4.8. RELIABILITY OF A SHAFT UNDER TORSION 203
4.8.2 RELIABILITY OF A SHAFT UNDER TORSION FOR A
DEFORMATION ISSUE
e shaft will be deformed due to an applied torque. e angle of twist between two cross-
sections of a shaft is a relative rotational angle in radian between the two cross-sections. e
following equation can calculate the angle of twist for a shaft under torque:
D
Z
L
0
T
.
x
/
dx
G J.x/
; (4.31)
where is an angel of twist in radian between the two cross-sections of the shaft with a length L.
T
.
x
/
is the resultant internal torque at the shaft axial coordinate x. J.x/ is the polar moment of
inertia of the shaft cross-section at the shaft axial coordinate x. G is the shear Young’s modulus.
When a shaft can be divided into several segments each of which is subjected to a constant
internal resultant torque and is with a constant cross-section, the angle of twist of such shaft can
be expressed as
D
X
T
i
L
i
GJ
i
; (4.32)
where T
i
, L
i
, and J
i
are the resultant internal torque, the length and the polar moment of inertia
in the ith segment. G is the shear Young’s modulus. When the angle of twist of a shaft is larger
than the specified value, the shaft will be treated as a failure. e limit state function of such
shaft under torsion for deformation can be expressed as:
g
.
G; L
i
; J
i
; T
i
/
D
X
T
i
L
i
GJ
i
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(4.33)
where is a specified maximum allowable angle of twist in radian, which is one of the design
specifications. is treated as a deterministic value. e rest of the parameters has the same
meaning as those in Equation (4.32).
e limit state function (4.33) can be used to calculate the reliability of the shaft under
torsion for a deformation issue.
One example will be shown to demonstrate how to calculate the reliability of a shaft under
torsion for a deformation issue.
Example 4.16
A schematic of a stepped shaft on an emergency braking is shown in Figure 4.13. e shear
Young’s modulus follows a normal distribution with a mean
G
D 1:117 10
7
(psi) and a stan-
dard deviation
G
D 2:793 10
5
(psi). e diameters for the shaft AB segment and the BCD
segment are d
1
D 0:75 ˙ 0:005
00
and d
2
D 1:25 ˙ 0:005
00
. e lengths for the segment AB, BC,
and CD are L
1
D 10 ˙ 0:010
00
, L
2
D 5 ˙ 0:010
00
, and L
3
D 5 ˙ 0:010
00
. Section D is treated as
a fixed end during the emergency braking. e torque at section A is T
A
D 560 ˙ 80 (lb/in). e
204 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
T
A
L
1
A B
C
D
L
2
L
3
T
C
Figure 4.13: e schematic of a shaft.
torque at section C is T
C
D 1380 ˙ 120 (lb/in). e maximum allowable angle of twist for the
shaft between section A and the section D is D 0:030 (radian). Use the Monte Carlo method
to calculate the reliability of the shaft under torsion for this deformation issue.
Solution:
1. e angle of twist between section A and section D.
e shaft shown in Figure 4.13 can be divided into three segments: AB segment, BC seg-
ment, and CD segment. According to Figure 4.13, the resultant internal torques for the
AB segment and the BC segment will be the same and is equal to T
A
. e resultant internal
torques for the CD segment is T
A
C T
C
. So, the angle of twist of the shaft between the
section A and the section D per Equation (4.31) is:
D
X
T
i
L
i
GJ
i
D
T
A
L
1
GJ
AB
C
T
A
L
2
GJ
BC
C
.T
A
C T
C
/L
3
GJ
CD
D
32T
A
L
1
Gd
4
1
C
32T
A
L
2
Gd
4
2
C
32.T
A
C T
C
/L
3
Gd
4
2
: (a)
2. e limit state function.
e limit state function of this shaft under torsion per Equation (4.33) is:
g
.
G; d
1
; d
2
; L
1
; L
2
; L
3
; T
A
; T
C
/
D
32T
A
L
1
Gd
4
1
32T
A
L
1
Gd
4
2
32.T
A
C T
C
/L
3
Gd
4
2
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(b)
e limit state function (b) has eight random variables. Geometric dimensions can be
treated as normal distributions. eir means and standard deviations can be determined
by Equation (4.1). e distribution parameters of eight random variables are listed in Ta-
ble 4.32.
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