
Retaining Wall Design Problem 3  Check Sliding Resistance:

Given the proposed retaining wall from problem 1 below, using the lateral force calculated in problem 1, 14,248.737 lb/ft, check the resistance to sliding. The thickness of concrete wall and base is 36", and the width of the base is 12 feet. The density of concrete is 150 lb/ft^3.


1. From the retaining walls submenu open the resistance to sliding  no key calculation, RSlidingWedge.aspx. Input the "Title" of the calculation, the "Designer/Checker" information and the variables.
To find the shear reactant V which occurs along the structural wedge boundary AB, set the sum of the moments equal to zero around a point "C" on the base along the inside face of the retaining wall and the outside face of the structural wedge opposite from AB.
Mc = 0 = 14248.737(6)  V(8); solving for V = 10686.55 lb/ft. W = weight of soil, Ws and weight of concrete, Wc, of the concrete wall and base. Ws = 15(8)120 + 8(2)120/2 = 15,360 lb/ft; Wc = 150(3)(12) + 150(15)3 = 12,150 lb/ft. Therefore W = 15,360 lb/ft + 12,150 lb/ft = 27,510 lb/ft
The sum of the vertical resultants, Vsum = 10,686.55 lb/ft + 27,510 lb/ft = 38,196.55 lb/ft. Click on the "Calculate" button. Read the results. The resistance to sliding is 14,701.856 lb/ft > 14,248.737 lb/ft therefore the design is valid for sliding resistance.


2. Summary

The calculated resistance to sliding is 14,701.856 lb/ft with a Factor of Safety, FS = 1.5. The resistance to sliding of 14,701.856 lb/ft is greater than the lateral force on the wall of 14,248.737 lb/ft, therefore the proposed design is valid for sliding resistance.

