Question

A parking garage is designed for two levels of cars. To make more money, the owner decides to double the size of the garage in each dimension (length, width, and the number of levels). For the support columns to hold up four floors instead of two, how should he change the columns' diameter?

(a) Double the area of the columns by increasing their diameters by a factor of 2

(b) Double the area of the columns by increasing their diameters by a factor of \(\sqrt 2 \)

(c) Quadruple the area of the columns by increasing their diameters by a factor of 2

(d) Increase the area of the columns by a factor of 8 by increasing their diameters by a factor of \(2\sqrt 2 \)

(e) He doesn't need to increase the diameter of the columns

Short answer

The correct option is (d).

Step-by-step solution

Concepts

The building is made so that the columns can take the maximum stress due to the weight of the building. For this problem, the stress is the same for both situations.

Explanation

Let A be the initial area of the columns, W be the weight of the garage, and d be the initial diameter of the column.

The initial stress on the column is \({\rm{stress}} = \frac{W}{A}\).

Let \(A'\) be the area of the columns and \(d'\) be the columns' diameter after changing the garage's dimension.

Now, when the length, width, and number of levels are doubled, the volume of the garage becomes eight times the initial volume of the garage.

Then, there will be eight times more cars, i.e., the weight of the garage is now eight times more (8W).

To maintain the balance of the force, the stress on the columns should remain unchanged. Therefore, the stress on the columns is

\(\begin{aligned}{c}\frac{W}{A} = \frac{{8W}}{{A'}}\\A' = 8A\\\pi \frac{{{{\left( {d'} \right)}^2}}}{4} = 8 \times \pi \frac{{{{\left( d \right)}^2}}}{4}\\d' = 2\sqrt 2 d\end{aligned}\).

Hence, the area of the columns should be doubled, and the diameter should be increased by a factor \(2\sqrt 2 \).