Environmental Engineering Reference
In-Depth Information
on which offers less resistance to sliding. The bearing capacity failure mechanism would
involve shearing within the clay foundation.
The deterministic calculations were performed using the CGPR Retaining Wall Stability
Computation Sheet 2.05 (Yang and Duncan 2002, converted to the SI system for these
examples). The factors of safety against sliding on the silty sand layer beneath the footing,
sliding in the clay foundation, and bearing capacity in the clay foundation are summarized
briefly in the following paragraphs.
3.6.2 Factor of safety against sliding on top of the silty sand layer
The factor of safety against sliding on the silty sand layer beneath the footing is given by the
following formula:
N
E h
µ
F =
(3.20)
in which N is the normal force on base (N/m), μ the base friction coefficient, which is
the tangent of base friction angle, and E h the horizontal earth pressure force on vertical
plane through heel of wall (N/m). The magnitudes of N , μ, and E h are
N = 536,769 N/m,
μ = tan δ = tangent of the base friction angle = 0.5,
E h = 191,767 N/m, and
F = 1.40 against sliding on the silty sand.
3.6.3 Factor of safety against sliding on the clay foundation
Meyerhoff's method (Meyerhoff 1953) is used to account for the fact that the footing is
eccentrically loaded. A reduced effective footing width is used, such that the reduced width
is centrally loaded. The reduced footing width is 2X = (2)(1.56) = 3.12 m. The factor of
safety against sliding on a clay layer can be calculated using this expression:
Sx
E
()
2
u
F
=
(3.21)
h
where S u is the undrained shear strength of the clay and X is the distance from the front of
the footing to the normal force on the base of the retaining wall. As shown in Figure 3.16:
S u = 120 kPa
X = 1.56 m
E h = 191,767 N/m
F = 1.95 against sliding at the top of the clay
3.6.4 Factor of safety against bearing capacity failure
The factor of safety against bearing capacity failure in the clay foundation is found from the
following equation:
NS
q
cu
F
=
(3.22)
 
 
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