The end column along a property line is connected to an interior column by a trapezoidal footing. The following data are given with reference to Fig. 14.3(b):

Column Loads: $Q_{1}=2016 kN , Q_{2}=1560 kN$ .

Size of columns: 0.46 x 0.46 m.

$L_{c}=5.48$ m.

Determine the dimensions a and b of the trapezoidal footing. The net allowable bearing pressure $q_{n a}=190$ kPa.

Question

The end column along a property line is connected to an interior column by a trapezoidal footing. The following data are given with reference to Fig. 14.3(b):

Column Loads: [latex]Q_{1}=2016 kN , Q_{2}=1560 kN[/latex].

Size of columns: 0.46 x 0.46 m.

[latex]L_{c}=5.48[/latex] m.

Determine the dimensions a and b of the trapezoidal footing. The net allowable bearing pressure [latex]q_{n a}=190[/latex] kPa.

Accepted Answer

Determine the center of bearing pressure [latex]x_{2}[/latex] from the center of Column 1. Taking moments of all the loads about the center of Column 1, we have

[latex](2016+1560) x_{2}=1560 imes 5.48[/latex]

[latex]x_{2}=\frac{1560 imes 5.48}{3576}=2.39 m[/latex]

Now [latex]x_{1}=2.39+\frac{0.46}{2}=2.62 m[/latex]

Point O in Fig. 14.3(b) is the center of the area coinciding with the center of pressure. From the allowable pressure [latex]q_{a}=190[/latex] kPa, the area of the combined footing required is

[latex]A=\frac{3576}{190}=18.82 sq \cdot m[/latex]

From geometry, the area of the trapezoidal footing (Fig. 14.3(b)) is

[latex]A=\frac{(a+b) L}{2}=\frac{(a+b)}{2}(5.94)=18.82[/latex]

or [latex](a+b)=6.34 m[/latex]

where, [latex]L=L_{c}+b_{1}=5.48+0.46=5.94 m[/latex]

From the geometry of the Fig. (14.3b), the distance of the center of area [latex]x_{1}[/latex] can be written in terms of a, b and L as

[latex]x_{1}=\frac{L}{3} \frac{2 a+b}{a+b}[/latex]

or [latex]\frac{2 a+b}{a+b}=\frac{3 x_{1}}{L}=\frac{3 imes 2.62}{5.94}=1.32 m[/latex]

but [latex]a+b=6.32 m ext { or } b=6.32-a[/latex]. Now substituting for b we have,

[latex]\frac{2 a+6.34-a}{6.34}=1.32[/latex]

and solving, a = 2.03 m, from which, b = 6.34 - 2.03 = 4.31 m.

Question 14.5: The end column along a property line is connected to an inte...

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