
#55
Feb2912, 09:33 PM

P: 63

nvn,
Also, attached, I have my bucket forces for position 3. Please look this over and tell me if it is correct and also if my signs in my math are correct. Thank you. 



#56
Mar112, 01:06 AM

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P: 2,110

grandnat_6: In your file shear_momentforcesp2.pdf in post 54, M_37.9665 and M_37.9665+ are wrong. Follow carefully the example I gave you in shearmomentforces12.png in post 49. Also, in your M_57.8547 equation, you should not have two 1007.50 values. Carefully proofread everything you type. File bucketforcesp3.pdf in post 55 currently looks correct.




#57
Mar112, 11:10 PM

P: 63

nvn,
I was wondering why M 57.8547 did not come out when I punched in numbers by the diagram and then used the equation I typed and they did not come out exactly the same. I double checked it, and thought it was something to do with my calculator. Instead of panning around on my little 13" laptop screen I should probably print it out so i can get a full view of it. I believe my shear_momentforcesp2 should be correct now. I also attached armforces3. This should be correct too. Thank you. 



#58
Mar212, 12:39 AM

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P: 2,110

grandnat_6: In shear_momentforcesp2.pdf, you computed M_37.9665 wrong. Did you doublecheck your calculation? Secondly, did you draw the 801.554 vector backwards? Check whether it should be up or down. File armfocesp3.pdf currently looks correct.




#59
Mar312, 10:28 PM

P: 63

nvn,
You are right, I did miscalculate M37.9665, and got the vector of 801.554 backwards. I think I have it fixed now in Shear_momentforcesp2.pdf. I also showed on the plot what the maximium bending moment is for beam 2. I have a question on the attached armvectorforces3.pdf. I could not get it to balance out. I spent a good amount of time checking my numbers and my math to make sure it was correct. I did notice on triangle GEH, I used the angle in green (46.6413) I found out if I use the angle 43.3287. it balances out. I think this is the only thing I did wrong. Is there a rule for which angle you are suppose to use? I'm surprised this did not come up before. The attached shear_momentforcesp3 shows my shear diagram and moment diagram. The shear diagram does not cross the neutral axis. I have shown on the plot the maximium bending moment of beam 2. If this is all correct, should we start sizing up the beams or move on the the uprights? If sizing up the beams, is MC/I going to be the only thing used, or will we be using F/A±MC/I? If the later, then I don't really know where to start, but will give it a try. Thank you. 



#60
Mar412, 12:48 PM

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P: 2,110

grandnat_6: In shear_momentforcesp3.pdf, you computed M_32.1404 wrong. Did you doublecheck your calculation? Always do. M_37.9665 is again wrong (same mistake you made in post 54); see post 56. Ironically, M_32.1404+ and M_37.9665+ are correct. I think this mistake would also make your maximum bending moment on beam 2 wrong. The M_57.8547 equation has the wrong sign on one term; correct the wrong term. Ironically, the righthand side of the M_57.8547 equation has the correct answer, except it should be 0.143462, not positive.
You should use 46.6713 deg. Actually, 46.6713 deg, to be exact. This did not come up before, because you happened to do it correctly. Use the angle you are rotating the structure. You rotated the structure 46.6713 deg; i.e., 46.6713 deg counterclockwise. Generally, use (P/A) +/ (M*c/I), unless P/A is negligible. For beam 1, for position 3, P = FEx = 388.0, from point E to D; and from point D to G, P = 388.0  1335.0 = 947.0. 



#61
Mar412, 04:51 PM

P: 63

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 23 day break and try not to think about it, and come back fresh. Regards. 



#62
Mar412, 07:32 PM

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P: 2,110

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? Doublecheck that.




#63
Mar512, 10:44 PM

P: 63

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 1557.692504.66= 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. 



#64
Mar612, 09:44 AM

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P: 2,110

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.




#65
Mar812, 11:07 PM

P: 63

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 



#66
Mar912, 05:57 PM

P: 63

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. 



#67
Mar1112, 06:41 AM

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P: 2,110

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 = crosssectional 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. 



#68
Mar1112, 07:56 PM

P: 63

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 whats 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. 



#69
Mar1212, 01:55 PM

P: 1

i thank you all very much! Okeke




#70
Mar1312, 09:58 PM

P: 63

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, 



#71
Mar1412, 12:08 AM

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P: 2,110

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 text books 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 aswelded (not annealed), normalized, DOM, or DOM stressrelieved, 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 aswelded, depending on the SAE grade designation. However, A500, grade B, on the other hand, specifically defines a strength. 



#72
Mar1412, 09:49 PM

P: 63

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, 


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