4 x 1250 bd squared
W = ------------,
L

[TeX:  $W = \frac{4 \times 1250 bd^2}{L}$]

or

5000 bd squared
W = --------. 
L

[TeX:  $W = \frac{5000 bd^2}{L}$]

But, to reduce the load to the proper working strain, we must divide this equivalent by 4, the factor of safety, and we shall have

5000 bd squared
W = --------.
4L

[TeX:  $W = \frac{5000 bd^2}{4 L}$]

Let us apply the formula—­

Case I. Given a span of 14 feet,
a breadth of 8 inches,
a depth of 14 inches.

Required the safe load.

5000 bd squared
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

----------------
/ 18000 x 168 x 4      ------
d =  /  --------------- =   / 268.8 = 16, inches nearly.
\/      5000 x 9       \/

[TeX:  $d = \sqrt{\frac{1800 \times 168 \times 4}{5000 \times 9}} = \sqrt{268.8} = 16$]

Case III.  Given the safety load 22,400 lbs.
the depth 18 inches.
the length 14 feet.

Required the breadth. 
Deriving b from the foregoing, we have,

W x 4L
b = ----------
5000 x d squared

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

substituting

22400 x 4 x 168
b = --------------- = 9.3 inches nearly.
5000 x 324

[TeX:  $b = \frac{22400 \times 4 \times 168}{5000 \times 324} = 9.3$]

For a cast iron beam or girder—­Mr. Hodgkinson found from numerous carefully conducted experiments that, by arranging the material in the form of an inverted T—­thus creating a small top flange as well as the larger bottom one, the resistance was increased, per unit of section, over that of a rectangular beam, in the ratio of 40 to 23.

In this beam the areas of the top and bottom flanges are inversely proportional to the power of the iron to resist compression and extension.  Mr. Hodgkinson’s formula for the dimensions of his girder, is

26 ad
W = ------
L

[TeX:  $W = \frac{26 ad}{L}$]

The factor of safety being 6 for cast iron beams—­the formula for the working load will be,

26 ad
W = ------
6 L

[TeX:  $W = \frac{26 ad}{6 L}$]

and, to find area of lower flange, we shall have

      6 WL
  a = ——­
      26 d

[TeX:  $a = \frac{6 WL}{26 d}$]

The general proportions of his girders are as follows: