Sample Problem 8-4: Determining wall thickness of a box

The outside dimensions of a rectangular box (bottom and four sides, no top), made of aluminum, are 24 by 12 by 4 inches. The wall thickness of the bottom and the sides is x. Derive an expression that relates the weight of the box and the wall thickness x. Determine the thickness x for a box that weighs 15 lb. The specific weight of aluminum is 0.101 lb/in.³.

Solution

The volume of the aluminum VAl is calculated from the weight W of the box by:

where γ is the specific weight. The volume of the aluminum based on the dimensions of the box is given by

V_Al = 24 · 12 · 4 −(24 − 2x)(12 − 2x)(4 − x)

where the inside volume of the box is subtracted from the outside volume. This equation can be rewritten as

(24 – 2x)(12 – 2x)(4 – x) + V_Al −(24 · 12 · 4) = 0

which is a third-degree polynomial. A root of this polynomial is the required thickness x. A program in a script file that determines the polynomial and solves for the roots is:

Note in the second to last line that in order to add the quantity V_Al −(24 · 12 · 4) to the polynomial Vin it has to be written as a polynomial of the same order as Vin (Vin is a polynomial of third order). When the program (saved as Chap8SamPro4) is executed, the coefficients of the polynomial and the value of x are displayed: