Each brace is subject to compression, and is liable to fail if too small, in the same manner as the rafter. Its size is to be ascertained, therefore, in the manner described for the rafter; which need not be here repeated, except, perhaps, as to the liability to fail by flexure; for in this case Ave have the breadth given, and need to find the thickness. The breadth of the brace is fixed by the thickness of the rafter, for it is usual to have the two pieces flush with each other. Rule XI. (Art. 114) is to be used, but with this difference, namely: instead of the thickness, use the breadth as one of the factors in the divisor. Thus -

t = Wa(I+3/2 er2)/Cb. (105.)

In working this rule, it is required, in order to get the value of r, the ratio between the height and thickness, to assume the thickness before it is ascertained; and after computation, if the result shows that the assumed value was not a near approximation, a second trial will have to be made. Usually the first trial will be sufficient.

For example, the brace D E is about 9 2/3 feet or 116 inches long-. As the strain in it is only 5200 pounds, the thickness will probably be not over 3 inches. Assuming it at this, we have r = l/t = 116/3 = 38 2/3; the square of which is about 1495.

Therefore, we have -

3/2 x 0.00109 x 1495 = 2.4445

add unity = 1 3.4445 The equation reduces, therefore, to this -

t = 5200 x 10x3.4445/9500x6 = 3.1414;

or, the required thickness of the brace is 3 1/7 inches, or the brace should be, say, 3 1/4 x 6 inches. In this case the result is so near the assumed value, a second trial is not needed.

For the second brace, we have the length equal to about 12 1/4 feet or 147 inches; and the strain equal to 6575 pounds (Art. 225). The ratio, therefore, may be obtained by assuming the thickness, say, at 4. With this, we have -

r =l/t = 1474 = 36.75; the square of which is 1350 9/16,

With this value of r2 -

3/2 x .00109 x 1350 9/16= 2.208l

3.2081 Then -

t = 6575x10x3.2081/9500x6=3.7006.

Comparing this result with the assumed value of t = 4, we find the difference so great as to require a second trial.

As the value of r was taken too low, the result obtained is correspondingly low. The true value is somewhere between

3 . 7 and 4. Assume it now, say, at 3 .9. With this value, we have -

r = l/t = 147/3.9 = 37.692; the square of which is 1420.7.

With this value of r2 -

3/2 x .00109 x 1420.7 = 2.32282 add unity = 1 .

3.32282

Then -

t = 6575 x10x3.32282/9500x6 = 3.833.

This result is a trifle less than the assumed value, 3.9. The true value is between these, and probably is about 3.86. This is quite near enough for use. This brace, therefore, is required to be 3 . 86 x 6 inches, or, say, 4x6 inches.