RETAINING WALL STABILITY CALCULATION SAMPLE BASICS AND TUTORIALS

RETAINING WALL DESIGN EXAMPLE - STABILITY TUTORIALS
How To Design Retaining Wall


EXAMPLE

Determine the factor of safety (FS) against sliding and overturning of the concrete retaining wall in Figure. The concrete weighs 150 lb/ft3 (23.56 kN/m3, the earth weighs 100 lb/ft3 (15.71 kN/m3), the coefficient of friction is 0.6, and the coefficient of active earth pressure is 0.333.

Calculation Procedure:



1. Compute the vertical loads on the wall

Select a 1-ft (304.8-mm) length of wall as typical of the entire structure. The horizontal pressure of the confined soil varies linearly with the depth and is represented by the triangle BGF in Fig. 10

Resolve the wall into the elements AECD and AEB; pass the vertical plane BF through the soil. Calculate the vertical loads, and locate their resultants with respect to the toe C. Thus:

W1 = 15(I)(ISO) = 2250 Ib (10,008N);
W2 = 0.5(15)(5)(150) = 5625;
W3 =0.5(15XS)(IOO) = 3750

Then ^W = 11,625 Ib (51,708 N). Also, Jc1 = 0.5 ft; X2= 1 + 0333(5) = 2.67 ft (0.81 m); Jc3 = 1 + 0.667(5)-433 ft (1.32m).

2. Compute the resultant horizontal soil thrust

Compute the resultant horizontal thrust T Ib of the soil by  applying the coefficient of active earth pressure.

Determine the location of T. Thus BG = 0.333(1S)(IOO) = 500 Ib/lin ft (7295 N/m); T = 0.5(15)(500) = 3750 Ib (16,680 N); y = 0.333(15) = 5 ft (1.5m).

3. Compute the maximum frictional force preventing sliding

The maximum frictional force Fm = where M = coefficient of friction. Or Fm = 0.6(11,625) - 6975 Ib (31,024.8N).

4. Determine the factor of safety against sliding 

The factor of safety against sliding is FSS = FJT = 6975/3750 = 1.86.

5. Compute the moment of the overturning and stabilizing forces

Taking moments with respect to C, we find the overturning moment = 3750(5)
= 18,750 lb-ft (25,406.3 N-m).

Likewise, the stabilizing moment = 2250(0.5) + 5625(2.67) + 3750(4.33) = 32,375 lb-ft (43,868.1 N-m).

6. Compute the factor of safety against overturning

The factor of safely against overturning is FSO = stabilizing moment, lb-ft (N-m)/overturning moment. lb-ft (N-m) = 32,375/18,750 = 1.73.

What is the function of shear keys in the design of retaining walls?

SHEAR KEYS ON RETAINING WALL TUTORIALS
Civil Engineering Tutorials


In determining the external stability of retaining walls, failure modes like bearing failure, sliding and overturning are normally considered in design. In considering the criterion of sliding, the sliding resistance of retaining walls is derived from the base friction between the wall base and the foundation soils.

To increase the sliding resistance of retaining walls, other than providing a large self-weight or a large retained soil mass, shear keys are to be installed at the wall base. The principle of shear keys is as follows:

The main purpose of installation of shear keys is to increase the extra passive resistance developed by the height of shear keys.

However, active pressure developed by shear keys also increases simultaneously. The success of shear keys lies in the fact that the increase of passive pressure exceeds the increase in active pressure, resulting in a net improvement of sliding resistance.

On the other hand, friction between the wall base and the foundation soils is normally about a fraction of the angle of internal resistance (i.e. about 0.8 ) where is the angle of internal friction of foundation soil. When a shear key is installed at the base of the retaining wall, the failure surface is changed from the wall base/soil horizontal plane to a plane within foundation soil.

Therefore, the friction angle mobilized in this case is instead of 0.8 in the previous case and the sliding resistance can be enhanced.