Torsional Rigidity Measurement...warning pseudoscience involved

[Walt Fricke]

John - how about you get someone to look at your digital levels while you put more and more of your own weight out on the end? Just to get that qualitative look before you start quantifying.

Should give you an idea of how things are going, though the question you raise is a good one, since you have a fulcrum in the middle of chassis.

[1QuickS]

It does not matter where the supports are, torque applied to chassis will result in the same rotation about the centroid axis.

[Flieger]

Originally Posted by JohnJL

Thanks guys.

Wouldn't it be 75 lbs though, since the deflection is across the plane? Half the twist is "up" on the far side of the beam and half is on the same side as the beam?

....Banker doing physics.....

You need to think of using a torque wrench (or any wrench). If you have a deep socket or extension you need to apply a good amount of pushing force on the socket to counteract your other hand pulling on the handle. If you apply a force of 1 pound to the end of a wrench the wrench just moves linearly, falling off the nut. If you apply a pull of 1 pound at the handle and a push of 1 pound at the socket and the handle is 1 foot long, then you are applying a couple (two opposing forces separated by a distance) of magnitude 1 foot-pound of torque.

Firstly, we assume the chassis is close enough to call bilaterally symmetric (left and right are mirrored). Now, a lever arm of length L (from the center of the chassis to the weight) with weight W at the end is added. You have two mounting points for the lever on the chassis, and in this case the mounting points are also bilaterally symmetric. They are separated by a distance D (so the total length of your beam is L+D/2).

This is a simple statics problem. We want the sum of the forces and the sum of the moments acting on the lever to be zero. We have a moment about the end of the lever (at the chassis centerline) L x W and a force W. Therefore, the two reaction forces at the chassis (call them R1 for the left side and R2 for the right) must cancel the applied force and moment.

Now, so that I only need to type scalars we assume that a tension force in the bolt is a positive force, and a positive moment is one that turns the end of the lever towards the ground.

R1 + R2 = W

LW = R1*D/2 - R2*D/2

Now, some algebra

R2 = W - R1
LW = (R1-R2)*D/2
LW = (R1-(W-R1))*D/2
LW = (R1-W+R1)*D/2
LW = 2*R1*D/2 - W*D/2
LW + W*D/2 = R2*D

W*(L/D+0.5) = R1 This is a positive quantity as I have defined my vectors, so the left side bolt is in tension with a force significantly greater than the applied weight

R2 = W-R1

R1 > W, therefore R2 is negative and the right side bolted joint is in compression.

The bolt on the side where the weight is is doing most of the work.

There are quicker ways to get this answer but this is a little more to the book.

[Flieger]

Originally Posted by 1QuickS

...One data point is not a curve so like Flieger said you will want a few data points to plot a stiffness response curve...

Even more basic than that, I was just pointing out that if he only gets a deflection equal to one unit of the resolution of the device, then he has +/- 50% uncertainty (error bars). That's hardly measuring something. 2 units he has +/- 25%, 3 units +/- 16.7%, 4 units +/- 12.5%, etc. So the bigger the deflection, the smaller the uncertainty.

[540964]

Originally Posted by 911pcars

Oh yeah. Let's order up a SPMM4000. Can you quote us a price.... out the door?

Sherwood

About 1.2M GBP.
Or you can rent time on one. Morse Measurements

[Walt Fricke]

The reinforcing cross member is held to the chassis with an M12-8.8 bolt. The first 8 indicates that the tensile strength of the steel is 800 Newtons per square millimeter. That's 4568 psi, or 20,340 pounds tensile strength.

So I don't think you will break the bolt.

[TCracingCA]

Hey Boss, how much torque do you want me to apply?

[Flieger]

Originally Posted by Walt Fricke

The reinforcing cross member is held to the chassis with an M12-8.8 bolt. The first 8 indicates that the tensile strength of the steel is 800 Newtons per square millimeter. That's 4568 psi, or 20,340 pounds tensile strength.

So I don't think you will break the bolt.

You are correct that an 8.8 will have an ultimate tensile strength of at least 800 MPa (Mega Pascals = Pascals * 10^6 = 10^6 N/m^2 = N/mm^2). The yield will be 80% of that.

However, 800 MPa is 116,000 psi. 6061-T6 aluminum is around 30 ksi (30,000 psi). 4130 steel (normalized) is about 63 ksi. 4.5 ksi is about the strength of cheese .

The minor diameter of an M12 bolt is 9.6 mm minimum, which gives an area of 72.4mm^2, corresponding to an ultimate tensile load of 13,476 pounds. You want to stay well clear of yield however, which means more like 6,500 pounds.

[Walt Fricke]

I knew I should have found one of those charts which shows strength by bolt size and grade directly, which would have kept me from overlooking the minor diameter. And maybe kept me from confusing UTS with the elastic limit.

[Walt Fricke]

But 200 pounds (a guy sitting on the lever) would need 32.5 feet to apply a force of 6,500 lbs, would it not?

A lot of force may be necessary. Remember the photo, in Puhn I think, of a bunch of guys standing on the end of a rather long lever to do this kind of measurement?

[Flieger]

Originally Posted by Walt Fricke

But 200 pounds (a guy sitting on the lever) would need 32.5 feet to apply a force of 6,500 lbs, would it not?

A lot of force may be necessary. Remember the photo, in Puhn I think, of a bunch of guys standing on the end of a rather long lever to do this kind of measurement?

If the lever arm was 32.5 feet (not 32.5 feet from weight to opposite side crossmember mount) then there would be 6500 ft-lbs of moment on there, so I would expect the chassis to deflect .5 degrees or less based on the stiffnesses given in post #34. The tension force in one bolt would be about 3350 pounds (I assumed a 24" bolt spacing, I fon't know what it really is). The other bolted joint would have 3150 pounds of compressive force in it, so no bolt necessary.

[JohnJL]

Hi guys,
Do keep discussing, just letting you know im away on holiday without access to the car for 10 days. I will resume next saturday.

Regarding the discussion on the steering rack bolts, isnt it a bit of a moot point? If it takes more force to twist the chassis than the steering rack bolts can withstand, and the bolts didnt break or pull before, then the car wasnt being subjected to enough twisting force for it to do so? So it doesnt matter? I guess it matters if my experiment subjects the car to more force than it encounters while driving and i were to break something. But that is an interesting question for me...if i have to subject the car to so much force that it exceeds normal use to get a measurable deflection then my results dont really matter for the purposes of seeing if my chassis mods impact handling?

[1QuickS]

The rack mount does not experience much force in tension or compression during operation but does react lateral loading. Its design was probably based upon loads induced by bumping a curb or bouncing through a chuck hole and not from normal loads due to tire friction with the road such as from braking or cornering. The large bolt is simply large due to the requirement of the lug welded into the chassis needing enough shear strength for the lateral loads. As such, the lug is probably adequate for reasonable tension/compression loading per your requirements.

Based upon your setup I calculate:

• 90 pounds generates 590 ft-lbs of torsion to chassis
• assuming torsional spring rate of 15k ft-lbs/degree your chassis twisted 0.04 degrees
• tension load on cross bar bolt (closet to the applied weight) is 298 lbs
• compressive load on opposite mount is 208 lbs
• deflection of your loading bar is 2.25" assuming 1 3/4" x 1 3/4" x 3/16" angle
• compressive stress in leg of angle is 40.5ksi which is 80% of yield strength of your angle

[JohnJL]

Originally Posted by 1QuickS

The rack mount does not experience much force in tension or compression during operation but does react lateral loading. Its design was probably based upon loads induced by bumping a curb or bouncing through a chuck hole and not from normal loads due to tire friction with the road such as from braking or cornering. The large bolt is simply large due to the requirement of the lug welded into the chassis needing enough shear strength for the lateral loads. As such, the lug is probably adequate for reasonable tension/compression loading per your requirements.

Based upon your setup I calculate:

• 90 pounds generates 590 ft-lbs of torsion to chassis
• assuming torsional spring rate of 15k ft-lbs/degree your chassis twisted 0.04 degrees
• tension load on cross bar bolt (closet to the applied weight) is 298 lbs
• compressive load on opposite mount is 208 lbs
• deflection of your loading bar is 2.25" assuming 1 3/4" x 1 3/4" x 3/16" angle
• compressive stress in leg of angle is 40.5ksi which is 80% of yield strength of your angle

You just blew my mind.

Before I left I did bolt the crossbar back on and removed the ruler/placed a bolt on the end of the lever bar so I can just stack the weights on the end without them slipping off. I am going to take some collective advice here and;

- I took the straps off the torsion bars and will hook the body directly to the lift arms - no bushings there.

- Will use the bar bolted to the steering member bolts only for loading with a bunch more brake discs.

- Will measure the change in angle across the rack rather than the change in height of the end of the bar

Still not intending this to be lab-level data but I am listening to suggestions here.

I als realize earlier I used the term "gravity-based" in reference to my dial indicators. What I should have said is that they are not spring-loaded. I have typically used them in a vertical, upright position such that gravity pulls the measuring pin down rather than any spring.

[911pcars]

"....I als realize earlier I used the term "gravity-based" in reference to my dial indicators. What I should have said is that they are not spring-loaded. I have typically used them in a vertical, upright position such that gravity pulls the measuring pin down rather than any spring.
"

I've never seen dial indicators without some mechanism to apply some tension to the arm/plunger. That would require a precise low friction mechanism to accurately measure movement. And other than used in a vertical axis, it wouldn't be accurate if used to measure movement at any other angle (e.g. horizontal and thereabouts). You sure the plunger only uses its weight to maintain contact with the part to measure?

Sherwood

[JohnJL]

Yep. One I got from Pelican about 10 years ago when I did my first engine build. I forget where I got the other.

[LJ851]

Originally Posted by JohnJL

Yep. One I got from Pelican about 10 years ago when I did my first engine build. I forget where I got the other.

Maybe they are just broken...

[1QuickS]

Forgot to add deflection at end of angle due to chassis twist which is an additional 0.06". Note that deflection of angle is 2 1/4" and deflection due to chassis twist is 1/16"; therefore, the angle as a loading beam is too limber and is close to yielding/buckling already.

Recommendations:

• support front with digital scales under loading beam
• use 2" x 3" x 3/16" wall rectangular tube to load chassis; orient so 3" face is vertical
• create anchor point in floor and use turnbuckle to load up the beam
• attachments to beam to be through the 2" wide wall which would be horizontal
• place digital level on cross member between rear shock mounts
• place digital level on horizontal member in front of gas tank

Dial indicators will record movement at the strapped mounting points as twist which is an error. Also, if dial indicators are not isolated from the supporting structure then they will move with the supports which is an error, they must be mounted to the shop floor to help assure they are as isolated as possible from motion of the chassis and its supports.

[Walt Fricke]

Paul
His goal is not to that level of precision.

For the stiffer beam suggested he'd have to conjure up a way of mounting the beam. With horizontal bolting he'd need to weld a small tube in to hold the horizontal bolt to replace the strength lost by drilling (though it could be done in the neutral axis to reduce that effect). That bolt could be larger than stock so its max shear would approach max tensile in the current arrangement. And he'd have to weld up some kind of fitting which would bolt into the stock location, and then stick down on both sides of the beam for the cross bolt.

He has said he doesn't want to come up with floor stands for dial indicators.

I am pretty sure he isn't going to want to install something on his garage floor for a one off experiment, and getting a pull out strength for some kind of drill a hole and insert a fastener system may not even be possible, as those are usually set for shear. I was happy enough using old 1/16" bolts climbing the Nose of El Cap, but I was very careful not to place an outward pull on them 30 years ago.

And won't digital levels, placed as suggested, be about as good as dial indicators? Perhaps we are not concerned with deflections of less than 0.1 degrees? Yes, given the fairly modest weights he proposes using, or even just 200 lbs of largish male, if he could be very precise then extrapolation would be more valid, I think.

The levels will compensate for the non-rigid attachment at the rear, won't they? Wasn't that your idea earlier?

What he may prove to himself is that the chassis deflection he records is so far down in the noise that it won't tell him if various strengthening measures he anticipates taking are doing much.

But it looks like a worthwhile endeavor, and he's pretty much got it all set up. Nothing much lost if his angle arm buckles.

[1QuickS]

Right.