MECH2400 5400 Mechanical Design Quiz 3 2014 Solution

MECH2400 5400
Mechanical Design 1
SID
Quiz 3 Solution
This quiz is worth 15% of your assessment in this subject
THIS IS A CLOSED BOOK EXAM
Part 1 Multiple-choice (0.25 marks each)
Answer all questions
Please circle the most correct answer
1. In relation to Gear Geometry, which of the following statements is false?
1. The addendum circle is the equal to the P.C.D. plus two addendums.
2. The involute profile is not defined below the base circle.
3. The line of contact is defined as the P.C.D. divided by the number of teeth.
4. The dedendum circle is equal to the P.C.D. minus two dedendums.
5. The contact ratio is a measure of tooth load sharing.
2. Which of the following shaft design cases are not able to be accurately calculated using AS14032004?
1. Shaft design cases where multiple stress concentrations are present.
2. Shaft design cases where stepped shafts need to be considered.
3. Shaft design cases where cyclic vibrations are present.
4. Shaft design cases that involve an axial load.
5. Shaft design cases that involve hollow shafts.
3. In terms of Gear Geometry why is minimizing the access or approach action on a meshing set of
gears more advantageous than increasing the access or approach action?
1. Minimising the access or approach action minimisise the noise generated by meshing gears.
2. Minimising the access or approach action minimisise the heat generated by meshing gears.
3. Minimising the access or approach action minimisise the effects of using a large pressure angle.
4. Minimising the access or approach action minimisise the effects of having a large base diameter.
5. Minimising the access or approach action minimisise the effect of the tip of the tooth wiping away lubrication.
4. When selecting a grade of steel for a shaft design what is the main reason that it is advantageous
to use a softer or lower grade of steel?
1. To minimize the likelihood of corrosion.
2. To minimize the Factor of Safety required.
3. To take advantage of the better surface finish that is possible using softer or lower grades of steel.
4. Because hollow shaft designs are not possible using harder or higher grades of steel.
5. To minimize the effects of stress concentration factors.
5. Which of the following is true when comparing Toothed belt drives to Gear Drives.
1. Toothed belt drives are able to take higher loads than gear drives.
2. Toothed belt drives are more precise in the rotational accuracy than gear drives.
3. Toothed belt drives have a variable pressure angle unlike gear drives that have a set pressure angle.
4. Toothed belt drives require more lubrication consideration than gear drives.
5. Toothed belt drives have less rotating inertia than gear drives.
6. Which of the following statements is true when comparing Rubber Bush Couplings against Chain
Couplings?
1. Rubber Bush Couplings are able to absorb shock loads better than Chain Couplings.
2. Rubber Bush Couplings rotate in either direction compared to Chain Couplings which can only rotate in one direction.
3. Rubber Bush Couplings do not allow slippage compared to Chain Couplings which allow for a 10% RPM reduction.
4. Rubber Bush Couplings require grease compared to Chain Couplings which require no grease.
5. Rubber Bush Couplings are smaller in size compared to Chain Couplings for the same load carrying capacity.
7. What advantage does a Woodruff key have over plain square key?
1. Woodruff keys are self-aligning compared to a plain square key.
2. Woodruff keys induce a lower stress concentration factor on the shaft when compared to a plain square key.
3. Woodruff keys are hidden from view.
4. Woodruff keys are able to take larger loads than a plain square key.
5. Woodruff keys are available in more sizes.
8. In shaft design, which of the following comparisons regarding interference fits vs conventional
keyways as a method of transmitting power is true?
1. Interference fits do not generate a stress concentration factor unlike conventional keyways.
2. Interference fits are able to be used on hollow and solid shafts unlike conventional keyways which can only be used
on solid shafts.
3. Interference fits take less axial distance than a conventional keyway.
4. Interference fits do not require machining into the finished shaft unlike conventional keyways.
5. All of the above.
9. What is the main use of dowel pins?
1. To transmit power between a two adjacent shafts.
2. To replace a set screw which has a damaged thread.
3. To act as a mechanical fuse between a shaft and coupling.
4. To accurately position or align two pieces together.
5. To replace a tapped hole arrangement which has a damaged thread.
10. What is one main advantage of using a Scotch Key?
1. Scotch Keys are able to take major shock loads.
2. Scotch Keys are able to resist axial and rotational movement.
3. Scotch keys have the highest torque carrying ability of any key.
4. Scotch keys enable the hub to move in the axial but not rotational direction.
5. Scotch keys enable the hub to move in the radial but not the axial direction.
Part 2 Written Solutions (1.0 mark each)
Answer all questions
1. What is one major advantage of using a Sliding Block (Oldham) coupling over a rigid or flexible
coupling?
Sliding Block (Oldham) couplings allow for lateral misalignment of two shafts.
(Or words to the effect)
(1 mark for the first correct answer. The type of misalignment may be stated or sketched)
2. Name two advantages that V-Belt Drives have over Flat Belt Drives?
Increased friction or power carrying ability or belt tension is reduced, closer centre distances
are possible or negligible slip,
(0.5 mark for each correct answer. Only the first two answers are considered)
3. If a mechanical designer wishes to use a very small drive pulley (20mm Dia.) on a low power
flexible belt drive arrangement what type of flexible belt drive is recommended? (Hint: Motion
synchronization is not required)
A flat belt.
(1.0 mark for correct answer. Only the first answer is to considered)
4. What advantage would a mechanical designer have by using a large pressure angle gear set?
e.g. 25° instead of a 14.5° pressure angle.
Stronger teeth or the teeth are thicker at the dedendum.
(1.0 mark for correct answer. Only the first answer is to considered)
5. What advantage would a mechanical designer have by using a lower pressure angle gear set?
e.g. 14.5° instead of a 25° pressure angle.
Smaller separating forces generated between gears, (and as a consequence, smaller diameter
shafts and bearings are required), quieter gears or more efficient gears.
(1.0 mark for correct answer. Only the first answer is to considered)
6. In Gear Geometry Design, what is the main purpose of the Base Circle?
To generate the involute.
(1.0 mark for correct answer. Only the first answer is to considered)
7. What is an advantage of using gears with an involute profile instead of gears with a cycloidal
profile?
Involute profile gears continue to function even if the centre distance varies or
the involute profile is more common or the manufacture is covered by standards or
easier to manufacture than cycloidal gears.
(1.0 mark for correct answer. Only the first answer is to considered)
8. Does welding onto a shaft demonstrate good mechanical design practice? Why?
No (0.5 marks for correct answer)
Reduced fatigue performance or reduced mechanical properties
(0.5 marks for correct answer. Only the first answer is to considered)
9. Give one reason why backlash is important in a gear set?
Backlash allows space for lubrication or expansion and contraction due to thermal effects.
(1.0 mark for correct answer. Only the first answer is to considered)
10. Rubber couplings are able to take misalignment in which two directions?
Angular and Axial misalignment
(0.5 mark for each correct answer. Only the first two answers are considered)
Part 3 Design Application (2.5 marks in total)
Answer all questions
1.0 Figure 1.0 shows an in line-shaft arrangement having two universal couplings with an input power of 1kW
at 400 RPM. The loss in power from the input to the output shaft under the current arrangement is
approximately 12%.
Output
Input
Figure 1.0
1. Sketch and roughly dimension an alternate arrangement that still uses the two universal couplings but
minimizes the loss in power. The 0.5m and 1.6m dimensions that locate the two universal couplings may
be adjusted by up to +/- 50% but only one coupling may be moved. The link between the two couplings
remains fixed in length. (0.25 marks).
1.63m
0.25m
8.7°
8.7°
Reducing the vertical displacement by 50% to 0.25m brings the two shafts more in line this is confirmed
by looking at Graph 1. If the horizontal dimension was reduced by 50% the misalignment and efficiency
would be worse. Recalculating using trigonometry / Pythagoras theorem aligns the couplings to within
approximately 8.7°.
Refering to Graph 1 at the end of this quiz, estimate the efficiency of the alternate arrangement that you
have designed. (0.25 marks)
98% * 98% = 96% (0.25 marks) Answer may vary slightly due to reading from graph.
2.0 Figure 2.0 shows a steel shaft that has been heat treated to minimum yield strength of 525 MPa and
has a diameter of 36mm. The shaft rotates at 600 RPM and transmits 30kW through a gear.
1. Use Table 1 to determine values for “b” and “h” as noted in Figure 2.0 below. (0.25 marks)
b = 10mm
h = 8mm (0.25 marks for both) (- 0.25 marks if the units mm are missing from any question)
F
F
Figure 2.0
2. Determine the force “F” acting along the surface of the shaft. (0.5 marks)
Angular speed ω = 600(2)π/60 = 62.8 rad/s
τ = 30 000/62.8 = 478 Nm
(0.25 marks up to this step)
F = τ/r = 478/0.018 = 26,556 N
(0.25 marks up to this step)
3. Using steel for the key with a yield strength (Sy) of 455MPa and the distortion energy theory,
determine the maximum shear strength (Ssy) of the key. (0.25 marks)
Ssy = 0.577Sy = (0.577)(455) = 262.5 MPa
(0.25 marks)
4. Using a factor of safety of 2.8 calculate the length “L” of key required to resist failure by shear in
mm (0.5 marks)
(262.5 * 106) / 2.8 = 26556 / (0.01 * L)
L = 0.0283m = 28.3mm (0.5 marks for both)
5. Calculate the length “L” of key required to resist failure by crushing in mm (0.5 mark)
(455.0 * 106) / 2.8 = 26556 / (0.01 * (L/2))
L = 0.0327m = 32.7mm (0.5 marks for both)
Some Basic Formulas and Table 1
ω = 2πN/60
P=τω
τ = Fr
Ssy = 0.577Sy
(Ssy / n) = (F / bL)
(Sy / n) = (F / (bL/2))
n = Factor of Safety
b = width of key
L = Length of Key
3
Graph 1
Approx. 99%
The efficiency of a needle bearing universal joint
Ref: Repco Universal Joints Co.
THIS IS THE END OF QUIZ 2
DO NOT TURN OVER
THIS QUIZ UNTIL YOU
ARE TOLD TO DO SO
THIS IS A CLOSED
BOOK EXAM

Download Report