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Single load placed on two separate beams

  1. Sep 29, 2011 #1
    I've got a situation where I'll be placing a load (a vacuum pump) on two separate beams.

    Basically, the pump attaches to a machine with a bolted flange. I'm wondering what would happen if we removed all the bolts except for the top and bottom one and let the pump hang...what the bending loads on the beams would be.

    At first I was tempted to use the parallel axis theorem to find the resulting moment of inertia of the two beams. But then I thought that since the beams are independed of each other, and the load is free to slide on them then we simply must add the two beams' moments of inertia. However, that also doesn't seem 100% right as the beams are not 100% percent independent...they share a common load.

    Does anybody know how to approach this problem?

  2. jcsd
  3. Sep 29, 2011 #2


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    If you assume the structure is stiff compared with the flexibiltiy of the beams, then both beams must deflect the same amount, and the combined stiffness will be twice the stiffness of one beam.

    Of course that is only approximate, because the structure will not be perfectly rigid, and there will be some geometric mis-matches between the parts when you assemble it, but it will probably be good enough.

    If the flexibility of the structure and the beams are the same order of magnitude, you would need to make a FE model of the complete assembly to find how the loads are distributed.
  4. Sep 29, 2011 #3
    I see. So it seems that the fact that the two beams are about half a meter apart vertically doesn't help them much in this situation....they might as well be half a meter apart horizontally.

    So I basically figure out how much one beam would be stressed if we hung the load on it, and multiply it by 2.
  5. Sep 30, 2011 #4
    Hold on, I just want to make sure I understand this.

    You have these two horizontal beams spaced vertically about half a meter, yes? And a load is hung between them. The load is supported by both of these beams, and nothing else?

    Just divide the load in half and consider it a point load at the center of your hole. Then just do normal deflection calcs

    Or, you could consider the load's center of mass and calculate the bending moments as well as the vertical loads, if you want to get a more precise answer.
  6. Sep 30, 2011 #5
    Yes, exactly.

    Thanks both for the help. There's just one thing that I can't get out of my head: the fact that the beams are spaced half a meter apart. I can't help but feel that this should somehow contribute to their load capability. Almsot like they should behave as one beam of the same cross-section area but half a meter wide, giving it a much greater moment of inertia.

    I think about it once and this makes sense, then I think about it again and it doesn't make sense. Does anyone know where I'm coming from? Can you help me understand what's going on?
  7. Sep 30, 2011 #6
    they aren't connected at their deflecting surfaces (except by the load at the point of deflection) so they wont support eachother as, say, an I-beam would. They do support eachother in the fact that they divide the load, but the real help comes from the moment generated by the hanging load.

    I think what you are trying to visualize is the fact that if the same load were supported by a single beam at the center, it would bend it more. You would be right. The fact that there are two beams mean that each reduces the force on the other, thus reducing the bending moment generated; effectively eliminating it.

    Also if, say, the pump was only supported by the lower beam, as the beam deflected, the top of the pump would tilt outward. This would move the center of mass out further from the load center on the beam, and would thereby increase the bending moment. The top beam keeps the top of the pump in more or less the same horizontal position (and even brings it in as the beam deflects, lowering the bending moment somewhat).

    that help?
    Last edited: Sep 30, 2011
  8. Oct 1, 2011 #7


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    Lsos: Do you really have that large gap between the pump base plate and the fixed plate, as shown in your diagram? If so, it is a weak connection. If you do not have this gap, then why did you draw it with a gap? If there is no gap, then local moment on each bolt is generally insignificant. Please let us know.
  9. Oct 3, 2011 #8
    Travis_King, I think I'm understanding it better now.

    Basically, my situation is a combiantion of the two following scenarios:



    In the first scenario, the load simply gets split evenly betwen the two beams, and the distance between them doesn't help anything. However, in the second scenario the distance DOES help out. Alot. Essentially, the distance between the beams minimizes the effect of the second scenario. Is this correct?

    Is this correct?

    Normally the gap doesn't exist because the pump is bolted to the base plate (flange). However, the idea was to have a couple sliding pins to help guide the pump during installation and removal, which is the situation I'm trying to illustratate. I wanted to do some calculations to figure out how long the pins can be and how many need to be used.
  10. Oct 3, 2011 #9
    Maybe not a perfect mathematical model but I would find where the approximate center of mass in the pump is, take this dimension and find the moment from the end of the beam then use this to calculate deflection. Dividing the load by the number of beams.

  11. Oct 5, 2011 #10


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    Lsos: Do you mean the pump base plate bolt holes are vertically slotted, and the pump slides up and down, vertically, in your diagram, in which case there is no gap between the pump base plate and the fixed plate? Or do you mean the pump is completely free to slide left and right in your diagram, in which case it is not really bolted to the fixed plate?
  12. Oct 5, 2011 #11
    Is the bolded part true? I was thinking that this component of the load is NOT divided by two, because the distance between the beams works to their advantage.

    The pump is completely free to slide left and right. It is NOT bolted...sorry if I didn't make this clear. I definitely can see how this part was not understood.

    The pump normally IS attached by means of a bolted joint, but my diagram shows what happens during mounting/ dismounting. All the bolts/ studs are removed except maybe 2 on which the pump is free to slide on.
  13. Oct 6, 2011 #12


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    Lsos: Your explanation in post 11 now clarifies your assembly. There will be no cantilever tip applied moment, shown as the second picture in post 8. There will also be no bolt (pin) axial load.

    The pump will slightly rotate, due to gravity and the clearance holes. But we might be able to ignore the rotation, depending on your exact dimensions, weight, and CG.

    The load on the lower bolt will not be reliably significant; therefore, it should be assumed zero. Therefore, the only load applied to your system is a cantilever tip load, P, applied to the upper bolt. Therefore, the distance between the upper and lower bolt does not matter much. You can ignore transverse shear stress; it will be insignificant. Therefore, the bending moment on the upper bolt thread, at the fixed plate midplane, is M1 = (0.5*t1 + gp + 0.5*t2)*P, where P = pump total weight (or possibly slightly greater), t1 = fixed plate thickness, gp = gap between pump base plate and fixed plate, t2 = pump base plate thickness, and M1 = bending moment applied to upper bolt thread.

    The bending stress on the bolt thread shank is sigma = M1*(0.5*d1)/I1 = 32*M1/(pi*d1^3), where pi = 3.141 593, and d1 = bolt tensile stress area diameter. For your M8 x 1.25 bolt, d1 = 6.8273 mm. You could probably use factors of safety of FSu = 2.0, FSy = 1.70.
  14. Oct 7, 2011 #13
    You mean zero tip moment, or do you mean that the effect of the tip moment will be greatly reduced?

    The rest I understand...except ignoring the second bolt. Under my conditions one bolt would snap like a twig (exaggeration), but I can't help but feel that multiple bolts might be able to do the job.
  15. Oct 8, 2011 #14


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    Lsos: I mean zero tip moment.

    Upon further investigation, I found that my post 12 phrase "or possibly slightly greater" is an understatement. I now found that P is much greater than I originally expected. Here is the solution to your given problem.

    (1) h1 = s1 + dn.
    (2) h2 = s1 + d2.
    (3) theta = asin{[-h1*t2 + h2*(t2^2 + h2^2 - h1^2)^0.5]/(t2^2 + h2^2)}.
    (4) a2 = t2*cos(theta) + h2*sin(theta).
    (5) a3 = (a1 - t2)*cos(theta) - 0.5*h2*sin(theta).
    (6) P2 = (a3/a2)*W.
    (7) P1 = W + P2.​

    where dn = bolt nominal diameter (mm).
    d2 = pump base plate bolt hole diameter (mm).
    t2 = pump base plate thickness (mm).
    s1 = center-to-center spacing between upper and lower bolt (mm).
    theta = pump base plate rotation angle (positive clockwise).
    a1 = perpendicular distance from pump base plate mating surface to pump CG (mm).
    a2 = horizontal distance from force P1 to force P2.
    a3 = horizontal distance from force P1 to force W.
    W = pump total weight (N).
    P1 = cantilever tip transverse (vertical) load, applied by pump flange to upper bolt (positive downward).
    P2 = cantilever transverse (vertical) load, applied by pump flange to lower bolt (positive upward).

    Notice, P2 will always be less than P1. Therefore, you only need to compute P1. Force P1 is called P in post 12. Take P1, and plug it in for P in post 12, then perform the computations listed in post 12, to obtain the upper bolt bending moment, M1, and the upper bolt bending stress, sigma.

    If you post your exact input values (dn, d2, t1, t2, gp, s1, a1, W), then we can check your math, if you use the above formulation. Give it a try.
  16. Oct 11, 2011 #15
    Took me a while to run through and understand these calculations :) Thanks for the help.

    dn = 10
    d2 = 10.5
    t2 = 28
    s1 = 439.8
    a1 = 200
    W = 1030

    Plugging in the values, I got:

    h1 = 449.8
    h2 = 450.3
    theta = .907579
    a2 = 35.126
    a3 = 168.412
    P2 = 4938
    P1 = 5968

    I didn't look deeply into how you derived the formula for theta. I understand it comes from the geometry of the flange and the clearance between the bolt/ bolt hole.

    Then I guess we have some distances, we do a sum of the moments, and we get the resulting forces.

    Does this capture the whole problem though? I can't help but feel we have overcomplicated it, but in turn missed some fundamental aspect.

    For instance, the thing that immediately strikes me is that P1 is now much, much greater than the weight of the pump. The implication of these calculations is that putting the pump on 2 bolts, is worse than putting it on 1.

    Taking this a few steps further, what happens if we reduce the amount that the pump is allowed to rotate? Reduce the clearance between the pump flange and the bolts while making the pump flange thinnner....and P1 tends towards infinity.

    Did I miss something? Am I going deeper into this now then these calculations intended?
    Last edited: Oct 11, 2011
  17. Oct 11, 2011 #16


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    Lsos: Yes, the solution in post 14 is applicable only for significant pump flange rotation angles, theta. As theta becomes very small (perhaps less than 1 or 2 deg?), the upper portion of the lower bolt hole might no longer significantly lose contact with the lower bolt, thereby somewhat changing the system of forces, with load sharing occurring between the bolts. Also, the solution in post 14 neglects friction (which is arguably negligible, and unreliable). As theta becomes very small, friction might become significant, to help resist the pump overturning moment.

    Nonetheless, one way or another, you must resist the pump overturning moment. If there is a small moment arm (a2) between the resisting forces, and you neglect friction, then P1 and P2 must go to very high values; but they stop increasing. (They do not approach infinity when theta = 0 deg.) If you have a small a2 value, and you neglect friction, then the bolts are squeezed together with a large mechanical advantage, like pliers.

    If you just put the pump on one bolt, and the pump were free to rotate, say, 45 deg, then yes, the pump would only apply its weight to the upper bolt. But the pump cannot rotate that much, due to the bolt hole. Instead, if you place the pump on one bolt, then P1 = W, but you would also have an applied cantilever tip moment, as shown in post 8. It would not be too difficult to develop this one-bolt solution, to check the bolt stress, if you are really interested in a one-bolt scenario.

    Do you really want to develop a two-bolt solution that includes friction, or develop a small rotation angle solution yourself? It might be tedious, even if you use FEA, with surface contact and friction. Near the cantilever tip, are the bolts greased, or dry? Are the bolts fully-threaded rods?

    If the bolts deflect a large amount when loaded, this might increase the pump flange rotation angle, which might need to be taken into account, in which case the pump rotation angle might not be very small.
  18. Oct 13, 2011 #17


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    Lsos: (1) What are the input values for gp and t1? (2) Near the cantilever tip, are the bolts greased, or dry? (3) Are the bolts fully-threaded M10 x 1.50 rods? Or do the bolts (studs) have a solid shank, and if so, at what distance from the fixed plate mating surface do the bolt threads terminate and begin again? (4) What are the materials of the fixed plate and pump flange? (5) What is the material of the bolts (steel property class 4.8, 5.8, 8.8, or 10.9)?
  19. Oct 14, 2011 #18
    Thank you for taking interest in this problem. I'm starting to understand that it is more complex than it looked at first glance....no wonder I had trouble with it :) I tried to mess around with some FEA analyeses, but a combination of not much experience with the program we have here and/or the many variables involved make the models constantly fail before they give any result.

    gp= 100mm, but this is one of the variables I'm trying to determine.
    t1= 28mm

    The bolts are dry. This is all supposed to happen in a clean room. I was planning on giving them only as much thread as they need to screw into the base plate, and leaving the rest smooth. Of course, I can make them whatever I want. That said, I was thinking of making them at least 10.9 class. The materials of the flanges are 400 series stainless steel.
  20. Oct 14, 2011 #19


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    Lsos: For initial, ballpark sizing of the system, let's initially ignore the applied moment, and let P1 = 0.5*W = 515 N (as others have mentioned), then compute gp for each bolt size, keeping in mind, when we later include the applied overturning moment, it will probably significantly decrease gp.

    Stu = 1040 MPa for steel bolt property class 10.9. FSu = 2.0. Using the post 12 sigma equation, and using M1 = (0.25*t1 + gp + 0.5*t2)*P1, and Ru = FSu*sigma/Stu = 1.0, then simplifying, gives gp = (0.099 128*d1^3) - (21 mm). This currently gives the following list (all values mm), where d1 is defined in post 12.

    dn = 10, d1 = 8.5927, gp = 41.89.
    dn = 12, d1 = 10.3582, gp = 89.17.
    dn = 14, d1 = 12.1236, gp = 155.6.
    dn = 16, d1 = 14.1237, gp = 258.3.
    dn = 20, d1 = 17.6544, gp = 524.5.​

    Pick out which gp value you prefer, from the above list, and let us know soon, keeping in mind, when we later include the applied overturning moment, it will probably significantly decrease gp.

    By the way, always leave a space between a numeric value and its following unit symbol. E.g., 28 mm, not 28mm. See the international standard for writing units (ISO 31-0).
  21. Oct 15, 2011 #20


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    Lsos: I went ahead and assumed you prefer the ballpark value gp = 155.6 mm, corresponding to dn = 14 mm (M14 x 2.0). I also assumed d2 = 14.7 mm. But now we need to check if gp = 155.6 mm is anywhere close to correct for dn = 14 mm.

    I now found that the rigid-body mechanics solution in post 14 is inapplicable to dn = 14, d2 = 14.7, and gp = 155.6, because the pump flange supposedly rotates far enough for both sides of each bolt hole to hit the bolts, thus generating a cantilever tip applied moment, M2. And I found the cantilevers, with these dimensions, are flexible enough such that P1 = P2 = 0.5*W. (And P2 is currently in the same direction as P1.)

    I also found you can use an effective coefficient of friction (COF) of mu = 0.2336. Even though COF is, say, mu = 0.15, using mu = 0.2336 compensates for the fact that the bolt hole corners slightly "grab on" to each bolt, because the pump flange rotates about 1.70 deg further than the cantilever tips. Here is the current solution for dn = 14, d2 = 14.7, and gp = 155.6. Even though the attached diagram also shows the cantilevers, it is only a free-body diagram (FBD) of the pump, not the cantilevers. The forces on the cantilevers would be the reverse of the forces on the pump.

    z2 = cantilever tip vertical deflection (positive downward).
    theta = pump flange rotation angle (positive clockwise).
    phi2 = cantilever tip rotation angle (positive clockwise).
    Fxf2 = axial force on cantilever due to friction.
    L2 = cantilever length.

    z2 = 5.361 mm
    theta = 4.772 deg
    phi2 = 3.071 deg
    h2 = s1 + d2 = 439.8 + 14.70 = 454.50 mm
    a2 = t2*cos(theta) + h2*sin(theta) = 65.71 mm
    a3 = (a1 - t2)*cos(theta) - 0.5*h2*sin(theta) = 152.50 mm
    P1 = P2 = 0.5*W = 515 N
    mu = 0.2336
    Fxf2 = mu*P2 = 0.2336*515 = 120.30 N
    L2 = 0.25*t1 + gp + 0.5*t2 = 0.25*28 + 155.6 + 0.5*28 = 176.6 mm

    From the attached FBD, summation of moment about point A follows.

    summation(M_A) = 0 = s1*Fxf2 - a2*P2 - a3*W + 2*M2.

    Solving for M2 gives, M2 = 69 000 N*mm. Therefore, the bending moment on the upper bolt thread is M1 = L2*P1 + M2 = 176.6*515 + 69 000 = 159 950 N*mm. Hence, the bending stress on the bolt thread shank is sigma = 32*M1/(pi*d1^3) = 32*159 950/(pi*12.1236^3) = 914.30 MPa. The bolt allowable stress is Sta = Stu/FSu = (1040 MPa)/2.0 = 520 MPa. The stress level is Ru = sigma/Sta = 914.30/520 = 175.8 %, which exceeds 100 %, which therefore means the bolt is currently overstressed.

    Therefore, what value of gp would supposedly give Ru = 100 %? Solving the above for L2, with Ru = 100 %, gives, L2 = 42.66 mm. Therefore, gp = L2 - 0.25*t1 - 0.5*t2 = 42.66 - 21 = 21.66 mm. However, a gp value this low might put you in a different stiffness category, closer to the rigid-body mechanics solution in post 14 (?).

    What would you want to do now? Go to dn = 16 mm? Or what?

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