If the soil exerts a trapezoidal distribution of load on the bottom of the footing, determine the intensities $w_1$ and $w_2$ of this distribution needed to support the column load... If the soil exerts a trapezoidal distribution of load on the bottom of the footing, determine the intensities $w_1$ and $w_2$ of this distribution needed to support the column loadings.
Understand the Problem
The question asks to determine the intensities $w_1$ and $w_2$ of the trapezoidal distribution of load exerted by the soil on the bottom of the footing, given the column loadings. This problem involves principles of statics and soil mechanics to ensure equilibrium.
Answer
$w_1 = 30.3 \text{ kN/m}$ $w_2 = 17.2 \text{ kN/m}$
Answer for screen readers
$w_1 = 30.3 \text{ kN/m}$ $w_2 = 17.2 \text{ kN/m}$
Steps to Solve
- Calculate the total vertical load
The total vertical load is the sum of the three column loads: $V = 60 \text{ kN} + 80 \text{ kN} + 50 \text{ kN} = 190 \text{ kN}$
- Calculate the total length of the footing
The total length of the footing is: $L = 1 \text{ m} + 2.5 \text{ m} + 3.5 \text{ m} + 1 \text{ m} = 8 \text{ m}$
- Express the area under the trapezoidal load distribution
The area under the trapezoidal load distribution represents the total vertical force exerted by the soil, which must equal to the total column loads. The area of a trapezoid is given by: $A = \frac{1}{2} (w_1 + w_2) L $ So, $\frac{1}{2} (w_1 + w_2) \times 8 = 190$ $4(w_1 + w_2) = 190$ $w_1 + w_2 = \frac{190}{4} = 47.5$ This gives us the first equation: $w_1 + w_2 = 47.5 \hspace{1cm} (1)$
- Calculate the location of the centroid of the applied loads
Taking the moment about the left edge of the footing: $\sum M = 60 \text{ kN} \times 1 \text{ m} + 80 \text{ kN} \times (1 \text{ m} + 2.5 \text{ m}) + 50 \text{ kN} \times (1 \text{ m} + 2.5 \text{ m} + 3.5 \text{ m})$ $\sum M = 60 + 80 \times 3.5 + 50 \times 7 = 60 + 280 + 350 = 690 \text{ kN.m}$
The location of the centroid is: $\bar{x} = \frac{\sum M}{V} = \frac{690}{190} = \frac{69}{19} \approx 3.632 \text{ m}$
- Express the location of centroid of the trapezoidal load distribution
The location of the centroid of a trapezoid is given by: $\bar{x} = \frac{L}{3} \times \frac{2w_2 + w_1}{w_2 + w_1}$ So, $3.632 = \frac{8}{3} \times \frac{2w_2 + w_1}{w_2 + w_1}$ $3.632 \times 3 = 8 \times \frac{2w_2 + w_1}{w_2 + w_1}$ $10.896 = 8 \times \frac{2w_2 + w_1}{w_2 + w_1}$ $\frac{10.896}{8} = \frac{2w_2 + w_1}{w_2 + w_1}$ $1.362 = \frac{2w_2 + w_1}{w_2 + w_1}$ $1.362(w_2 + w_1) = 2w_2 + w_1$ $1.362w_2 + 1.362w_1 = 2w_2 + w_1$ $0.362w_1 = 0.638w_2$ $w_1 = \frac{0.638}{0.362} w_2$ $w_1 = 1.762 w_2 \hspace{1cm} (2)$
- Solve the system of equations
Substitute (2) in (1): $1.762w_2 + w_2 = 47.5$ $2.762w_2 = 47.5$ $w_2 = \frac{47.5}{2.762} = 17.2 \text{ kN/m}$
Substitute $w_2$ in (1): $w_1 + 17.2 = 47.5$ $w_1 = 47.5 - 17.2 = 30.3 \text{ kN/m}$
$w_1 = 30.3 \text{ kN/m}$ $w_2 = 17.2 \text{ kN/m}$
More Information
The intensities $w_1$ and $w_2$ represent the soil's reaction pressure at the edges of the footing. The distribution is assumed to be trapezoidal, which is a common simplification in soil mechanics.
Tips
A common mistake is not correctly calculating the centroid location, either for the applied loads or for the trapezoidal distribution itself. Also, forgetting to convert units or making algebraic errors when solving the system of equations are potential pitfalls.
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