# Help with Statics and Strengths of Materials

by grandnat_6
Tags: materials, statics, strengths
 HW Helper Sci Advisor P: 2,010 grandnat_6: Some dimensions you listed in some of your files do not match some dimensions and angles in other files. E.g., one angle I tried to check should perhaps be 17.5924 deg (?), but some of your files list 18.0000 deg. We cannot necessarily expect to obtain a force balance on your moment summation, unless you use accurate numbers throughout the entire problem. Generally always maintain at least four (preferably five) significant digits throughout all your dimensions, angles, and intermediate calculations. Because your dimensions do not seem to match from file to file, over the next week, I will first need to go back through all your files, and see if I can figure out the exactly correct dimensions and angles, before you rounded them. You cannot change dimensions or angles of your assembly, midstream, during the calculations, and necessarily expect to get a force balance. In the meantime, if you could post a diagram showing accurate dimensions and angles on your original bulldozer system (before you rotated beam 1 horizontal), that would be great.
 P: 63 nvn, How about this. It would be good practice for me to start over. After review, I think I should also put more of an angle on the 3 degree ram. If I work below grade, the ram will be below horizontal on the D pin and that will not make for a good design. I am also contemplating if I should also be using the weight of the rams in my calculations. The ram used for pins C,B weight is published about 20.51#'s and ram F,D weights 27.56. They are both 2" cylinders. My estimate on the beams 1 and 2 are also about 30lbs each. I choose to ignore them, but would like any input if I should add these to the diagram otherwise I will leave them off. I was planning on using a safety factor of 5 in yeild. In a way it will be more like 7 due to using all the force on the bucket to on one arm. I will round to the forth place decimal on my weights and angles. Please give me a week to post the new diagrams and math, if everything goes well I might be able to post them earlier. nvn, how does this sound? I don't want to make you spend all your time on my work. I can do it, again, good practice for me. :)
 HW Helper Sci Advisor P: 2,010 grandnat_6: That sounds fine. I would say, add 60 % of the weight of your hydraulic cylinders and beams to the vertical load applied to your bucket tip. That should be adequate, while avoiding making the computations significantly more complicated. I want to make sure you know what the term "significant digits" means. See the first four bullets under the above link. "Significant digits" is not the same as "decimal places." The general rule is for significant digits, not decimal places. (Decimal places vary, depending on the problem.) See my advice in post 19 regarding significant digits. For your particular problem right now, it means you can probably use one decimal place for forces and weights, three decimal places for angles, and three (or sometimes four) decimal places for dimensions.
P: 63
Thanks for the link for the sigificant digits. I think I understood it.

I have attached my new work. Please note dimensions have changed. I got up to the same point we left off. My moment diagram did not seem correct again because it does not return to zero. I did some reading in my statics book, I have not read it in the book, but I think it is probably correct since pin E and D are basicly holding the beam and the forces on pin B act as a cantiliver. My book does not show an example of a couple and a force acting on the same point, but I have noticed if I add the force of the couple of 65200.00"# acting on pin B to my maximium moment of -65193, I'm within 7lbs.

My book says not to add the force going though the distance you are taking moments from, but does not say anything about a couple. So I'd assume the couple does not get added either?

Thanks.
Attached Files
P: 63
nvn,

I don't know why it was blurry. All I do is select DWG to PDF for the printer. As far as colors I do not know how to change that. I only use 5 colors in CAD. Sorry for the inconvenience.

I could not read your blue text either. Do you mean I forgot the force of FBy and not FAy?

I have added the other 3 dimensions asked for in the attachments.

I forgot to mention, I left the 500lbs acting on the bucket. Loader buckets that were made for this tractor, the literature says it can lift 350lbs. Adding another 100lbs for the bucket and 60% of the weight for the rams brings it to 478.84lbs. So I left it 500lbs because the oil in the cylinders will add weight and also some of the hydraulic lines. Reasonable?

Thanks.
Attached Files
 BUCKETFORCES.pdf (31.8 KB, 12 views) ARMFORCES.pdf (33.0 KB, 3 views) 3.pdf (22.8 KB, 1 views)
 HW Helper Sci Advisor P: 2,010 grandnat_6: What do you mean you could not read my blue text? Can you not zoom in? Was that blue color faded, or what? If you cannot read my shade of blue text, what color do you want me to use for my mark-up? Black, or dark blue, or what? No, in your first file in post 24, when summing moments about point B, you forgot to include the moment about point B caused by FAy. As shown in my post 23 file, when you do this, you will obtain FA = 3155.30 lbf.
 P: 63 nvn, At lunch today I read your reply. The blue text used to mark up my drawings is good. I just can't make it out in the last file you marked up. Probably due to the blurriness. Last night I tried three different computer programs to zoom in. I also printed it out and used a magnifying glass both on the print out and the screen. I could not make it out. I worked on trying to get the FA=3155.30lbf. I could not get it. I knew in the back of my mind since pin A is not in line with pin B it probably would not be worked the same. I think the problem is in finding FAy. The only thing that makes sense to me is to take the moment of 500lbs*20.563" and then divide by the distance of .57" = 18038.0lbs I then took that and added it to my 13681.5lbsf obtained in my original problem. Then dividing by 7.4375" to obtain 4264.8lbsf. finally dividing 4264.8lbf by the cos of 58.841 which equals 8242.5lbf. Please take the time to mark up my bucket forces in post 24. Thank you.
HW Helper
P: 2,010
grandnat_6: Here is a larger image of something I posted in post 23. See the last paragraph of posts 23 and 25.

Aside 1: I just now noticed, there is a new, strange bug in PF (Physics Forums). Currently, when I upload a .png file, PF erroneously changes it to a .jpg file, which is wrong and corrupts the image, and bloats a 15 KB .png file to 51 KB. This is why you could not read my post 23 file. PF ruined it. That is not the file I posted. Notice, this PF bug was not occurring when I posted my first file, in post 6. My post 6 file is a .png file, and is perfectly sharp. That is how all my files really are. But notice, in posts 9 and 23, PF ruined my .png files, changing them to garbage .jpg, which is not what I uploaded. The .jpg format is only for photographs, and is a completely wrong format for line graphics.

Aside 2: Here is a temporary work-around for this PF bug. After you download my attached file, remove the .txt extension. Now you will see the real .png file I uploaded, which is sharp. Similarly, I also attached the post 23 file, below. Notice, it is now sharp.
Attached Files
 analysis05.png.txt (15.2 KB, 10 views) analysis04.png.txt (85.4 KB, 3 views)
 P: 63 nvn, there is no attachment in post 27.
P: 21,719
 Quote by nvn I just now noticed, there is a new, strange bug in PF
It is not a bug, it is a feature. When you upload attachments you are presented a list of max widths. Images larger than the max are resized and saved as jpg. That's both to save space on the screen and to save space on the server.

Workaround - upload the (too large) image somewhere else and link to it.
 P: 63 Borek, Thanks for the info. I think nvn, thinks I was nuts or something! maybe still does. nvn, Thanks for telling me what FA is. I think I figured it out in the attached bucket forces.pdf. Re-cap, Since the pins of A and B are in line on the first bucket, the force going though pin A in the Y direction does not produce a moment because the force goes though pin B and you can not use a force going though a point you are summing moments from. Thus pin b in the X direction can, so all the forces acting on it must go through the X direction to be balanced reaction force. On the second bucket since pin A is offset from pin B in both the X and Y direction a moment is produced in both X and Y of pin B. Since FA has two directions they must be included. I do have a few questions about this. What happens if the bucket was pined to a wall by pins A and B in both situations? Is there only an X directed force in both cases on pin B or will the reacting forces be the same as in our first and second bucket analysis? Are there any rules to follow when doing this? The equation in black on the bucket forces pdf; what is the rule to cancel out the X, and Y to obtain FA? I mean, FAx and FAy are the same forces but they are in different directions so how does the X and Y cancel out? I also attached my arm forces. I am hoping for a good report. Thank you.
 HW Helper Sci Advisor P: 2,010 grandnat_6: I do not yet see attachments in post 30. You can hit the Edit button on post 30, to make corrections, if you wish. Regarding your question, I show you how to perform the calculation with FAx and FAy in my analysis05.png file in post 27. If the bucket were pinned to a wall (with no roller support), the analysis would be different (because the constraints are different); and therefore, the pin A and B forces would be different. There would be x and y reaction forces at pin A and pin B, which would be four unknowns, which is statically indeterminate (potentially very difficult problem to solve; don't go there). Borek: Thank you very much for the tip. That clears it up for me.
P: 63
Strange, they were there last night. They are well below the .pdf limit for file size. Here they are again.

Thank you.
Attached Files
 BUCKETFORCES.pdf (32.7 KB, 3 views) ARMFORCES.pdf (34.2 KB, 2 views)
 HW Helper Sci Advisor P: 2,010 grandnat_6: By the way, numbers less than 1 must always have a zero before the decimal point. E.g., 0.57, not .57. See the international standard (ISO 31-0); or see any credible text book. Regarding your question in post 30, the black equation is incorrect. It should not have x and y subscripts. See the blue text in my analysis06.png file to see why, and to see the correct algebra. All answers in both attached files currently appear to be correct. Could you also provide dimensions xC, yC, xD, yD, xG, and yG, shown in the attached file? You can just list them; you do not necessarily need to draw them again. Attached Thumbnails
P: 63
nvn,

DOH! The math for pin A makes perfect sense now! I don't know why I didn't see that FA*cos/sin = x/y direction respectively.

I've also noticed you've been removing a lot of my + signs and making them - sign because I have the force negative in the equation. I had it in my head that since we are summing moments; meant that we are adding all positive and negative forces together. I looked back in my book and have noticed some problems have been done both ways. Must have depended on who was working the problem at that time.

I have read the provided links, I'll try to apply this moving forward. Thank you.

Attached I have replaced my old forces with the new correct forces in arm vector forces.pdf, and have plotted a new shear force diagram in shear_momentforces.pdf.

In added dimensions.pdf I have added the dimensions requested. It was easiest for me to add them in. I'm assuming these are being used to check my angles?

Thank you.
Attached Files
 added dimensions.pdf (13.5 KB, 3 views) ARMVECTORFORCES.pdf (43.6 KB, 2 views) SHEAR_MOMENTFORCES.pdf (38.5 KB, 1 views)
 HW Helper Sci Advisor P: 2,010 grandnat_6: I have assumed your dimensions in my analysis08.png file, below, are accurate, and that your rotated structure should correspond to it. Therefore, any value in analysis09.png that does not exactly correspond is rewritten in blue. The moment diagram is shown in analysis10.png. See the last paragraph of post 16. M_57.855 = +2.034 inch*lbf because you might have used four significant digits for a few values, but even if you did not, you would need to use six significant digits throughout all calculations to obtain a moment summation accurate to five significant digits. Hence, our summation is off in the fifth significant digit (relative to some of the maximum moments). But this is close enough to call it balanced. Could you state dimensions xG and yG to six significant digits, so I can get the exact rotation angle? Attached Thumbnails
P: 63
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. Im not quite sure, if we are taking moments around D; are figuring for two seperate moments for D? It appears, (because of the significant digit error) that maybe you are summing only the forces to the left in one equasion 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.