δ = L. 0,03 or δ = 30. 0,03 = 0,9. From Table IV we have for spruce:

Calculation of Keyed girder.

e = 850000

From Table I, Section No. 17, we have for' the weakest section of the girder, which would be through a key, and as shown in Fig. 141, i = b/12 (d3 -d13)

= 10/12 (243 -63) = 11340, therefore inserting these values in the transposed Formula(40) w = 0,9.48.850000. 11340 3603

= 8925

Or the safe centre load, not to crack plastering, would be, say 9000 pounds.

Table XVI Value Of Y In Formula 78 100224

Fig. 141.

Now let us try the keys. We first take the greatest horizontal shearing, which will be at the end keys.

The vertical shearing at- these keys will be equal to the reaction (see Table VII, or Formula 11.)

As the load is central, each reaction will, of course, be one-half the load, or 4500 pounds, therefore the vertical shearing strain at end key, will be (a little less than) x = 4500 Now from Formula (13) we know that the horizontal shearing strain at the same point is: 3/2. x/a For the area we take the full area of cross-section or a= 10.24 = 240, therefore horizontal shearing strain:

3.4500/2.240 = 28,125 pounds per square inch. The amount of this strain that will act on each key is, of course, equal to the area at the neutral axis from centre to centre of key, or 40. 10 = 400 square inches multiplied by the strain per square inch, or

400. 28,125 = 11250 pounds. To resist this we have a key 12" X 10" = 120 square inches, being sheared across the grain. From Table IV we know that the safe shearing stress of Georgia pine across the grain or fibres, is:

(g/f) = 570 pounds so that the key could safely stand an amount of horizontal shearing

= 570. 120 = 68400 pounds or more than six times the actual strain. Had we, however, placed the grain of the key horizontal, the shearing would be with the grain or along the fibres; the safe shearing stress this way for Georgia pine (Table IV) is only 50 pounds per square inch, so that the key would only have resisted

= 50.120 = 6000 pounds, or it would have been in serious danger of splitting in two.

Now take the Key A B immediately to the right of the weight. The bending-moment at A will be (Table VII) mA= 4500.166 = 747000 and at B mB= 4500.154 = 693000 Now at A and B, the girder being uncut, the moment of resistance will he (Table I, Section No. 2) r = 10. 242/6 = 960

Dividing the bending moment by the moment of resistance (Formula

18 transposed) gives the extreme fibre strains, at A = 747000/960 = 778 pounds.

and at B = 693000/960 = 722 pounds.

Now the centres (x and y, see Fig. 139) of each side of key will be 1 1/2" from neutral axis, the extreme fibres being, of course, 12" distant from neutral axis, therefore average strain on side of key at A. (Or x, Fig. 139) = 1 1/2/12. 778 = 97 pounds, and at B (or y, Fig. 139) = 1 1/2/12 . 722 = 90 pounds.

The extreme compression, will, of course, be on the lower edge of key, at A and will be = 2.97 = 194 pounds per square inch.

From Table IV we find that Georgia pine will safely stand a pressure of 200 pounds per square inch, across the fibres, so that we are just a little inside of the safety mark. We now have to consider our key as a cantilever with cross-section 10" wide and 12" deep.

projecting 3" beyond the support and loaded uniformly with a weight equal to 90 pounds per square inch, or u = 90.3.10 = 2700 pounds.

Now the bending moment at support is, (Formula 25.) m = 2700.3/2 = 4050 pounds.

Central Keys.

The moment of resistance (Table I, Section 2) is r = 10.122/6 = 240

Therefore (Formula 18) the extreme fibre strains on key = m/r = 4050/240 = 17 pounds Or not enough to be even considered seriously.

Another method of combining and strengthening wooden girders, is to cut them with saw-shaped notches, as shown in (Fig. 142) and fit the teeth closely together, firmly bolting the two parts together, so as to force them to act together as one girder. Sometimes the top surface slants towards each end, and iron bands are driven on towards the centre, till they are tight. But bolts are more reliable, and not likely to slip; where the girder is broad, they should be doubled, that is, placed in pairs across the width of girder. The distance between bolts should not exceed twice the depth of girder. Great care must be taken to get the right side up.1 Many text-books even being careless in this matter. It must be remembered that the upper fibres are in compression, crowding towards the centre, while the lower ones, in tension, are pressing away.

Notched girders.

Notched girders

Fig. 142.

The girder must therefore be placed, as shown, so that the two sets of fibres will meet at the short joints and oppose each other. The girder is easily calculated similarly to the former example. The crushing on C D or A B can be found, and also the stress on their extreme edges; this must not exceed the safe stress of the material for compression along the fibres. Then D B or C A must have area sufficient to resist the horizontal shearing strain.

In all these girders the most careful fitting of joints is necessary; then, too, the ends must have sufficient bearing not to crush under the load. Thus, take the former example, the reaction was 4500 pounds; the safe resistance of spruce to crushing across the fibres is (Table IV) = 75 pounds.

We need therefore an area = 4500/75 = 60 square inches, and as the girder is 10" broad it should bear on each support 60/10 = 6

Size of


1 The reasons for placing the notched girder in the position shown in Figure 142 have been fully stated by me in a letter published May 10, 1890, in the American Architect, Vol. XXVIII, No, 750.