Combined mat footing design for two columns A combined footing design is required for the column loads shown in Figure 10.28. Columns C1 and C2 are 18 ft apart and column C1 is 2 ft from the edge of the footing. Both columns are 18 in × 18 in. The thickness of the footing is assumed to be 24 in.

Question

Combined mat footing design for two columns
A combined footing design is required for the column loads shown in Figure 10.28. Columns C1 and C2 are 18 ft apart and column C1 is 2 ft from the edge of the footing. Both columns are 18 in × 18 in. The thickness of the footing is assumed to be 24 in.

Accepted Answer

Step 1: Locate centroid of loads by creating a free-body diagram and balance forces with the resultant (Figure 10.29):
[latex]P_{C1} = 80 + 60 = 140^k\P_{C2} = 160 + 120 = 280^k\R = ΣP_i = P_{C1} + P_{C2} = 420^k\R\bar{x} =∑P_i x_i = P_{C1}x_1+P_{C2}x_2[/latex]
Solve for the centroid
[latex]\bar{x}=\frac{P_{C_1}x_1+P_{C_2}x_2}{R}[/latex]
= [latex]\frac{140^(2^\prime)+280^k(20^\prime)}{420^k}[/latex]
= 14′
Step 2: Size footing such that the footing is centered about [latex]\bar{x}[/latex]:
Length of footing (L)[latex] = 2\bar{x} =[/latex] 2(14′) = 28′
Step 3: Determine the required width, B, of the footing:
The effective bearing capacity, [latex]q_e[/latex], is determined by subtracting the weight of the footing (Figure 10.30).
[latex]q_e = q_a − hw_c[/latex] = 3000 lb/ft² − 2′(145 lb/ft³) = 2710 lb/ft² = 2.7 ksf
Setting the effective bearing capacity equal to the total load, [latex]P_{C1}+P_{C2}[/latex], over the area of the footing, BL,
[latex]q_e=\frac{P_{C_1}+P_{C_2}}{BL}[/latex], rearrange terms, and solve for B,
[latex]B=\frac{P_{C_1}+P_{C_2}}{q_eL}=\frac{140^k+280^k}{2.7(k/ft^2)(28 ft)}=[/latex] 5.56 ft
Round up, the dimension, to the next foot and use 6′−0″ for the width of the footing.
We can now create the design shear and moment diagram for the combined footing as follows.
The factored loads are as follows:
[latex]P_{uc_1}=1.2(80)+1.6(60)=192[/latex] k
[latex]P_{uc_2}=1.2(160)+1.6(120)=384[/latex] k
and the uniform bearing capacity, [latex]w_u[/latex], is calculated as
[latex]w_u=\frac{P_{uc_1}+P_{uc_2}}{L}=\frac{192 k+384 k}{28 ft}=20.57\frac{k}{ft}[/latex]
Using equations of statics, the shear and moment diagrams can be created as shown in Figure 10.31. For simplicity of design, the moments are taken, from the left, at 2′, 9.33′, and 20′, which are the centerlines of the columns and the point of zero shear between the columns.
[latex]+ ⤸ M_{2^\prime}= 20.57(2.0)(1)=41.14^{k\cdot ft}\+ ⤸ M_{9.33^\prime}=20.57(9.33)\frac{9.33}{2}−192(7.33)=− 512.1^{k\cdot ft}\+ ⤸ M_{20^\prime}=20.57(20^\prime)\frac{20}{2}−192(18)=664.58^{k\cdot ft}[/latex]

Question 10.6: Combined mat footing design for two columns A combined footi...

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