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