Pile loads due to shear wall loading Given the wall and pile cap configuration in Figure 10.37, determine the loading on piles due to the shear wall loads. Note that the shear wall is asymmetrical to the pile cap and the piles are not evenly spaced (Table 10.1).

Question

Pile loads due to shear wall loading
Given the wall and pile cap configuration in Figure 10.37, determine the loading on piles due to the shear wall loads. Note that the shear wall is asymmetrical to the pile cap and the piles are not evenly spaced (Table 10.1).

Table 10.1 Location of center of gravity of pile group
Pile	Area (ft²)	[latex]X_i[/latex] (ft)	[latex]A_iX_i[/latex] (ft³)
1,2	2 (0.785) = 1.57	3	4.71
3,4	1.57	8	12.56
5,6	1.57	14	21.98
7,8	1.57	19	29.83
9,10	1.57	23	36.11
	7.85		105.19

Accepted Answer

The solution type of problem can be completed in three steps:
Step 1. Find the center of gravity of the pile group
Step 2. Determine the design moment about the center of gravity
Step 3. Determine the load on each of the piles using
[latex]Q=\frac{P}{n}\pm\frac{\sum m d_i}{\sum d_i^2}[/latex]
Step 1.
Locating the center of gravity is best done in a tabulated form as follows:
Area of pile [latex]=\frac{\pi(1^{ft})^2}{4}=0.785[/latex] ft²
[latex]\bar{x}=\frac{\sum A_i x_i}{\sum A_i}=\frac{105.19}{7.85}=13.4[/latex] ft
Hence, the center of gravity is 13.4 ft from the left most edge of the pile cap.
Step 2.
Set up a free-body diagram to depict the loading on the pile cap about the center of gravity and sum moments (Figure 10.38).
A clockwise moment about the c.g. is shown here.
[latex]+ ⤸ M_{c,g} =100^k (19^{ft})-12^{k/ft}(15)\left[13.4-11.5\right]=1558^{k\cdot ft}[/latex]
The equivalent point load (R) of the uniform load is [latex]12^{k/ft}(15^\prime) = 180^k[/latex]. This load is located at the center of the wall and to the left of the c.g.
Step 3.
Quantifying the loads on each pile is done in tabulated form using the familiar equation as shown (Table 10.2).

Table 10.2 Loads on each pile
Pile	[latex]d_i[/latex] (ft)	[latex]d_i^2[/latex] (ft²)	[latex]\frac{P}{n}[/latex] (k)	[latex]\frac{\sum m d_i}{\sum d_i^2}[/latex]	[latex]P=\frac{P}{n}\pm\frac{\sum m d_i}{\sum d_i^2}[/latex]
1,2	13.4 − 3 = 10.4	2(108.16)	[latex]\frac{180}{10}=[/latex] 18	[latex]\frac{1558(10.4)}{522.4}=31.02[/latex]	= 18 − 31.02 = −13.02 uplift
3,4	13.4 − 8 = 5.4	2(29.16)	18	16.1	= 18 −16.1 = 1.9 uplift
5,6	14 − 13.4 = 0.6	2(0.36)	18	1.79	= 18 + 1.79 = 19.79 uplift
7,8	19 − 13.4 = 5.6	2(31.36)	18	16.7	= 18 + 16.7 = 34.7
9,10	23 − 13.4 = 9.6	2(92.16)	18	28.63	= 18 + 28.63 = 46.63
		522.4

[latex]Q=\frac{P}{n}\pm\frac{\sum m d_i}{\sum d_i^2}[/latex]
In summary, the minimum and maximum pile loads are at piles 1 and 2, which experience an uplift force of 13.02 kips, and piles 9 and 10, which experience a downward force of 46.63 kips. It is important to note that the weight of the footing was not accounted for in this problem and must be ultimately incorporated for a properly designed system.

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