Help with Statics and Strengths of Materials

[grandnat_6]

Hi nvn,

Sorry I did not reply last night. I was with my family yesterday.

The numbers requested are xG=30.9554" and yG=11.3586.

I have attached beam 2 vector analysis and beam2 shear_momentforces.

Since we need to use six significant digits, I choose to change the numbers you provided and move on. I believe my shear diagram is good, but is off slightly because of the signifiant digits. I also tried to follow your math for the moment diagram. I am not quite sure, if we are taking moments around D; are figuring for two separate moments for D? It appears, (because of the significant digit error) that maybe you are summing only the forces to the left in one equation and then plotting it. Then you are slightly moving to the right of D by .0001 and then summing again including the force and moment in D?

If you don't mind, and you have the exact numbers, could you please complete and show me how you figured out the maximium moment like in analysis10.png? This way we can use the numbers in the future. I would like to move forward and start analysising a new position where pins E and B are in the same horizontal plan with the bucket tipped so the bottom side is horizontal with the plane of pins E and B. I think this should be the next step. I'll use six significant digits both for force and dimensions from this point on.

Let me know what you think about this.

Thank you.

 

[nvn]

grandnat_6: I wanted only point G and the pin locations accurate to six significant digits, before the structure is rotated. Five significant digits is adequate for most of the other values. There is generally no need to use six significant digits for most of the other values. Since I forgot to ask before, could you give me dimensions xA, xB, yB, xC, yC, xD, yD, xF, and yF accurate to six significant digits, before the structure is rotated, before I reply to the other questions? (By the way, normally you would not use six significant digits. This exception is just so I find out the true, initial location of the pins and structure, instead of wasting time on irrelevant round-off error.)

All moments in analysis10.png are taken about the centerline of the members. Yes, in analysis10.png, the first moment equation is adjacent to point D (or an infinitesimal distance to the left of D). The second moment equation is an infinitesimal distance (0.000001) to the right of D. Neither equation includes FDy, because FDy causes no moment about the summation point.

 

[grandnat_6]

nvn,

I'm assuming you wanted the dimensions from pin E? Attached are those dimensions.

Thanks.

I've also looked over your moment diagram and calulations for beam 1 for a fifth time. I went ahead and attempted to calculate beam2. I think I have it, just that the slight differences in our dimensions might have to do with it.

I am also assuming in your moment calcuations the + sign means you are adding the moments produced in the Y direction?

On your moment calcuation of M57.855; why the reason for subtracting 19.258"-3.1980" for the moment? Also, why is the force of FEx being calcuated as a moment? How I understand it, we are checking for beam one, every dimension and force is being calcuated from the centerline of the beam. Since FEx is acting on the center line, how it is producing a force on beam 1. I must be missing something...?

Thanks.

 

[nvn]

grandnat_6: Are you sure you gave me accurate values for xG and yG? Could you double-check the accuracy of those two values? Point G defines the location and slope of both beams, and I'm wondering if it might be throwing a few of my numbers off. Also, what is the x dimension from E to the bucket tip (where the applied loads are applied)?

Regarding your questions in post 38, the plus (+) sign means, it is a moment summation at an infinitesimal distance to the right of the specified x coordinate. Regarding your second question, moments are computed from the centerline of beam 1, until we get to beam 2. Then moments are computed from the centerline of beam 2. The vertical distance from point B to FEx is 19.258. The vertical distance from point B to FDx is (19.258 - 3.1980).

 

[grandnat_6]

I see CAD did not set the x direction of pin B on the center mark of pin B. This might be the error. I've added some extra angles for reference.

Let me know if you need anything else.

Thanks.

 

[nvn]

grandnat_6: I think the dimensions in post 40 give an angle of 142.25971 deg between centerlines GE and GB, not 142.39913 deg. I called the load application point at the bucket tip point H. Post 40 changes the vertical distance between points A and B to 7.437320, and changes the vertical distance between points H and B to 4.010720, which supposedly changes an earlier bucket diagram. Is that what you want? And also, e.g., the 20.04246 dimension in beam2shear_momentforces.pdf currently does not correspond to post 40.

In post 40, the centerline length of beam 1 (EG) is 32.973505, and the centerline length of beam 2 (GB) is 31.463566.

 

[grandnat_6]

nvn,

This project is becoming a real bear for us. I went back to the start and found the bottom of my bucket was not quite horizontal.

Sorry, for the inconvenience about this.

I have re-drawn the bucket and have dimensioned all the x,y pin locations including the G point from pin E. Then I rotated the arm and have dimensioned all the x,y coordinates from pin E. If your calculations agree with mine, then we will be on the same page. I'll start over with the calculations and use six significant digits. I know you mentioned to only use six for the rotation, but I want to make sure it is accurate as possible for the forces and dimensions as well.

Regards,

Rob

 

[nvn]

grandnat_6: No problem. Good catch. All of my dimensions match post 42 to four decimal places, and most of my dimensions match to five decimal places.

Five significant digits for hand calculations is adequate (and excellent). You usually never need six significant digits for hand calculations. Usually, you use at least four significant digits for hand calculations, or you can use five significant digits when you want to be more precise. You can use six significant digits occasionally, as long as you know it is unnecessary, and is often not done.

 

[grandnat_6]

nvn,

Attached are my new bucket and arm calculations. I did use six significant figures. I noticed on the bucket after I was done finding all forces, there is an error of .003 in the Y direction. After finding all the forces on the arm. The numbers match perfectly.

I'll work on the rotated arm and shear/moment diagram tomorrow.

Thanks.

 

[nvn]

grandnat_6: After you obtain answers, you can check your work, yourself, by summing forces and moments, to ensure they balance to approximately zero. Your bucketforces.pdf file in post 44 looks correct. In your armforces.pdf file, 70 037.8 is wrong. If you try a moment summation check using your answers in armforces.pdf, you will see the moments do not balance to zero. The 70 037.8 is wrong. Try again.

 

[grandnat_6]

nvn,

I did check the forces in the x and y to make sure they were balanced. I even checked those calculations 3 times! err... Figures, I went back to check and get a different answer right off the bat. Maybe I'm working to hard...

Anyway. I have attached the fix, I also attached my vectorforces for rotated beams and the shear moment diagram. The moment diagram is about 7lbs short. I hope it is not wrong.

Let me know.

Thanks.

 

[nvn]

grandnat_6: Could you give me your armbucket_dimensions.pdf file in post 42, except make all dimensions and angles therein accurate to six decimal places? And also ensure point G is accurate. I want to double-check my initial geometry.

 

[grandnat_6]

nvn,

Attached are the requested dimensions to 7 decimal places.

 

[nvn]

grandnat_6: Corrections to your armforces.pdf file are highlighted in blue in armforces12.png, attached below. Study this file, so you will understand the mistake you are making with minus signs. Remember this important rule.

If a vector is drawn in the negative direction (or would cause a negative moment), then the term is preceded by a minus sign in the summation general equation.​

I made a mistake in a moment diagram calculation in a previous file, so I corrected my mistake in http://img12.imageshack.us/img12/7648/shearmomentforces12.png . Because moments are computed from the centerline of beam 2, the moment must be computed about point I, as shown in the file.

Due to round-off error, the moment balance is off by -3.49, which is close enough and good.

 

[grandnat_6]

nvn,

I think I understand what is going on with the moments. Since point G is the intersection of beam 1 and beam 2; anything from the right of point G has to be for beam 2. Likewise anything to the left of point G has to for beam1. Correct? If so, what happens if point C or D lands on intersection G? I'd assume it would work in with and be calculated for both beams?

I take it then our maximium moment for beam 1 is 45,350.3"# and beam two is 38,443.2"#?

So, at this point this position is done being calculated, next is to start with position 2? Then finally position 3? Then we can use the biggest bending moment of each beam to size up the beams?

I'll pay better attention to my signs.

Thank you.

 

[nvn]

grandnat_6: Paragraph 1 in post 50 is correct. If point C or D lands on point G, then let it be on beam 1. Or, you could say it is on both beams, if you wish. Either way you prefer.

Yes, your maximum moment on beam 1 in post 50 is correct. You can see, the maximum moment on beam 2 is directly below point G on your moment diagram, not at point I.

So far, your plan sounds good.

 

[grandnat_6]

nvn,

I understand what you are saying about point G and beam 2. The max moment for beam two is slightly lower than Beam 1 max moment.

I have attached the new position of the loader and bucket. I have found all forces acting on the bucket and also on the arm. I hope I improved this time around.

I did notice there is a slight error in the Y direction of the bucket, and also on the arm. I hope I did not make an error.

Thanks.

 

[nvn]

grandnat_6: All three files in post 52 are correct.

 

[grandnat_6]

nvn,

I've been keeping busy on this. Attached you will find my shear moment diagram for position 2.

I've also dimensioned position 3.

Let me know if I done anything wrong.

Thanks.

 

[grandnat_6]

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.

 

[nvn]

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.

 

[grandnat_6]

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.

 

[nvn]

grandnat_6: In shear_momentforcesp2.pdf, you computed M_37.9665 wrong. Did you double-check 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.

 

[grandnat_6]

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.

 

[nvn]

grandnat_6: In shear_momentforcesp3.pdf, you computed M_32.1404 wrong. Did you double-check 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 right-hand 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.

 

[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,