advertisement

ME317 Tolerance Stack-ups Problems due to tolerance stack-ups include: Failure to assemble Interference between parts Failure of parts to engage Failure to function as intended Overview: Tolerance stack-ups are accumulations of variations on drawings or in part assemblies. Example: Accumulation of tolerances on a drawing. 3.00±0.01 7.00±0.01 What is the effective dimension and tolerance between the two holes? In this case the tolerances add directly The furthest apart the two centers can be is 7.01-2.99=4.02 The closest is 6.99-3.01=3.98 Thus, the effective dimension and tolerance is 4.00±0.02 This addition of tolerances may make it hard to join with a mating part that has two pins that fit in those holes. What can we do as designers to improve this? The logical thing is to put the dimension and tolerance directly on the hole spacing if that is the location most important to function. 7.00±0.01 Example: Assembly of individual parts Suppose we have two blocks, A and B that are to fit into a slot in C, all with tolerances shown. A±a B±b C±c Gap Let’s look at the largest and smallest gap that we could have. Largest Gap = (C+c) – (A-a) – (B-b) = C – (A+B) + (a+b+c) Smallest Gap = (C-c) – (A+a) – (B+b) = C – (A+B) – (a+b+c) Consequently the dimension and tolerance of the gap is effectively: {C-(A+B)} ± (a+b+c) If we have 10 parts (say a clutch pack) that all have to fit into a housing, the addition of all those tolerances can be significant. If controlling the spacing of the clutches is important to function we have a problem. What can we do as designers? Solutions • Tighten up the tolerances on each component so the sum of the tolerances is lower. • Include a spacer that comes in different sizes to take up any slack resulting from the tolerance addition (similar to shimming). • Design so that the tolerance stacks are not relevant to function. • Consider that the variation in each part is likely to be statistically distributed. Example: 2-D Consider a swing weight governor similar to the one shown below Simplifying further, assume: Holes are perfectly located Pins and holes have zero clearance and move freely • Only length of links has tolerance L1±l1 H L2±l2 The height can be found using geometry (specifically, the Law of Cosines). H2 = L12 + L22 - 2l1l2cosα (where α is the angle between the legs) And Hmax2 = (L1+l1)2 + (L2+l2)2 – 2(L1+l1)(L2+l2)cosα Hmin2 = (L1-l1)2 + (L2-l2)2 – 2(L1-l1)(L2-l2)cosα Tolerance stack of swing weight governor L1 4 L1max 4.02 l1 0.02 L1min 3.98 L2 3 L2max 3.02 l2 0.02 L2min 2.98 Angle alpha Hmax Hmin Hnom +/- 180 7.04 6.96 7 0.04 170 7.014 6.934 6.974 0.040 160 6.935 6.856 6.896 0.039 150 6.805 6.728 6.766 0.039 140 6.624 6.549 6.587 0.038 130 6.394 6.322 6.358 0.036 120 6.117 6.048 6.083 0.035 110 5.795 5.730 5.763 0.033 100 5.431 5.370 5.401 0.030 90 5.028 4.972 5.000 0.028 80 4.590 4.539 4.564 0.025 70 4.120 4.075 4.098 0.022 60 3.625 3.586 3.606 0.019