Help with Statics and Strengths of Materials

i thank you all very much! Okeke

 nvn, I've been doing a lot of research the last few days. Can you help me clear the air on this? In terms of shear strength, tension yield strength, compression yield strength; I've been reading some use a value for shear of .75, .60, or .557 as the mulitplier for ultimate and yeild strength, to define the allowable shear strength. Is shear strength suppose to be a different value compared using a safety factor of 3 for use in yeild tension and yield compression? Is the value of .75,.60,.557 suppose to be the point where it shears, and then you have to add a safety factor on top of that? So far my fabricator told me A513 is the economical choice, 4130 tube is availible for higher strength. He can also price me some dome x100 or A514, but will have to make the beams. Thank you,

nvn,

I was unable to get any info from my supplier on the tubing, he seems like he does not want to share this with me. Since it was though email, I'll give him a few days, perhaps he is working on it.

I think I understand what you are saying in your last post. I have done an example, please see attached.

Beam 1 in position 1 has the most stress, therefore I only sized up the beam to fit position 1 as it will be more than enough for position 2 and 3. I have also found the yield shear strength, yield tension strength, and yield compressive strength. Since the material is isotropy, in this situation only the tension would need to be calculated. The allowable shear in yield is the most burdening factor, I would have to say the shear stress would be the only thing needed to be calculated. Once the beam meets the shear criteria, it will be well within the tension and compressive stress allowable stress.

I did not use FSu=FSy for now, because I don't understand what you mean by "for now".

Since the 500# and 850# force is double the load per arm, since each arm ideally carries the full load most times; I have divided the axial force and bending moment by 2 and have sized up another beam for that was well. Because there are instances where one arm may carry the majority of the load, I am unsure to use the safety factor of 3 for the double load or the single load.

Please let me know if my math and assumptions are correct.

Thank you,
Attached Files
 BEAM1SIZEP1.pdf (23.2 KB, 5 views)

 Recognitions: Homework Help Science Advisor grandnat_6: Forget about shear yield strength; you virtually never need to use it. Secondly, you do not need to compute shear stress for sizing the global cross section of beam 1 or 2; therefore, delete shear stress and shear strength from your current calculations. All you need to check for globally sizing beam 1 or 2 is the axial stress plus bending stress, and compare it to tensile yield strength, Sty. You can let the full load, in position 1, be applied to only one arm. And I think you can use FSy = 3. Let's say you have a 4 x 2.5 x 0.3125 rectangular tube for beam 1. Even though the axial force reduces immediately to the right-hand side of point D, let's use the axial force immediately to the left-hand side of point D, regardless, which is P = 2.871 kip. If Sty = 50 ksi, the calculations would be as follows. (1) sigma = (2.871/3.23) + [(45.350 kip*in)/3.06] = 15.709 ksi. (2) Ny = Sty/(FSy*sigma) = (50 ksi)/(3.0*15.709) = 1.0610 > 1.0; therefore, not overstressed.

nvn,

Thank you for the help, It didn't really make sense to use shear in yield anyway. I'd assume you would want to use Shear ultimate strength when you want to design something to shear like a key or a shear pin? You wouldn't want a safety factor for this either, correct?

I was unable to get pricing from my fabricator yet, but I decided to go ahead and make some beams out of 100XF. Attached is my work. on the left side of the shear/moment diagram I have calculated the beam size. The number I use for moment of inertia and area, was obtained directly from the CAD program. I subtracted the OD area and moment of inertia from the ID to obtain my numbers. I then made a beam to these numbers. I hope it is all correct.

To the left of the shear/moment diagram I have tried to make a tapper-ed beam. I started by using the beam selected from the left hand side and used that size for that particular moment. Moving to the left 10" from that moment I made a smaller beam to the same width and thickness. Calculating out a size for that moment, and plotting it on the shear/moment diagram it gave me the angle for the beam. Now on the end of the far left of the shear/moment diagram where zero is at, there I'd assume there is only the axial force of 2871# acting on that area. But it is so small the radii of the beam would not fit. According to a steel website that offers 100XF states to use a bend radii of 1.75 x T. At the point of zero there is no way the beam can be made that small keeping the minimum radii, so I was not able to check the 2871#F with the cross sectional area. So I would have to widen my tapper to a minimal length to satisfy the bending operation and to make sure the cross sectional area will fall within the safe allowable stress. The only other way to make this better is to manipulate the P/A +/- MC/I=St equation to solve for A,C, and I, while using St =32614psi from the previous equation. Almost impossible.

Does this all sound correct?

Thank you!
Attached Files
 BEAM1SIZEP1.pdf (28.0 KB, 6 views) OD3.5X2X.188.pdf (38.0 KB, 1 views) ID3.5X2X.188.pdf (38.3 KB, 1 views)

 Recognitions: Homework Help Science Advisor grandnat_6: Yes, you would use shear ultimate strength (Ssu) for pins and keys. And yes, use an ultimate factor of safety (FSu) for this. At the tapered end of beam 1 (point E), you have axial force, P = 2871, and shear force, V = 1063. I did not know you would make a tapered beam. Therefore, you need to check axial stress and shear stress at end E. Therefore, to check the general beam size at end E (not necessarily connection details), you can use the following. (1) Axial stress, sigma = P/A. Ensure Ny = Sty/(FSy*sigma) ≥ 1.0. (2) Shear stress, tau = 1.50*V/(2*h*t), where h = beam cross-sectional depth, and t = wall thickness. Ensure Ny = 0.60*Sty/(FSy*tau) ≥ 1.0.
 nvn, For the tapered beam, I'm keeping my options open, and never done one in school. So I would like to know how to make one. I understand where you are getting the axial stress from. It makes total sense. I have not seen the equation in #2 before. (2) Shear stress, tau = 1.50*V/(2*h*t), where h = beam cross-sectional depth, and t = wall thickness. Ensure Ny = 0.60*Sty/(FSy*tau) ≥ 1.0. I have done some google research. On wikipedia I found a formula for that is different, but I can tell it will not work in this situation because there is no moment. tau = VQ/It. Where does the 1.50 come from? What is the 2 from in (2*h*t)? For H, is that cross sectional depth from E to D? So, 32.1404? Or is it the distance from the centroid to the extreme fiber? Thank you.
 Recognitions: Homework Help Science Advisor grandnat_6: 1.50 comes from V*Q/(I*t) for a rectangular cross section. The 2 is because there are two webs (two sides) in a rectangular tube cross section. The h is two times the distance from the centroid to the extreme fiber.

nvn,

Attached I have done the math for Point E. I am kinda limited on this because of the bend radii. I think my fabricator will be able to make the bends, but I'll find out from him. It's not going to matter in this situation because I'm well below the safe shear value, but should I be adding P/A to 1.50*v/(2*h*t)?

A question that now arises, When I get the two halves bent; where should the seams be ideally? I want to say the seams should go on the YY section. They say the welding bead is stronger than the base metal, but I feel since XX is doing the main work, it might be possible to change the metalurgy of the 100XF metal and weaken it. What do you think?

In post 64 it was mentioned the axial force for beam 3 is P=2774#, could you please show that math for this? I can not figure out how you obtained that.

Thank you.
Attached Files
 BEAM1SIZEP1.pdf (31.2 KB, 6 views)

 Recognitions: Homework Help Science Advisor grandnat_6: Do not add P/A to 1.50*V/(2*h*t). You misunderstood dimension h; it is cross-sectional depth, not extreme fiber distance (c). Try rereading post 77. I am not sure which faces are best for a seam. I currently guess the side faces would be best, instead of the top and bottom faces. However, check a few existing electric-resistance-welded rectangular tubes, to see where the seam is located; I would currently guess it is on one side, not the top nor bottom face. The axial force in member BG is the axial component of forces FBx and FBy. Therefore, for position 3, P = -2504.66*cos(37.6009 deg) - 1294.11*sin(37.6009 deg) = -2774.00.
 nvn, If I read post 77 correctly now. h should equal 1.625". If this is correct, tau will equal 2610 psi. Beam 2 position 1 carries the most stress from the other two positions, the stress in beam 2 is very close to the stress in beam 1, and meets the minimium safety factor. Therefore I might as well make beam 1 and beam 2 with the same ends. Beam 2 will just have a slightly larger taper (2.19 degrees vs beam 1 1.67 degree) due to the difference in length betweeen beam 1 and beam 2. Thank you for showing me the math for position 3 beam 2, I was using the compliment angle of the 1294.11# force. Thank you for your help!

nvn,

Attached I have done the vector analysis of the upright in position 1. Vector hj is a link connection on the backside of the upright at point h and transfers force to a pin at the front of the tractor at pin j.

g is a connection point to the sub frame, and is colinear vertically to pin E.

Please ignore in my calcuations the missing # and * signs as for some reason either the lap top or the program would not allow me to enter those signs, even after a reboot.

Is this look correct?

Thanks.
Attached Files
 uprightp1.pdf (25.4 KB, 4 views)

 Recognitions: Homework Help Science Advisor grandnat_6: No, that is wrong. You did not reverse the direction of forces at point E. And, it seems you did not even bother to check whether or not your forces and moments (your answers) are in equilibrium. Check your signs. Also, try to not address general questions specifically to me (unless it is something specific to me), so you could possibly get help from anyone reading this forum.

yep, thought I forgot something.

It balances out now. This correct?

Thank you.
Attached Files
 uprightp1.pdf (25.8 KB, 3 views)

 Recognitions: Homework Help Science Advisor grandnat_6: Your answers in post 83 look correct.

Attached I have the shear/moment diagram, and have also sized up a beam. Is this correct?

Also, I have made an alternative beam. It's a U-shape. In this one, Pin E is offset from Pin g by 1" this makes a very minute change to the forces. I assume because of this it should be ok to use the shear moment force diagram from the rectangular tube to size up the U-shape beam. I do show a wall inside the beam that will close the U into a tube. One upright will have this to hold hydraulic oil to run the system the other will not.

Since the shape is not fully closed, I understand my tension and compressive will be different. Will this effect my moment of ineria for calcuation purposes?

Thanks.
Attached Files
 ALTERNATIVE UPRIGHT.pdf (32.5 KB, 5 views) SHEAR_MOMENTFORCESuprightp1.pdf (46.9 KB, 2 views)