Practical Design to Eurocode 2
Foundations
Eurocode 7
Eurocode 7 has two parts:
Part 1: General Rules
Part 2: Ground Investigation and testing
Limit States
The following ultimate limit states apply to foundation
design:
EQU: Loss of equilibrium of the structure
STR: Internal failure or excessive deformation of the
structure or structural member
GEO: Failure due to excessive deformation of the ground
UPL: Loss of equilibrium due to uplift by water pressure
HYD: Failure caused by hydraulic gradients
Categories of Structures
Category Description
Risk of
geotechnical
failure
Examples from
EC7
Small and relatively
simple structures
Negligible
None given
Conventional types of
structure no difficult
ground
No exceptional
risk
Spread
foundations
All other structures
Abnormal risks
Large or
unusual
structures
STR/GEO ULS
Unfavourable
Favourable
Leading
variable
action
Exp 6.10
1.35Gk
1.0Gk
1.5Qk
Exp 6.10a
1.35Gk
1.0Gk
Exp 6.10b
1.25Gk
1.0Gk
1.5Qk
1.50,iQk
1.0Gk
1.0Gk
1.3Qk
1.30,iQk
Permanent Actions
Accompanying variable
actions
Main
Others
Combination 1
1.50,iQk
1.50,1Qk
1.50,iQk
Combination 2
Exp 6.10
Notes:
If the variation in permanent action is significant, use Gk,j,sup and Gk,j,inf
If the action if favourable, Q,i = 0 and the variable actions should be ignored
Partial factors
Symbol
Combination 1
Combination 2
Angle of shearing
resistance
1.0
1.25
Effective cohesion
1.0
1.25
Undrained shear
strength
cu
1.0
1.4
Unconfined strength
qu
1.0
1.4
Bulk density
1.0
1.0
Spread Foundations
EC7 Section 6
Three methods for design:
Direct method check all limit states
Indirect method experience and testing used to
determine SLS parameters that also satisfy ULS
Prescriptive methods use presumed bearing
resistance (BS8004 quoted in NA)
Pressure distributions
EQU : 0.9 Gk + 1.5 Qk (assuming variable action is
destabilising e.g. wind, and permanent action is
stabilizing)
STR : 1.35 Gk + 1.5 Qk (6.10)
(6.10a or 6.10b could be used)
Strip and Pad Footings
(12.9.3) Plain concrete
hF
a
a
bF
0,85 hF
(9gd/fctd,pl)
a
gd is the design value of the ground pressure
as a simplification hf/a 2 may be used
Reinforced Bases
Check critical bending moment at column face
Check beam shear and punching shear
For punching shear
the ground reaction
within the perimeter
may be deducted
from the column load
Worked Example
Design a square pad footing for a 350 350 mm column
carrying Gk = 600 kN and Qk = 505 kN. The presumed
allowable bearing pressure of the non-aggressive soil is
200 kN/m2.
Category 2, using prescriptive methods
Base size: (600 + 505)/200 = 5.525m2
=> 2.4 x 2.4 base x .5m (say) deep.
Worked Example
Use C30/37
Loading = 1.35 x 600 + 1.5 x 505
= 1567.5kN
ULS bearing pressure =
1567.5/2.42
= 272kN/m2
Critical section at face of column
MEd = 2.72 x 2.4 x 1.0252 / 2
= 343kNm
d = 500 50 16 = 434mm
K = 343 x 106 / (2400 x 4342 x 30)
= 0.025
Worked Example
z = 0.95d
= 0.95 x 434 = 412mm
As = MEd/fydz
= 343 x 106 / (435 x 412) = 1914mm2
Provide H16 @ 250 c/c (1930mm2)
Beam shear
Check critical section d away from column face
VEd = 272 x (1.025 0.434) = 161kN/m
vEd = 161 / 434 = 0.37MPa
vRd,c (from table) = 0.41MPa => beam shear ok.
Worked Example
Punching shear
Basic control perimeter at 2d from face of column
vEd = VEd / uid < vRd,c
= 1, ui = (350 x 4 + 434 x 2 x 2 x ) = 6854mm
VEd = load minus net upward force within the area of the
control perimeter)
= 1567.5 272 x (0.352 + x .8682 + .868 x .35 x 4)
= 560kN
vEd = 0.188MPa; vRd,c = 0.41 (as before) => ok
Workshop Example
Pad foundation for a column taking Gk = 300kN, Qk =
160kN. Permissible bearing stress = 150kPa.
Work out size of base, tension reinforcement and any
shear reinforcement.
Retaining Walls
Ultimate Limit States
for the design of
retaining walls
General expressions
10
Calculation Model A
11
Calculation Model B
12
Partial factors
Symbol
Combination 1
Combination 2
Angle of shearing
resistance
1.0
1.25
Effective cohesion
1.0
1.25
Undrained shear
strength
cu
1.0
1.4
Unconfined strength
qu
1.0
1.4
Bulk density
1.0
1.0
Overall design
procedure
13
Initial sizing
Figure 6 for overall
design procedure
14
Panel 2
Overall design
procedure
15
Design against
sliding
(Figure 7)
16
Overall design
procedure
Design against Toppling
(Figure 9)
17
Overall design
procedure
Design against bearing failure
(Figure 10)
18
Overall design
procedure
19
Figure 13
Strut and tie methods
20
Which is the Stronger?
P1
P2 2P1
Bi-axial Strength of Concrete
compressive strength of
concrete with transverse
tension
fcu
fcu
fct
tensile stress in concrete
fct
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Strut and Tie Models
(6.5.2)
Struts
EC2 provides a simplified approach to limiting stresses in struts.
Where there is no transverse tension
Rd,max = fcd (ie 0,57fck)
Otherwise where there is transverse tension
about 0,3fck)
Rd,max = 0,6fcd (ie
where = 1-fck/250
Sizing nodes (6.5.4)
c-c-c
The size of a node is
(1-fck/250)fcd
determined by the limiting
stresses in the struts and the
(Exp (6.60))
anchorage length of bars
If the stresses in all struts
meeting at a node are the
same, the boundaries of the
c-c-t
node will be perpendicular to 0.85(1-fck/250)fcd
the axis of the strut and the
(Exp (6.61))
node will be in a state of
hydrostatic stress
Bars can be
anchored both
through and beyond
a node
Unless special confinement is
provided, the calculated
c-t-t
compressive stress in the node
0.75(1-fck/250)fcd
regions should not exceed:
(Exp (6.62))
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Strut and Tie Models Can Help
Us Understand .
1 - Anchorage of tension reinforcement at simple support
behaviour similar to deep beam at support
Even at a simple support there is tension
in the reinforcement
2 - Detailing of nib
bearing
strut needs to
be supported
on bar
bar to be anchored
beyond point of load
application
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3 - Anchorage of bars
4 - Lapping of bars
F tan
F tan
F/2
F/2
Forces from cranks reduced if laps in plane parallel to surface
but similar splitting forces from lap
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Deep Beams
w kN/m
Clause 9.7 suggests that strut and
tie methods will be used in the
design of deep beams, however
no advice is given on the shape or
limit to the height of the assumed
strut
parabolic curve
67.4
0.6 L
0.2 L
tension zone
Pile Cap Exercise
25
Pile Cap Exercise
fck = 30 MPa / fyk = 500 MPa
breadth = 1050
bottom cover = 100, bars 32mm
side cover
= 75
5.5 MN (Ult)
700 x 700 column
1. Set up an arrangement of
struts and ties
1650
2. Check the strength of the
struts 6.5.2 (2)
3. Check the strength of the
nodes 6.5.4 (4)
4. Calculate the areas of
reinforcement required
5. What other checks should
200
be made to EC2?
750
3400
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Pile Cap Exercise Solution (1)
5500 kN
1. Arrangement of struts and
ties
Angle of strut
= tan -1 (1530 / 1125)
= 53.7
700 x 700 column
Width of each strut
= 700 sin 53.7
= 564 mm
Compressive force in each
strut
= 2750 / sin 53.7
= 3412 kN
1530
3412 kN
53.7
564
Tensile force in tie
= 2750 / tan 53.7
= 2020 kN
120
2020 kN
1125
2750 kN
2750 kN
Pile Cap Exercise Solution (2)
700
2. Strength of struts
Stress in strut:
Projecting the rectangular geometry
from the column gives minimum area,
and maximum stress
= 3.412 / (0.564 x 0.7)
= 8.6 MPa
3412 kN
53.7
564
Cl 6.5.2 (2) : Strength of strut
sRd,max
= 0.6 fcd
= 0.6 (1-fck/250) fcd
= 0.6 (1 30 / 250)
0.85 x 30 / 1.5
= 8.98 MPa OK
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Pile Cap Exercise Solution (3)
5.5 MN (Ult)
3. Strength of nodes
700mm
Upper node:
Ed,1
Stress on hor. plane Ed,1
= 2.75 / (0.35 x 0.7)
= 11.2 MPa
Stress on vert. plane Ed,2
= 2.02 / (0.35 tan 53.7 x 0.7)
= 6.1 MPa
As these will be principle stresses (no
shear), the other plane is ok by inspection.
Ed,2
3412 kN
2.02MN
Cl 6.5.4 (4a) :
Strength of compression node without ties
= Rd,max
= (1-fck/250) fcd
= (1 30 / 250) 0.85 x 30 / 1.5
= 14.96 MPa OK
Pile Cap Exercise Solution (4)
3. Strength of nodes
Lower node:
The exact geometry of node is difficult to
define, so conservatively compare strut
stresses to node allowable stresses
Stress in strut Ed,1
= 8.6 MPa
Ed,1
Stress in pile Ed,2
= 2.75 / (0.752 / 4)
= 6.2 MPa
Cl 6.5.4 (4b) :
Strength of compression node with anchored ties
= Rd,max
= 0.85 (1-fck/250) fcd
= 0.85 (1 30 / 250) 0.85 x 30 / 1.5
= 12.7 MPa OK
Ed,2
Ed,1
Ed,2
750
750
2750 kN
2750 kN
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Pile Cap Exercise Solution (5)
4. Area of reinforcement
160 mm
Tension = 2020 kN
Area of steel required
= 2020 x 1000/ (500/1.15)
= 4646 mm2
6H32 (4825 mm2)
53.7
5. Other checks
112 mm
Anchorage length, lbd :
Cl 8.4.2 (2) : fbd
= 2.25 12fctd
= 2.25 x 1 x 1 x 1x 2/1.5 = 3.0 MPa
Cl 8.4.3 (2) : lb,rqd
= (/4)(sd/fbd)
= (32/4) x {2020,000 x 4/(6 x x322 x 3)}
= 1116.3 mm
Cl 8.4.4. (1) :
2750 / tan 53.7
= 2020 kN
2750 kN
1 =0.7; , 2 = (1-0.15(91-32)/32 = 0.72; 3 = 1; 4 = 1;
5 = 1 0.04 (2.75/(1.5x.75)) = 0.902; so 2 3 5 = 0.649 but Exp(8.5) 2 3 5 = 0.7
Hence lbd = 1
2 3 4 5Lb,rqd =
0.7 x 0.7 x 1116.3 = 547 mm
Assume bar fully stressed at centreline of pile
Standard mandrel diameter (Cl 8.3 (3) & Table 8.1N) m = 7 = 7 x 32 = 224 mm
Standard extension of bar beyond bend = 5 = 5 x 32 = 160 mm
Tying
www.eurocode2.info
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Tying systems (2)
Peripheral ties (9.10.2.2):
Ftie,per = (20 + 4n0) 60kN where n0 is the number of storeys
Internal ties (including transverse ties) (9.10.2.3):
Ftie,int = ((gk + qk) / 7.5 )(lr/5)Ft Ft kN/m
Where (gk + qk) is the sum of the average permanent and variable floor
loads (kN/m2), lr is the greater of the distances (m) between the centres of
the columns, frames or walls supporting any two adjacent floor spans in the
direction of the tie under consideration and
Ft = (20 + 4n0) 60kN.
Maximum spacing of internal ties = 1.5 lr
Horizontal ties to columns or walls (9.10.2.4):
Ftie,fac = Ftie,col (2 Ft (ls /2.5)Ft) and 3% of the total design ultimate vertical
load carried by the column or wall at that level. Tying of external walls is only
required if the peripheral tie is not located within the wall. Ftie,fac in kN per
metre run of wall, Ftie,col in kN per column and ls is the floor to ceiling height in
m.
Tying systems (3)
Vertical ties (9.10.2.5):
In panel buildings of 5 storeys or more, ties should be provided in
columns and/or walls to limit damage of collapse of a floor.
Normally continuous vertical ties should be provided from the
lowest to the highest level.
Where a column or wall is supported at the bottom by a beam or
slab accidental loss of this element should be considered.
Continuity and anchorage ties (9.10.3):
Ties in two horizontal directions shall be effectively continuous and
anchored at the perimeter of the structure.
Ties may be provided wholly in the insitu concrete topping or at
connections of precast members.
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