Help with Statics and Strengths of Materials

  1. Hi,

    I just signed up with the forum. I have a degree in Mechanical Design, have been working as a drafter. It's been about 10 years since I've done any statics and wanted to re-learn this stuff by making a bucket loader for my garden tractor. The last few months I've been reading my Statics and Strengths of Materials book. I think I might have confused myself.

    I'm currently working on the bucket and want to analysis the forces on the hydraulic ram pin A, and the solid arm at Pin B.

    I've tried to do this in different ways and keep getting different answers. It's frustrating when you don't have someone to tell you what you are doing right or wrong.

    The attachments below show a diagram and my forces of 500lbs to lift and also my 850#'s my tractor can push. Each step I show what I did along with the math.

    I'm not sure if I'm correct on this. I'm pretty sure I have my forces balanced in the X direction, but when I go though using SIN 30.47 degrees to sum forces in the Y direction, I get a much bigger, unrealistic number. I found out if I use the compliment angle SIN 59.63 it comes out correctly.

    This confuses me because if I use right angle trig I should be using 30.47degrees for to find the opposite leg.

    Can someone check my work and let me know if what I have so far is correct? If so; How do I go about finding the balancing forces in the Y direction?

    I appreciate any help.


    Thank you,
    Rob
     

    Attached Files:

  2. jcsd
  3. Hello Rob and welcome to Physics Forums.

    Each subject heading has a separate forum here. I have suggested to the moderators that this really belongs in one of the engineering forums - you will receive better help there as well.

    You really need to identify the forces acting as well as the geometry something like in the sketch I have shown.

    Yes this design will be sufficiently slow acting to use static equilibrium in the design,

    The bucket will have its own weight acting downwards at the centre of gravity of the bucket along with the load at its centre of gravity.
    You will need to combine these to one load at its centre of gravity and should consider different loads to see how this combination varies.

    I am assuming that your lifting flange has some sort standard arrangement with a pivot at the lower eye and a ram attached to the upper one.

    The pivot will only have vertical and horizontal forces acting, by definition.
    Depending upon how the ram is attached it will have either vert and horizontal forces if also via a pin. Or if it is rigidly attached then it will exert a moment.

    Oncle you have these forces you can use the fact for that a body in equilibrium under three forces - the three forces meet at a point.

    You can solve this by calculation or by drawing.

    You may also need to consider moment equilibrium.

    You pdfs show nicer drafting than mine by the way.

    go well
     

    Attached Files:

  4. Studiot,

    Thanks for placing my question in the proper location. I know the load should be at the center of gravity, but I choose the cutting edge as force placement since it will make a bigger bending moment. If you pick a corner of a piece of steel or concrete sheet, or a hole could be drilled in the cutting edge and a chain and clevis can be used to lift. These forces would be on the cutting edge.... Correct? The 500 lbs does include the bucket weight.

    You are correct with the configuration of the ram and pivoit.

    I used CAD to make my diagrams, you're hand written one looks much better than what my hand written one would be. :)

    Thanks,

    Rob
     
  5. A couple of things
    1. compliment angle of 30.47 is 59.53 degrees
    2 For your momnets, you have to include that from the 500 pound force acting at a distance of 20.50
    3. Resolve you forces into x and y compnents in a diagram - you already have the 500 and the 850 pounds so leave them as Fx and Fy at the tip as is - it is visually easier to see what is going on and not forget about a force.
    4. If pinned at both ends, which it would have to be, you can assume the force of the ram acts at the angle of the ram as you have done in your moment calculation ( include the 500# force moment) at the angle of 51 degrees.
    5. At the pivot you do not know the angle of the force - that you determine after you have sovled for the Fx and Fy at the pivot. You seem to assume that if you have some sort of arm at an angle of 54 degrees the force is transferred through the arm at 54 degree angle. That would be somewhat true if the pivot is pinned at both ends, which you readily can understand is impossible - the arm has to be fixed at the upper end to your tractor.
    6 Limit your decimal places to one digit - 500# and 850 # is resolved into a force of 986 # ( not 986.15). One decimal used just as a placeholder would be 986.2#. (0.15 is about 2 ounces - do you need that much accuracy )

    Positive x to the right, positive y up
    In the x direction : Fpivot_x + Fram_x = 850
    In the y direction : Fpivot_y + Fram_y = 500
    Moments : re-adjust your moment calculation
     
    Last edited: Jan 25, 2012
  6. I have attached a new PDF. Are the forces correct? I think they are, because they balance out. I also think i'm wrong because the forces on the pins A, and B. seem high.

    If this is correct. I take the forces and transfer them to the arm, but they will be opposite in direction. I can then sum forces around E to find D and then sum forces in the X/Y direction again.

    If am correct on this, I will have a few more questions once I get the arm figured out.

    Thanks for all the help!

    Rob
     

    Attached Files:

    Last edited: Jan 25, 2012
  7. nvn

    nvn 2,124
    Science Advisor
    Homework Helper

    grandnat_6: Regarding your pdf in post 5 ...

    (1) summation(Mb) = 0 = 20.563*500 + 6.25*850 - 6.50*Fa*cos(51 deg), where Fa = axial force in hydraulic cylinder AC. Therefore, solving for Fa gives, Fa = 3812.2 lbf (16.96 kN).

    (2) summation(Fx) = 0 = -850 - 3812.2*cos(51 deg) + Fbx. Therefore, Fbx = 3249.1 lbf (14.45 kN).

    (3) summation(Fy) = 0 = -500 + 3812.2*sin(51 deg) + Fby. Therefore, Fby = -2462.6 lbf (-10.95 kN).
     

    Attached Files:

    Last edited: Jan 26, 2012
  8. nvn has corrected your arithmetic, since you had somne geometry wrong (the perpendicular distance for the 850 force )

    Yes Ram forces are usually very high compared to loads. This translates into higher forces at the pivots.
    Hopefully making such connections strong enough is bread and butter to an mech engineering designer. Note the forces on the attachment of the lifting flange to the bucket (weld?) will also be very high.

    go well
     
  9. NVN and Studiot,

    It took me a few times last night to figure out how to forces went in the math. I think I understand it now.

    Attached I have analyzed the arm. The forces from the bucket have been transferred over. If I understand correctly; by rule, they act in the opposite direction.

    I used all dimensions perpendicular to pin E. I think this is correct. I also broke both hydraulic rams down in their x/y components and solved using theirs perpendicular distances, but had about 100 lbs difference. So I am not sure if I did this correct since I feel I should have been within 10lbs. Could be rounding error with distance and forces?



    Next thing I am unsure on how to set up the beam to check the beam for bending and shear stresses. The attached armforces pdf, shows how I rotated the beam section horizontal. (one beam will be welded to the other to give the angle).

    Do I rotate the forces with the rotation or do they stay as they are? I know I'll be using combined stress equations to size up the beams. I am confident in making the shear and moment diagrams. I plan on checking the beam(s) every 6" so as to make them tapper-ed. I just don't know how to go about setting up the forces or the beam for this. I think this will be the most difficult part so far.

    Again, Thank you for all the help so far. I'm not trying to to have you design my loader, but trying to learn as I go. I'd appreciate any in-site on how to start this.



    Studiot, I do plan on welding plates on the sides for the pins of the bucket, but I also thought about making it a quick attach feature, but I don't think it's really necessary for me. Thanks for the tip for high stress area. I'll keep that in mind as I go.



    Thanks,

    Rob
     

    Attached Files:

  10. nvn

    nvn 2,124
    Science Advisor
    Homework Helper

    grandnat_6: Your numbers might be slightly inaccurate, because you might be rounding your intermediate values too much. Do not round too much along the way. Otherwise, your equilibrium checks will not balance.

    I showed you a correct free-body diagram (FBD) in post 6. But there is something you seem to not have noticed yet, as follows.

    (1) A FBD must show the force vectors. A vector is a line with an arrowhead at one end. You cannot just write slanted numbers, and then omit the vector arrow. And you cannot just draw a line, and then omit the vector arrowhead.

    (2) Each force vector must have an arrowhead in the direction you assume. (If the vector magnitude is positive, then you assumed the correct direction. If the vector magnitude is negative, then the force acts in the direction opposite to the shown vector.)

    (3) Each force vector must be labeled with either a vector name (preferably), or its numerical value.​

    See the attached file, below, for an example of a correct FBD. I also marked and corrected your calculations therein. Notice, you should generally first write an equation using variable names; and then afterwards, fill in the numerical values.

    Regarding your armforces.pdf file, you cannot remove the bend in arm EDCB. If you do, you are changing the structure, which is not the structure you are analyzing. And you cannot change the hole C and D locations, relative to the members. You can rotate an arm segment only if you section-cut the arm and rotate all forces on that free body by the same amount. Or you can rotate the entire structure only if you rotate all forces on the structure by the same amount. But you cannot rotate one segment of a structure relative to another segment while keeping them connected together.
     

    Attached Files:

    Last edited by a moderator: Jan 29, 2012
  11. NVN,

    Thank for the help.

    I'll admit I was being lazy about my FBD. I did take the time and did it this time.

    Next step I believe is to make a shear and bending moment diagram. From your last post I have rotated the beams so beam 1 is horizontal. The forces were all rotated the same amount Please see attached arm vector addition PDF. (upper diagram)

    In the lower diagram, I have taken a short cut to make sure I'm correct on this. Since my forces are now not on a horizontal/verical direction I need to break them down individually into horizontal and vertical forces and add/subtract them together.

    I remember in combined stress I'll be using F/A for loads acting axially. This will be done for pin E.

    Mc/I for loads perpendicular to the axis. This will be for the vertical forces of pins E,C,D. I want to say Pin B vertical force is considered in this as well with both the shear/and bending moment diagrams having a length of 56.13".

    The horizontal forces of pins C,D, and B will be using Fec/I Since they are eccentrically loaded.

    Once these have been found I can add or subtract these values together along with the F/A value Also, if I remember correctly, F/A will either be +/- to Mc/I depending on checking for compression/tension and shear.

    Please let me know what I have correct and what I'm doing wrong, I'll make diagrams and caculations and post them of beam 1 when I get a chance.

    Thank you!

    Rob
     

    Attached Files:

  12. nvn

    nvn 2,124
    Science Advisor
    Homework Helper

    grandnat_6: Nice FBD, except a FBD typically must also have dimensions, like the example I posted in post 9.

    However, in your armvectorforces.pdf file, your summation(Fy) calculations, and FEy value, are currently incorrect. See the calculations I posted in post 9.

    So far, your plan sounds good.
     
  13. Ok,

    On the attached I have rotated the structure and it's forces so beam 1 is horizontal. I then made a chart to break these forces down in there horizontal x, and vertical y components.

    I then added the horizontal x, and vertical y components respectively per the pins they are acting on.

    When I did my shear diagram, none of my forces have crossed the zero line, nor did they balance out to zero.

    Were did I go wrong?

    Thank you
     

    Attached Files:

  14. nvn

    nvn 2,124
    Science Advisor
    Homework Helper

    grandnat_6: Your solution in post 12 is currently incorrect at points B and E, as follows. You have the wrong sign (wrong direction) on FBy. And you erroneously ignored the negative sign on FEx in my post 9 attached file. Try again.
     
  15. I found out what was wrong with the diagram in post 9. The math shows it's a negative force but the arrow head on the vector is facing positive. I have corrected this on my armvectorforces.pdf. I must have forgotten about the negative number since the vector was going positive.

    The shear force and bending moment diagram. I plotted them out in shear_momentforces.pdf. I know the shear part is correct now because the addition equals zero, so it is balanced. The moment diagram I am unsure if it is correct. shouldn't I be going from the 26842.3"# directly to zero instead of 52377.1"#?

    Since my shear force diagram passes though the zero line twice the maxium moment should be either of those two figures, and should be 29838.6"#'s since it is the largest number, if 52377.1"#'s is not correct.

    Thank you.
     
  16. I don't know why, but my files did not attach.
     

    Attached Files:

  17. nvn

    nvn 2,124
    Science Advisor
    Homework Helper

    grandnat_6: First, just a minor comment.
    If you draw a vector pointing toward the left, and the force points to the left, then the vector magnitude is positive, not negative; but its direction is negative. Likewise, if you draw a vector pointing downward, and the force is downward, then the vector magnitude is positive, not negative; but its direction is negative.

    Conversely, if you draw a vector pointing toward the right, and the force points to the left, then the vector magnitude is negative.

    In other words, because four vectors in your shear_momentforces.pdf file are drawn in the negative direction, then you should not put a negative sign on their magnitude. Remove the negative sign. You already know they are negative forces, because the vector is drawn in the negative direction.​

    Your shear force diagram currently appears correct. Nice work. Your bending moment diagram is correct from point E to D, but is currently incorrect from point D to C to B, because you did not include the moment of any horizontal force that is not on the same y coordinate as the location you are summing moments about. Try again.
     
  18. nvn,

    I thought you might say something about my negative signs. :smile: The only reason I put them there so I wouldn't forget they were negative. I did remove them since your point is clear.

    The attached I have added the moments of the forces in the Y direction. My statics book does not give an example of this. (with an offset force horizonal to the beam axis.) Also, all examples are positive moments and no negative moments.

    I still think I have done something wrong because my moment diagram does not return to zero on the right size. I have double checked my moments of pins C,D,B, and made sure my moments where in the right direction. I did find I made a mistake with the moment at B going CCW, but realized it is CW. I also used a vertical force of C as for the moment calculation and have corrected that using the horizontal force. If my moment diagram is correct, why does it not return to zero in this case?

    Thank you for the time your spending with me on this. It's been a little harder going than I though. If anything, I am re-learning a lot. I hope others have have been following this and it is helping them as well. I appreciate your time and knowledge.
     

    Attached Files:

  19. Indeed, but I have been standing back since nvn is doing such a grand job. I applaud the effort he has put into this. Also you deserve praise for your level of cooperation and your own efforts. Both are reaping just rewards.

    go well
     
  20. nvn

    nvn 2,124
    Science Advisor
    Homework Helper

    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.
     
  21. 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. :)
     
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