Practical Design to Eurocode 2
Week 5 - Foundations
Eurocode 7
Eurocode 7 has two parts: Part 1: General Rules Part 2: Ground Investigation and testing
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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 Negligible Examples from EC7
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Small and relatively simple structures
None given
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Conventional types of structure – no difficult ground All other structures
No exceptional risk
Spread foundations
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Abnormal risks
Large or unusual structures
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EC7 – ULS Design
EC7 provides for three Design Approaches
UK National Annex - Use Design Approach 1 – DA1
For DA1 (except piles and anchorage design) there are two sets of combinations to use for the STR and GEO limit states.
Combination 1 – generally governs structural resistance Combination 2 – generally governs sizing of foundations
STR/GEO ULS –
Unfavourable
Actions partial factors
Leading variable action Accompanying variable actions
Main Others
Notes: If the variation in permanent action is significant, use Gk,j,sup and Gk,j,inf If the action if favourable, γQ,i = 0 and the variable actions should be ignored
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Factors for EQU, UPL and HYD
Permanent Actions Unfavourable EQU UPL HYD 1.1 1.1 1.335 Favourable 0.9 0.9 0.9 Variable Actions Unfavourable 1.5 1.5 1.5 Favourable 0 0 0
Symbol Angle of shearing resistance Effective cohesion Undrained shear strength Unconfined strength Bulk density γφ γc’ γcu γqu γγ
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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)
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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)
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Strip and Pad Footings (12.9.3) – Plain concrete
hF a bF a
0,85 ⋅ hF ≥ √(3σgd/fctd,pl) a
σgd 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.
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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 = 272 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 ⇒ As = MEd/fydz 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. = 0.95 x 434 = 412mm = 343 x 106 / (435 x 412) = 1914mm2
⇒ Provide H16 @ 250 c/c (1930mm2)
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Worked Example
Punching shear Basic control perimeter at 2d from face of column vEd = βVEd / uid < vRd,c β = 1, ui = (350 x 4 + 434 x 2 x 2 x π) = 6854mm VEd = load minus net upward force within the area of the control perimeter) = 1567.5 – 272 x (0.352 + π x .8682 + .868 x .35 x 4) = 560kN vEd = 0.188MPa; vRd,c = 0.41 (as before) => ok
Workshop Problem 1
Pad foundation for a 300mm square column taking Gk = 600kN, Qk = 350kN. Permissible bearing stress = 225kPa. Concrete for base C30/37.
Work out • size of base, • tension reinforcement and • any shear reinforcement.
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Workshop Problem 1
Category 2, using prescriptive methods Base size: (Gk + Qk)/bearing stress = ______m2 ⇒ ____ x ____ base x ____mm deep (choose size of pad) Use C__/___ (choose concrete strength) Loading = γg x Gk + γq x Qk = _____kN ULS bearing pressure = ____/____2 = _____kN/m2 Critical section at face of column MEd = ____ x ____ x _____2 / 2 = _____kNm d = ___ – cover – assumed ø = _____mm K = M/bd2fck= ______
Workshop Problem 1
⇒ z = ____d = ____ x ____= ____mm ⇒ As = MEd/fydz = _____mm2 ⇒ Provide H__ @ ____ c/c (_____mm2) Check minimum steel 100As,prov/bd = _____ For C__/__ concrete As,min = ____ ∴ OK/not OK Beam shear Check critical section d away from column face VEd = ___ x _____= _____kN/m vEd = VEd / d = _____MPa vRd,c (from table) = ____MPa ∴ beam shear OK/not OK.
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Workshop Problem 1
Punching shear Basic control perimeter at 2d from face of column vEd = βVEd / uid < vRd,c β = 1, ui = = _____mm VEd = load minus net upward force within the area of the control perimeter) = _____ – ____ x ( = _____kN vEd = _____MPa; vRd,c = ______ (as before) => ok/not ok )
Retaining Walls
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Ultimate Limit States for the design of retaining walls
Calculation Model A
Model applies if bh≥ ha tan (45 - ϕ’d/2)
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General expressions
Ws = b sHγ k,c Wb = tbBγ k,c bh = B − b s − b t b L s = bt + s 2 B Lb = 2
For Calculation Model A
h = tb + H + bh tan β b tan β Wf = bh H + h γ k,f 2 b L f ≈ bt + bs + h 2 Ω=β L vp = B
Symbol Angle of shearing resistance Effective cohesion Undrained shear strength Unconfined strength Bulk density γφ γc’ γcu γqu γγ
Overall design procedure
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Initial sizing
bs ≈ tb ≈ h/10 to h/15 B ≈ 0.5h to 0.7h bt ≈ B/4 to B/3
Overall design procedure
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Figure 6 for overall design procedure
Panel 2
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Overall design procedure
Design against sliding (Figure 7)
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Overall design procedure
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Design against Toppling (Figure 9)
Overall design procedure
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Design against bearing failure (Figure 10)
Panel 4 – Expressions for bearing resistance
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Overall design procedure
Figure 13 – Structural design
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Piles
Flexural and axial resistance of piles
‘Uncertainties related to the cross-section of cast in place piles and concreting procedures shall be allowed for in design’ ‘In the absence of other provisions’, the design diameter of cast in place piles without permanent casing is less than the nominal diameter Dnom: • Dd = Dnom – 20 mm for Dnom < 400 mm • Dd = 0.95 Dnom for 400 ≤ Dnom ≤ 1000 mm • Dd = Dnom – 50 mm for Dnom > 1000 mm ICE SPERW B1.10.2 states ‘The dimensions of a constructed pile or wall element shall not be less than the specified dimensions’. A tolerance of 5% on auger diameter, casing diameter, and grab length and width is permissible.
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Flexural and axial resistance of piles
• The partial factor for concrete, γc, should be multiplied by a factor, kf, for calculation of design resistance of cast in place piles without permanent casing. • The UK value of kf = 1.1, therefore γc,pile = 1.65 • “If the width of the compression zone decreases in the direction of the extreme compression fibre, the value η fcd should be reduced by 10%”
Bored piles
Reinforcement should be detailed for free flow of concrete Minimum diameter of long. reinforcement = 16mm Minimum number of longitudinal bars = 6 BUT – BS EN 1536 Execution of special geotechnical work Bored Piles says 12 mm and 4 bars!
Pile cross section: Ac Ac ≤ 0.5 m2 0.5 m2< Ac≤ 1.0 m2 m2 Ac > 1.0 Min area of long. rebar, As,bpmin ≥ 0.5% Ac ≥ 2500 mm2 >1130 mm ≥ 0.25% Ac Pile diameters < 800 mm
What is Strut and Tie?
A structure can be divided into: • B (or beam or Bernoulli) regions in which plane sections remain plane and design is based on ‘normal’ beam theory, D (or disturbed) regions in which plane sections do not remain plane; so ‘normal’ beam theory may be considered inappropriate and Strut & Tie may be used
and •
D regions
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What is strut and tie?
Strut-and-tie models (STM) are trusses consisting of struts, ties and nodes. a) Modelling Imagine or draw stress paths which show the elastic flow of forces through the structure b) STM Replace stress paths with polygons of forces to provide equilibrium.
Conventionally, struts are drawn as dashed lines, ties as full lines and nodes numbered. 57 A deep beam
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At failure which is bigger P1 or P2?
P1
P2
Concept by R Whittle, drawn by I Feltham. Used with permission
At failure which is bigger P1 or P2?
P1
P2 ≈ 2P1
Concept by R Whittle, drawn by I Feltham. Used with permission
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What is strut and tie?
Strut and tie models are based on the lower bound theorem of plasticity which states that any distribution of stresses resisting an applied load is safe providing:
Equilibrium is maintained and Stresses do not exceed “yield”
What is strut and tie?
In strut and tie models trusses are used with the following components: • Struts (concrete) • Ties (reinforcement) • Nodes (intersections of struts and ties) Eurocode 2 gives guidance for each of these.
In principle - where non-linear strain distribution exists, strut and tie models may be used. e.g • Supports • Concentrated loads • Openings
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Struts
Where there is no transverse tension σRd,max = fcd = 0.85 fck /1.5 = 0.57 fck Otherwise, where there is transverse tension
σRd,max
Where:
= 0.6 ν’fcd ν’ = 1-fck/250
σRd,max
= 0.6 x (1-fck/250) x 1.0 x fck /1.5 = 0.4 (1-fck/250) fck
Bi-axial Strength of Concrete
compressive strength of concrete with transverse tension
fcu
fct
fcu
tensile stress in concrete
fct
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Biaxial stress relationship
σz
Reduction in compression strength
tension
σy
Strut & Tie Models
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Struts
Dimensions of the strut are determined by dimensions of the nodes and assumptions made there. The stress in struts is rarely critical but the stress where struts abut nodes is (see later). However . . . . .
Discontinuities in struts
Areas of non-linear strain distribution are referred to as “discontinuities” Partial discontinuity Full discontinuity
Curved compression trajectories lead to tensile forces
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Partial discontinuity
Tension in the reinforcement is T When b ≤ H/2 T = ¼ [(b – a )/b] F Reinforcement ties to resist the transverse force T may be “discrete” or can be “smeared” over the length of tension zone arising from the compression stress trajectories
T T
Full discontinuity
When b > H/2 T = ¼ (1 – 0.7a /h) F Reinforcement ties to resist the transverse force T may be “discrete” or can be “smeared” over the length of tension zone arising from the compression stress trajectories
T T
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Ties
Design strength, fyd = fyk/1.15 Reinforcement should be anchored into nodes
Nodes
Nodes are typically classified as:
CCC – Three compressive struts
CCT – Two compressive struts and one tie
CTT – One compressive strut and two ties
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CCC nodes
The maximum stress at the edge of the node:
σRd,max = k1 ν’fcd
Where: k1 ν’ = 1.0 = 1-fck/250
σRd,max = (1-fck/250) x 0.85 x fck /1.5
= 0.57 (1-fck/250) fck
Pile-cap example
Using a strut and tie model, what tension reinforcement is required for a pile cap supporting a 500 mm square column carrying 2 500 kN (ULS), and itself supported by two-piles of 600 mm diameter. fck = 30 MPa
2 500 kN (ULS)
150 2700
1400
Breadth = 900 mm
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Pile-cap example
STM
Angle of strut Width of strut*
=
tan-1(900/1300)
2 500 kN (ULS)
= 34.7° = 250/cos 34.7° = 304 mm Force per strut = 1250/cos 34.7° = 1520 kN Force in tie
500/2 = 250
34.7o 34.7o 866 kN
Pile-cap example
Detailing Detailed checks are also required for the following: • • • Small piles Determine local tie steel across struts (if req’d) Detailing of reinforcement anchorage (large radius bends may be required)
Anchorage starts from here
Strut dimensions
RE previous statement that calculated strut dimensions were “Conventional but simplistic - see later” For the CCT node:
Not used in previous calcs. Hence struts themselves rarely critical.
Similarly for the CCC node
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Pile-cap example
Comparison: design using bending theory MEd =2500 x 1.800/4 = 1125 kNm Assume: 25 mm φ for tension reinforcement 12 mm link d = h – cnom - φlink - 0.5φ = 1400 – 75 - 12 – 13 = 1300 mm
Worked example
K ' = 0.208 M Ed K = bd 2f ck
= 1125 × 10 900 × 13002 × 30 = 0.025 < K '
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As = 1125 x 106 / (435 x 1270) = 2036 mm2 Use 5 H25 (2454 mm2)
c.f. using S&T 1991 mm2 req’d and 5H25 provided
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Workshop problem 2
Using a strut and tie model, what tension reinforcement is required for a pile cap supporting a 650 mm square column carrying 4 000 kN (ULS), and itself supported by two-piles of 750 mm diameter. fck = 30 MPa
4 000 kN (ULS)
150 3300
Workshop problem 2 Model answer
Angle of strut = tan-1(____/_____) = _____° Width of strut (?) = ____/cos ____° = ____ mm Force per strut Force in tie = _____/cos ____° = _____ kN = _____ kN
325 Strut angle
1800
Breadth = 1050 mm
4 000 kN (ULS)
2000 kN (ULS) 2250
2000 kN (ULS)
(?)
100
= ____ tan _____°
1800
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Workshop problem 2 Model answer
Check forces in truss Stress in strut = _____x103/(____x650) = _____ MPa Strength of strut:
σRd,max
= 0.4 (1-fck/250) fck = _____ MPa
Area of steel required: As ≥ ____ x 103/435 ≥ ______ mm2 Use H
Workshop problem 2 Model answer
Nodes: bottom
From above
σRd,2 σRd,1
= _____ MPa as before = 2000 x 103/(3752 π) = _____ MPa
______ kN 2000 kN
σRd,max = 0.48 (1-fck/250) fck
= _____ MPa
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Workshop problem 2 Model answer
Nodes: top
2398 kN 2398 kN
Workshop problem 2 Model answer
Design using bending theory
MEd =4000 x 2.250/4 = _____ kNm Assume: 25 mm φ for tension reinforcement 12 mm link d = h – cnom - φlink - 0.5φ = ______ mm
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Workshop problem 2 Model answer
K’ = ______ K=M/bd2fck K= ______ z= ______ x d z = ______ mm As = MEd/fydz As = ______/ (____ x _____) = _____mm2 Use __ H____ (_____ mm2)