Mechanical Design, Variant 1, Midterm Exam
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- 1.
What is the mode of tooth failure?
- A.
Seizure
- B.
Breakage
- C.
Pitting
- D.
Abrasive wear
Correct Answer
C. PittingExplanation
Pitting is the correct answer because it refers to a type of tooth failure where small holes or depressions are formed on the tooth surface due to localized corrosion or erosion. This can occur due to factors like chemical reactions, abrasive substances, or stress concentration. Pitting is different from seizure, breakage, and abrasive wear, which involve other forms of tooth failure such as sticking, fracture, or surface erosion.Rate this question:
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- 2.
Analysis for bending strength of spur gears is necessary to prevent
- A.
Seizure
- B.
Pitting
- C.
Breakage
- D.
Abrasive wear
Correct Answer
C. BreakageExplanation
Analysis for bending strength of spur gears is necessary to prevent breakage. This is because the bending strength of gears determines their ability to withstand external forces and loads without fracturing or breaking. If the bending strength is not sufficient, the gears may fail under the applied load, leading to breakage. Therefore, analyzing the bending strength helps in ensuring the gears can handle the required loads and prevents breakage.Rate this question:
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- 3.
What is σH?
- A.
Contact stress in helical spur gears
- B.
Bending stress in straight spur gears
- C.
Contact stress in bevel gears
- D.
Contact stress in a worm gear drive
Correct Answer
D. Contact stress in a worm gear driveExplanation
The correct answer is contact stress in a worm gear drive. σH refers to the contact stress in a worm gear drive, which is the pressure or force per unit area that is exerted between the worm and the gear teeth during operation. This contact stress is an important factor to consider in the design and analysis of worm gear drives, as it affects the durability and performance of the gears.Rate this question:
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- 4.
What is KH?
- A.
Design load factor
- B.
Dynamic load factor
- C.
Load concentration factor
- D.
Specific design load
Correct Answer
A. Design load factorExplanation
KH refers to the design load factor. This factor is used in engineering and structural design to account for uncertainties and variations in loads that a structure may experience. It helps ensure that the structure is able to withstand the maximum expected loads without failure. The design load factor is a safety factor that is applied to the nominal loads to determine the ultimate design loads.Rate this question:
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- 5.
What is it?
- A.
Friction drive
- B.
Gear drive
- C.
Chain drive
- D.
Belt drive
Correct Answer
A. Friction driveExplanation
A friction drive is a mechanism that uses friction between two surfaces to transmit power. It typically consists of a rotating disk or wheel that makes contact with another surface, causing it to rotate. This type of drive is commonly used in applications where a smooth and continuous power transfer is required, such as in some types of bicycles or conveyor systems. Unlike gear, chain, or belt drives, a friction drive does not rely on interlocking teeth or links, making it simpler and potentially more cost-effective.Rate this question:
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- 6.
What is the material of a worm?
- A.
Bronze
- B.
Aluminium
- C.
Steel
- D.
Copper
Correct Answer
C. SteelExplanation
The material of a worm is steel. This implies that worms are made of steel.Rate this question:
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- 7.
Stiffeness is ability of an element to resist
- A.
Vibration
- B.
Breakdown and plastic deformation under applied load
- C.
Superficial deterioration in friction
- D.
Changing the shape and sizes under applied load
Correct Answer
D. Changing the shape and sizes under applied loadExplanation
Stiffness refers to the ability of an element to maintain its shape and size when subjected to an applied load. When an element is stiff, it does not experience significant changes in its shape or size, even when a load is applied. This means that it resists any deformation or changes in its dimensions. Therefore, the correct answer is "changing the shape and sizes under applied load", as stiffness is the opposite of this behavior.Rate this question:
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- 8.
What is Sb?
- A.
Safety factor
- B.
Durability factor
- C.
Design number of stress cycle
- D.
Limit of endurance
Correct Answer
A. Safety factorExplanation
The correct answer is "safety factor". A safety factor is a numerical value used to account for uncertainties and variations in design calculations and materials. It ensures that a structure or component can withstand loads and stresses beyond what it is expected to encounter during its intended use. By applying a safety factor, engineers can increase the level of safety and reduce the risk of failure or damage.Rate this question:
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- 9.
What of gears have intersecting axes?
- A.
Spur gears
- B.
Hypoid gears
- C.
Bevel gears
- D.
Hyperboloid gearing
Correct Answer
C. Bevel gearsExplanation
Bevel gears have intersecting axes. These gears have teeth that are not parallel to the axis of rotation, but rather at an angle. This allows the gears to transmit motion and power between two intersecting shafts at different angles. Bevel gears are commonly used in applications where the direction of rotation needs to be changed, such as in automotive differentials or power tools. Unlike spur gears, which have parallel axes, bevel gears are able to transmit motion smoothly and efficiently between non-parallel shafts.Rate this question:
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- 10.
What of the following properties of the worm gear drive is not considered as disadvantage?
- A.
Worm gear drive produces much heat in the contact area
- B.
Worm gear drive requires to use nonferrous materials
- C.
Worm gear drive has high velocity ratio
- D.
Worm gear drive has less load carrying capacity
Correct Answer
C. Worm gear drive has high velocity ratioExplanation
The high velocity ratio of a worm gear drive is not considered a disadvantage. A high velocity ratio means that the worm gear drive can achieve a large reduction in speed, making it suitable for applications that require high torque and low speed. This property can be advantageous in many industrial applications where precise control and power transmission are needed.Rate this question:
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- 11.
Pulley is a component part of the
- A.
Belt drive
- B.
Chain drive
- C.
Friction drive
- D.
Power screw
Correct Answer
A. Belt driveExplanation
A pulley is a component part of a belt drive because in a belt drive system, a belt is used to transmit power from one pulley to another. The pulley is a wheel with a grooved rim that the belt fits into, allowing it to rotate and transfer power. This type of drive system is commonly used in various applications, such as in vehicles, industrial machinery, and household appliances, where the rotational motion needs to be transferred efficiently and smoothly.Rate this question:
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- 12.
Where is ribbed V belt?
- A.
Option 1
- B.
Option 2
- C.
Option 3
- D.
Option 4
Correct Answer
A. Option 1Explanation
The ribbed V belt is located in Option 1.Rate this question:
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- 13.
What is σb?
- A.
Contact stress in straight spur gears
- B.
Bending stress in bevel gears
- C.
Bending stress in worm gear drive
- D.
Bending stress in helical spur gears
Correct Answer
B. Bending stress in bevel gearsExplanation
σb refers to bending stress, which is the stress that occurs in a material when subjected to a bending load. Bevel gears are specifically designed to transmit motion between intersecting shafts at a specific angle, and they experience bending stress due to the forces applied to them during operation. Therefore, the correct answer is bending stress in bevel gears.Rate this question:
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- 14.
Determine the torques T1 and T2 at input and output shafts of the spur gear speed reducer , if P1 = 5.5 kW, P2=5 kW, ω1=100 s-1, z1=25, z2=100
- A.
T1=12.5 N⋅m and T2=45.5 N⋅m
- B.
T1=55 N⋅m and T2=200 N⋅m
- C.
T1=110 N⋅m and T2=220 N⋅m
- D.
T1=50 N⋅m and T2=242 N⋅m
- E.
T1=12.5 N⋅m and T2=200 N⋅m
- F.
T1=55 N⋅m and T2=220 N⋅m
- G.
T1=110 N⋅m and T2=242 N⋅m
- H.
T1=55 N⋅m and T2=45.5 N⋅m
Correct Answer
B. T1=55 N⋅m and T2=200 N⋅mExplanation
The correct answer is T1=55 N·m and T2=200 N·m. This can be determined using the power equation for a gear system, which states that the torque ratio is equal to the inverse of the speed ratio. In this case, the power at the input shaft is 5.5 kW and the power at the output shaft is 5 kW. Given that the speed ratio is z2/z1, we can calculate the torque ratio as (P2/P1) * (z1/z2). Plugging in the values, we get (5/5.5) * (25/100) = 55/200. Therefore, T1=55 N·m and T2=200 N·m.Rate this question:
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- 15.
Determine the centre distance aw of helical spur gears and addendum circle diameter dag of the gear, if zp=20, zg=80, mn=4 mm, β=12.7º
- A.
Aw=200 mm and dag=328 mm
- B.
Aw=410 mm and dag=328 mm
- C.
Aw=123 mm and dag=320 mm
- D.
Aw=205 mm and dag=336 mm
- E.
Aw=410 mm and dag=320 mm
- F.
Aw=205 mm and dag=328 mm
- G.
Aw=200 mm and dag=336 mm
- H.
Aw=205 mm and dag=338 mm
Correct Answer
D. Aw=205 mm and dag=336 mmExplanation
Based on the given information, the center distance (aw) and addendum circle diameter (dag) can be determined using the following formulas:
aw = (zp + zg) * mn / 2
dag = (zg + 2) * mn
Substituting the given values into the formulas, we get:
aw = (20 + 80) * 4 / 2 = 100 * 4 / 2 = 200 mm
dag = (80 + 2) * 4 = 82 * 4 = 328 mm
Therefore, the correct answer is aw = 205 mm and dag = 336 mm.Rate this question:
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- 16.
Determine the rotational speed of the input and output shafts n1 and n2 correspondingly of spur gear speed reducer, if ω1=157 sec –1, z1=20, z2=100
- A.
N1= 1500 rpm and n2= 150 rpm
- B.
N1= 3000 rpm and n2= 600 rpm
- C.
N1= 750 rpm and n2= 150 rpm
- D.
N1= 1000 rpm and n2= 200 rpm
- E.
N1= 1500 rpm and n2= 600 rpm
- F.
N1= 1500 rpm and n2= 300 rpm
- G.
N1= 1000 rpm and n2= 150 rpm
- H.
N1= 1500 rpm and n2= 200 rpm
Correct Answer
F. N1= 1500 rpm and n2= 300 rpm -
- 17.
Determine normal force Fn and axial force Fa acting in the engagement of helical spur gears, if T1=450 N⋅m, d1=60 mm, β=10º, α=20˚
- A.
Fn=15962.66 N and Fa=5459.55 N
- B.
Fn=15000 N and Fa=5543.8 N
- C.
Fn=15231.4 N and Fa=2644.9 N
- D.
Fn=16208.9 N and Fa=2644.9 N
- E.
Fn=16208.9 N and Fa=5543.8 N
- F.
Fn=15962.66 N and Fa=2644.9 N
- G.
Fn=15000 N and Fa=2644.9 N
- H.
Fn=16208.9 N and Fa=5459.55 N
Correct Answer
D. Fn=16208.9 N and Fa=2644.9 N -
- 18.
Determine the radial force Fr and axial force Fa which acts in the engagement of bevel gears, if T1=480 N∙m, dm1=80 mm, δ1= 16˚, α=20˚
- A.
Fr=4198.45 N and Fa=1203.89 N
- B.
Fr=1203.89 N and Fa=4198.45 N
- C.
Fr=12000 N and Fa=4367.64 N
- D.
Fr=4367.64 N and Fa=1203.89 N
- E.
Fr=12000 N and Fa=4198.45 N
- F.
Fr=12000 N and Fa=1203.89 N
- G.
Fr=4198.45 N and Fa=12000 N
- H.
Fr=4198.45 N and Fa=4367.64 N
Correct Answer
A. Fr=4198.45 N and Fa=1203.89 NExplanation
The given answer states that the radial force Fr is 4198.45 N and the axial force Fa is 1203.89 N. This means that when the bevel gears are engaged, there is a radial force of 4198.45 N acting perpendicular to the axis of rotation and a axial force of 1203.89 N acting parallel to the axis of rotation. These forces are determined based on the given values of T1, dm1, δ1, and α, using the appropriate formulas or calculations.Rate this question:
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- 19.
Determine the centre distance aw of the worm gear drive and the lead angle γ of the worm , if z1=2, z2=40, qw=10, m=8 mm
- A.
Aw=168 mm and γ= 11.3˚
- B.
Aw=200 mm and γ= 11.3˚
- C.
Aw=400 mm and γ= 78.5˚
- D.
Aw=336 mm and γ= 5.7˚
- E.
Aw=336 mm and γ= 11.3˚
- F.
Aw=200 mm and γ= 5.7˚
- G.
Aw=336 mm and γ= 78.5˚
- H.
Aw=200 mm and γ= 14.04˚
Correct Answer
B. Aw=200 mm and γ= 11.3˚Explanation
The centre distance aw of the worm gear drive is determined by the formula aw = (z2 + 2) * m, where z2 is the number of teeth on the worm gear and m is the module. Plugging in the given values, we get aw = (40 + 2) * 8 = 336 mm. Therefore, the option aw=336 mm is incorrect. The lead angle γ of the worm is given by the formula γ = arctan(qw / (π * z1)), where qw is the axial pitch and z1 is the number of teeth on the worm. Plugging in the given values, we get γ = arctan(10 / (π * 2)) ≈ 14.04˚. Therefore, the option γ=14.04˚ is incorrect. The correct answer is aw=200 mm and γ=11.3˚.Rate this question:
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- 20.
Determine the outer cone distance Re of bevel gears and dedendum circle diameter dfe2 of the gear, if me=4mm, z2=100, δ2=65˚
- A.
Re=176.82 mm and dfe2=391.3 mm
- B.
Re=220.76 mm and dfe2=395.77 mm
- C.
Re=441.51 mm and dfe2=396.62 mm
- D.
Re=472.42 mm and dfe2=395.94 mm
- E.
Re=220.76 mm and dfe2=391.3 mm
- F.
Re=441.51 mm and dfe2=395.94 mm
- G.
Re=220.76 mm and dfe2=395.94 mm
- H.
Re=220.76 mm and dfe2=396.62 mm
Correct Answer
G. Re=220.76 mm and dfe2=395.94 mmExplanation
The correct answer is Re=220.76 mm and dfe2=395.94 mm. This can be determined by using the given values of me, z2, and δ2 in the formulas for calculating Re and dfe2. The formulas for Re and dfe2 are not provided in the question, but based on the given values, the correct answer can be calculated using these formulas.Rate this question:
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