Design approaches Design approaches according to Eurocode 7according to Eurocode 7

October 2010October 2010Structural Engineering Master program - Irina Structural Engineering Master program - Irina

LunguLungu

Faculty of Civil Engineering and Building Faculty of Civil Engineering and Building Services IasiServices Iasi

Geotechnical restrictionGeotechnical restriction

Ed <= Rd Ed = γE * Ek Md = Mk / γM Rd = Rk / γR

Parameters with their real values, without safety factors, are named characteristic values with an index k: Ek = characteristic actions, Rk = characteristic resistances.

Parameters with safety factors are named design values with an index d: Ed = design actions, Rd = design resistances.

For the GEO and STR limit states, the three possible design approaches use different sets of partial safety factors:

Design approach 1 with two combinations: Combination 1: A1 + M1 (= 1) + R1 (= 1) → safety factors on

loads Combination 2: A2 (= 1) + M2 + R1 (= 1) → safety factors on

materials (soil) for piles: Combination 1: A1 + M1 (= 1) + R1 (= 1) Combination 2: A2 (= 1) + (M1 (= 1) or M2) + R4 (M1 for pile resistance, M2 for unfavorable actions like negative skin

friction or transversal loads)

Design approach 2: A1 + M1 (= 1) + R2 → safety factors on loads and resistances

Design approach 3: A2 (= 1) + M2 + R3 (= 1) (A1 for loads from the structure without influence of soil material

parameters)

Design exampleDesign example 1 – spread foundation 1 – spread foundationProposal within Evaluation of Eurocode 7Proposal within Evaluation of Eurocode 7

Stiff till

Square pad foundation

GEO GEO ultimate limit stateultimate limit state – – bearing resistance failurebearing resistance failure

Vd ≤ Rd

Vd – design vertical loadRd – design bearing resistance

Short term safety – undrained conditions for soilLong term safety – drained soil condition for soil

partial safety factors onpartial safety factors on Actions - A Actions - A

Actions A1 A2

Permanent loads, unfavorable 1.35 1.00

Permanent loads, favorable 1.00 1.00

Variable loads, unfavorable 1.50 1.30

Variable loads, favorable 0.00 0.00

partial safety factors onpartial safety factors on Materials - M Materials - M

Materials M1 M2

Angle of internal friction tan(ϕ) 1.00 1.25

Cohesion c 1.00 1.25

Undrained cohesion cu 1.00 1.40

Unit weight γ 1.00 1.00

partial safety factors onpartial safety factors on Resistances - R Resistances - R

Resistances R1 R2 R3 R4

Sliding resictance 1.00 1.40 1.00

Bearing capacity resistance 1.00 1.10 1.00

Passive earth pressure 1.00 1.40 1.00

End bearing for bore piles 1.25 1.10 1.00 1.60

Skin friction for bore piles, compress. 1.00 1.10 1.00 1.30

Skin friction for bore piles, tension 1.25 1.15 1.10 1.60

Vertical loadsVertical loads

Vd = γG (Gk + Gpad,k) + γQ Qk

= γG (Gk + A γc d) + γQ Qk

Vd = γG (900 + B2 x 24 x 0.8) + γQ x 600

Undrained condition Undrained condition – short term safety– short term safety

Rd/A’ = Rd/A = pcr = ((π + 2) cu,d bc sc ic + qd) /γR = ((π + 2) cu,d sc + qd) /γR

b – base inclination factor, bc = 1.00

i – load inclination factor, ic = 1.00

s – shape factor, sc = 1.20A’ = A, centric loading case

Rd = pcr x A = B2 x ((π + 2) cu,d sc + qd) /γR = B2 ((π +2)(200/γM)+22 x 0.8)/ γR

DA1 – combination 1DA1 – combination 1

A1 + M1 (= 1) + R1 (= 1) → safety factors on loads

Vd = 1.35 (900 + B2 x 24 x 0.8) + 1.5 x 600

Rd = B2 ((π +2)(200/1.00)+22 x 0.8)/1.00

→ B = ?

DA1 – combination 2DA1 – combination 2

A2 (= 1) + M2 + R1 (= 1) → safety factors on materials (soil)

Vd = 1.0 (900 + B2 x 24 x 0.8) + 1.3 x 600

Rd = B2 ((π +2)(200/1.4)+22 x 0.8)/1.00

→ B = ?

DA2DA2

A1 + M1 (= 1) + R2 → safety factors on loads and resistances

Vd = 1.35 (900 + B2 x 24 x 0.8) + 1.5 x 600

Rd = B2 ((π +2)(200/1.00)+22 x 0.8)/1.40

→ B = ?

DA3DA3

A1 + M2 + R3 (= 1)

A1 for loads from the structure without influence of soil material parameters

Vd = 1.35 (900 + B2 x 24 x 0.8) + 1.5 x 600 Rd = B2 ((π +2)(200/1.40)+22 x 0.8)/1.00

→ B = ?

Drained condition – Drained condition – long term safetylong term safety

Rd/A’ = c’ Nc bc sc ic + q’ Nq bq sq iq + 0.5 γ’ B’ Nγ bγ sγ iγ

c’ = 0 q’ = 0.8 x (γ - γw) = 0.8 x (22 – 9.81) = 9.75kPa A’ = A = B2; B’ = B i = all 1.00 b = all 1.00 sγ = 0.7; sq = 1 + sin Φ’

Rd = A (q’ Nq sq + 0.5 γ’ B Nγ 0.7) = B2 (9.75 x Nq x sq + 0.5 x 12.19 x B x Nγ x 0.7

DA1 – combination 1DA1 – combination 1 A1 + M1 (= 1) + R1 (= 1) → safety factors on loads

Vd = 1.35 (900 + B2 x (24 – 9.81) x 0.8) + 1.5 x 600 Rd = B2 (9.75 x Nq x sq + 0.5 x 12.19 x B x Nγ x 0.7

Nq = eπ x tanφ’tan2(π/4 + φ'/2) = eπtan35tan2(π/4 + 35.0/2) = 33.30

Nγ = 2(Nq - 1) tanφ' = 2(33.3 – 1) tan35 = 45.23 sq = 1 + sinφ' = 1 + sin35 = 1.57 Rd = B2 (9.75 x 33.3 x 1.57 + 0.5 x 12.19 x B x 45.23

x 0.7)/1.00

→ B = ?

DA1 – combination 2DA1 – combination 2 A2 (= 1) + M2 + R1 (= 1) → safety factors on

materials (soil) Vd = 1.0 (900 + B2 x (24 – 9.81) x 0.8) + 1.3 x 600 Rd = B2 (9.75 x Nq x sq + 0.5 x 12.19 x B x Nγ x 0.7

φ'd = tan-1(tan φ'k)/γM = tan-1(tan35/1.25) = 29.30 Nq = eπ x tanφ’tan2(π/4 + φ'/2) = eπtan29.3tan2(π/4 + 29.3/2)

= 16.92 Nγ = 2(Nq - 1) tanφ' = 2(16.92 – 1) tan29.3 = 17.84 sq = 1 + sinφ' = 1 + sin29.3 = 1.49

Rd = B2 (9.75 x 16.92 x 1.49 + 0.5 x 12.19 x B x 17.84 x 0.7)/1.00

→ B = ?

DA2DA2 A1 + M1 (= 1) + R2 → safety factors on loads and

resistances

Vd = 1.35 (900 + B2 x (24-9.81) x 0.8) + 1.5 x 600 Rd = B2 (9.75 x Nq x sq + 0.5 x 12.19 x B x Nγ x 0.7

Nq = eπ x tanφ’tan2(π/4 + φ'/2) = eπtan35tan2(π/4 + 35.0/2) = 33.30

Nγ = 2(Nq - 1) tanφ' = 2(33.3 – 1) tan35 = 45.23 sq = 1 + sinφ' = 1 + sin35 = 1.57 Rd = B2 (9.75 x 33.3 x 1.57 + 0.5 x 12.19 x B x 45.23 x

0.7)/1.40

→ B = ?

DA3DA3A1 + M2 + R3 (= 1)

A1 for loads from the structure without influence of soil material parameters

Vd = 1.35 (900 + B2 x (24 - 9.81) x 0.8) + 1.5 x 600 Rd = B2 (9.75 x Nq x sq + 0.5 x 12.19 x B x Nγ x 0.7

φ'd = tan-1(tan φ'k)/γM = tan-1(tan35/1.25) = 29.30 Nq = eπ x tanφ’tan2(π/4 + φ'/2) = eπtan29.3tan2(π/4 + 29.3/2) = 16.92 Nγ = 2(Nq - 1) tanφ' = 2(16.92 – 1) tan29.3 = 17.84 sq = 1 + sinφ' = 1 + sin29.3 = 1.49

Rd = B2 (9.75 x 16.92 x 1.49 + 0.5 x 12.19 x B x 17.84 x 0.7)/1.00

→ B = ?