Dimensioning of a hole from tolerances

Hi
I hope you guys can help.

I have two parts that should be bolted together with four M6 bolts. The tolerances of the bottom part are quite big and the holes for this part are fixed to Ø6. The tolerances of both parts are fixed due to manufacturing and lowering of costs.
I need to determine the size of the holes of the top part in order for them to fit together. I have attached a photo showing the tolerances of both parts.

I am in doubt how to calculate the size of the holes. For instance for the left side, is this the correct way?
Diameter = Delta[sub]dim[/sub] + Tol[sub]dist[/sub] + D[sub]shaft[/sub]= (211.5-210)+[sub][/sub]3.5 + 6 = 1.5 + 3.5 + 6 = Ø11

Thank you in advance


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Replies

  • Ramani Aswath
    Ramani Aswath
    It is not clear if the dimension in the drawing is centre to centre or edge to edge.The sketch indicates edge to edge. Assuming so the minimum at the bottom will be 209 mm (?), the maximum at the top will be 218.5. Difference will be 9.5mm. Add the 3mm nominal for the radius of the hole and you end up with 9.5+3 = 12.5 as dia of the lower left hole for the top plate, which is too large for fixing with an M6 bolt unless a large washer is used. There will be considerable play also in use especially if vibrations are involved.

    Also it may be better to do all your calculations granting that the top left holes of both plates match and then calculate the other three hole diameters with this as reference.

    Others may offer a better solution. Let us wait for responses.
  • Peter Hansen
    Peter Hansen
    A.V.Ramani
    It is not clear if the dimension in the drawing is centre to centre or edge to edge.The sketch indicates edge to edge. Assuming so the minimum at the bottom will be 209 mm (?), the maximum at the top will be 218.5. Difference will be 9.5mm. Add the 3mm nominal for the radius of the hole and you end up with 9.5+3 = 12.5 as dia of the lower left hole for the top plate, which is too large for fixing with an M6 bolt unless a large washer is used. There will be considerable play also in use especially if vibrations are involved.

    Also it may be better to do all your calculations granting that the top left holes of both plates match and then calculate the other three hole diameters with this as reference.

    Others may offer a better solution. Let us wait for responses.
    Thank you for your reply.

    I am sorry I did not make it more clear, the dimensions are centre to centre. How did you get 218.5?
  • Ramani Aswath
    Ramani Aswath
    I shall have to recalc for centre to centre.
    In the meantime, the normal practice is to make a slot in the second piece with slot axis on the line joining centres of the holes and length to accommodate the worst case differences in tolerance plus the 6mm hole diameter. By doing that the play in the assembly will be avoided. Slot width will have to accommodate the bolt dia.
    We have to consider the locations when the top plate has max tolerance and the bottom minimum, as well as the reverse. the final slot (or hole if you still want a round hole) will have to accommodate both deviations.
  • Ramani Aswath
    Ramani Aswath
    Let us see if this can be simplified.
    For the bottom plate the min and max dist bet centres = 208 and 215 mm
    For the top plate the min and max dist bet centres = 209 and 211 mm
    Since the bottom plate's variation is bigger than that of the top plate this can be ignored. The hole in the bottom plate must be 3+3.5+3 = 9.5 Radius (19 dia) if the other hole remains 6 dia. Or the difference can be split as 12.5 dia for both holes.

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