This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.
A Flitch-plate girder of 20-foot span consists of two Georgia pine beams each 6" X 16" with a sheet of plate-iron 16" deep bolted between them. The girder carries a load of 13000 pounds at its centre; of what thickness should the plate be? The girder supports a plastered ceiling.
From Table XIII we find that a Georgia pine beam 6" X 16" of 20-foot span will safely carry without cracking plaster 7080 pounds uniform load, or 3990 pounds at its centre (See Case(6) Table VII,) so that the two wooden beams together carry 7980 pounds of the load, leaving a balance of 5020 pounds for the iron plate to carry. The deflection of a 20-foot span Georgia pine beam 6" X 16" with 3990 pounds centre load will be, Formula (40)
δ = 1/48. 3990.2403/e. i.
. δ = 1/48. 3990. 2403/1200000. 2048 = 0,47"
Strength of wooden part.
We now have a wrought-iron plate which must carry 5020 pounds centre load, of a span of 20 feet, 16" deep, and must deflect under this load only 0,47". Inserting these values in Formula (40) we have:
0,47 = 1/48. 5020.2403/e. i.
From Table IV we have for wrought-iron e = 27000000
While for i, we have (Table I. Section No. 2) i = b.d3/12 = b.163/12 = 341.b
Inserting these values and transposing we have: b= 5020. 2403/48. 27000000. 341. 0,47 = 0,33
Or the plate would have to be 1/3" X 16". Now to make sure that this deflection does not cause too great fibre strains in the iron, we can calculate these from Formulae (18) and (22). The bending moment at the centre will be (22) m = 5020. 240/4 = 301200
The moment of resistance will be (Table I. Section No. 2) r = b.d2 /6 = 0,33.162/6 = 14
And from (18) m/(k/f) = r, or transposing and inserting values,
(k/f) = 301200/14 = 21514 pounds.
As the safe modulus of rupture of wrought-iron is only 12000 pounds (Table IV) we must increase the thickness of our plate. Let us call the plate 5/8" X 16", we should then have r = 5. 162/8.6 = 26,67 and
(k/f) = 301200/26.67 = 11256
So that the plate would be a trifle too strong. This would mean that both plate and beams would deflect less. The exact amount might be obtained by experimenting, allowing the beams to carry a little less and the plate a little more, until their deflections were the same, but such a calculation would have no practical value. We know that the deflection will be less than 0, 47" and further, that plastering would not crack, unless the deflection exceeded 3/6 of an inch (Formula 28) as
20.0,03 = 0, 6"
In regard to the bolts, the best position for them would, of course, be along the neutral axis, that is, at half the height of the beam. For here there would be no strain on them. But to place them with sufficient frequency along this line would tend to weaken it too much, encouraging the destruction of the beam from
Size of Bolts.
longitudinal shearing along this line. For this reason the bolts are placed, alternating, above and below the line, forming two lines of bolts, as shown in Fig. (137). The end bolts are doubled as shown; the horizontal distance, a-b, between two bolts should be about equal to the depth of the beam. If we place the bolts in our example, say 3" above and 3" below the neutral axis, we can readily calculate the size required. Take a cross-section of the beam (Fig. 138) showing one of the upper holts. Now the fibre strains along the upper edge of the girder, we know are (k/f ) or 1200 pounds per square inch, for the wood, and we just found the balance of the load coining on the iron would strain this on the extreme upper edge = 1125(3 pounds per square inch. As the centre line of the bolt is only 3" from the neutral axis or 3/8 of the distance from neutral axis to the extreme upper fibres, the strains on the fibres along this line will be, of course, on the wood § of 1200, or 450 pounds per square inch: and on the iron 3/8 of 11256=4221 pounds per square inch. Now, supposing the bolt to be 1" in diameter. It then presses on each side against a surface of wood = 1" X 6" or = six square inches. The fibre strain being 450 pounds per square inch, the total pressure on the bolt from the wood, each side, is:
C.450 = 2700 pounds. On the iron we have a surface of 1" X 5/8" = 5/8 square inches. And as the fibre strain at the bolt is 4221, the total strain on the bolt from the iron is = 5/8. 4221 = 2638 pounds. Or, our bolt virtually becomes a beam of wrought-iron, circular and of 1" diameter in cross-section, supported at the points A and B, which are 6 5/8" apart, and loaded on its centre C with a weight of 2G38 pounds.
Therefore we have, at centre, bending-moment (Formula 22) m = 2638.6 5/8/4 = 4369,
From Table I, Section No. 7, we know that for a circular section, the moment of resistance is, r = 11/14. r3 = 11/14. (1/2)3 = 0,098
Now for solid circular bolts, and which are acted on really along their whole length it is customary to take ( k/f) the safe modulus of rupture rather higher than for beams. Where the bolts or pins have heads and nuts at their ends firmly holding together the parts acting across them they are taken at 18000 pounds for steel and at 15000 pounds for iron. We have therefore transposing (Formula 18) for the required moment of resistance r= - - = 0,291. Inserting this value for r in the 15000 above we have for the radius of bolt, 11/14. r3 = 0,291 and