The formula W = --------
4L

[TeX: $W = \frac{5000 bd^2}{4 L}$]

becomes, by substitution,

5000 x 8 x 196
W = -------------- = 11.666 lbs.
4 x 8

[TeX: $W = \frac{5000 \times 8 \times 196}{4 \times 168} = 11,666$ lbs.]

Case II. Given the safety load 18000 lbs.
the breadth 9 inches,
the length 14 feet.

Required the depth.
From the above formula we have

-------
/ W X 4L
d = / ------
\/ 5000 b


[TeX: $d = \sqrt{\frac{w \times 4L}{5000 b}}$]

substituting