2 II II

where

_a_ = the area of steel per linear foot,
_x_{II}_ = the distance from the center of the steel to the outer
fiber, and
_S_{II}_ = the strength of the steel in tension.

Then for a beam, 12 in. wide,

2 _d_
_d_ _S_ = _a_ _S_ ( ----- - _x_ ) ,
II 2 II

or

2
_d_ _S_
_a_ = --------------------- .
_d_
_S_ ( ----- - _x_ )
II 2 II

Carrying this to its conclusion, we have, for example, in a beam 12 in.
deep and 12 in. wide,

_S_ = 500,
_S_{II}_ = 15,000,
_x_{II}_ = 2-1/2 in.
_a_ = 1.37 sq. in. per ft.

The writer has used this formula very extensively, in calculating new
work and also in checking other designs built or to be built, and he
believes its results are absolutely safe. There is the further fact to
its credit, that its simplicity bars very largely the possibility of
error from its use. He sees no reason to introduce further complications
into such a formula, when actual tests will show results varying more
widely than is shown by a comparison between this simple formula and
many more complicated ones.

GEORGE H. MYERS, JUN. AM. SOC. C. E. (by letter).--This paper brings out
a number of interesting points, but that which strikes the writer most
forcibly is the tenth, in regard to elaborate theories and complicated
formulas for beams and slabs. The author's stand for simplicity in this
regard is well taken. A formula for the design of beams and slabs need
not be long or complicated in any respect. It can easily be obtained
from the well-known fact that the moment at any point divided by the
distance between the center of compression and the center of tension at
that point gives the tension (or compression) in the beam.

The writer would place the neutral axis from 0.42 to 0.45 of the
effective depth of the beam from the compression side rather than at the
center, as Mr. Godfrey suggests. This higher position of the neutral
axis is the one more generally shown by tests of beams. It gives the
formula _M_ = 0.86 _d_ _As_ _f_, or _M_ = 0.85 _d_ _As_ _f_, which the
writer believes is more accurate than _M_ = 5/6 _d_ _As_ _f_, or
0.83-1/3 _d_ _As_ _f_, which would result if the neutral axis were taken
at the center of the beam.

_d_ = depth of the beam from the compression side to the center