Drive chains, also called roller chains, fill the gap between drive belt/pulley systems and drive gears. These chains convey drive force over long distances like belts, yet provide positive engagement like gears, increasing torque and velocity in inverse proportion. Chains are used in engines, machines, conveying systems, and transportation applications such as bicycles. The primary considerations involved in calculating chain tension requires implementing an accurate analysis of the forces acting on the chain.

Define the bicycle chain application. An adult bicycle has a 52-tooth front chain sprocket with an 8.281-inch pitch diameter. A 200-pound rider stands on and applies his full weight to the right pedal which is level with the pedal crank shaft, and centred seven inches from it, while both hand brakes are fully applied. With this information you can calculate the static chain tension at the instant before the brakes are released to start forward motion.

Calculate the mechanical gain of the pedal arm and front sprocket. Since the pedal arm represents an effective radius of seven inches, it is longer than front sprocket's pitch radius = 8.281 inches/2 = 4.1405 inches, so the force on the chain is 7 inches/4.1405 inches = 1.69 times pedal force.

Calculate pedal force at the instant downward pedal motion starts. With the 200-pound rider standing on the right pedal, and with the pedal arm perpendicular to downward gravity, then the rider's full 200-pound weight is being applied to the seven inch pedal arm as a force by gravity alone.

Calculate the instantaneous chain tension at the starting moment by multiplying the 200-pound downward force by 1.69, the mechanical gain ratio of the pedal-sprocket lever system. Substituting values, 90.7kg. X 1.69 = 153kg.

Define the conveyor chain application. A free-hanging number 40 roller chain weighing 0.181kg./foot is used to convey finished dry cleaning on hangers between the hanging area and the pick-up counter that is 80 feet away from it. A single group of items weighing 7.71 Kilogram is hanging motionless at the halfway point. The chain is two feet lower where the clothes are hanging than at its endpoints. From this information you can calculate the tension in this conveyor chain.

Define the identical right triangles created by the depressed chain. Because the weight is hanging exactly midway, the right triangle long legs are each 80 feet/2, or 40 feet long. The short sides are two feet long. A right triangle calculator shows that the small angle between the diagonals and horizontal is 2.86 degrees.

Resolve all the forces acting on the chain. The total weight of the chain is 80.01 feet times 0.181kg./foot = 14.5kg. + 7.71kg. pulling down on the middle = 22.2kg. downward. The forces on each end support are equal to the angular downward and inward force equal to chain tension.

Solve for chain tension by dividing the total downward force, 22.2kg. by 2 X (sin of the 2.86-degree angle). This yields 22.2kg./(2 X 0.0499) = 49/0.0998 or 223kg. tension in the chain.

#### Tip

Inspect drive chains regularly to make sure they can sustain the tension forces required.

#### Warning

Over stretching chains by overloading them just one time can require complete replacement to avoid harmful or injurious breakage.

#### Tips and warnings

- Inspect drive chains regularly to make sure they can sustain the tension forces required.
- Over stretching chains by overloading them just one time can require complete replacement to avoid harmful or injurious breakage.

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