By the end of this chapter, you will be able to:
Mastering these skills will help you confidently analyze and design retaining walls that are safe and reliable in real-world construction projects.
Compiled by: John Wachira
Determining forces on retaining walls is essential for ensuring their structural stability and safety in building construction. Retaining walls resist lateral earth pressures and other loads, and understanding the distribution of these forces requires knowledge of the wall’s center of gravity and stability conditions. This chapter provides mathematical methods to locate the center of gravity for various retaining wall sections, a critical step in analyzing overturning, sliding, and bearing capacity. Accurate calculations of these parameters help building technologists design safe retaining structures in Kenyan construction projects.
Retaining walls must be stable against overturning, sliding, and excessive bearing pressure to prevent failure. The center of gravity (CG) of the wall and the resultant force from soil pressures are fundamental in assessing these stability criteria. This section covers how to calculate the CG for uniform and non-uniform wall sections and use it to evaluate stability.
The center of gravity is the point where the entire weight of the retaining wall can be considered to act. It is crucial for determining the moment arms for overturning analysis.
The formula for the coordinate of the CG, \( \bar{x} \), along a horizontal axis is:
$$ \bar{x} = \frac{\sum (A_i x_i)}{\sum A_i} $$
where \(A_i\) is the area of part \(i\), and \(x_i\) is the distance of the centroid of part \(i\) from the reference axis.
Example 1: Calculate the center of gravity (CG) of a rectangular retaining wall section 2.5 m wide and 4 m high.
Given: \(b = 2.5\, m\), \(h = 4\, m\)
$$ \bar{x} = \frac{2.5}{2} = 1.25\, m $$$$ \bar{y} = \frac{4}{2} = 2\, m $$ Answer: CG is at 1.25 m from one side horizontally and 2 m from base vertically
Example 2: For a composite wall consisting of two rectangles: base 3 m wide by 0.5 m thick (\(A_1 = 1.5\, m^2\), centroid \(y_1 = 0.25\, m\)), and stem 0.5 m wide by 3 m high (\(A_2 = 1.5\, m^2\), centroid \(y_2 = 0.5 + 1.5 = 2\, m\)). Find the vertical CG from base.
Total area:$$ A = 1.5 + 1.5 = 3\, m^2 $$ Vertical CG:$$ \bar{y} = \frac{(1.5 \times 0.25) + (1.5 \times 2)}{3} = \frac{0.375 + 3}{3} = \frac{3.375}{3} = 1.125\, m $$ Answer: CG is 1.125 m above the base
Stability involves ensuring that the resultant force from all acting loads passes through a safe zone within the base to prevent overturning or sliding.
The resultant force acting on the retaining wall combines the wall’s weight and the lateral earth pressure.
The shape and dimensions of the retaining wall affect the CG location and stability.
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Create a free accountThis chapter covers the fundamental principles of determining forces on retaining walls, beginning with the concept of center of gravity and its role in ensuring wall stability. It explains how to calculate the center of gravity for both uniform and non-uniform wall sections, which is crucial for assessing stability and the risk of overturning. Various types of pressures acting on retaining walls are discussed, including active, passive, and at-rest earth pressures, along with methods for calculating hydrostatic and earth pressures. The chapter then explores lateral thrust and stability analysis, presenting different methods for calculating lateral forces with a focus on Rankine's and Coulomb's theories. Finally, it addresses stability checks and design considerations tailored to gravity and cantilever retaining walls, emphasizing the importance of proper analysis to prevent failure and ensure structural integrity.
A rectangular retaining wall section has a base width of 3 m and height of 4 m. Calculate the center of gravity from the base. (Easy) (2 marks)
Determine the center of gravity of an L-shaped retaining wall section composed of a vertical stem 0.5 m thick and 4 m high, and a base slab 3 m wide and 0.5 m thick. (Medium) (3 marks)
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