ENT345 Mechanical Components Design - UniMAP...
Transcript of ENT345 Mechanical Components Design - UniMAP...
ENT345 Mechanical Components Design
Dr. Haftirman Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis UniMAP Page 1
1) LOAD AND STRESS ANALYSIS
i. Principle stress
ii. The maximum shear stress
iii. The endurance strength of shaft.
1) Problem 3- 71
A countershaft carrying two-V belt pulleys is shown in the figure. Pulley A
receives power from a motor through a belt with the belt tensions shown.
The power is transmitted through the shaft and delivered to the belt on
pulley B. Assume the belt tension on the loose side at B is 15 percent of
the tension on the tight side.
a) Determine the tensions in the belt on pulley B, assuming the shaft is
running at a constant speed.
b) Find the magnitudes of the bearing reaction forces, assuming the
bearings act as simple supports.
c) Draw shear-force and bending moment diagrams for the shaft. If
needed, make one set for the horizontal plane and anther set for the
vertical plane.
d) At the point of maximum bending moment, determine the bending
stress and the torsional shear stress.
e) At the point of maximum bending moment, determine the principal
stresses and the maximum shear stress.
Solution
Assume the belt tension on the loose side at B is 15 percent of the tension on the
tight side. T2 = 0.15 T1
ENT345 Mechanical Components Design
Dr. Haftirman Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis UniMAP Page 2
a) Determine the tensions in the belt on pulley B, assuming the shaft
is running at a constant speed.
βπ» = π
(πππ β ππ)π΅(πππ)ππ + (π»π β π»π)π΅(πππ)ππ = π
ππππππ΅ππ + (π. πππ»π β π»π)π΅(πππ)ππ = π
ππππππ΅ππ β πππ. ππ»π = π
ENT345 Mechanical Components Design
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π»π = ππππ΅ππ On pulley B
π»π = (π. ππ)ππππ΅ππ = ππ. π π΅ππ
π» = π»π + π»π = πππ + ππ. π = πππ. π π΅ππ
b) Find the magnitudes of the bearing reaction forces, assuming the
bearings act as simple supports.
Y 31.82 N 45 N
31.82N T2=0.15 T1
X 212.132 N 300 N
A 45Λ B
Z 212.132 N
T1
βπππ¦ = 0
ππππππππΒ°(πππ) β πππ. π(πππ) + πΉπͺπ(πππ) = π
πΉπͺπ = βπππ. π π΅
βπΉπ§ = 0
πΉπΆπ β ππππππππΒ° + πππ. π β πππ. π = π
πΉπΆπ = πππ. π π΅
βπππ§ = 0
ππππππππΒ°(πππ) + πΉπͺπ(πππ) = π
πΉπͺπ = βππ. ππ π΅
ENT345 Mechanical Components Design
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βπΉπ¦ = 0
πΉπΆπ + ππππππππΒ° β ππ. ππ = π
πΉπΆπ = βπππ. π π΅
y 243.952
O A B C X
300 mm 400 mm 150 mm
-157.9 86.1
ENT345 Mechanical Components Design
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Z 243.952 150.7 N
O A B C X
300 mm 400 mm 150 mm
107.2 287.5
c) Draw shear-force and bending moment diagrams for the shaft. If
needed, make one set for the horizontal plane and anther set for
the vertical plane.
y 243.952
O A B C X
300 mm 400 mm 150 mm
-157.9 86.1
V (N)
86.1
O
-157.9
ENT345 Mechanical Components Design
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M (Nm)
O x
-47.37
Z 243.952 150.7 N
O A B C X
300 mm 400 mm 150 mm
107.2 287.5
V (N)
136.8
O
-107.2 150.7
ENT345 Mechanical Components Design
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M (Nm) 22.56
O x
-32.16
d) At the point of maximum bending moment, determine the bending
stress and the torsional shear stress.
The critical location is at A where both planes have the maximum
bending moment. Combining the bending moments from the two
planes;
π = β(β47.37)2 + (β32.16)2 = 57.26 ππ
The torque transmitted through the shaft from A to B is
π = (300 β 45)(0.125) = 31.88 ππ
The bending stress and the torsional stress are both maximum are on
the outer surface of a stress element.
π =ππ
πΌ=
32π
ππ3=
32(57.26)
π(0.020)3= 72.9π₯106ππ = 72.9πππ
π =ππ
π½=
16π
ππ3=
16(31.88)
π(0.020)3= 20.3π₯106ππ = 20.3πππ
ENT345 Mechanical Components Design
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e) At the point of maximum bending moment, determine the principal
stresses and the maximum shear stress.
π1, π2 =ππ₯
2Β± β(
ππ₯
2)2
+ (ππ₯π¦)2
=72.9
2Β± β(
72.9
2)2
+ (20.3)2 =
π1 = 78.2πππ
π2 = β5.27πππ
ππππ₯ = β(ππ₯
2)2
+ (ππ₯π¦)2
= β(72.9
2)2
+ (20.3)2 = 41.7 πππ
ENT345 Mechanical Components Design
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2) Problem 3-73
A gear reduction unit uses the countershaft shown in the figure. Gear A
receives power from another gear with the transmitted force FA applied at
the 20Λ pressure angle as shown. The power is transmitted through the shaft
and delivered through gear B through a transmitted force FB at the pressure
angle shown.
a) Determine the force FB, assuming the shaft is running at a constant
speed.
b) Find the magnitudes of the bearing reaction forces, assuming the
bearings act as simple supports.
c) Draw shear-force and bending-moment diagrams for the shaft. If
needed, make one set for the horizontal plane and another set for the
vertical plane.
d) At the point of maximum bending moment, determine the bending
stress and the torsional shear stress.
e) At the point of maximum bending moment, determine the principal
stresses and the maximum shear stress.
ENT345 Mechanical Components Design
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FB
Y FA
20Λ 25Λ
Z
a) Determine the force FB, assuming the shaft is running at a constant
speed.
βπ» = π
βπππππ(πππππΒ°)(πππ) β ππ©(πππππΒ°)(πππ) = π
ππ© = πππππ π΅
b) Find the magnitudes of the bearing reaction forces, assuming the
bearings act as simple supports.
βπππ§ = 0
βπππππ(πππππΒ°)(πππ) β πππππ(πππππΛ)(πππ) + πΉπͺπ(ππππ)
= π
πΉπͺπ = ππππ π΅
βπΉπ¦ = 0
πΉπΆπ β πππππ(πππππΒ°) β πππππ(πππππΛ) + ππππ = π
πΉπΆπ = ππππ π΅
ENT345 Mechanical Components Design
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βπππ¦ = 0
πππππ(πππππΒ°)(πππ) β πππππ(πππππΛ)(πππ) β πΉπͺπ(ππππ) = π
πΉπͺπ = βπππππ π΅
βπΉπ§ = 0
πΉπΆπ β πππππ(πππππΒ°) + πππππ(πππππΛ) β πππππ = π
πΉπΆπ = πππ π΅
y
8319
O A B C X
400 mm 350 mm 300 mm
5083 3762 9640
ENT345 Mechanical Components Design
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Z 10337 10830
O A B C X
400 mm 350 mm 300 mm
494 20673
c) Draw shear-force and bending-moment diagrams for the shaft. If
needed, make one set for the horizontal plane and another set for
the vertical plane.
d)
V (N)
5083
1321
O
-8319
M (Nm)
2496 x
2033
-47.37
V (N) 9843
ENT345 Mechanical Components Design
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O
-494
-10830
M (Nm) 3249
O
-198
d) At the point of maximum bending moment, determine the bending
stress and the torsional shear stress.
The critical location is at B where both planes have the maximum
bending moment. Combining the bending moments from the two
planes;
π = β(2496)2 + (3249)2 = 4097 ππ
The torque transmitted through the shaft from A to B is
π = 11000(πππ 20Β°)(0.3) = 3101 ππ
The bending stress and the torsional stress are both maximum are on
the outer surface of a stress element.
ENT345 Mechanical Components Design
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π =ππ
πΌ=
32π
ππ3=
32(4097)
π(0.050)3= 333.9π₯106ππ = 333.9πππ
π =ππ
π½=
16π
ππ3=
16(3101)
π(0.050)3= 126π₯106ππ = 20.3πππ
e) At the point of maximum bending moment, determine the principal
stresses and the maximum shear stress.
π1, π2 =ππ₯
2Β± β(
ππ₯
2)2
+ (ππ₯π¦)2
=333.9
2Β± β(
333.9
2)2
+ (126.3)2
=
π1 = 376πππ
π2 = β42.4πππ
ππππ₯ = β(ππ₯
2)2
+ (ππ₯π¦)2
= β(333.9
2)2
+ (126.3)2 = 209 πππ
f) The endurance strength of the shaft:
Material: AISI 1040 CD with Sut = 590 MPa
Diameter of shaft: d = 50 mm
Machined or cold-drawn a = 4.51 and b = - 0.265
Sπβ² = 0.5(590) = 295 πππ
ππ = πππ’π‘π = (4.51)(590)β0.265 = 0.83157
Surface factor ka =0.83157
The loading situation is rotating bending.
ππ = (π
7.62)β0.107
= (50
7.62)β0.107
= 0.8176
πΊπ = πππππΊπβ² = (π. πππππ)(π. ππππ)(πππ) = πππ. ππππ΄π·π.
The endurance strength is Se = 200.568 MPa
ENT345 Mechanical Components Design
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3) Problem 11-34
A gear-reduction unit uses the countershaft depicted in figure.
Find the two bearing reactions. The bearings are to be angular-contact ball
bearings, having a desired life of 40 kh when used at 200 rev/min. Use 1.2
for the application factor and a reliability goal for the bearing pair of 0.95.
Select the bearings from Table 11-2
Solution
ENT345 Mechanical Components Design
Dr. Haftirman Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis UniMAP Page 16
FB
Y FA
20Λ 25Λ
Z
b) Determine the force FB, assuming the shaft is running at a constant
speed.
βπ» = π
βππππ(πππππΒ°)(πππ) β ππ©(πππππΒ°)(πππ) = π
ππ© = ππππ. πππ π΅
c) Find the magnitudes of the bearing reaction forces, assuming the
bearings act as simple supports.
βπππ§ = 0
βππππ(πππππΒ°)(πππ) β ππππ. πππ(πππππΛ)(πππ) + πΉπͺπ(ππππ)
= π
πΉπͺπ = πππ. πππ π΅
βπΉπ¦ = 0
πΉπΆπ β ππππ(πππππΒ°) β ππππ. πππ(πππππΛ) + πππ. πππ = π
πΉπΆπ = πππ. πππ π΅
ENT345 Mechanical Components Design
Dr. Haftirman Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis UniMAP Page 17
βπππ¦ = 0
ππππ(πππππΒ°)(πππ) β ππππ. πππ(πππππΛ)(πππ) β πΉπͺπ(ππππ)
= π
πΉπͺπ = βππππ. ππ π΅
βπΉπ§ = 0
πΉπΆπ β ππππ(πππππΒ°) + ππππ. πππ(πππππΛ) β ππππ. ππ = π
πΉπΆπ = ππ. πππ π΅
ENT345 Mechanical Components Design
Dr. Haftirman Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis UniMAP Page 18
y
807.7
O A B C X
400 mm 350 mm 300 mm
493.543 365.27 935
Z 10337 10830
O A B C X
400 mm 350 mm 300 mm
48.405 20673
πΉπ = β(πΉππ)π + (πΉππ)
π = β(πππ. πππ)π + (ππ. πππ)π
πΉπ = βππππππ. πππ = πππ. πππ π΅
πΉπ = β(πΉππ)π + (πΉππ)
π = β(πππ. π)π + (βππππ. ππ)π
πΉπ = βπππππππ. ππ = ππππ. πππ π΅
ENT345 Mechanical Components Design
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ππ«πΆ = (π. π)(πππ. πππ) = πππ. πππ π΅
ππ«πͺ = (π. π)(ππππ. πππ) = ππππ. ππ π΅
ππ« =π³π«ππ«ππ
π³πΉππΉππ=
(πππππ)(πππ)(ππ)
(πππ)= πππ
Realibility for RA and RB are βR = β0.95 = 0.975
(πΆ10)π = 595.093
[
480
0.02 + 4.439 [ππ (1
0.975)]
11.483
]
13
= 6351.465 π
Bearing Type is angular contact bearing at O, series number 02- 12
Bore : 12 mm
OD : 32 mm
Width : 10 mm
Fillet radius : 0.6 mm
Shoulder diameter: ds = 14.5 mm, dH= 28 mm
(πΆ10)π = 1590.97
[
480
0.02 + 4.439 [ππ (1
0.975)]
11.483
]
13
= 16980.522 π
Bearing Type is angular contact bearing at C, series number 02- 30
Bore : 30 mm
OD : 62 mm
Width : 16 mm
Fillet radius : 1.0 mm
Shoulder diameter: ds = 35 mm, dH= 55 mm