Chapter 2: Statistical Considerations

Chapter 2: Statistical
Considerations
BAE 417 417-Design Design of Machine Systems
Fall 2006

Limits, Fits, and Tolerances
(Sections 22-6
6 and 2-7) 2 7)
Capital letters refer to holes; lowercase letters are used for shafts

Limits and Fits
Basic Size Size- size to which limits or deviations are
assigned
Deviation Deviation- algebraic difference between size and
corresponding basic size
Upper deviation deviation- indicates maximum difference
Lower deviation deviation- indicates minimum difference
Fundamental deviation deviation- which of the above is closer
to the basic size

Limits and Fits
Tolerance Tolerance- difference between max. and min.
size limits
International tolerance grade numbers (IT)-
designated groups of tolerances that vary
depending on basic size
Hole basis basis- system of fits corresponding to hole
sizes (H is the fundamental deviation)
Shaft basis basis- system of fits corresponding to shaft
size (h is the fundamental deviation)
See Table 22-8
8 for a Description of Preferred Fits Using the Basic Hole System

Conventions and Variable
Definitions
D- basic hole size
d- basic shaft size
δ- deviation
δ δu-- upper deviation
δl- lower deviation
δF- fundamental deviation
ΔD- tolerance grade for hole
Δd- tolerance grade for shaft

For holes
D max = D + ΔD D (Eq. 22-38)
38)
D min = D
For shafts with clearance fits c, d, f, g, and h
d max = d + δ F (Eq. 22-39)
39)
d min = d + δF – Δd
For shafts with interference fits k, n, p, s, and u
d min = d + δ F (Eq. 22-40)
40)
d max = d + δF + Δd

Example
Find the shaft and hole dimensions for a loose
running fit with a 34 34-mm mm basic size.

Solution
From Table 22-8
8 we find the ISO symbol to be
34H11/c11
From Table AA-11
11 using the IT tolerance grade
of 11 (IT11) and a basic size of 34 (34 mm) we
get ΔD D = Δd= d= 0.160 mm
Using Eq. 22-38
38 we get:
D max = D + ΔD D = 34 + 0.160 = 34.160 mm
D min = 0.160 mm

Solution
Using Table AA-12
12 we find the fundamental
deviation ( (δF) ) for the 34c11 shaft to be
δF= = -0.120 0.120 mm
Using Eq. (2 (2-39) (2 (2-39) 39) we get:
d max = d + δF = 34 – 0.120 = 33.880 mm
d min = d + δF – Δd d = 34 - 0.120 – 0.160 = 33.720 mm

Dimensions and Tolerances
Nominal size - size for speaking of an element (i.e., ½”
bolt or 3” pipe)
Basic size – exact theoretical size
Limits – stated maximum and minimum dimensions
Tolerance – difference between two limits
Bilateral – 1.005 ± 0.002 in
+ 0.
004
Unilateral – 1.
005 in
−0.
000
Natural tolerance – plus and minus three standard
deviations

Dimensions and Tolerances
Clearance – difference in dimensions (diametral
or radial)
Used only when the internal member is smaller than
the external member
difference in dimensions
Used when the internal member is larger than the
external member
Interference – difference in dimensions
Allowance – minimum clearance or maximum
interference

Tolerance Stacking
The “gap”, “grip”, or “interference” of an assembly of mating parts is
defined as:
w
= ∑ x −∑
y i j
where x’s are right tending and y’s are left tending displacements

Homework Assignment
Do problems 22-25,
25, 2-27, 2 27, and 2-32 2 32 for next
Tuesday.