m³,

k
’= 35°, c
k
’= 0 kPa;

cv,k
’= 30°,
Q
hk
= 300 kN
Q
vk= 600 kN
h = 10m
2 m

concrete
= 24.5 kN/m³

18. Thank you for your attention!