Note on equivalent diameter dequiv and A95
A95,standard
A95,standard
0.95 · d
d = 8mm
Figure 1: Illustration of A95,standard
A95,standard is the area in a standard test specimen (diameter Ø3in = 8mm) that is exposed to stresses σ in
the interval 0.95σmax ≤ σ ≤ σmax when the specimen is loaded in rotating bending. The area is illustrated in
figure 1 and it is computed as:
(d2 − (0.95 · d)2 ) · π
4
(1 − 0.952 ) · π 2
=
·d
4
= 0.0766 · d2
A95,standard =
= 0.0766 · (8mm)2
= 4.90mm2
(1)
1
A95,specimen
x
x
h
0.95 · h
A95,specimen
b
Figure 2: Illustration of A95,specimen for a cross section exposed to non-rotating bending around the xx-axis.
A95,specimen is the area in the actual cross section that with the actual loading is exposed to stresses σ in the
interval 0.95σmax ≤ σ ≤ σmax . For the example in figure 2 where the cross section is loaded with non-rotating
bending the A95,specimen is computed as:
A95,specimen = b · (h − 0.95 · h)
= 0.05 · b · h
(2)
dequiv
A95,specimen
dequiv
Figure 3: Illustration of dequiv
dequiv is defined as the diameter of a circular cross section that has the same A95 as A95,specimen when loaded
in rotating bending. The relation can be written as:
A95,specimen = 0.0766 · d2equiv
or
dequiv =
r
A95,specimen
0.0766
(3)
2
As additional explanation to the numbers in Norton is here shown the computation of A95 for a circular cross
section loaded with non-rotating bending. The area is shown in figure 4.
A95 = 2 ·
= 2·
= 4·
=
=
=
=
=
=
=
Z R
0.95·R
Z R
2 · b dy
2·
0.95·R
Z R p
0.95·R
p
R2 − y 2 dy hint: pythagoras is used to express b
R2 − y 2 dy integration by looking up in an equation collection
#
R
R2 − y 2 R2
−1 y
4·
+
· sin
2
2
R
0.95·R
p
√
✟
!
✟
2
2
2 −(0.95·R)2
2
2
✟
0.95·R·
R
R· R✟−R
R
R
R
0.95·
R
4·
−
+
·sin−1
−
·sin−1
✟✟2
2
2
2
R
R
✟
p
4
· R2 · sin−1 (1) − 0.95 · R2 · 1 − 0.952 − R2 · sin−1 (0.95)
2
p
2 · R2 · sin−1 (1) − 0.95 · 1 − 0.952 − sin−1 (0.95)
2 p
d
2·
· sin−1 (1) − 0.95 · 1 − 0.952 − sin−1 (0.95)
2
p
1 d2 · · sin−1 (1) − 0.95 · 1 − 0.952 − sin−1 (0.95)
2
2
d · 0.0104615
(4)
"
y·
p
b
dy
y
0.95 · R
R
A95
x
d
Figure 4: Illustration of the area with minium 95% of the maximum stress when loaded with non-rotating bending. This is used in the
computation of the correction factor Csize .
3
Example
A round shaft with diameter d = 70 mm is exposed to non-rotating bending. The correction facor Csize is
computed in the following. The A95 is computed as:
A95,specimen = d2 · 0.0104615
= (70 mm)2 · 0.0104615
= 51.26 mm2
(5)
The equivalent diameter is computed as:
dequiv =
r
A95,specimen
0.0766
r
51.26 mm2
0.0766
= 25.869 mm
=
(6)
Thereby the correction factor Csize is computed based on equation (6.7b) in Norton:
Csize = 1.189 · d−0.097
equiv
= 1.189 · (25.869 mm)−0.097
= 0.8672
(7)
The fatigue strength is reduced by the factor Csize = 0.8672 due to size of the cross section and the loading
being nonrotating bending in stead of rotating bending.
4