Help with Statics and Strengths of Materials

[grandnat_6]

nvn,

Please excuse my inability to comprehend this. Trust me when I say I'm working really hard at this, (maybe too hard) I did double check my work, but of course once I post and then go back to it after your reply I get a different calculation. There are so many numbers, I won't doubt if i forgot something or typed something in error.

The M57.8547, I wrote that by hand and got the right answer except I made a typo on the sign in the term when I copied it to the computer. I also completely ignored the minus sign in the answer. I believe I have the correct bending moments now for position 3.

I was unsure if the shear moment force for position 2 was also incorrect , I did notice I mistakenly made a typo and put 2654.52# instead of 2662.52#.

Once I get these bending moment diagrams correct, I think I'll step away from this and take a 2-3 day break and try not to think about it, and come back fresh.

Regards.

 

[nvn]

grandnat_6: In shear_momentforcesp2.pdf, M_37.9665 is now correct. You computed M_37.9665+ slightly inaccurately; and you did not update it on the moment diagram yet. And the peak value on the moment diagram has a misplaced decimal point. In shear_momentforcesp3.pdf, for the maximum moment on beam 2, I currently got -24 101.7. Are you sure -24 150.4 is correct? Double-check that.

 

[grandnat_6]

nvn,

Your right, I did screw up the diagrams again. err.. The attached should be good now.

I think I see how you get P. So I take it for Beam 2 it's C,E 1557.69-2504.66= -946.97?

Thanks for all your help and sticking with this far. I'll post one of the beam sizes in about 3 day.

Thanks again.

 

[nvn]

grandnat_6: Both files in post 63 look correct. Your axial force in beam 2 in the post 63 text is incorrect. I would currently say, for simplicity, assume the axial force in beam 2, from point B to G, is the axial component of forces FBx and FBy. Therefore, for beam 2, for position 3, P = -2774.00, from point B to G.

 

[grandnat_6]

Hi nvn,

Tonight I worked on sizing up beam 1 p3. The attached is my work. The safety factor is wrong, that I know, I know I mentioned in an earlier post I was going to use a safety factor of 7 but it looks like it is going to be too high, this is from comparing to what is in brochures for similar loaders. The beams for the factor of safety 7 would be way too large.(I actually used 5, the +2 comes from the doubled force of each arm) My statics book says to use a safety factor in yeild for steel shock load 5, varying load 3, and steady load 2. When using these in school, I do remember the beams seemed very large and unrealistic from what you would find out in the around us. Do you happen to have an opinion on what should be used? Maybe I wasn't suppose to use the 850# force? It seems to make sense to use it, since the tractor can apply that force when driving into a pile of dirt or gravel.

The beam I used for this is a 2.5x1.5x3/16" rectangular tube. The information I found on the website is http://www.cim.mcgill.ca/~paul/HollowStruct.pdf. I also used the information for the strength from http://en.wikipedia.org/wiki/ASTM_A500.

I'm sure there are going to be a lot of corrections on this. I also could not come up with the same axial force -2774.00 for beam 2.

Regards,
grandnat_6

 

[grandnat_6]

The attached I have sized up the pins for pin E, and B.

I also sized up the connection plate that will be welded to the arms for both pins C, and D.

 

[nvn]

grandnat_6: I am not yet sure what yield factor of safety, FSy, is customary for bulldozer arm beams, but I am currently thinking FSy = 3 might be adequate, because the load is varying. But if you want to voluntarily use FSy = 4 or 5, that is OK. We should probably research what is customary for arm beams. Does anyone else reading this know? I am currently leaning toward FSy = 3, at the moment, without having thought it over much yet.

Your yield factor of safety for the pins, FSy = 5, sounds good, because they will be subjected to shock loading. Using the 850 lbf applied load currently seems correct. You can easily hit a large tree root, a large buried rock, or a buried concrete pier.

I am not yet quite understanding your comment about a beam yield factor of safety of 5 + 2. If you have one beam, a 2.5 x 1.5 x 0.1875 is severely overstressed. Do you have one arm in your assembly, or two arms? If you have one arm, and use FSy = 3, it would be as follows for beam 1, position 3, for a 3.5 x 2.5 x 0.1875 inch steel A500B rectangular tube; A = cross-sectional area = 1.89 in^2, Sx = section modulus = Ix/c = 1.76 in^3, Sty = tensile yield strength = 46 ksi.

(1) Normal stress, sigma = (P/A) +/- (M/Sx) = [(-0.947 kip)/(1.89 in^2)] - [(24.360 kip*in)/(1.76 in^3)] = -14.34 ksi.

(2) Yield safety factor, Ny = Sty/(FSy*sigma) = (46 ksi)/(3.0*14.34 ksi) = 1.069 > 1.0; therefore, not overstressed.

 

[grandnat_6]

nvn,

Glad to hear from you, I thought maybe you gave up on me. I've done an Internet search and have attached some brochure pages for a loader bucket and the tractor I'm going to put it on, I do own a model of tractor specified in the brochure.

I have also attached a page out of the machinery's hand book. I see their multipliers are different that what i used for my calculations. (75% versus 60% I read on line some time ago on line, of course though, what I read was for bolts.) The best thing at this point, I think I should do is talk to my fabricator and have him call his supplier and see what materials are available for steel tubing. Seems like Internet searches came up with A500B or A513. Also have to note, just because they carry a 3x2.5x.1875 beam in A500B does not always mean they carry it in A513 either. So I will have him find out what's available and also what the shear, tension, and compression values are.

Why I mentioned about a safety factor of 5+2, the loader will have two arms, the 500# and 850# is what is acting on the whole assembly. I chose to use those values for one arm, because it is possible to pick something up with the one side of the bucket and not the other, which will end up putting more stress on one arm. So one of the arms have a safety factor of 2 already built into it due to the forces, then adding another safety of 5 would make it a safety factor of seven. Maybe this was not necessary to do?

I also researched, the machinery's hand book which shows different safety factors. According to them, it looks as if a factor of 4 is the highest. From it's reading it appears its due to the reliability of material being used, how severe the loading is, and the environmental conditions.

 

[Okeke O. Inno]

i thank you all very much! Okeke

 

[grandnat_6]

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]

grandnat_6: I have almost always used Ssu = shear ultimate strength = 0.60*Stu, and Ssy = shear yield strength = 0.577*Sty, where Stu = tensile ultimate strength, and Sty = tensile yield strength. Most textbooks claim the above. I currently do not know why Machinery's Handbook instead says Stu = 0.75*Stu.

The above values are shear strength, and tensile strength, the point where the material shears or ruptures. These material strength values do not include a safety factor. They are material strength values, not allowable stress.

I am still currently leaning toward a yield factor of safety of FSy = 3 for your arm beams, and perhaps FSy = 5 for the pins. The factor of safety is the same in tension and compression. The allowable tensile (or compressive) stress is Sta = Sty/FSy. The allowable shear stress is Ssa = Ssu/FSu = 0.60*Stu/FSu, where FSu = ultimate factor of safety. Because your current FSy values are so high, you can just use FSu = FSy, for now.

We do not yet know the tensile yield strength (Sty) of your A513 steel tubes, because you did not state an SAE steel grade designation yet. A513 covers a lot of SAE steel grade designations (SAE 1008, 1010, 1020, 4130, 4140, just to name a few). You (your supplier) must state the SAE steel grade designation, before we can look up the strength of your A513. If no SAE grade designation for A513 tubes is stated, then we would be forced to assume SAE 1008.

And, your supplier must state whether the A513 steel tube thermal condition is as-welded (not annealed), normalized, DOM, or DOM stress-relieved, before we can look up the strength. If no thermal condition for A513 tubes is stated, then we would be required to assume normalized, or perhaps as-welded, depending on the SAE grade designation. However, A500, grade B, on the other hand, specifically defines a strength.

 

[grandnat_6]

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,

 

[nvn]

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.

 

[grandnat_6]

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!

 

[nvn]

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.

 

[grandnat_6]

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.

 

[nvn]

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.

 

[grandnat_6]

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.

 

[nvn]

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.

 

[grandnat_6]

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!

 

[grandnat_6]

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.

 

[nvn]

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.

 

[grandnat_6]

yep, thought I forgot something.

It balances out now. This correct?

Thank you.

 

[nvn]

grandnat_6: Your answers in post 83 look correct.

 

[grandnat_6]

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.

 

[nvn]

grandnat_6: alternative_upright.pdf in post 85 looks correct. In your second file in post 85, you omitted M_11.5700+ in your moment diagram, and moment diagram calculations. Your axial force P is wrong, and is not the maximum axial force in the vicinity of your maximum moment.

Regarding your second question, here you still use the moment of inertia of the cross section, in a similar manner, along with the correct extreme fiber distances. Check textbooks, and example problems in textbooks.

 

[grandnat_6]

SolidElast,

Thanks for the link, I'll have to play around with it some day.
_____________________________________________________________________________

I fixed my moment diagram and calculations for the upright. I have also done the shear moment diagram for the alterative upright.

Is P going to be zero, or is it the product of Fgy and FEy? If it is neither. How is its force calculated? Would this be the same for the alterative beam since point g and E are not in colinear?

Thanks.

 

[nvn]

grandnat_6: P = Fgy + FFy, in the vicinity of a point F section cut. It is roughly the same even if points g and F are noncollinear.

 

[grandnat_6]

The attached I have sized up a beam for upright1. I also sized up a beam for the alternative upright beam. This was done using sections where the pins are located.

A few notes, I didn't size the beam though pin g because I am planning on welding a formed C-channel to the bottom of it. I know this beam could be a lot more narrow, but I wanted to make room to use it as a hydraulic tank.

I didn't check for shear (tau)yet because I did not know if the value in the equation would be different due to the U-shape compared to the 1.50 used for the rectangular tube. If it is different what should the value be?

What factor of safety is recommended for the bearing, tension, shear loads on the 3/16" thick walls for the pins?

Thanks.

 

[nvn]

grandnat_6: You cannot pretend the beam cross section shear force is a bending moment, and use it for the bending stress formula. You must use bending moment in the bending stress formula.

In the bending stress formula, use maximum extreme fiber distance, not minimum extreme fiber distance. You currently used minimum extreme fiber distance, which is incorrect.

You can use the 1.50 factor for shear stress for the U-shaped cross section.

For the pin material, you can use a yield factor of safety of 3, and an ultimate factor of safety of 5. For the beam material, you can use a yield factor of safety of 3.

File shear_momentforcesuprightp1.pdf in post 89 currently looks correct.