Page:Cassells' Carpentry and Joinery.djvu/103

Rh $$\begin{align} \therefore \ & d = \sqrt[3]{875} = 9.5 \ \text{nearly,} \\ \text{and}\ & b = .6d = .6 \times 9.5 = 5.7\ \text{in.} \end{align}$$

Therefore, a binder 9·5 in. x 5·7 in. would carry a floor weighing 1 cwt. per ft. super, over a span of 10 ft. The following rule is given by Tredgold:—

$$D = \sqrt[3]{\frac{L^2}{B}} \times 3.42.$$

In this case, again, the breadth must assumed. Let this be taken as 5·5 in.,

$$\begin{align} \text{then}\ D &= \sqrt[3]{\frac{10 \times 10}{5.5}} \times 3.42 \\ \therefore \ d &= \sqrt[3]{18} \times 3.42 \\ \therefore \ d &= 2.7 \times 3.42 \\ &= 9.2\ \text{nearly,} \end{align}$$

which corresponds very nearly with the first case.

Girders 10 ft. apart from centre to centre carry a floor weighing 1 cwt. per ft. super. Required, the breadth and depth for strength; span 20 ft. (1) The total load carried by the girder is 20 x 10 x 1.25 = 250 cwt.—that is, the length multiplied by half the bay on either side multiplied by the load per ft. super. (2) Let B.W. = 7 times the safe load = 250 x 7 = 1,750 cwt. (3) Let breadth be .6 d. (4) Let c the constant be 4 cwt.

Then

$$\begin{align} \text{B.W.} &= \frac{2cbd^2}{L} \\ 1750 &= \frac{2 \times 4 \times .6d \times d^2}{20} \\ \therefore \ d^3 &= \frac{1750 \times 20}{2 \times 4 \times .6} = 7262.5 \\ \therefore \ d &= \sqrt[3]{7262.5} = 19.25 \  \text{in. nearly,} \\ \therefore \ b &= .6d = .6 \times 19.25 = 11.550 \  \text{in.} \end{align}$$

Therefore, the breadth and depth of a suitable girder for the required purpose must be 11.5 in. wide and 19.25 in. deep. It is needless to remark that a wooden girder 20 in. deep is impracticable, and a wrought-iron girder would be substituted for it; but as the above is merely an illustrative example, the construction of the girder need not be discussed. Tredgold's rule for fir girder is:

$$D = \sqrt[3]{\frac{L^2}{B}} \times 4.2$$

Let the breadth (which must be assumed) be 12 in. Then:—

$$\begin{align} D &= \sqrt[3]{\frac{20 \times 20}{12}} \times 4.2 \\ &= \sqrt[3]{\frac{400}{12}} \times 4.2 \\ &= \sqrt[3]{33.3} \times 4.2 \\ &= 3.2 \times 4.2 \\ &= 14 \ \text{in. nearly.} \end{align}$$

It is evident from this that a girder 20 in. deep is by far too large, or that a girder of 14 in. is much too small. If the formulæ in each case are examined it will be found that the first is based on the strength of a small beam determined by trial, while the second is doubtful. It is certain, however, that a girder, 12 in. by 14 in., and 20 ft. long, is not capable of carrying a load of 250 cwt., as determined by the recognized formulæ. It may be mentioned further that the loads are considered as distributed loads, while in reality they are loads placed at certain fixed points, namely, the points where the binders are connected to the girder; consequently the dimensions obtained by the formulæ are slightly less than they ought to be.

Thus

$$\begin{align} B.W. &= \frac{2 c b d^2}{L} \\ &= \frac{2 \times 4 \times 11.5 \times 19.25 \times 19.25}{20} \\ &= 1704.58 \ \text{cwt.} \end{align}$$

which is less than the actual breaking weight calculated for, namely, 1,750 cwt. The strongest floor, for the quantity of timber used, is given in the first case, while the apparent strength shown in the second and third cases results in actual weakness. But single floors should not be used for spans exceeding 16 ft.; and though they are sometimes used for spans up to 24 ft., in such cases deflection is considerable, resulting in cracked ceilings, etc. It may, nevertheless, be stated that each floor has its advantages and its disadvantages. The above