80 E L²

[TeX: $\frac{WL^2}{80 E} = {bd^3}$, whence $W = \frac{{bd^3} x 80 E}{L^2}$]

now, having the weight given, and assuming the dimensions of
the cross-section--we shall have

-----
/ WL² WL²
d = ³/ -----, and b = ------
\/ 80 Eb 80 Ed³

[TeX: $d = \sqrt[3]{\frac{WL^2}{80 EB}}$, and $b = \frac{WL^2}{80 ED^3}$]

in the above formulæ,

W = weight in pounds.
L = length in feet.
E = a constant.
b = breadth in inches.
d = depth in inches.


=Transverse Strains.= The strain caused by any weight, applied
transversely, to a beam supported at both ends, is directly as the
breadth, and square of the depth, and inversely as the length. It
causes the beam to be depressed towards the middle of its length,
forming a curve, concave to the horizontal and below it. In assuming
this form--the fibres of the upper part of the beam are compressed,
and those of the lower part are extended--consequently there must be