Thread: Stiff Members

Sorry about the title it just came out that way.

This discussion started on another thread and instead of derailing the other thread completely I have transferred the posts on to this new thread. I hope the discussion can continue to with the assistance of the real engineers as I am not a mech eng, just an amateur.

I think the tech talk started here thanks to Bush65

Originally Posted by Bush65

Simple structural lesson follows to try and clear-up some of the misconceptions in previous posts.

From a structural approach, under normal conditions the axle housing is a simple beam subjected to bending loads. These loads are called bending moments a moment (similar to a torque), is force applied at a distance from the section of the beam that we are evaluating for strength.

The area moment of inertia and sectional modulus (moment of inertia / radius from neutral axis to extreme fibre), together with the material yield strength are the properties of beam sections that must be determined to find how great a bending moment can be carried by a beam.

The value of a bending moment at any particular section is the vector sum of all applied forces x the distance of each force from the section, taken only for one side from the particular section. To see how this works consider the front axle with the vehicle stationary on a level bit of road and looking at it from the front:

For the section where the swivel bolts to the axle tube flange. The bending moment at the bolted joint is Mbolt = RH side wheel load x distance from the RH wheel to bolted joint. Note: the direction of the moment is clockwise.

For the section at the spring perch, the value of the moment will be greater because the distance to the load at the RH wheel is greater. Its direction is still clockwise.

For any section between the LH and RH spring perches, we have 2 forces of nearly equal magnitude, one vertically up at the wheel and the other vertically down at the spring. Here we have to calculate both of these moments then find the vector sum. As before we have a clockwise vector from the wheel load. The moment due to the spring force is anti-clockwise and smaller magnitude because the distance from the section under consideration is smaller. As a convention we take anti-clockwise moment as negative values - then the vector sum is M = Mwheel - Mspring.

Without going into the proof, it turns out that for this example, the bending moment remains constant for every section between the spring perches. Also note that the bending moment is greatest at the spring perch.

Edit: to continue from where I left off:

If however the diff pumpkin is hung-up on a rock with a wheel off the ground, the bending moment will be greatest at the section at the rock and will have the opposite sign. The moment will then reduce as the section being considered is closer to the spring perch.

Once the bending moment has been evaluated, we can determine the strength or bending stress at the section under consideration. The bending stress in the section increases the further away from the neutral axis (providing the material has not buckled or yielded. So we are interested in the stress at the extreme fibre and bending stress = M / Z (where Z is the sectional modulus and Z = I /y where I is area moment of inertia and y is radius from neutral axis to extreme fibre).

If we want to modify the axle housing to increase its strength, we need to increase the sectional modulus Z. This is achieved by adding material further from the neutral axis. When a beam is subjected to bending the fibres in tension increase in length and those in compression reduce in length. There is a position where there is no change in length and this is called the neutral axis. For a symmetrical section, such as the stock axle housing, the neutral axis is along the centreline, and the distance to the extreme fibre on the tension side is equal to the distance to the extreme fibre on the compression side. For equilibrium (necessary physics), the sum of all the moments of the stress in every fibre on the compression side must be equal to the sum of all the moments of the stress in every fibre on the tension side - do not confuse this statement with the discussion about bending moment of forces, it concerns the properties of the section resisting bending moments.

If more material is added on one side, making the section asymmetrical, the neutral axis is shifted toward the side where the material was added. This somewhat complicates finding the section modulus.

An 'I' section is commonly used for structural members subjected to bending. These have a weight advantage for a required sectional modulus because most of the material is located furthest from the neutral axis. The ratio of the flange width to thickness is chosen to avoid buckling of the flange outstanding from the web.

From the above, it should be seen that the most advantage from any strengthening exercise will result if we add most material as far as practical from the neutral axis. Using material which is too thin can result in buckling.

Then came this

Originally Posted by slug_burner

for the visual leaner, not exact but I think this will help. I hope this does not confuse you even more.







The product of the blue lines multiplied by the red line is greater in the top image in the third pic (even if I subtract the product of that same amount of metal from where it was previously multiplied by a shorter blue line to the old position of the top plate) than the method you suggest by using a wider C section but keeping the top plate in the same place as the second image in first pic. You are adding only a small amount of metal so only marginally stiffer.

Then came this, after that I think we have caught up and will add via new posts

Originally Posted by uninformed

Thanks SB. Am I the only non-engineer type on this site

So it would be marginaly stiffer...but doesnt matter. Im guessing in the real world you can only go so high due to clearance issues, and the taller you make it the sides are more prone to buckling.

So if the modified housing is trying to bend, the top of the "C" section is in compression and the bottoms of the sides of it are in tension, which would be reacting against the top of the axle tube that is trying to be in compression....is this why the neutral point is raised?

If so, with a round hollow section, would there be a "sweet spot" or best place to locate the sides of the added "C" section, to best engage with the top of said tube to try and battle the compression forces.

Would I also be correct-ish in saying that if you have a 80mm OD round tube and to change the top section of it being in compression during bending to being in neutral, you would have to add a "C" section close to 80mm high above it.....with the same wall thickness as the round tube.

Bush engineering at best....WAG really.

what about different profile rather than the standard "C" section that Ben used? I have seen some nice desert racing housings.....mmmm $$$$

Does the "C" section on the lower front of the front axle housing in Series vehicles offer any housing strengthening?

Sorry to derail your thread Ben, great work on the truck so far.....even if you did put air bags in it

again, thanks John. You did a fine job in your description, it is just my comprehension that is questionable.