We now lay out, Figure 269, the pin, and find that we have a double lever arrangement ; the fulcrum being the 2 inch wide 67000 pounds strain. To the left of this we have a lever 1 3/4 inch wide loaded with two loads, one 28000 pounds one inch wide and one beyond 15000 pounds 3/4 inch wide. To the right we have lighter loads, the heaviest bending-moment will therefore be on the left side, or: m = 1/2. 28000+ 1 3/8.15000 = 34625 pounds-inch. The safe-bending-moment on a 2 15/16 inch pin is: m = 11/14.(1 15/32)3.15000 = 37500 pounds-inch. The pin is therefore safe. The same result could have been read off directly from Table XXXIX.

Table Of Strains Of Roof 200134

Fig 269.

We next design the main rafter; as this is to be in one length, we design it only for lower panel (91500 pounds-compression) and the other panels will be, of course, too strong. Besides the compression we will have a transverse strain or bending-moment on rafter, per square foot, as follows :

Size of Main





3 inch blocks,






Iron tees (say)





69,4 pounds,

or allowing for weight of rafter, say, 70 pounds per foot. The rafter lengths being 17 feet and 17 1/2 feet apart, we have total uniform load u = 17.17 1/2.70 = 20825 or say 21000 pounds. We now draw anywhere along rafter a vertical line a b = 21000 pounds; then draw a c normal to rafter; it scales 13000 pounds, therefore the actual transverse load on rafter is 13000 pounds.

The required moment of resistance to resist this load will be Formula(18) r = 13000.(17.124)/8.12000 = 27,7

By reference to Table XXI, we find we require one 12 1/2 inch

80 pounds channels to take care of the transverse strain. We also note that the area of channel is 8 square inches, which is about what we need additional for resisting the compression. We will decide then to use two 12 1/4 inches 80 pounds channels.

The square of the radius of gyration isTable Of Strains Of Roof 200135 =21,10

The area of the two will be 16 square inches, but as just one-half of their total moment of resistance r is needed to resist transverse strain, we will have only 8 square inches left of the total area to resist compression.

The column is 17 feet or say 204 inches long. One end will be a pin end, the other a milled or planed end with fair bearing. We have then for safe compressive load Formula (3) : w= 8.12000/1+2042.0,000033/21,1 = 90300 pounds, or near enough to pass as safe, If the transverse strain had required say a moment of resistance of r -35 we should have had left to resist compression of the total area 16 square inches only,

Transverse strain on rafter.

Compression on rafter.

= 16 - (35/2.27,7).16

= (about) 6 square inches instead of 8, which (in such a case), would not be enough and would require a heavier channel. The thickness of our channel web we see from Table XXI, is 0,39 inches, or two webs = 0,78. The safe bearing on pin will therefore be = 0,78.2 15/16.12000= 28200 pounds. The web of channel will therefore need thickening at the shoe pin, but not at the two others. At the apex the pin does not bear on channels, but is connected indirectly by a 2-inch thick plate.

At the shoe the channels have planed ends and rest directly on the shoe, the strain therefore against their webs, will be 60000 pounds, due to the rod trying to tear the pin out. The webs take care of 28200 pounds of bearing, leaving 60000 - 28200 = 31800 pounds, to be transferred by rivets to thickening or reinforce plates. Of this each side takes care of

31800/2 = 15900 pounds.

The thickness required for each plate, is therefore

15900/2 15/16.12000= 0.42 or say 1/2 inch each.

The number of rivets required must next be settled. We use 5/8 inch rivets:

The channel being thinner than the plate determines the bearing value for each, or with its share of the 14100 pounds carried by each channel web, the total bending-moment will therefore be from Formula (25).

= 0,39.5/8.12000 = 3000 pounds, or

15900/3000= 5,3 or say six rivets required for bearing.

The rivets are in single shear, their area = 0,3068 and their value

= 0,3068.8000= 2454 pounds, and we require 15900/2450 = 6,5 or say seven rivets.

For bending-moment each rivet is a single lever held by the half-inch plate, projecting 0,39 inches and uniformly loaded on free end m = 14100.0,39/2 = 2750 pounds-inch.

Bearing of rafter webs on pins.

Reinforce plates to web.

Rivets in reinforce plates.

The safe bending-moment on a 5/8 inch rivet we previously found to be 360 pounds-inch, and require therefore

2750/360 = 7,7 or say eight rivets.

The disposition of rivets around the pin, however, requires nine rivets.

We now go to the apex point. We use here a 2-inch plate, its bearing on pin must be all right, as we made the other end of vertical rod 2 inches thick. The upper end of rod we make forked, each side 1 inch thick, and 2 inches between to admit plate.

The plate is so large that there is evidently no danger of the pin shearing out.

For the rivets we can see it will require a large number and therefore decide to use larger or say 7/8 inch rivets.

The bearing value on two webs of each rivet will be = 7/8.2.0,39.12000 = 8190 pounds. The (double) shearing value of each rivet will be

= 2.0,6013.8000 = 9620 pounds. The safe bending-moment on each rivet will be

= 11/14(7/16)3.15000 = 987 pounds-inch.

If now we consider that the two rafters have planed ends and butt fairly against each other, taking up the thrust from each, the rivets need only take care of the vertical down pull 67000 pounds, all of the rivets taking a share. In that case we should need for bearing

67000/9360= 7,2 rivets, for shearing less, and for total bending-moment, remembering that the channel backs are 2 inches apart and that the rivets are beams supported at both ends and uniformly loaded, from