Managing risk and achieving reliable geotechnical designs using Eurocode 7 - Risk and Reliability in Geotechnical Engineering - page 415

using φ k

′

= 28 2 . ° for which K a;k = 0.309. Hence, the characteristic wall frictional resistance

against uplift is

R5 K = 0 .

; σδ

′

tanA = 0 . 53918 × 00 . × ( . 0 − . 81 ) × 6 × tan 282 . × 216 = 8 79 kN

k a kv k

The overall factor of safety on the wall frictional resistance against uplift is obtained by
dividing the characteristic resistance by the factored resistance:

OFS = R k /R d = 879/858 = 1.02

Therefore in this situation factoring φ ′ and reducing it from 28.2° to 23.2° has provided
virtually no margin of safety or degree of reliability on the wall frictional resistance. The
reason for this is because, while applying the γ M to φ ′ reduces tanδ k , at the same time it
increases K a;k with the result that there is very little safety.
The φ ′ value above was obtained by assuming that a lower φ ′ value is conservative and
dividing the inferior tan φ ′ value by γ M . Eurocode 7 states in §2.4.5.2(5) that a character-
istic value may be an upper value, that is, greater than the mean value. If that approach is
adopted in this example, the superior φ ′ is given by

φ

′

= tan

− 1

(tan ξ φ

′

) = tan

− 1

(. 1 0 75 × tan) 3 18 0 = . °

k mean

and the φ ′ value is

φ

′

= tan( tan) tan(.

− 1

γ φ

′

=

− 1 125

× tan 318 .) = 37 8 . °

d Mk

Using this superior value for φ ′ gives K a;d = 0.203, so that the design wall resistance is
now

R5 K = 0 .

; σδ

′

tan A = 0 . 52318 × 00 . × (. 0 − .) 81 × 6 an37 8 × . × 216 = 83 6 kN

d a dv d

For the upper φ k

′

= 31 8 ., K a;k = 0.265 so that the characteristic wall resistance is

R5 K = 0 .

; σδ

′

tan A5 = 0 . × 0 . 265 × (. 18 0 − 9816 .) × × tan. 31 8 × 216 = 8 72 kN

k a kv k

This R k is very similar to the R k calculated using the lower φ ′ value. Using these resistance
values based on the superior φ ′ , the overall factor of safety on the wall resistance against
uplift is now

OFS = R k /R d = 872/836 = 1.04

The above calculations show that, for this particular example, adopting the superior φ ′
value gives a similar overall factor of safety to that obtained when using the inferior φ ′ value.
This example demonstrates that in certain situations, such as when determining the wall resis-
tance on buried structures, using factors to increase or decrease the ϕ′ value may not provide
the required level of safety and hence care is needed when determining the design parameter