lecture 5 foundations

8/9/2019 Lecture 5 Foundations

1/30

Practical Design to Eurocode 2

Foundations

Eurocode 7

Eurocode 7 has two parts:

Part 1: General Rules

Part 2: Ground Investigation and testing


2/30

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 thestructure 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

Large or

unusual

structures

Abnormal risksAll other structures3

Spread

foundations

No exceptional

risk

Conventional types of

structure no difficult

ground

2

None givenNegligibleSmall and relatively

simple structures

1

Examples from

EC7

Risk of

geotechnical

failure

DescriptionCategory


3/30

STR/GEO ULS

1.30,iQk1.3Qk1.0Gk1.0GkExp 6.10

Notes:

If the variation in permanent action is significant, use Gk,j,sup and Gk,j,infIf the action if favourable, Q,i = 0 and the variable actions should be ignored

Combination 2

1.50,iQk1.5Qk1.0Gk1.25GkExp 6.10b

1.50,iQk1.50,1Qk1.0Gk1.35GkExp 6.10a

1.50,iQk1.5Qk1.0Gk1.35GkExp 6.10

Combination 1

OthersMainFavourableUnfavourable

Accompanying variableactions

Leadingvariableaction

Permanent Actions

Partial factors

1.01.0Bulk density

1.41.0quUnconfined strength

1.41.0cuUndrained shear

strength

1.251.0cEffective cohesion

1.251.0Angle of shearing

resistance

Combination 2Combination 1Symbol


4/30

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)


5/30

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)


6/30

a a

bF

hF

(9gd/fctd,pl)h

a

F0,85

gd is the design value of the ground pressure

as a simplification hf/a 2 may be used

Strip and Pad Footings(12.9.3) Plain concrete

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


7/30

Worked Example

Design a square pad footing for a 350 350 mm columncarrying 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


8/30

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


9/30

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


10/30

Ultimate Limit States

for the design ofretaining walls

General expressions


11/30

Calculation Model A


12/30

Calculation Model B


13/30

Partial factors

1.01.0Bulk density

1.41.0quUnconfined strength

1.41.0cuUndrained shear

strength

1.251.0cEffective cohesion

1.251.0Angle of shearing

resistance

Combination 2Combination 1Symbol

Overall design

procedure


14/30

Initial sizing

Figure 6 for overall

design procedure


15/30

Panel 2

Overall design

procedure


16/30

Design against

sliding

(Figure 7)


17/30

Overall design

procedure

Design against Toppling(Figure 9)


18/30

Overall design

procedure

Design against bearing failure(Figure 10)


19/30

Overall design

procedure


20/302

Figure 13

Strut and tie methods


21/302

P1

P2 2P1

Which is the Stronger?

fcu

fcu

fct

fct

compressive strength of

concrete with transverse

tension

tensile stress in concrete

Bi-axial Strength of Concrete


22/302

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 Rd,max = 0,6fcd (ieabout 0,3fck)

where = 1-fck/250

Struts

Strut and Tie Models(6.5.2)

Sizing nodes (6.5.4)

The size of a node is

determined by the limiting

stresses in the struts and the

anchorage length of bars

If the stresses in all struts

meeting at a node are the

same, the boundaries of the

node will be perpendicular to

the axis of the strut and the

node will be in a state ofhydrostatic stress

Bars can be

anchored boththrough and beyond

a node

c-c-c(1-fck/250)fcd

(Exp (6.60))

Unless special confinement is

provided, the calculated

compressive stress in the node

regions should not exceed:

c-c-t0.85(1-fck/250)fcd

(Exp (6.61))

c-t-t0.75(1-fck/250)fcd

(Exp (6.62))


23/302

1 - Anchorage of tension reinforcement at simple support

Even at a simple support there is tension

in the reinforcement

behaviour similar to deep beam at support

Strut and Tie Models Can HelpUs Understand .

strut needs tobe supported

on bar

bar to be anchored

beyond point of load

application

2 - Detailing of nib

bearing


24/302

3 - Anchorage of bars

F

F/2 F/2

F tan

F tan

Forces from cranks reduced if laps in plane parallel to surface

but similar splitting forces from lap

F F

4 - Lapping of bars


25/30


26/302

5.5 MN (Ult)

7503400

1650

fck = 30 MPa / fyk = 500 MPa

breadth = 1050

bottom cover = 100, bars 32mm

side cover = 75

1. Set up an arrangement of

struts and ties

2. Check the strength of the

struts 6.5.2 (2)

3. Check the strength of thenodes 6.5.4 (4)

4. Calculate the areas of

reinforcement required

5. What other checks should

be made to EC2?

700 x 700 column

200

Pile Cap Exercise


27/302

1. Arrangement of struts andties

Compressive force in eachstrut

= 2750 / sin 53.7

= 3412 kN

Tensile force in tie

= 2750 / tan 53.7

= 2020 kN

Width of each strut

= 700 sin 53.7

= 564 mm

Angle of strut

= tan -1 (1530 / 1125)

= 53.7

1125

2750 kN

1530

120

53.7

5500 kN

700 x 700 column

2750 kN

2020 kN

3412 kN

564

Pile Cap Exercise Solution (1)

2. Strength of struts

Stress in strut:Projecting the rectangular geometryfrom the column gives minimum area,

and maximum stress= 3.412 / (0.564 x 0.7)= 8.6 MPa

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

53.7

3412 kN

564

700



28/302

Upper node:

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 (noshear), the other plane is ok by inspection.

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

3. Strength of nodes

3412 kN

700mm

2.02MN

Ed,2

Ed,1

5.5 MN (Ult)


3. Strength of nodes

Lower node:The exact geometry of node is difficult todefine, so conservatively compare strutstresses to node allowable stresses

Stress in strut Ed,1= 8.6 MPa

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-f

ck/250)f

cd

= 0.85 (1 30 / 250) 0.85 x 30 / 1.5

= 12.7 MPa OK

2750 kN2750 kN

Ed,2

Ed,2

750 750

Ed,1Ed,1



29/302

4. Area of reinforcement

Tension = 2020 kN

Area of steel required

= 2020 x 1000/ (500/1.15)

= 4646 mm2

6H32 (4825 mm2)

5. Other checks

Assume bar fully stressed at centreline of pile

Standard mandrel diameter (Cl 8.3 (3) & Table 8.1N)m = 7 = 7 x 32 = 224 mmStandard extension of bar beyond bend = 5 = 5 x 32 = 160 mm

Cl 8.4.4. (1) :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 235 = 0.649 but Exp(8.5) 235 = 0.7

Hence lbd = 12 345Lb,rqd = 0.7 x 0.7 x 1116.3 = 547 mm

53.7

2750 / tan53.7= 2020 kN

2750 kN

112 mm

160mm

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


Tying

www.eurocode2.info


30/30

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 verticalload 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 (2)

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 andanchored at the perimeter of the structure.

Ties may be provided wholly in the insitu concrete topping or at

connections of precast members.

Tying systems (3)

lecture 5 foundations

Documents