Practical Design to Eurocode 2
23/11/16
Practical Design to Eurocode 2 The webinar will start at 12.30
Course Outline Lecture Date
Lecture 10: Foundations
Speaker
Title
1
21 Sep Charles Goodchild Introduction, Background and Codes
2 3
28 Sep Charles Goodchild EC2 Background, Materials, Cover and effective spans 5 Oct Paul Gregory Bending and Shear in Beams
4
12 Oct Charles Goodchild Analysis
5
19 Oct Paul Gregory
6
26 Oct Charles Goodchild Deflection and Crack Control
7
2 Nov
Paul Gregory
Detailing
8
9 Nov
Jenny Burridge
Columns
9
16 Nov Jenny Burridge
Fire
10
23 Nov Jenny Burridge
Foundations
Slabs and Flat Slabs
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Practical Design to Eurocode 2
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Foundations Lecture 10 25th November 2015
Model Answers Lecture 9 Exercise: Fire resistance of Column
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Fire resistance of column Using Equation 5.7, work out the fire resistance of a 250 x 750 column with an axial capacity of 3750kN and an axial load in cold conditions of 3500kN. The column is on the ground floor of a three storey building and the length is 4.5m. The cover is 30mm, main bars are 20mm and the links are 10mm diameter.
Design Exercise Answers = 120(
+
+ + 120
+
)
.
μfi = 0.7 x 3500/3750 = 0.65
Rηfi = 83(1 - 0.65) = 28.8
a = 30 + 10 + 10 = 50mm
Ra = 1.6(50 - 30) = 32
l0,fi = 0.5 x 4.5 = 2.25m
Rl = 9.6(5 - 2.25) = 26.4
b’ = 1.2 x 250 = 300mm
Rb = 0.09 x 300 = 27
n>4
Rn = 12
R = 120((Rηfi + Ra + Rl + Rb + Rn)/120)1.8 = 131 minutes
Lecture 10: Foundations
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Foundations
Outline – Week 10, Foundations We will look at the following topics: • Eurocode 7: Geotechnical design –
Partial factors, spread foundations.
• Pad foundation – Worked example & workshop • Retaining walls • Piles
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Eurocode 7 6
(p43 et seq)
Eurocode 7 has two parts: Part 1: General Rules
Plus NA
Part 2: Ground Investigation and testing
Plus NA
Eurocode 7 How to…… 6. Foundations The essential features of EC7, Pt 1 relating to foundation design are discussed.
Note: This publication covers only the design of simple foundations, which are a small part of EC7. It should not be relied on for general guidance on EC7.
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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
Lecture 10: Foundations
Risk of geoExamples from technical failure EC7
1
Small and relatively simple structures
Negligible
None given
2
Conventional types of structure – no difficult ground
No exceptional risk
Spread foundations
3
All other structures
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 –
Actions partial factors
Permanent Actions Unfavourable
Favourable
Leading variable action
Accompanying variable actions
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
1.5ψ0,iQk
1.0Gk
1.0Gk
1.3Qk
1.3ψ0,iQk
Main
Others
Combination 1 1.5ψ0,iQk 1.5ψ0,1Qk
1.5ψ0,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 is favourable, γQ,i = 0 and the variable actions should be ignored
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Factors for EQU, UPL and HYD
Limit state
Permanent Actions Unfavourable
Favourable
Unfavourable
Favourable
EQU
1.1
0.9
1.5
0
UPL
1.1
0.9
1.5
0
HYD
1.35
0.9
1.5
0
Partial factors –
Parameter
Variable Actions
material properties
Symbol
Combination Combination 1 2
EQU
Angle of shearing resistance
γφ
1.0
1.25
1.1
Effective cohesion
γc’
1.0
1.25
1.1
Undrained shear strength
γcu
1.0
1.4
1.2
Unconfined strength
γqu
1.0
1.4
1.2
Bulk density
γγ
1.0
1.0
1.0
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Geotechnical Report The Geotechnical Report should: • be produced for each project (if even just a single sheet) • contain details of: • the site, • interpretation of ground investigation report, • geotechnical recommendations, • advice
Foundation design recommendations should state: • bearing resistances, • characteristic values of soil parameters and • whether values are SLS or ULS , Combination 1 or Combination 2 values
Spread Foundations EC7 Section 6 Three methods for design: • Direct method – check all limit states: • Load and partial factor combinations (as before) • qult=c’Ncscdcicgcbc + q’Nqsqdqiqgqbq +γ’BNγsγdγiγgγbγ/2 where – – – – – – – – –
Lecture 10: Foundations
c = cohesion q = overburden γ = body-weight Ni = bearing capacity factors si = shape factors di = depth factors ii = inclination factors gi = ground inclination factors bi = base inclination factors
“We just bung it in a spreadsheet” Settlement often critical See Decoding Eurocode 7 by A Bond & A Harris, Taylor & Francis
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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).
Used in subsequent slides).
Spread Foundations Design procedures in:
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Fig 6/1 (p46)
Procedure for depth of spread foundations
Pressure distributions
SLS pressure distributions
ULS pressure distribution
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Load cases
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 (Using (6.10). Worse case of Exp (6.10a) or (6.10b) could be used)
Plain Concrete Strip Footings & Pad Foundations: Cl. 12.9.3, Exp (12.13)
0,85⋅ hF ≥ √(3σgd/fctd,pl) a
hF
where: σgd is the design value of the ground pressure • as a simplification hf/a ≥ 2 may be used
a
a bF
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Plain Concrete Strip Footings & Pad Foundations allowable pressure σgd 50 100 150 200 250
70 140 210 280 350
C16/20
C20/25
C25/30
C30/37
hF /a
hF /a
hF/a
hF /a
0.65 0.92 1.12 1.29 1.45
0.60 0.85 1.04 1.21 1.35
0.55 0.78 0.95 1.10 1.23
e.g. cavity wall 300 wide carrying 80 kN/m onto 100 kN/m2 ground: bf = 800 mm a = 250 mm hf = say assuming C20/25 concrete 0.85 x 250 = 213 say 225 mm
0.52 0.74 0.90 1.04 1.17
hF a
a bF
Reinforced Concrete Bases • Check critical bending moments at column faces • Check beam shear and punching shear
For punching shear the ground reaction within the perimeter may be deducted from the column load
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Pad foundation Worked example
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.
Answer: Category 2. So using prescriptive methods: Base area: (600 + 505)/200 = 5.525m2 => 2.4 x 2.4 base x 0.5m (say) deep.
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Worked Example Loading = 1.35 x 600 + 1.5 x 505 = 1567.5 kN ULS bearing pressure = 1567.5/2.42 = 272 kN/m2 Critical section at face of column MEd
= 272 x 2.4 x 1.0252 / 2 = 343 kNm
d
= 500 – 50 – 16 = 434 mm
K
Use C30/37 concrete
= 343 x106/(2400 x 4342 x30) = 0.025
Worked Example ⇒ z = 0.95d
= 0.95 x 434 = 412mm
⇒ As = MEd/fydz
= 343 x 106 / (435 x 412) = 1914mm2
⇒ Provide 10H16 @ 250 c/c b.w (2010 mm2) (804 mm2/m) Beam shear: Check critical section d away from column face VEd = 272 x (1.025 – 0.434) = 161kN/m vEd = 161 / 434 = 0.37MPa ρ = 2010/ (434 x 2400) = 0.0019 = 0.19% vRd,c (from table) = 0.42MPa
6/Table 6 (p47) Concise Table 15.6
=> beam shear ok.
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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 = 0.188 MPa; vRd,c = 0.42 (as before) => ok
Retaining Walls Chapter 9
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Ultimate Limit States for the design of retaining walls
Calculation Model A
Rankine theory Model applies if bh≥ ha tan (45 - ϕ’d/2)
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Calculation Model B
Inclined ‘ virtual’ plane theory Model applies to walls of all shapes and sizes
General expressions General
Ws = b sHγ k,c
h = tb + H + bh tanβ
Wb = tbBγ k,c
b tanβ Wf = bhH + h γk,f 2 b Lf ≈ bt + bs + h 2 Ω=β
bh = B − b s − b t b L s = bt + s 2 B Lb = 2
Lecture 10: Foundations
Model A
Model B
Lvp = B
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9 (Figure 4) Overall design procedure
Initial sizing bs ≈ tb ≈ h/10 to h/15 B ≈ 0.5h to 0.7h bt ≈ B/4 to B/3
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9 (Figure 4) Overall design procedure
9 (Figure 6) Figure 6 for overall design procedure
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Soil Densities
Ex Concrete Basements
Design value of effective angle of shearing resistance, φ’d tan φ’d = tan (φ’k/γφ) where φ’k = φ’max for granular soils and = φ’ for clay soils, φ’max and φ’ are as defined as follows γφ = 1.0 or 1.25 dependent on the Combination being considered.
Ex Concrete Basements
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Angle of shearing resistance Granular Soils Estimated peak effective angle of shearing resistance, φ’max = 30 + A + B + C Estimated critical state angle of shearing resistance, φ’crit = 30 + A + B, which is the upper limiting value.
Ex Concrete Basements
Clay soils Long term ≡ Granular Soils
Ex Concrete Basements
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Calcs – Material properties & earth pressures
9 (Panel 2)
9 Panel 2
(p71)
2
’
9 (Figure 4) Overall design procedure
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9 (Figure 7)
Design against sliding (Figure 7)
Sliding Resistance 9 (Panel 3)
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9 (Figure 4) Overall design procedure
Design against Toppling 9 (Figure 9)
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9 (Figure 4) Overall design procedure
Design against bearing failure 9 (Figure 10)
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Expressions for bearing resistance 9 (Panel 4,Figure 11)
9 (Figure 4) Overall design procedure
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Structural design 9 (Figure 13)
Remember: Load and Partial Factor Combinations Parameter
Symbol
Comb. 1
Comb. 2
γG,unfav
1.35
1.0
γG,fav
1.00
1.00
γQ
1.50
1.30
Angle of shearing resistance
γφ
1.0
1.25
Effective cohesion
γc’
1.0
1.25
Undrained shear strength
γcu
1.0
1.4
Unconfined strength
γqu
1.0
1.4
Bulk density
γγ
1.0
1.0
Actions Permanent action : unfavourable Permanent action: favourable Variable action Soil Properties
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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 Specification for piling and embedded retaining walls (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!]
Minimum areas: Pile cross section: Ac Ac ≤ 0.5 m2 0.5
m2
1.0 m2
Lecture 10: Foundations
m2
Min area of long. rebar, As,bpmin
Pile diameters
≥ 0.5% Ac
< 800 mm
≥ 2500
mm2
≥ 0.25% Ac
>1130 mm
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Minimum area of reinforcement, As,bpmin (mm2)
Minimum reinforcement 5000 4500 4000 3500 3000 2500 2000 1500 1000 500 0 0
200
400
600
800
1000
1200
1400
1600
Pile diameter, mm
Workshop Design a 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 Category 2, using prescriptive methods Base size: (Gk + Qk)/bearing stress = _____ __ _____ __ = __ __ __ _ _m2 ⇒ ______ x ______ base x ______mm deep (choose size of pad) Use C 30/37 (concrete) 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 ⇒ z = ____d ⇒ As = MEd/fydz =
= ______ x ______= ____mm
Table 15.5
_____mm2
⇒ Provide H____ @ ______ c/c (_______mm2) Check minimum steel 100As,prov/bd = _______ For C30/37 concrete As,min = ____ ∴ OK/not OK
12.3.1
Beam shear Check critical section d away from column face VEd = ______ x ________= ________kN/m vEd = VEd / d = ________MPa ρ = _____/ (______ x ______) = _____ = _____ % vRd,c (from table) = ____MPa
Lecture 10: Foundations
∴ beam shear OK/not OK.
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Workshop Problem 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
End of Lecture 10 (and the course!)
Emails:
[email protected] [email protected] [email protected]
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