Eurocode 7 - Elements of Soil Mechanics - page 151

Europe-wide adoption of the Standard and offers designers in different nations an approach most relevant
to their needs. The UK National Annex to EN 1997-1 states that Design Approach 1 is to be used in the
UK and worked examples in the following chapters illustrate the use of this method.
As mentioned earlier, the choice of partial factors to be used is dependent on the design approach being
followed (for the GEO and STR limit states). For each design approach, a different combination of partial
factor sets is used to verify the limit state. For Design Approach 1 (for retaining walls and shallow footings),
two combinations are available and the designer would normally check the limit state using each combina-
tion, except on occasions where it is obvious that one combination will govern the design (the combination
of partial factor sets for Design Approach 1 is different for pile foundations - see Chapter 10) .

Design Approach 1: Combination 1: A1 + M1 + R1
Combination 2: A2 + M2 + R1
Design Approach 2: A1 + M1 + R2
Design Approach 3: A* + M2 + R3
( Note . A*: use set A1 on structural actions, set A2 on geotechnical actions).

The sets for actions (denoted by A), material properties (denoted by M) and ground resistance (denoted
by R) for each design approach are given in Table 5.1. Also given in the table are the partial factors for
the EQU limit state.

Example 5.4: Design approaches; design actions

A concrete foundation is to be cast into a soil deposit as shown in Fig. 5.13. The
foundation has a representative self weight, W of 50 kN.
During a check for bearing resistance (see Chapter 9) , the vertical representative
actions V G; k and V Q; k are considered as unfavourable . Determine the design values of
each action, for each Design Approach.

Solution:

The design values of the actions are achieved by multiplying the representative actions
by the appropriate partial factors of safety from Table 5.1.
e.g. Design Approach 1, Combination 1 (DA 1-1)

G = × = G γ 600 1 35 × . = 810 kN

d k G

V G; d (kN) V Q; d (kN) W G; d (kN)

DA 1-1 γ G = 1.35

γ Q = 1.5

810 225 67.5

DA 1-2 γ G = 1.0

γ Q = 1.3

600 195 50

DA 2 γ G = 1.35

γ Q = 1.5

810 225 67.5

DA 3 γ G = 1.35

γ Q = 1.5

810 225 67.5