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Bowles on Strip Loads
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Transcript of Bowles on Strip Loads
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Figure 11-16 Identification of terms used in the Boussin esq equation [Eq. (11-20)] for lateral pressu re.
Equation (11-20) can also be written as
P \3rh 1 - 2 / x 1
ar
=
2^ [ ^ - - R (RT7) \
( U
-
2
a )
where the several terms including 0, z, r, andR are identified on Fig. 11-16. This form of the
equation is particularly suitable for prog ram min g on a small calculator, since the point load
P
is usually fixed with given
x, y
coordinates and we want to vary
z
to obtain the wall pressure
profile.
The computer programming of this equation allows one to solve any of the given backfill
surcharge loads of Fig. 11-17 defined as follows:
1. Point load. Use the equation in the given form.
2.
Line load. Treat as either one load or a series of concentrated loads along a line of uni
width acting on unit areas.
3 . Strip load. Treat as a series of parallel line loads acting on strips of some un it wid th.
4 . Loaded area. Treat as a series of parallel line loads acting on strips of finite leng th.
We can easily analyze a constant uniformly loaded area (say, the interior part of an em-
bankment) or one with a linear varying load (say, the embankment side slopes). In either of
these cases the loaded area is divided into strips with some load intensity
q
and some small
unit width B, on the order of 0.25 to 0.5 m. Th ese strips are then subd ivided into unit
areas of some lengthL also on the order of 0.25 to 0.50 m. The se un it areas are treated a
a series of point loads of
Q qBL
acting at the center of each of the unit areas. The severa
unit area contributions mak ing up the total loaded area are then sum m ed to obtain the tota
lateral pressure acting at some point at the depth of interest (either in the soil or on a wall)
This is the procedure used in program SM BL Pl (B-8) on your program diskette.
The general validity of using a form of Eq. (11-20) for surcharges was established in sev-
eral publications, including the work of Spangler (193 6), Spangler and Mickle (1956), Re hn-
man and Broms (1972), and others.
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Figure 11-17
Surcharge loads that can be used with the computer program SMBLPl (B-8) on your program
diskette. NSQW, NSQL = number of unit elements away from wall and parallel to wall, respectively, as used in
the computer program.
The early work of Spangler and Spangler and Mickle introduced an error into the general
application of the equations; however, that can be avoided by direct use of Eq. (11-20) and
an appropriate value for Poisson's ratio
/ J L .
When the work of Spangler was first published, he used /J L = 0.5 [and later in Span-
gler and Mickle (1956)], which substantially simplifies Eq. (11-20)but may not be correct
Spangler's work consisted of trying to measure the lateral pressure against a 1.829 m (2.134
m total height) high X 4.572 m wide retaining wall with a constant stem thickness of 0.150
m. He used metal ribbons (since earth-pressure cells were not readily available in the early
1930s) and simply dumped a granular
(WL =
17.5,
Wp =
13.2%) backfill behind the wall
with no compaction at all to produce an extremely loose state. After a time, he had a truck
backed onto the loose backfill so that the rear wheels could simulate two concentrated loads.
To simulate a line load he laid a railroad crosstie parallel to the wall, onto which the rear
wheels (a single axle with dual tires) of a loaded truck were backed. Since the wall was only
4.572 m long and a railroad crosstie is about 3 m long, a strip model was not very likely to
have been produced.
From these efforts Sp angler (both references) found that the meas ured lateral pressure w as
about twice that predicted by Eq. (11-20) with /J L
=
0.5. From the reported results, Mindlin
(1936a), in discussing Spangler's (1936) work suggested that the factor of 2 could be ex-
plained by a rigid wall producing the effect of a mirror load placed sym me trically in front of
the wall (Fig. 11-ISa) . The author began looking at this problem more closely and decided
that the mirror load is not an explanation. As shown in Fig. 11-18&, a mirror load on a rigid
a) Concentrated load
NSQW = NSQL - 1
b)
Line load
NSQW = 1
NSQL = 1 (or as req'd)
(c) Strip load
NSQW =
sufficient for width
NSQL = l (or as req'd)
d) Linear varying load.
e)
Linear varying load.
/ )
Foot ing
load oflimited
BxL.
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(a) Mirrored loads on rigid wall.
Figure 11-18 Th e case for lateral pressu re on a rigid wall.
wall would simply cancel the lateral shear stresses in the wall and certainly not double the
horizontal pressure. A flexible wall could possibly double the lateral pressure but would have
to be extrem ely flexible (and hav e the loads a pplied sequ entially). Re ferring to Fig. 11-18Z?,
we see that the horizontal pressure ah produced by P
0
(applied first) acting on a rigid area of
dydz
would develop shear resistance
r
such that
rt(2dz+2dy) = a^
W hen mirror load
P
0
is
applied, a second shea r stress r w ould deve lop on the elem ent but in the opposite direction, so
the shear stress would cancel and we would simply have on each side of the wall a horizontal
stress ah (not2ah). If the wall is flexible and little shear stress develops, the element would
(if the loads were applied sequentially) displace laterally toward the
-a
direction to prod uce
a resisting soil stress on the
a
side of
a
h
= ah r(2dz
+
2dy).
Since this would become
locked in when P
o
is applied, the stress on the left would become ah + a
h
, and since a
h
is
transmitted through the wall, the right side would also have the existing cr^+ transmitted a
h
value from load
P
0
.
If the loads were applied simultaneously the stresses would simply be
ah
on each side (and not
2ah).
Mindlin got around this complication by inserting a statement that the wall was rigid but
could not carry shear. There is no such wall type known to the author.
Because the Spangler work was done in the early 1930s, it is difficult to speculate on
the cause of the high stresses except to note that the wall had rather finite dimensions. The
surcharge load w as caused b y a truck backing onto the backfill. Wh en it stopped at the desired
position it would have produced an inertial force that was amplified because the fill was not
well compacted. The backfill probably was of limited extent, so that it is also possible some
type of arching occurred that increased the lateral pressure.
More recently Rehnman and Broms (1972) showed (using modern earth-pressure cells)
that when the soil behind the wall was dense the lateral pressure from point loads was much
less than when the soil was loose. They also found that gravelly backfill produced larger
lateral pressures than finer-grained materials. This observation indicates that both soil state
and Poisson's ratio are significant parameters.
Rigid wall
(b) Horizontal stresso
h
on differential
aready dz.
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3 . Con centrated loads well outside the Ra nkin e zone contribute toP
a
in the trial wedge case
leading one to the opinion that the trial wedge is not correct, but conservative.
4 . Since Eq . (11-20) gives small lateral pressures w hen the load is very close to the wa ll, thi
result may mean either that the surcharge load is being carried downward by vertical wall
friction rather than by lateral pressure or that Eq. (11-20) is not valid for a load close to
such a massive discontinuity in the elastic half-space.
5 . Wall pressure s comp uted by Eq. (11-20) are rather small once the wall-to-load d istance i
greater than the Rank ine zone.
Computer Program for Lateral Pressure
Computer program SMBLPl (B-8) on your program diskette can be used for all the lateral
pressure problems shown on Fig. 11-17. By superposition, almost any conceivable surcharge
load can be analyzed, quite rapidly in most cases.
From above table at 1.3 m from wall:
Trial wedge = 65 kN
Boussine sq JR+ Coulomb P
a
gives
Coulomb:
Total
Inside
Outside
TABLE 11-7
Com parison of trial wedge and Bou ssinesq wall forces compu ted using Eq. 11-20).
Also shown is the Coulomb active pressure force.
2.217 m
Load position
from wall, m
Trial wedge, kN
0 0.3
65.0
0.6
65.0
1.0
65.0
1.3
65.0
1.6
64.8
2.0
63.4
2.3
61.6
2.6
59.3
3.0 m
55.5
CoulombP
a
= 17.30) 4)
2
0.276) = 38.2 kN vs. 38.1 of trial wedge)
Boussinesq
fi =
0.3
0.5
0.7
1.0
8.8
13.1
18.0
25.3
8.3
11.5
14.8
19.9
5.3
7.5
9.7
13.0
Inside
4.1
5.8
7.5
10.0
3.3
4.6
6.0
8.1
2.5
3.6
4.6
6.2
Outside
2.1
3.0
3.9
5.2
1.8
2.5
3.3
4.5
1.4
2.1
2.7
3.7