Design approaches according to Eurocode 7

Structural Engineering Master program

Geotechnical restriction

Ed

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 example 1 spread foundation

Stiff till

Square pad foundation

GEO ultimate limit state bearing resistance failure

Vd Rd

Vd design vertical load

Rd design bearing resistance

Short term safety undrained conditions for soil

Long term safety drained soil condition for soil

partial safety factors on 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 on Materials - M

Materials M1 M2

Angle of internal friction tan() 1.00 1.25

Cohesion c 1.00 1.25

Undrained cohesion cu 1.00 1.40

Unit weight 1.00 1.00

partial safety factors on 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 loads

Vd = G (Gk + Gpad,k) + Q Qk= G (Gk + A c d) + Q Qk

Vd = G (900 + B2 x 24 x 0.8) + Q x 600

Undrained condition short term safety

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

b base inclination factor, bc = 1.00

i load inclination factor, ic = 1.00

s shape factor, sc = 1.20

A = A, centric loading case

Rd = pcr x A = B2 x (( + 2) cu,d sc + qd)

/R = B2 (( +2)(200/M)+22 x 0.8)/ R

DA1 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 (( +2)(200/1.00)+22 x 0.8)/1.00

B = ?

DA1 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 (( +2)(200/1.4)+22 x 0.8)/1.00

B = ?

DA2

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 (( +2)(200/1.00)+22 x 0.8)/1.40

B = ?

DA3

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 (( +2)(200/1.40)+22 x 0.8)/1.00

B = ?

Drained condition long term safety

Rd/A = c Nc bc sc ic + q Nq bq sq iq + 0.5 B N b s i

c = 0

q = 0.8 x ( - w) = 0.8 x (22 9.81) = 9.75kPa

A = A = B2; B = B

i = all 1.00

b = all 1.00

s = 0.7; sq = 1 + sin

Rd = A (q Nq sq + 0.5 B N 0.7) = B2 (9.75 x Nq x sq +

0.5 x 12.19 x B x N x 0.7

DA1 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 N x 0.7

Nq = e x tantan2(/4 + '/2) = etan35tan2(/4 + 35.0/2) =

33.30 N = 2(Nq - 1) tan' = 2(33.3 1) tan35 = 45.23 sq = 1 + sin' = 1 + sin35 = 1.57 Rd = B

2 (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 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 N x 0.7

'd = tan-1(tan 'k)/M = tan

-1(tan35/1.25) = 29.30 Nq = e

x tantan2(/4 + '/2) = etan29.3tan2(/4 + 29.3/2) = 16.92

N = 2(Nq - 1) tan' = 2(16.92 1) tan29.3 = 17.84 sq = 1 + sin' = 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 = ?

DA2

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 N x 0.7

Nq = e x tantan2(/4 + '/2) = etan35tan2(/4 + 35.0/2) =

33.30 N = 2(Nq - 1) tan' = 2(33.3 1) tan35 = 45.23 sq = 1 + sin' = 1 + sin35 = 1.57 Rd = B

2 (9.75 x 33.3 x 1.57 + 0.5 x 12.19 x B x 45.23 x 0.7)/1.40

B = ?

DA3A1 + 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 N x 0.7

'd = tan-1(tan 'k)/M = tan

-1(tan35/1.25) = 29.30 Nq = e

x tantan2(/4 + '/2) = etan29.3tan2(/4 + 29.3/2) = 16.92 N = 2(Nq - 1) tan' = 2(16.92 1) tan29.3 = 17.84 sq = 1 + sin' = 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 = ?

Design widths for different Design

Approaches and Design conditions

Design Approach Foundation width (m) OFS = Ruk/VkUndrained Drainedcondition condition

DA1 (Combination 1)

DA1 (Combination 2)

DA2

DA3

Design example 2

GEO ultimate limit state bearing resistance failure

Vd RdVd design vertical load

Rd design bearing resistance

Short term safety undrained conditions for soil

Long term safety drained soil condition for soil

partial safety factors on 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 on Materials - MMaterials M1 M2

Angle of internal friction tan() 1.00 1.25

Cohesion c 1.00 1.25

Undrained cohesion cu 1.00 1.40

Unit weight 1.00 1.00

partial safety factors on Resistances - R

Resistances R1 R2 R3

Sliding resictance 1.00 1.10 1.00

Bearing capacity resistance 1.00 1.40 1.00

about vertical loads

Nq = etan ' tan2(45 + '/2)

Nc = (Nq - 1) cot'

N = 2 (Nq - 1) tan

sq = 1 + sin' for a square or circular shape

s = 1 - 0.3(B'/L') for a rectangular shape

s = 0.7 for a square or circular shape

sc = (sqNq - 1)/(Nq - 1)

ic = iq (1 - iq)/N c tan'

iq = [1 - H/(V + A c cot')]m

i = [1 - H/(V + A c cot')]m + 1

m = [2 + (B'/L')]/[1+(B'/L')] when H acts in the direction of B'

about resistances

Values to determine, based on the B value