What are the Abutments? Know Its Types & Functions with suitable example

An abutment is a substructure that supports one end of a bridge’s superstructure while also supporting the embankment, which acts as the bridge’s approach, laterally. Additionally, the abutment shields the embankment from scouring the stream on a river bridge. Masonry, regular concrete, or reinforced concrete can be used to construct bridge abutments.

The three separate structural components listed below often make up an abutment:

1. The breast wall which directly supports the dead and live loads of the superstructure, and retains the filling of the embankment in its rear,
2. The wing walls, act as extensions of the treast wall in retaining the fill though not taking any loads from the superstructure.
3. The back wall (also known as dirt wall), which is a small retaining wall just behind the brid go scat preventing the flow of material from the fill on to the bridge seat.

In abutment design, the forces to be considered are:

1. Dead load due to superstructure.
2. Live load on the superstructure.
3. Self-weight of the abutment.
4. Longitudinal forces are due to tractive effort and braking, temperature variation, and concrete shrinkage.
5. Thrust on the abutment due to retained earth and the effect of live loads on the fill at the rear of the abutment. The latter effect is considered in design as an equivalent surcharge. The Bridge Code (Clause 714.4) requires all abutments to be designed for a live load surcharge of 1.2 m height of earth fill.

Of the above forces, earth pressure is the most difficult to compute correctly. The magnitude of earth pressure varies with the character of the material used for backfill and its moisture content. In abutment construction, it is important to place the fill material carefully and arrange for its proper drainage. A good drainage system may be secured by placing rock fill immediately behind the abutment.

The braking force is usually larger than the tractive effort and is taken as 0.2 of the weight of the design vehicle. The other longitudinal forces due to temperature variation and concrete shrinkage at the bearing level may be conservatively assumed as 10% of the dead load from the superstructure.

The design of an abutment is performed by assuming preliminary dimensions of the abutment section depending on the type of superstructure, substructure, and foundations, and checking for stability against overturning, base pressures, and sliding. The factor of safety against overturning should be greater than 2.0. Further, the eccentricity of the resultant of all forces on the abutment should lie within one-sixth of the base width so that there is no tension at the base. The maximum stress should be less than the safe bearing capacity of the soil. The safety factor against sliding should be more than 1.5. The required calculations are indicated in detail in the example.

For masonry abutments, it is usual to provide a batter of about 1 in 25 to 1 in 12 for the front face of the breast wall. The rear batter is adjusted to get the width required to resinc the net pressures within the prescribed limits. When reinforced concrete abutments are adopted, it would be permissible to have vertical faces both in front and at the rear faces of the breast wall. The toe and the heel portions of the base slab are so proportioned that the eccentricity of the resultant is limited to one-sixth of the base width.

Wing walls will normally have sections similar to those shown in Fig. 4. A wing wall can be cast monolithically with the abutment breast wall to form a single monolithic structure. It is often desirable to provide a construction joint between the abutment and wing walls when these are of stone masonry or mass concrete, especially if the levels of foundations are different. Wings can be splayed or made perpendicular to the breast wall depending on the site conditions.

Typical forms of reinforced concrete abutments are shown. The wing walls have been cantilevered without extending the base of breast wall for support, as would have been necessary for masonry abutments. The length of the cantilever return where adopted may be restricted to 4.0m. The slop of the bottom edge of the wing should be such as to have this edge below the level of the revetment of the embankment. A gravity type breast wall is used

A bridge abutment may fail in several ways as below, and the final design should be checked to avoid these failures. The breast wall may fail by tensile cracks, crushing or shear. The all say tilt forward due to excessive overturning moment due to earth pressure. The wall may dide forward due to earth pressure if the vertical forces are inadequate. Though the wall may le structurally strong, failures may occur along a curved surface by rupture of the soil due to inadequate shear resistance.

Example of Design of Abutment:

The angle of internal friction of soil on the wall, z = 17.5°
Approach slab: R.C. slab 300 mm thick, adequately reinforced.
Load from superstructure per running foot of abutment wall:
Dead load = 119 kN/m
Live load = 85 kN/m
(The above two values are to be obtained from the calculations for superstructure, and are taken to act over a width of 8.5 m).

Bearings: Neoprene pads of overall size 320 x 500 × 65 mm, embedding 5 plates of 3 mm thickness and 6 mm clearance in plan. G = 1 kN/mm*.

It is required to check the adequacy of the required section. The reinforcement details sre
not computed here.
(c) Self-weight of abutment: Treating the section as composed of 4 elements as shown in Fig. 6, the weight of each element and moment about the point O on the front toe are computed.
(d) Longitudinal forces:

1. Force due to braking:

Force due to 70R wheeled vehicle = 0.2 x 1000 = 200 kN
This force acts at 1.2 m above the road level (Clause 214.3).
Force on one abutment wall = 100 kN
Horizontal force per m of wall = 100/8.5 = 11.8 kN/m

2. Force due to temperature Variation and Shrinkage:

assuming moderate climate, variation in temperature is taken as (+/-) 17^o C as per Clause11.8.5 of Bridge Code.

Coefficient of thermal expansion = 11.7*10^6
Strain due to temperature variation= 17*11.7*10^{-6} = 1.989*10^{-4}
From Clause 220.3, strain due to concrete shrinkage = 2.0*10^{-4}
Total Strain due to temperature and shrinkage = (1.989+2.0)10^{-4} = 3.989*10^{-4}
Horizontal deformation of the deck due to temperature and shrinkage affecting one abutment = 3.989*10^{-4} * 17060/2 =3.44mm

Strain in bearing = \left(\frac{Deformation}{Elastomer \: thickness}\right)

\left(\frac{3.44}{65-(5*3)}\right) = 0.069

Assuming G = 1 N/mm^2

Horizontal force due to strain in longitudinal direction at bearing level (computed as per Clause 707.2.3)

=4.0KN/m

(iii) Vertical reaction due to bearing

Vertical reaction at one abutment = \left(\frac{200(1.2+1.6)}{16.10\times 8.5}\right) = 4.1KN/m

(d) Earth pressure:

Active earth pressure P=0.5wh^2.k_a
where K_a is obtained from Equation.
Here \theta =90^o , \Phi = 35^o , z= 17.5^o , \delta = 0^o
Substituting the value in the equation, we get Ka =0.496
Height of backfill below approach slab = 5.6m
Active earth pressure = 0.5\times 18\times 5.6^2 \times 0.496 = 140.0kN/m
Height above the base of the center of pressure = 0.42*5.6 =2.35m
Passive pressure in front of toe slab is neglected.

(e) Live load surcharge and approach slab:

The equivalent height of earth for live load surcharge as per Clause 714.4 is 1.20m
Horizontal force due to L.L surcharge = 1.2x18x0.496×5.6 =60.0kN/m
Horizontal force due to Approach slab = 0.3x24x0.496×5.6 =20.0kN/m
The above two forces act at 2.8m above the base
Vertical load due to L.L surcharge and approach Slab = (1.2×18+0.3×24)2.6 =74.9kN/m

(f) Weight of earth on heel slab:

Vertical load = 18(5.6-0.75)2.6 = 227kN/m

(g) Check for stability-overturning

The force and their position are shown in Fig.
The forces and moments about the point O at the toe on the base are tabulated. Two cases of loading conditions are examined

• Span loaded condition
• Span unloaded condition

(Case 1) Span loaded condition

Overturning Moment about toe = 623.1 kN.m
Restoring moment about toe = 1740.9 kN.m
The factor of Safety against overturning = 1740.9/623.1 = 28>2.0 Safe.
Location of resultant from O

Eccentricity of resultant

(Case 2) Span unloaded condition

Overturning moment about toe = 572.4 kN.m
Restoring moment about toe = 1607.2 kN.m
The factor of safety against overturning = 1607.2/572.4 = 2.8>2.0 safe.

e= 0.68m < 0.80 m

(h) Check for stresses at the base

For span-loaded condition
Total downward forces = 691.4 kN

Extreme Stresses at base = \frac{6914}{4.8\times 10}10 \pm (\frac{6\times0.78}{4.8})

= 284.5 of 3.6kN/m2

Maximum Pressure = 284.5kN/m2 < 350kN/m2 Permissible
Minimum Pressure = 3.6kN/m2 > 0 (No tension)

Hence Safe.

(i) Check for sliding

Sliding force = 235.8 kN

Force resisting sliding = 0.6 x 691.4 = 414.8kN

Factor of Safety against sliding = 414.8/235.8 = 1.76>1.5 Safe.

(j) Summary

They assumed section of the abutment is adequate

What are the Abutments? Know Its Types & Functions with suitable example