Geology and Soil Mechanics, UW-Stout
Forces and Pressures on Retaining Walls
To examine whether a retaining wall can remain stable for a simple geometry is an exercise in
static equilibrium. When you estimate forces and pressures on walls, take care not to confuse the
three quantities: force (F), force/length (F/L), and force/area or pressure (P). The basic
objective is to apply the conditions for static equilibrium, which are:
1. All the forces in the horizontal direction must add to zero.
2. All the forces in the vertical direction must add to zero.
3. The clockwise moments (or torques) must equal the counter-clockwise moments.
(In actuality, the forces must balance in 3-dimensions and the moments must balance in two perpendicular
directions. But many simple geometries can be effectively analyzed in 2-dimensions.)
Example #1, Gravity Wall
Lets begin by working through an example using a simple gravity wall. To calculate the red (destabilizing) F/L vector in figure 1B, one needs to first determine the lateral soil pressure at the
base of the wall, P1.
P1 = Koh
P1 = (0.8)(15.1 kN/m3)(6 m)
P1 = 72.5 kN/m2.
Now the F/L can be determined by an analysis of a distributed load:
F/L = (1/2)(Koh)h = (1/2)Koh2 = (1/2)P1h
F/L = (1/2)(72.5 kN/m2 )(6 m)
F/Lsoil = 217.5 kN/m.
To determine the F/L due to the weight of the concrete, we need to first determine the total
weight in 1 m length (in and out of the page):
W = c(Volume) = (25 kN/m3)(6m)(2.6m)(1m)
W = 390 kN.
You can express this as a total force of 390 kN or a (F/L)c of 390 kN/m, where the length is in
and out of the page.
Figure 1
This wall is stable in regards to tipping over per unit of length if Mcl > Mccl. Determining Mcl:
Mcl = (F/L)cd = (390 kN/m)(1.3m) = 507 kN (or 507 kN.m/m),
Mccl = (F/L)soild = (217.5 kN/m)(2m) = 434 kN (or 434 kN.m/m)
The wall is stable. The soil will exert a "reactive force" onto the base of the wall to ensure Mcl =
Mccl if Mcl > Mccl .It is interesting to note that the minimum thickness necessary to keep the wall
stable from tipping is about 2.4 m keeping the rest of the conditions the same. Will the soil have
sufficient bearing capacity? The pressure underneath the concrete is calculated by
P = (Wtotal)/(area)= c(Volume)/(area) = (25 kN/m3)(6m)(2.6m)(20m)/( (2.6m)(20m) )
P = (25 kN/m3)(6m) = 150 kN/m2. Now, to calculate the soil's bearing capacity we need to first
determine a foundation geometry. The closest geometry is probably a continuous wall footing
with zero depth. (This is an estimate because the soil on the right side of the wall is pushing on
the soil mass directly underneath the wall. This could significantly weaken the soil's ability to
resist shearing and squishing out towards the left. So one needs to use some personal judgement
in evaluating the applicability of a continuous wall footing calculation.)
qult = cNc + DfNq + 0.5BN
To error on the safe side, lets consider the angle of internal friction, , to equal zero. Therefore,
Nc=5.14, Nq=1.0, N =0, and Df=0. Thus,
qult = cNc = (65 kN/m2)(5.14) = 334 kN/m2.
The soil will not experience a bearing capacity failure based upon this estimate (150 kN/m2 <
334 kN/m2).
Example #2, Tie-Back
Determining the forces and pressures acting upon a tie-back wall is quite similar to the procedure
for a simple gravity wall (from above). We need to first determine the degree to which the soil is
pushing on the structure (red vector).
Figure 2
The pressure at the base, P1, is determined by
P1 = Koh
P1 = (0.5)(13.0 kN/m3)(11.6 m)
P1 = 75.4 kN/m2.
Now the F/L can be determined by an analysis of a distributed load:
F/L = (1/2)(Koh)h = (1/2)Koh2 = (1/2)P1h
F/L = (1/2)(75.4 kN/m2 )(11.6 m)
F/Lsoil = 437.3 kN/m.
Now we need to apply the condition that Mcl = Mccl to determine the magnitude of the tension
in the tie-back cable or rod. Choosing a pivot point where (F/L)soil-base (bottom green vector in
figure 2B) is being applied, we get:
0 = Mcl - Mccl)
0 = (F/L)soil-base(d) + (F/L)tie(d) - (F/L)soil(d)
0 = (F/L)soil-base(0 m) + (F/L)tie(10.3 m) - (437.3 kN/m)(3.87 m - 0.3 m),
solving this equation we find (F/L)tie = 151.7 kN/m. Thus, if the tie-back rods are placed every
meter in length (in and out of the page), the tension on one rod is F=(F/L)tie(1m)=151.7 kN. If
the rods are placed every 3 meters, then F= (F/L)tie(3m)=455.1 kN. The first condition of
equilibrium can now be applied to solve for (F/L)soil-base . Such that,
0 = Fx
0 = (F/L)soil-base + (F/L)tie - (F/L)soil
0 = (F/L)soil-base + 151.7 kN/m - 437.3 kN/m
and
(F/L)soil-base = 285.6 kN/m.
This assumes the soil is strong enough to provide such a force for this given geometry. If it is
not strong enough the wall will fail due to the soil getting pushed out at the base.
Example #3, Tie-Back With a High Water Table
Tie-Back with a high water table
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