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Stress Analysis of Shaft

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AME 30363 Mechanical Engineering DesignHomework 1
·
Solutions
6.8 The shaft shown in sketch
c
is subjected to tensile, torsional, and bending loads. Determinethe principal stresses at the location of stress concentration.
Ans.
σ
1
= 52.99 MPa,
σ
2
= 0,
σ
3
= -12.27 MPa.
150 mm120 mm
D
= 45 mm
d
= 30 mm
r
= 3 mm100 N-m1000 N500 N
Sketch
c
, for Problem 6.8
Notes:
This problem can be easily solved through the principal of superposition. The stressconcentration factors are obtained from Figures 6.5 (a) and (b).
Solution:
The rod will see normal stresses due to axial loads and bending, and a shear stressdue to torsion. Note that the shear stress due to shear is zero at the extreme ﬁbers wherethe stresses are largest. The critical location is at the bottom where the bending and axialstresses are both tensile. Assign the
x
-axis to the rod axis. The normal stress is given by:
σ
c
=
K
c
1
P A
+
K
c
2
McI
= (1
.
9) 1000 N
π
4(0
.
03 m)
2
+ (1
.
65)(500 N)(0
.
120 m)(0
.
015 m)
π
64(0
.
030 m)
4
= 40
.
0 MPawhere the stress concentration factors of 1.9 and 1.65 are obtained from Figure 6.5 (a) and(b). The shear stress is
τ
=
K
c
TcJ
= (1
.
4)(100 Nm)(0
.
015 m)
π
32(0
.
030 m)
4
= 26
.
45 MPaEquation (2.16) gives the principal stresses.
σ
1
,σ
2
=
σ
x
+
σ
y
2
±
τ
2
xy
+
σ
x
−
σ
y
2
2
= 40
.
0 MPa2
±
(26
.
45 MPa)
2
+
40
.
0 MPa2
2
or
σ
1
= 53
.
0 MPa and
σ
2
=-12.7 MPa. Note that the shear stress is very small compared tothe normal stress; we could have taken
σ
x
as a principal direction.6.12 Two tensile test rods are made of AISI 4340 steel tempered at 260
◦
C and aluminum alloy2024-T351. The dimensionless geometry correction factor
Y
= 1. Find how high a stress eachrod can sustain if there is a crack of 2-mm half-length in each of them.
Ans.
AISI 4340:
σ
=631 MPa.1
Notes:
The fracture toughness for these materials is obtained from Table 6.1 on page 232.The nominal stress that can be sustained is then given by Eq. (6.4).
Solution:
From Table 6.1 on page 232,
K
ci
for AISI 4340 is 50.0 MPa-m
1
/
2
. For Al 2024-T351,
K
ci
is 36 MPa-m
1
/
2
. Therefore, the stress in the steel is given by Eq. (6.4) as:
K
ci
=
Y σ
nom
√
πa
→
σ
nom
=
K
ci
Y
√
πa
= 50
.
0 MPa(1)
π
(0
.
002)= 631 MPaFor the Al 2024-T351,
K
ci
=
Y σ
nom
√
πa
→
σ
nom
=
K
ci
Y
√
πa
= 36
.
0 MPa(1)
π
(0
.
002)= 454 MPa6.22 A machine element is loaded so that the principal normal stresses at the critical location fora biaxial stress state are
σ
1
= 20 ksi and
σ
2
= -15 ksi. The material is ductile with a yieldstrength of 60 ksi. Find the safety factor according to(a) The maximum-shear-stress theory (MSST)
Ans.
n
s
= 1.714(b) The distortion-energy theory (DET)
Ans.
n
s
= 1.97
Notes:
Equation (6.6) is used to obtain the safety factor for the MSST. Equation (6.11)gives the safety factor for the DET after the von Mises stress is calculated from Eq. (6.9). If the stress is biaxial, then one principal stress is zero.
Solution:
First of all, since the stress state is biaxial, then one normal stress is zero. There-fore, the three principle stresses are properly referred to as
σ
1
= 20 ksi,
σ
2
= 0 and
σ
3
=
−
15ksi, since
σ
1
≥
σ
2
≥
σ
3
. For the maximum shear stress theory, Eq. (6.6) gives the safetyfactor as:
σ
1
−
σ
3
=
S
y
n
s
→
n
s
=
S
y
σ
1
−
σ
3
= 60 ksi(20 ksi + 15 ksi) = 1
.
714The von Mises stress is obtained from Eq. (6.9) as:
σ
e
= 1
√
2
(
σ
2
−
σ
1
)
2
+ (
σ
3
−
σ
1
)
2
+ (
σ
3
−
σ
2
)
2
1
/
2
= 1
√
2
(
−
20 ksi)
2
+ (
−
35 ksi)
2
+ (15 ksi)
2
1
/
2
or
σ
e
= 30
.
4 ksi. Therefore, the safety factor is, from Eq. (6.11),
σ
e
=
S
y
n
s
→
n
s
=
S
y
σ
e
= 60 ksi30
.
4 ksi = 1
.
976.32 The shaft shown in sketch
j
is made of AISI 1020 steel. Determine the most critical sectionby using the MSST and the DET. Dimensions of the various diameters shown in sketch
j
are
d
= 30 mm,
D
= 45 mm, and
d
2
= 40 mm.
40 mm40 mm40 mm
Dd r
= 6 mm
r
= 9 mm
T
= 50 N-m10 kN1 kN
d
2
Sketches
j
, for Problem 6.322
Notes:
This problem requires the incorporation of stress concentration eﬀects into the com-ponent stresses before determining the principal stresses.
Solution:
(a)
Fillet.
First, considering the location of stress concentration 80 mm from the wall:
J
=
π
32
d
4
=
π
32(0
.
040 m)
4
= 7
.
95
×
10
−
8
m
4
I
=
J
2 = 3
.
98
×
10
−
8
m
4
A
=
πd
2
/
4 = 7
.
07
×
10
−
4
m
2
Also, from statics,
V
= 1 kN,
M
= 40 Nm,
N
= 10 kN,
T
= 50 Nm. The bottom locationis critical, since the bending and tensile stresses are additive at this location. Also, thereis no shear stress due to shear at the extreme location. The stress concentration due tobending is obtained from Fig. 6.5(b) as 1.4, while for tension it is
K
c
= 1
.
55 from Fig.6.5(a). The stress concentration for torsion is
K
c
= 1
.
2 from Fig. 6.5(c). Therefore,
σ
1
= 458 MPa and
σ
2
=
−
28 MPa.(b)
Groove
For the location 40 mm from the wall,
N
= 10 kN,
M
= 80 Nm
T
= 50Nm,
K
c
(bending) = 1
.
45,
K
c
(tension) = 1
.
5,
K
c
(torsion) = 1
.
2 (all from Fig. 6.6).Therefore,
σ
1
= 31
.
2 MPa and
σ
2
=
−
0
.
76.This means that the critical location is 80 mm from the wall, as the stresses are higher.7.3 The shaft shown in sketch
a
rotates at high speed while the imposed loads remain static. Theshaft is machined from ground, high-carbon steel (AISI 1080). If the loading is suﬃcientlylarge to produce a fatigue failure after 1 million cycles, where would the failure most likelyoccur? Show all necessary computations and reasoning.
1 in.1 in.2 in.2 in.3 in.3 in.
1 in.1 in.
r
=in.
1 ––16
r
=in.
1 –81 –4
r
=in.
1 –8
PP
/2
P
/2
y x
Sketch
a
, for Problem 7.3
Notes:
: The failure is expected to be due to bending, since the stresses are completelyreversing. One must therefore investigate all likely locations to determine the maximumnormal stress, including the eﬀects of stress concentrations.
Solution:
Note the locations a, b, c, and d in the ﬁgure, which are the probable failurelocations, either because of stress concentration (a, c, d) or because the loading is largest(b). Upon further inspection, the diﬀerence between shoulder a and d is the radius. Since3
the radius is smaller for the left shoulder than the right one, it will have a higher stressconcentration factor, and the right shoulder doesn’t have to be considered. In terms of theanalysis to be followed, there are no diﬀerences between the locations in the shaft in termsof manufacturing method, size eﬀects, temperature eﬀects, reliability, etc, except that thefatigue stress concentration factor must be calculated. Also, from the inside front cover, AISI1080 steel has an ultimate strength of 89 ksi.(a)
Left Shoulder.
Taking a section at point a, the moment is
M
= (
P/
2)(1 in.), withthe diameter equal to 1 in. The stress concentration is obtained from Fig. 6.5b, using
D/d
= 1
.
25 in.
/
1 in. = 1
.
25 and
r/d
= (1
/
16 in.)
/
1 in. = 0
.
0625 so that
K
c
= 1
.
75.From Fig. 7.9,
q
n
is about 0.75. Therefore, from Eq. (7.19),
K
f
= 1 + (
K
c
−
1)
q
n
= 1 + (1
.
75
−
1)(0
.
75) = 1
.
56The maximum stress at a is:
σ
=
K
f
McI
= (1
.
56)(0
.
5
P
)(1 in.)(0
.
5 in.)
π
64(1 in.)
4
= (7
.
94 in.
−
2
)
P
(b)
Midpoint of Shaft.
At the midpoint, the bending moment is
M
= (
P/
2)(3 in.) =(1
.
5 in.)
P
. The diameter of the shaft is 1.25 in. at this location. There is no stressconcentration here, so the maximum stress is given by Eq. (4.48) as:
σ
=
McI
= [(1
.
5 in.)
P
](1
.
25 in.
/
2)
π
64(1
.
25 in.)
4
= (7
.
82 in.
−
2
)
P
(c)
Groove.
Taking a section at the groove, the moment is
M
= (
P/
2)(2 in.) = (1 in.)
P
.Although the diameter is not speciﬁed at this location, it is reasonable to approximateit as 1 in.
K
c
is obtained from Fig. 6.6b, using
D/d
= 1
.
25 in.
/
1 in. = 1
.
25 and
r/d
= (1
/
8 in.)
/
1 in. = 0
.
125, so that
K
c
= 1
.
7. From Fig. 7.9,
q
n
is around 0.85.Therefore, from Eq. (7.19),
K
f
= 1 + (
K
c
−
1)
q
n
= 1 + (1
.
7
−
1)(0
.
85) = 1
.
595The maximum stress is
σ
=
K
f
McI
= (1
.
595)[(1 in.)
P
](0
.
5 in.
π
64(1 in.)
4
= (16
.
2 in.
−
2
)
P
(d)
Summary.
Since the stress is largest at the groove, this is the most likely fatigue failurelocation.7.4 For each of the AISI 1040 cold-drawn steel bars shown in sketches
b
and
c
determine(a) The static tensile load causing fracture,(b) The alternating (completely reversing) axial load
±
P
that would be just on the verge of producing eventual fatigue failure.4

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