Wing walls are essentially retaining walls adjacent to the abutment. The walls can be independent or integral with the abutment wall.
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 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:
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
q = angle of wall to road under bridge
f = skew angle ( -ve if < 90°)
Lw = L1 + 1/x [(L+K)Sinq]
Lw = L2 - [
(L+K) Cosq +
(L+K) SinqTanf - V/Cos f ] Cos f / Y
L + K = [ X Y (L2 - L1) + V X ] / [ X Cos (q - f ) + Y Sinq] ..................eqn.(1)
For minimum length of wall dL/dq = 0
ie. Tanq = Tanf + Y / X Cos f
For known lengths of wall (L+K) two values of q can be obtained from eqn.(1).
From eqn.(1)   -(A+B)Tan2(q/2) + 2(C+Y)Tan(q/2) + (B-A) = 0
 ==================================
Where A = [XY(L2 - L1) + VX] / [L + K] , B = X Cosf , and C = X Sinf
Example
 
Wall
 
 
 
L1
 
 
 
L2
 
 
 
V
 
 
 
X
 
 
 
Y
 
 
 
f
 
 
Level
at
top of
wall
 
(L+K)
 
 
Range
of
q
 
 
q
 
 
 
Lw
 
 
 
Kmax
 
 
 
N/E
 
 
56.63
 
 
63.56
 
 
2.2
 
 
2.0
 
 
2.0
 
 
10.5
 
 
63.01
 
 
10.97
 
68.08
to
32.42
 
32.5
 
 
59.57
 
 
4.08
 
S/E
56.91
64.21
2.0
2.0
2.0
-27.0
63.55
15.89
31.5
31.5
61.05
10.84
 
S/W
 
 
57.21
 
 
64.42
 
 
0.8
 
 
2.59
 
 
2.0
 
 
27.7
 
 
63.76
 
 
12.19
 
89.30
to
18.50
 
30.0
 
 
59.57
 
 
3.80
 
Minimum Lengths :-
N/E Wall : L+K = 10.437 @ q = 50.25°
S/E Wall : L+K = 15.889 @ q = 31.5°
N/E Wall : L+K = 9.996 @ q = 54.37°