= 0,718"

Or the diameter of boll should be 1,436" or say 17-16". But 1" will be quite ample, as we must remember that the strains calculated will come only on the one bolt at the centre of span of beam; and that, as the beam remains of same cross section its whole length the extreme fibre strains decrease rapidly towards the supports, and therefore also the strains on the bolts. The end bolts are doubled however, to resist the starting there of a tendency to longitudinal shearing. We might further calculate the danger of the holt crush-ing the iron plate at its bearing against it; or crushing the wood each side; or the danger of the iron bolt being sheared off by the iron plate between the wooden beams; or the danger of the iron bolt shearing off the wood in front of it, that is tearing its way out through the wood; but the strains are so small, that we can readily see that none of these dangers exist.

Another method of adding to (he sum of the separate strengths of the girders is to place one under the other making a straight joint between the two parts and to drive in hard wood keys, as shown in Fig. 140.

The keys can either be made a trifle thicker than the holes and the beams then firmly bolted together so as to take hold of keys securely; or, keys can be shaped in two wedged-shaped pieces to each key, and driven into the hole from opposite sides, after the beams are firml y bolted. In the latter case, care must be taken that the joint between the opposite wedges is slanting or diagonal, and not horizontal or else, of course, the keys would be useless. Either method allows, for tightening up, after shrinkage has taken place. Iron bands are frequently used in place of bolts but they are more clumsy, less liable to all fit exactly, and besides do not allow for tightening up so easily as with bolts. Where beams are very wide, however, the bands are very advantageous. Tredgold says the keys should be twice as wide, as high; and that the sum of all their heights should equal one and a third times the depth of girder. They can be easily calculated, however. As the main strain on them is a horizontal shearing strain, and the stress or resistance to shearing is greatest across the grain the keys should of course, be placed with their grain running as nearly as possible vertically. Of course, as the greatest horizontal shearing exists near the supports, the end wedges should be the strongest; it is customary, however, to make them all of the same size for convenience of execution. The amount of the horizontal shearing is found by Formula (13).

Keyed Cirders.

Keyed Cirders

Fie. 139.

Besides the horizontal shearing strain there will also be a crushing strain on the sides of wedges, which will be greatest, where the greatest fibre strains exist. This of course, is at the point of the greatest bending moment on the beam. Let us consider the wedge at A-B Fig. (140) which has been drawn enlarged in Fig. (139.)

The lower half of the girder, being in tension, in trying to stretch its fibres meets with the resistance of the wedge along E F, therefore tends to crush or compress this surface. The amount of this compression, per square inch, will be equal to the average fibre strain between E and F. Now the fibre strain at A can be readily found, by finding the "bending moment" at A and dividing this by the moment of resistance of the girder (see Table I) and Formula (18). This gives the fibre strains at A. The average fibre strain on E F will be to the strain at A as the distance of x from the neutral axis, is to the depth of half the beam; x being the centre of E F, or:

Table XVI Value Of Y In Formula 78 100223

Fig. 140.

Extreme fibre strain at A: average fibre strain on wedge = E A: E x. The amount of the compression on E F will of course equal the area of wedge at E F (that is E F multiplied by the breadth of girder), and this area multiplied by the average fibre strain on E F. The greatest compression on E F will of course be at F, and equal to twice the average fibre strain, as E F = 2. E x.

In the same way, we find that the upper half of the girder, being in compression, is forcing its fibres towards the centre causing compression on the surface D C. The amount of this compression is found similarly as for that on E F, the only difference being in the difference in bending moments at B and A. The key therefore becomes virtually a cantilever, the built-in part being between E F and C, and the load applied on the free end C D. the load being a uniform one and equal to the amount of the compression on CD. The weakest point of the girder itself will be either at the point of greatest bending moment, or at key nearest to it. where, of course the girder will not be of full section, being weakened in the part cut away for key. An example will more fully illustrate all of the foregoing.

Weakest Point of Girder.

Example. A spruce girder (Fig. 1 1") of 30-foot clear span is built up of two girders 1"" X 12" each, making the whole section 10" X 24". Georgia pine keys are used, each 6" X 12" (and, of course) 10" across girder; they are placed with grain vertically 3'4" between centres and same distance from centre of last key to support. The girder helps support a plastered ceiling. What is the safe centre load on girder?

The girder is (d = ) 24" deep and (L =) thirty feet long; now 1 1/6. L would be 35, therefore d is less than 1 1/6 L, and from rule contained in Table VIII for spruce we must calculate for deflection, not rupture, in order to be safe. Formula (40) gives the rule for deflection of a centre load on a girder or beam. It is:

δ = 1/48. w.13/e.i or transposing, w = δ.48.e.i/l3 where w would be the sate centre load, in pounds. Now in order not to crack plastering, we have from Formula (28)