Fig. 50. Brake Levers and Air Brake.
For double-truck cars the brake rigging is necessarily more complicated, as it must be arranged to give an equal pressure on all eight wheels of the car. Brake shoes are sometimes placed between the wheels of a truck and sometimes outside. The arrangement of brake shoes between wheels is apparently finding most favor, as when the shoes are applied in this position there is less tendency to tilt the truck frame when the brakes are applied, and this adds to the comfort of passengers in riding. [Fig. 50] shows one form of arrangement of brake levers common on a double-truck car equipped with air brakes, with inside-hung brake shoes.
Brake Leverages and Shoe Pressure. The levers between the air cylinder and the brake shoes are usually so proportioned that with an air pressure of 70 lbs. per sq. in. in the brake cylinders the total of the brake shoe pressures on the wheels will be equal to about 90 per cent of the weight of the car. The diagram [Fig. 51] has shoe pressures and strains in the several rods marked on shoes and rods.
The following example, based on the diagram, will explain the lever proportioning. Only round numbers are given on the diagram.
Assume a four-motor car weighing 40,000 pounds. A brake cylinder 7 inches in diameter is used. This gives 38.5 square inches and at 70 pounds air pressure a total force on the piston rod of 2,695 pounds. The weight of the car is 40,000 pounds. Taking 90 per cent of this gives a total of 36,000 pounds to be exerted by the brake shoe when an emergency stop is made. Each of the eight shoes will press against the wheels with a force of 4,500 pounds.
The dimensions of the truck are such that the “dead levers,” those fixed at one end and which carry shoes, cannot be over 13 inches long. The shoe will be hung three inches from one end, making the proportions 10 to 3, and the pressure on the strut rod between shoes will be 4,500 × ¹⁰⁄₁₃ or 3,461 pounds. To clear the truck frame the live lever extends 14 inches above the point of application of the brake shoe. To obtain 4,500 pounds pressure on the shoe, the distance between the brake shoe and the strut rod, which we will call “x,” will be found by regarding the upper end of the lever as fixed and the power applied at the lower end.
4500 = 3461 × 14 + x14 or
x = 4.2 inches.
Now to obtain the force required in the rod to the truck quadrant, the bottom end of the live lever must be regarded as the fulcrum. The equation is
x = 4500 × 4.218.2 = 1038 pounds.