Material Strength

I am okay with the calculations, need help finding the formulas.

Assume a section of 4130 Tubing 2"OD x .049" is sufficient in strength for a given application with concerns only about loads in two directions 90 degrees to each other and axial loads applied at each end.

I'm not concerned with the point at which the material will fail, only how much it will yield under a certain force.

How can I determine what size 6061-T6 tubing within a 2-3" diameter range will give the same "strength"?

How about mild steel in a 1 to 1 1/2" diameter?

If you are feeling really generous Ti and Carbon Fiber are also options.

Thanks for your help.

This should be close, but I'm sure a Mech Engineer will be closer than this Elec Engineer...

Vmax for steel = (-5wl^4 ) / (384EI)


where w = load in lbs/foot
l=length in whatever unit (for comparison this will be the same for both pieces)
E= modulus of elasticity (4130 = 29,700 KSI, 6061-T5 = 10,000 KSI)
I= moment of inertia

For a tube take the moment for ID and subtract it from the moment of OD

Formula is I= 1/4 * PI * r^4 for both calculations

You will then have to transpose for a tube diameter and available wall thickness.

It's 4:50 on a Friday and it's too late to do that.......


good luck

Excellent.

Thank you much.

I can google to find out the modulus of elasticity for all the materials I am considering.

So, is the amount a material will twist directly proportional to its' modulus of elasticity (given same MOI and length) ? OR is Vmax giving me the amount the material will yield in a horizontal/vertical plane?

Lets see if I have the problem correct.

You have a round tube of know diameter, wall thickness, and length.

The applied loads are torsional (twisting at both ends) and one (just for concept) vertical load and at some point between the ends and one horizontal load at some point between the ends.

Correct so far?

The next question is how is the tube supported?

Since this is a combined load you will need to solve in the following order:

1. Draw Free Body Diagram
2. Find Reactions
3. Shear Diagram
4. Bending Moment Diagram
5. Find bending stress
6. Find vertical shear stress
7. Find Critical Points
8. Using Mohr's circle find max Principal stress
9. Failure Analysis using max shear stress theroy or the distortion energy theory
10. Then check for deflection

This is for a ductile material under a static load - if its brittle or under a cyclical load then step 9 changes.

All in the day of the life of a ME

See if I can simplify/eliminate variables. In all cases the loads will be identical in location, magnitude, areas of support and length of tube.

The only variables would be the OD, ID and material.

Assume somebody much more educated than I has already done all the calculations and determined that 2" OD X .049" 4130 is adequate for the intended purpose.

200# static load in the center of a 48" long tube supported at each end.

axial load is 34 ft lbs with a 1ft lever arm.

Assume that whatever amount of bending and twisting of the tube is the maximum acceptable for the application.

48" long tube with a 200lbs transverse load applied at the 24" point; how are the ends supported?? knife edges?? clamped with one or more set screws?? embedded in blocks??

how is the 34 ft-lb torque applied?? 1 ft = ???

if this is a simple materials stiffness trade life is easier. E2 x I2 ≥ E1 x I1 takes care of bending stiffness where E is Young's Modulus for the material and I is the moment of inertia about the neutral axis of the section. and G2 x J2 ≥ G1 x J1 takes care of torsional stiffness where G is the shear modulus for the material and J is the polar moment of inertia of the section about the axis of torsion. be careful of the units..........................

if human safety is involved, have a Professional Mechanical Engineer or Structural Engineer do the combined loading analysis for your application.

Art
artesmith@earthlink.net

the suspense is killing me, what are you trying to build???

If the design goal is really just to match the stiffness of the existing tube then it's easy. To first order, the stiffness in both bending and torsion is directly proportional to material stiffness multiplied by (OD^4 - ID^4).

For example, aluminum is about 1/3 as stiff as steel, so a 3.00 OD x .049 aluminum tube would be slightly stiffer (because (2.00^4 - 1.902^4) < (3.00^4 - 2.902^4)/3). And it would weigh about half as much since aluminum is also about 1/3 as dense as steel.

In practice, it's quite likely that the stresses at the attachment points for the load or supports will dictate the size of the tubing rather than the pure stiffness of the section.

If the constraint is maximum OD of 3.00, then the above mentioned aluminum tube would be the lightest metallic solution. Carbon fiber makes it look silly, but you'd have to worry about fiber orientation and the stiffness calculation is a lot more difficult.

Nathan

Quote:

---

Originally Posted by Quickshoe

See if I can simplify/eliminate variables. In all cases the loads will be identical in location, magnitude, areas of support and length of tube.

The only variables would be the OD, ID and material.

Assume somebody much more educated than I has already done all the calculations and determined that 2" OD X .049" 4130 is adequate for the intended purpose.

200# static load in the center of a 48" long tube supported at each end.

axial load is 34 ft lbs with a 1ft lever arm.

Assume that whatever amount of bending and twisting of the tube is the maximum acceptable for the application.

---

So a static load of 34 lbs will be applied to the end of a 12" bar attached perpendicular to a round hollow section (tube). Basically half of a torsion bar setup, correct?

Quote:

---

Originally Posted by Art Smith

if this is a simple materials stiffness trade life is easier. E2 x I2 ≥ E1 x I1 takes care of bending stiffness where E is Young's Modulus for the material and I is the moment of inertia about the neutral axis of the section. and G2 x J2 ≥ G1 x J1 takes care of torsional stiffness where G is the shear modulus for the material and J is the polar moment of inertia of the section about the axis of torsion. be careful of the units..........................

---

Thank you Art and others. That will be perfect.

Quote:

---

Originally Posted by ArtSmith

if human safety is involved, have a Professional Mechanical Engineer or Structural Engineer do the combined loading analysis for your application.

---

Where is the fun in that :D:D?

Quote:

---

Originally Posted by provamo

the suspense is killing me, what are you trying to build???

---

Nothing terribly exiting. Custom bicycle frame. Sorry for the anti-climactic answer. Lots of intelligent people here, so I knew I'd get my answer, instead of somebody telling me they made theirs out of muffler tubing or fence post.

Quote:

---

Originally Posted by nulrich

If the design goal is really just to match the stiffness of the existing tube then it's easy. To first order, the stiffness in both bending and torsion is directly proportional to material stiffness multiplied by (OD^4 - ID^4).

For example, aluminum is about 1/3 as stiff as steel, so a 3.00 OD x .049 aluminum tube would be slightly stiffer (because (2.00^4 - 1.902^4) < (3.00^4 - 2.902^4)/3). And it would weigh about half as much since aluminum is also about 1/3 as dense as steel.

In practice, it's quite likely that the stresses at the attachment points for the load or supports will dictate the size of the tubing rather than the pure stiffness of the section.

If the constraint is maximum OD of 3.00, then the above mentioned aluminum tube would be the lightest metallic solution. Carbon fiber makes it look silly, but you'd have to worry about fiber orientation and the stiffness calculation is a lot more difficult.

Nathan

---

Thanks again everybody.

How about this homemade job: (not mine)
http://4.bp.blogspot.com/_atq6Cg90KP...00-h/KA+LD.jpg

Quote:

---

Originally Posted by LJennings

So a static load of 34 lbs will be applied to the end of a 12" bar attached perpendicular to a round hollow section (tube). Basically half of a torsion bar setup, correct?

---

Not static at all. Just trying to keep things simple. I'm fairly certain there is enough safety margin in the selection of the baseline set up that I'm just looking for a simple, first order type comparison.

Thanks again to everybody. You've been a great help.