Determing torque so that I may determine the diameter of my axle

[blinddog22]

I am a little rusty at the moment but let's say I have an axle with two loads being applied to it. Both 500 lb. I am trying to figure out the various stresses on it so i can determine a suitable diameter for the axle connecting the tires. The two tires and axle are all one piece. The bearing locations are indicated by teh boxes and 500 lb forces.
I made a FBD and calculated reaction forces, moment, shear, and bending. However, I am stuck with the torsion or torque. I don't know which one it is. I believe it is torque since the whole axle is one piece. How do i determine the torque given
diameter of wheel = 20in
axle is driven at a constant speed of 3 ft/s
friction coeff = 0.3

I will be using von-mises afterwards to determine the proper diameter of the axle.

Please see attached picture.

Thank you.
Can't seem to get picture to work, but here is an imjur link. http://imgur.com/a/HHmgn

 

[haruspex]

axle is driven at a constant speed of 3 ft/s

Ft/s is not a rotation rate. Do you mean the vehicle moves at that speed? On the level or is this going uphill?

friction coeff = 0.3

Do you mean rolling resistance? 0.3 is very high for that. Friction between tyre and road is not relevant.

 

[blinddog22]

Ft/s is not a rotation rate. Do you mean the vehicle moves at that speed? On the level or is this going uphill?

Do you mean rolling resistance? 0.3 is very high for that. Friction between tyre and road is not relevant.

I was given that the velocity was 3 ft/s, this is of the vehicle.

I believe it is rolling resistance. I was told to assume around 0.2, as long as I state my assumption. So i just choose 0.3 arbitrarily.

 

[haruspex]

I was given that the velocity was 3 ft/s, this is of the vehicle.

I believe it is rolling resistance. I was told to assume around 0.2, as long as I state my assumption. So i just choose 0.3 arbitrarily.

Ok. If the coefficient is 0.3, what is the magnitude of it as a force?

 

[blinddog22]

Ok. If the coefficient is 0.3, what is the magnitude of it as a force?

To be honest, I don't understand what you are asking.

This is for a trailer that is holding a 1000 pound load. The load is bearing on the axle at the indicated positions. I am not even sure where the 0.3 friction coeff comes into play calculation wise, unless it is needed to find the torque.

 

[haruspex]

To be honest, I don't understand what you are asking.

This is for a trailer that is holding a 1000 pound load. The load is bearing on the axle at the indicated positions. I am not even sure where the 0.3 friction coeff comes into play calculation wise, unless it is needed to find the torque.

Rolling tesistance acts , like friction, to oppose relative motion. If the axle load is 500 lb and the coefficient of rolling resistance is 0.3, how hard would you have to push horizontally to get that wheel turning?

 

[blinddog22]

Rolling tesistance acts , like friction, to oppose relative motion. If the axle load is 500 lb and the coefficient of rolling resistance is 0.3, how hard would you have to push horizontally to get that wheel turning?

You would have to push with a force greater than the friction force multiplied by the normal force.

 

[haruspex]

You would have to push with a force greater than the friction force multiplied by the normal force.

Rolling resistance coefficient multiplied by normal force.
So what is the strength of force you have to overcome, and where does it act?

 

[blinddog22]

I suppose in this case, it would be 1000lb since there is two 500 lb forces acting on the axle, and it would be acting on the axis of the axle. So you would need to overcome 1000(0.3) = 300 lb?

 

[haruspex]

I suppose in this case, it would be 1000lb since there is two 500 lb forces acting on the axle, and it would be acting on the axis of the axle. So you would need to overcome 1000(0.3) = 300 lb?

Each wheel gets its own share.
And where does the resisting force act?

 

[blinddog22]

Oh, ok. So then the force that needs to be overcome is 500(0.3) = 150 lb. Since each wheel gets its own share.
The resisting force acts between the wheel and the ground.

 

[haruspex]

Oh, ok. So then the force that needs to be overcome is 500(0.3) = 150 lb. Since each wheel gets its own share.
The resisting force acts between the wheel and the ground.

Right, so what torque does it exert on the axle?

 

[blinddog22]

150 lb x 9in (radius of tire) = 1350 lb-in?

 

[haruspex]

150 lb x 9in (radius of tire) = 1350 lb-in?

Yes.

 

[blinddog22]

Alright, then that is my torque applied to the axle so that I may calculate my maximum shear stress? As shown in my initial picture.

 

[haruspex]

Alright, then that is my torque applied to the axle so that I may calculate my maximum shear stress? As shown in my initial picture.

Yes.

 

[blinddog22]

Excellent. Thank you very much.

If you look at my diagram, the maximum moment i calculated was 5000 lb-in. Does that seem correct? When i calculate my sigma and tau and use the von mises criterion, I feel like the value seem awfully big. In the end I ended up with a diameter a little shorter than 1.42 inches. Which seems kind of skinny.

 

[haruspex]

Excellent. Thank you very much.

If you look at my diagram, the maximum moment i calculated was 5000 lb-in. Does that seem correct? When i calculate my sigma and tau and use the von mises criterion, I feel like the value seem awfully big. In the end I ended up with a diameter a little shorter than 1.42 inches. Which seems kind of skinny.

I'm not an engineer, so I don't know about von Mises' work.
As far as I am aware, you have calculated the minimum diameter under ideal conditions. Real axles have to cope with bumps in the road, wear and tear, and a safety margin.

 

[blinddog22]

Alright, no worries.

Ok, I suppose that makes sense then. A 1.5 diameter thick beam of 1018 steel sound like it should handle a 1000lb load.

Also, I just thought of something during my calculation. When i plug my torque into my shear stress equation to determine the torsion, I only need to plug in 1350 lb-in, correct? Not 1350 lb-in x 2 for the other wheel as well? Do i worry about the torque from the other wheel?

 

[haruspex]

Alright, no worries.

Ok, I suppose that makes sense then. A 1.5 diameter thick beam of 1018 steel sound like it should handle a 1000lb load.

Also, I just thought of something during my calculation. When i plug my torque into my shear stress equation to determine the torsion, I only need to plug in 1350 lb-in, correct? Not 1350 lb-in x 2 for the other wheel as well? Do i worry about the torque from the other wheel?

Somewhere along the axle must be the drive (differential?). The torque is between there and the wheels. The two sides are independent.

 

[blinddog22]

Its just a straight axle with 2 wheels. Meant for a trailer to carry a load. Treating it as one solid piece essentially. Same concept?

 

[haruspex]

Its just a straight axle with 2 wheels. Meant for a trailer to carry a load. Treating it as one solid piece essentially. Same concept?

In that case there is no torque.

 

[blinddog22]

Very well. No shear stress then.

 

[haruspex]

Very well. No shear stress then.

Not so sure about that. I thought that the vertical loading in beams constituted both shear and bending stress - http://portal.ku.edu.tr/~cbasdogan/Courses/MDesign/course_notes/ShearStressInBeamsNotes.pdf.
But as I wrote, I'm not the engineer here.

 

[blinddog22]

Alright, I will review that.

I have another question if you are willing and able. Do bolts affect the shear and moment force diagrams for beam loading/deflection at all? I have dug through 4 of my textbooks and scoured the internet but haven't found anything to explain it.

See attached pic if you don't mind. http://imgur.com/a/MuJTc

There are 4 beams, the two smaller ones on top of the two longer ones and bolted together using two bolts per connection. In the center of each beam there is a load acting on it. How are the shear/moment diagrams affected, or are they?

 

[haruspex]

Alright, I will review that.

I have another question if you are willing and able. Do bolts affect the shear and moment force diagrams for beam loading/deflection at all? I have dug through 4 of my textbooks and scoured the internet but haven't found anything to explain it.

See attached pic if you don't mind. http://imgur.com/a/MuJTc

There are 4 beams, the two smaller ones on top of the two longer ones and bolted together using two bolts per connection. In the center of each beam there is a load acting on it. How are the shear/moment diagrams affected, or are they?

There are downward forces at the midpoints of the two upper beams, and equal upwards forces at the midpoints of the two lower beams, right?
The bolts are, to a first approximation, redundant. They only come into play as the bending of the beams twists the corners from the horizontal, no?

 

[blinddog22]

there is a 1000lb load standing on four legs. So there is a 250 lb force acting on the midpoints of all four beams.

Honestly, I tried my hardest searching through mechanics of materials, design, and Beam deflection books trying to assure myself of the effect the bolts have on my shear and bending moment diagrams. I did not think the bolts matters when it came to the diagrams. I am not sure how to tackle them.

 

[haruspex]

there is a 1000lb load standing on four legs. So there is a 250 lb force acting on the midpoints of all four beams.

Honestly, I tried my hardest searching through mechanics of materials, design, and Beam deflection books trying to assure myself of the effect the bolts have on my shear and bending moment diagrams. I did not think the bolts matters when it came to the diagrams. I am not sure how to tackle them.

Are the legs at the corners? Under the bolts?

 

[blinddog22]

No, the 4 beams bolted together form the rectangular frame of a trailer. Something negligible will be placed on top of it to form a flat surface. Then a 1000lb load will be placed on it. The load has 4 legs that will all perfectly apply the loads to the 4 centers of the beams. The 4 squares with the X's through them signify the centers of the beams, which is also where the loads will be applied.

 

[blinddog22]

Perhaps this will make it more clear. I drew the frame at an angle with the 4 loads acting on it.

http://imgur.com/a/fBvpE

 

[blinddog22]

A side view looking at either of the left/right beams. At eye level.

http://imgur.com/a/MTemv

 

[haruspex]

No, the 4 beams bolted together form the rectangular frame of a trailer. Something negligible will be placed on top of it to form a flat surface. Then a 1000lb load will be placed on it. The load has 4 legs that will all perfectly apply the loads to the 4 centers of the beams. The 4 squares with the X's through them signify the centers of the beams, which is also where the loads will be applied.

Ok, but something must be supporting the frame from below. Where does that act?

 

[blinddog22]

The axle from earlier will be. Looking at the picture, http://imgur.com/a/fBvpE , it will be located on the dashed line, bisecting the two longer plates on the bottom of the assembly.

 

[haruspex]

The axle from earlier will be. Looking at the picture, http://imgur.com/a/fBvpE , it will be located on the dashed line, bisecting the two longer plates on the bottom of the assembly.

Then you can simplify the picture. You have 250 down and 500 up at each of those points, so the net is 250 up.

 

[blinddog22]

So the two points over the axle will be 250 up and the two points on opposite sides of the axle will be 250 down.

This is what I officially have for the moment. http://imgur.com/a/NUGdm
For the right/left beam that is supporting the downward 250 lb load. I took the moment about point O. I included both bolts on the right side using the summation of moments I solved for the F the bolts are taking. This resulted in the 950 lb force. Near the bottom I added that force per bolt on acting downward. 2 bolts on either end of the beam.
I feel like this is very very wrong.