# Homework Help: Verify my calculation, bending moment torque

1. Sep 28, 2007

### WhiteHawk

I made a diagram with the calcs on it, should be fairly simple, I just need to make sure i'm doing it right or if i'm way off.

These are the beginnings of my homemade engine dynamometer...I thank you for your assistance.

2. Sep 28, 2007

### WhiteHawk

where did this moved?

3. Sep 28, 2007

### TVP45

I'm sorry but I don't understand the diagram. Is one corner pinned? What does the 300 ft-lb represent? What does the force represent?

I believe, however, you have a mistake. You have ft-lb * ft = lb and that is not true.

I'm not sure what sort of dynamometer you're making, but are you aware of prony brakes?

4. Oct 1, 2007

### WhiteHawk

The square is real (mechanical object with no deflection). The one corner labled "bending moment" is on a pivot bearing, the square would rotate around that point. There is 300 ftlbs of torque being applied to the other point. The system would not actually move (with this design I would use a transducer to measure the force). Basically I would need a way to calculate the torque by measuring how much force (lbs or N) applied where I have the "?", so I can calculate HP with RPM.

As far as prony brakes, I think with the high output of a V8 engine wouldn't be practical for that design. Basically any way of efficiently braking the engine under wide open throttle to hold at specific RPMs would be a dynamometer. The most popular designs are a water pump that you restrict the outlet causing resistance or using a very large generator to create resistance. For my poor mans dyno idea, I would use a rear axle out of a truck, with large disc brakes to create the resistance.

So this theoretical torque I mention above is coming from the input shaft of the rear axle, and the axle would be mounted at that pivot point 6" from the center of the input shaft.

5. Oct 1, 2007

### TVP45

If I understand your diagram correctly (and I may not, so correct me), you have a rigid structure (the square) with a pivot at the lower left corner. There is a 300 ft-lb torque applied ABOUT the upper left corner (This is where I may not understand you. I read the curved arrow to indicate that the centerline of the input shaft is perpendicular to the square and points right at that upper left corner. Is that OK). You want to resist (and measure) the torque at the lower right corner.

If that is correct, the structure is probably not free to rotate about the pivot. The pivot will actually oppose the turning torque and there will be little if anything to measure. The exception to this is if the motor and shaft are actually mounted on the square and free to turn with it; in that case, it may destroy itself or not but that seems a difficult way (and dangerous one) to do the measurement.

Can you give more details? Thanks.
Tom

6. Oct 1, 2007

### WhiteHawk

I drew another diagram, with how I could maybe represent the scenario a little better.

Blue circle is the pivot.

Green circle is input torque.

Tan box is force measuring device.

The Important item, is the Magical Red lever to signify the torque, is this lever valid? See how I drew it standing vertical, it would seem as though the system would want to pivot now.

New Diagram

7. Oct 1, 2007

### TVP45

I still can't see clearly, but it looks like it doesn't quite work. Is the green circle connected to something which is connected to the earth? Or very heavy?

8. Oct 1, 2007

### WhiteHawk

The green circle is floating and can "move" with the system. A driveshaft with universal joints would be driving the green circle (like a car suspension can flex while the driveshaft is still able to turn).