Earth pressures and stability of retaining walls - The Mechanics of Soils and Foundations - page 396

and, solving by trial and error, or otherwise,

d = 40 m

With this value of d we have P a = 405 kN and P p = 320 kN. Hence, resolving

horizontally, from Eq. (24.15) the force in the prop is,

P = 405 − 320 = 85 kN

Example 24.3: Depth of a cantilever wall If the wall in Fig. 24.18(a) is not propped
it acts as a cantilever and the forces on the wall are shown in Fig. 24.18(c). From
Eq. (24.16), taking moments about the toe where the force Q acts,

d ) 2

1
3 (5

20 d 2

1
3 d

5(5 + × + d ) = ×

and, solving that by trial and error, or otherwise,

d = 8.5 m

To provide sufficient length to mobilize the force Q , the wall depth should be increased
by 20% so the required depth of penetration is about 10 m.

Further reading
Atkinson, J. H. (1981) Foundations and Slopes , McGraw-Hill, London.
Clayton, C. R. I., J. Milititsky and R. I. Woods (1993) Earth Pressure and Earth Retaining
Structures , Spon Press, London.
Heyman, J. (1972) Coulomb's Memoir on Statics , Cambridge University Press, Cambridge.
Kerisel, J. and E. Absi (1990) Active and Passive Pressure Tables , Balkema, Rotterdam.
Padfield, C. J. and R. J. Mair (1984) Design of Retaining Walls Embedded in Stiff Clays , CIRIA,
Report 104, London.