You are here: Home » Bridge Design » Design Notes » Wing Walls
Wing walls are essentially retaining walls adjacent to the abutment. The walls can be independent or integral with the abutment wall.
Providing the bridge skew angle is small (less than 20°), and the cutting/embankment slopes are reasonably steep (about 1 in 2), then the wing wall cantilevering from the abutment wall is likely to give the most economical solution.
Splayed wing walls can provide even
more of an economy in material
costs but the detailing and fixing
of the steel reinforcement is more
complicated than the conventional wall.
Design Considerations
Loads effects to be considered on the rear of the wall are:
- Earth pressures from the backfill material.
- Surcharge from live loading or compacting plant.
- Hydraulic loads from saturated soil conditions.
The stability of the wall is generally designed to resist 'active' earth pressures (Ka); whilst the structural elements are designed to resist 'at rest' earth pressures (Ko). The concept is that 'at rest' pressures are developed initially and the structural elements should be designed to accommodate these loads without failure. The loads will however reduce to 'active' pressure when the wall moves, either by rotating or sliding. Consequently the wall will stabilise if it moves under 'at rest' pressures providing it is designed to resist 'active' earth pressures.
Geometry for splayed wing walls
Plan on Wing Wall
X = slope to road under bridge
Y = slope from road over bridge
L = length of sloping wall
K = length of horizontal wall
V = verge width to end of wall
L1 = level at bottom of embankment
L2 = level at back of verge on road over bridge
Lw = ground level at end of wall
θ = angle of wall to road under bridge
φ = skew angle ( -ve if < 90°)
Lw = L1 + 1/x [(L+K)Sinθ]
Lw = L2 - [(L+K) Cosθ +
(L+K) SinθTanφ - V/Cosφ ] Cosφ / Y
L + K = [ X Y (L2 - L1) + V X ] / [ X Cos (θ - φ ) + Y Sinθ] ..................eqn.(1)
For minimum length of wall dL/dθ = 0
ie. Tanθ = Tanφ + Y / X Cosφ
For known lengths of wall (L+K) two values of θ can be obtained from eqn.(1).
From eqn.(1) -(A+B)Tan2(θ/2) + 2(C+Y)Tan(θ/2) + (B-A) = 0
==================================
Where A = [XY(L2 - L1) + VX] / [L + K] , B = X Cosφ , and C = X Sinφ
Example
Wall
L1
L2
V
X
Y
φ
Level
at
top of
wall
(L+K)
Max
&
Min
θ
θ
Lw
Kmax
N/E
56.6
63.6
2.2
2.0
2.0
10.5
63.0
11.0
68.1
to
32.4
32.5
59.6
4.1
S/E
56.9
64.2
2.0
2.0
2.0
-27.0
63.6
15.9
31.5
31.5
61.1
10.8
S/W
57.2
64.4
0.8
2.6
2.0
27.7
63.8
12.2
89.3
to
18.5
30.0
59.6
3.8
Minimum Lengths :-
N/E Wall : L+K = 10.437 @ θ = 50.25°
S/E Wall : L+K = 15.889 @ θ = 31.5°
S/W Wall : L+K = 9.996 @ θ = 54.37°
David Childs B.Sc, C.Eng, MICE