Formula (21) m = 67000.2/8 = 16750 pounds-inch.

Therefore number of rivets required to resist bending-moment

= 16750/987= 16,9 or say nine each side of joint. The plate, however, requires ten each side for even distribution.

Pin plates at apex.

Number of rivets.

If the channels do not butt fairly the plate will have to transfer the thrust from one to the other. This thrust will be equal to the horizontal resultant of compression on rafter at apex which is 54500 pounds. Its horizontal projection measures 34000 pounds. The rivets each side of joint must take care of this strain and the plate be large enough not to crush under it. We need not calculate any in this case, however, as this strain is just about one-half of the strain for which the total number of rivets were proportioned.

We next design the shoe. It is a flat cast-iron plate 2 inches thick, and 28 inches by 24 inches with two flanges each 3 inches thick to receive the pin. As the channels bear directly on the plate, the only strain on the flanges will be due to the pin trying to shear its way out, the strain being 60000 pounds. Resisting this there are two areas to each of the two flanges, each area 3 inches x 3 inches, or a total area = 4.3.3 = 36 square inches, the actual stress is therefore

60000/36 = 1667 pounds per square inch, which is safe.

The wall must carry the whole load of truss, each reaction is 5 7900 pounds due to vertical load. To this must be added the vertical resultant (or projection) of the largest wind reaction 21700 pounds. Its vertical projection measures about 14000 pounds making the total vertical reaction 71900 pounds. The area of plate is 28x24=672 square inches, therefore compression per square inch on brick-work

=71900/672 = 108 pounds, which is safe.

Had this truss been of larger span, we should have had to place one shoe either on a rocking saddle, or else on rollers. If the latter there should be sufficient rollers, and they should be of large enough diameter not to indent the plate, and to roll back and forth freely. It is usual to put upward flanges all around the bottom plate and downward flanges all around the upper plate to hold the rollers in place between them. The flanges should be less than radius of rollers, so as not to meet. The foot of girder must be secured against yielding sideways. The size of rollers is determined by the following formula:

Designing shoe plate.

Bearing area.

Include wind reaction.

Rockers or rollers.

Formula for rollers.

(132)

Where w = the safe load, in pounds, on each roller.

Where l = the length, in inches, of each roller.

Where d = the diameter in inches, of each roller, if of steel or wrought-iron and rolling between cast-iron plates. If between wrought-iron plates add 25 per cent to w.

Rollers should be used, (under one shoe only,) where trusses indoors are over eighty feet span, or out-doors if span is over sixty-five feet.

Example.

A truss of one hundred feet span has a reaction at each end of 95000 pounds. The shoe-plate is 20 inches wide the long way of rollers, and rollers are 1 inch in diameter. How many rollers are required?

Each roller will safely carry from Formula (132)

we shall require therefore

95000/15000 = 6,3 or say 7 rollers.

Of course, where rollers are used some arrangement must be made in the cornice to allow for the movement due to expansion and contraction of the truss ; or if the roof is continuous a slip-joint must be provided in the roof itself, the detail of which will depend upon the local circumstances.