Engineering anomalies

[epanzella]

Actually, no. The loads are only slightly different, according to the shaft diameter, since you can't put any more torque on the shaft via the handwheel than the pinion can react (plus the torque from the return spring, which is minor by comparison).

Essentially you have a shaft with a 3/4" lever at one end and about a 9 inch lever at the other end. The short end enjoys a 10 to 1 mechanical advantage.

 

[Codered741]

Essentially you have a shaft with a 3/4" lever at one end and about a 9 inch lever at the other end. The short end enjoys a 10 to 1 mechanical advantage.

While the gear and handle act as a second class lever, which will give you a mechanical advantage (from your example actually a 12:1 advantage), the forces on the shaft are the same. However, the shaft is acting on the pins at different radii, so the forces on the pins are slightly different.

Because the shaft is acting as the fulcrum in the lever, it encounters torque developed by the handwheel/return spring. This torque is expressed in Ft/Lbs, metric N/m, and is calculated by multiplying the force applied perpendicular to the handle, to the distance at which it is applied. Eg. you have a 1' long handle attached to a shaft, and apply 10 lbs of force to the handle, you will create 10 ft/lbs of torque on the shaft that it is connected to. (1ft * 10lbs = 10ft/lbs)

If you were to exert force closer to the shaft, say at 8in from the center, you would develop less torque. 8in * 10lbs = 80 IN/lbs / 12 = 6 2/3 Ft/Lbs. This torque is then transmitted undiminished, assuming the shaft is sturdy enough to handle the loads, for the entire length of the shaft.

If you want to find out the force that would be acting on the gear, you simply run the calculation in reverse. Continuing with our example, 10Ft/lbs, we will say that the gear OD is 3/4" (you would actually calculate this with the pitch diameter). Therefore, 10Ft/Lbs * 12 = 120IN/Lbs / 0.75IN = 90Lbs.

This is the number that you would use as the force when calculating the necessary cross-section for shear strength. The shear strength of mild steel is classified by ASTM as a minimum of 36,000 psi, and say our pin is 0.125" dia. The cross-sectional area of this pin is 0.012in^2. This number will be doubled, as the pin passes through the shaft twice, allowing twice the area to resist the force. Therefore,
0.024 in^2 * 36,000psi = 864Lbs.

This is the double shear strength of the pin. However, we have one more step to take in our calculations, as the pin is reacting to torque, not a direct force. To find the maximum allowable force on the pin, we need to calculate what torque would be developed on the shaft, in order to develop 864lbs at the shear plane.

First we need the diameter of the shear plane, say 0.375in (3/8" bore on a 3/4" gear seems reasonable). So we just do our calculation again, using 864lbs at 0.1875in, calculating the force at 1ft.

.1875in * 864lbs = 162 in/Lbs / 12 = 13.5lbs.

This is how much force you will need to apply to the handle to break the pin.

As you can see, when the shear diameter decreases, the forces acting on the part increase. I am not saying that these numbers are correct, as I do not have all of the dimensions of the actual shaft, or the actual material types used. If the pin was an alloy of steel, say 4130, even in its normalized and annealed state, its shear strength will increase to 70,000psi, making the applicable force 1680lbs, force at the handle being about 26lbs.

As for the reason for the mis-matched pins, I would guess it was probably for ease of assembly, as the gear was made in a machine shop, with craftsmen who dealt with those pins all the time, and the handle was put on by an assembly line worker, or even the consumer, to make it easier for assembly.

-Cody

 

[Dranreb]

Those are some interesting thoughts on this subject folks, thanks all for taking the time to reply!

Cody, your detailed explanation is just what I was after, much appreciated, although I had to Google 'second class lever'

I'm quite surprised at just how strong the taper pin is, it's nice to know that I can hang on the handle like gorilla a when a drill bit starts to screech and turn blue without it failing on me...:biggrin:

Bernard