Design of Gears MCQ Quiz - Objective Question with Answer for Design of Gears - Download Free PDF
Last updated on Jun 4, 2025
Latest Design of Gears MCQ Objective Questions
Design of Gears Question 1:
Face of a tooth in spur gear is:
Answer (Detailed Solution Below)
Design of Gears Question 1 Detailed Solution
Explanation:
Face of a Tooth in Spur Gear
- In the context of spur gears, the "face" of a tooth refers to the surface of the tooth that lies between the pitch circle and the top land. This definition is crucial for understanding the geometry of gear teeth and their role in transmitting motion and power in mechanical systems.
- The face of a tooth in a spur gear is specifically defined as the part of the tooth's profile that extends from the pitch circle to the top land. The pitch circle is an imaginary circle that represents the effective diameter at which the gear teeth engage with a mating gear. The top land, on the other hand, is the flat or slightly rounded surface at the very top of the tooth. The face of the tooth plays a critical role during the engagement of gears as it is part of the contact surface where power transmission occurs.
Explanation of the Gear Tooth Terminology:
- Pitch Circle: The pitch circle is an imaginary circle that defines the size of the gear. It is the circle where the teeth of two mating gears make contact and effectively transmit motion and force. This circle is fundamental in determining the gear ratio and speed ratio in gear systems.
- Top Land: The top land is the flat or rounded surface at the top of the gear tooth. It is the highest point of the tooth profile and is not involved in the engagement with the mating gear teeth.
- Face: The face of the tooth is the portion of the tooth profile that extends from the pitch circle to the top land. This surface interacts with the mating gear teeth during the operation of the gear system.
Importance of the Face:
- The face of the tooth is crucial in power transmission as it makes contact with the mating gear's teeth during engagement.
- The surface finish, shape, and accuracy of the face directly influence the efficiency, noise, and wear characteristics of the gear system.
- Proper design and manufacturing of the face ensure smooth and reliable operation of the gear system, reducing the chances of failure or excessive wear.
Design of Gears Question 2:
The tooth thickness of the 4 mm module gear measured at the pitch circle line is
Answer (Detailed Solution Below)
Design of Gears Question 2 Detailed Solution
Concept:
Tooth Thickness Calculation: The tooth thickness (t) at the pitch circle is given by the formula:
\( t = \frac{\pi \times m}{2} \)
where \( m \) is the module of the gear.
Calculation:
Given:
- Module (m) = 4 mm
Substituting the value of the module into the formula:
\( t = \frac{\pi \times 4 \, }{2} \)
\( t = \frac{3.1416 \times 4 \, }{2} \)
\( t = \frac{12.5664 \, }{2} \)
\( t = 6.2832 \, \text{mm} \)
Rounding off to one decimal place, we get:
\( t \approx 6.3 \, \text{mm} \)
Design of Gears Question 3:
A spur gear transmitting power is connected to the shaft with a key of rectangular section. The type(s) of stresses developed in the key is/are -
Answer (Detailed Solution Below)
Design of Gears Question 3 Detailed Solution
Explanation:
Key:
- A key is a mechanical component used to connect a rotating machine element to a shaft. The key prevents relative rotation between the two parts and can also transmit torque.
- When a spur gear transmits power through the shaft, the key experiences both shear and bearing stresses.
- Shear stress occurs because the key resists the relative rotational motion between the shaft and the gear, and the torque applied generates a force that causes shearing along the length of the key.
- Bearing stress occurs at the contact surfaces between the key and the keyway (slots) in the shaft and gear. These contact surfaces bear the load, causing compressive stress.
Design of Gears Question 4:
The form factor of a spur gear tooth depends on
A. Circular pitch
B. Pressure angle
C. No. of teeth
D. both A&B
Answer (Detailed Solution Below)
Design of Gears Question 4 Detailed Solution
Design of Gears Question 5:
Gear burnishing is a process for
Answer (Detailed Solution Below)
Design of Gears Question 5 Detailed Solution
Explanation:
Explanation:
Gear finishing processes:
- Conventional finishing process for gears:
- Gear shaving
- Gear grinding
- Gear honing
- Gear lapping
- Gear burnishing
- Gear skiving
- Advanced finishing processes for gears:
- Gear finishing by electrochemical honing.
- Gear finishing by electrochemical grinding.
- Gear finishing by abrasive flow finishing (AFF).
Gear hobbing:
- It is the process of generating gear teeth by means of a rotating cutter referred to as a “hob”
- Hobbing can be used to produce spur, Helical, & worm gears, as well as splines in almost all material (ferrous and non-ferrous metals & plastics), but not bevel or internal gears.
Top Design of Gears MCQ Objective Questions
A gear having a diameter of 40 cm transmits 10 kW at 250 r.p.m. If the pressure angle is 20°, the force on the tooth of the gear is:
Answer (Detailed Solution Below)
Design of Gears Question 6 Detailed Solution
Download Solution PDFConcept:
Spur gear:
- Spur gears are those gears in which teeth are straight and parallel to the axis of the shaft.
- These types of gear are noisy, more fatigue and have less life.
- Spur gears are used for very low power transmission.
Force analysis in Spur gear:
- Total force (F) acts on the common normal of the mating gears.
- Total force can be divided into two components.
1.Tangential component (Ft) = Fcosϕ
2. Radial component (Fr) = Fsinϕ
where ϕ is pressure angle
- Ft is used to produce the power required and Fr provides thrust.
- Torque produced can be calculated as:
T = Ft × R (where R is pitch radius)
- The power produced can be calculated as:
P = T × ω (where ω is the angular velocity of gear)
Calculation:
Given:
N = 250 rpm, D = 0.4 m, P = 10 kW, ϕ = 20°
Power, P = Fcosϕ × R × ω
\(P=Fcos \phi \times R\times \frac{2\pi N}{60}\)
\(10 \times10^3 =Fcos 20 \times 0.2\times \frac{2\pi \times 250}{60}\)
The total force on tooth is, F = 2032.42
In the formulation of Lewis equation for toothed gearing, the load WT acts on:
Answer (Detailed Solution Below)
Design of Gears Question 7 Detailed Solution
Download Solution PDFConcept:
In the Lewis analysis, the gear tooth is treated as a cantilever beam. The tangential component (WT) causes the bending moment about the base of the tooth.
The Lewis equation is based on the following assumptions:
- The effect of the radial component (Wr), which induces compressive stresses, is neglected.
- It is assumed that the tangential component (WT) is uniformly distributed over the face width of the gear. This is possible when the gears are rigid and accurately machined.
- The effect of stress concentration is neglected.
- It is assumed that at any time, only one pair of teeth is in contact and takes the total load.
In Lewis equation for bending stress, gear tooth is considered as
Answer (Detailed Solution Below)
Design of Gears Question 8 Detailed Solution
Download Solution PDFConcept:
In the Lewis analysis, the gear tooth is treated as a cantilever beam. The tangential component (WT) causes the bending moment about the base of the tooth.
The Lewis equation is based on the following assumptions:
- The effect of the radial component (Wr), which induces compressive stresses, is neglected.
- It is assumed that the tangential component (WT) is uniformly distributed over the face width of the gear. This is possible when the gears are rigid and accurately machined.
- The effect of stress concentration is neglected.
- It is assumed that any time, only one pair of teeth is in contact and takes the total load.
Lewis form factor is given by
\(y = \frac{{{t^2}}}{{6hm}}\)
- It is a dimensionless quantity
- It is independent of tooth size
- It is only a function of the shape
The analysis of the Bending stress in the gear tooth is done by Lewis equation.
Lewis's equation in spur gears are used to find the
Answer (Detailed Solution Below)
Design of Gears Question 9 Detailed Solution
Download Solution PDFConcept:
In the Lewis analysis, the gear tooth is treated as a cantilever beam. The tangential component (WT) causes the bending moment about the base of the tooth.
The Lewis equation is based on the following assumptions:
- The effect of the radial component (Wr), which induces compressive stresses, is neglected.
- It is assumed that the tangential component (WT) is uniformly distributed over the face width of the gear. This is possible when the gears are rigid and accurately machined.
- The effect of stress concentration is neglected.
- It is assumed that any time, only one pair of teeth is in contact and takes the total load.
Lewis form factor is given by
\(y = \frac{{{t^2}}}{{6hm}}\)
- It is a dimensionless quantity
- It is independent of tooth size
- It is only a function of the shape
The analysis of the Bending stress in the gear tooth is done by Lewis equation
A 20º full depth involute spur pinion of 4 mm module and 21 teeth is to transmit 15 kW at 960 rpm. Its face width is 25 mm. Given that the tooth geometry factor is 0.32 and the combined effect of dynamic load and allied factors intensifying the stress is 1.5; the minimum allowable stress (in MPa) for the gear material is
Answer (Detailed Solution Below)
Design of Gears Question 10 Detailed Solution
Download Solution PDFConcept:
Power transmitted by the gear (P) = Torque × Angular speed ⇒ ωT
Torque (T) = Tangential force × Radius ⇒ FT × R
The force transmitted by the gear is given by,
\({F_T} = \frac{{{\sigma _b}\;×\;b\;×\;Y\;×\;m}}{{{f_s}}}\)
where
σb = Minimum allowable stress for gear material
Y = Tooth geometry factor
fs = Combined effect of dynamic load and allied factors intensifying the stress.
b = Face width of the tooth
m = Module of the gear
Calculation:
Given:
m = 4 mm, Tp = 21, P = 15 kW ⇒ 15 × 103 W, N = 960 rpm, b = 25 mm, Y = 0.32
\(R=\frac{mT_p}{2}\Rightarrow\frac{4\;\times\;10^{-3}\;\times\;21}{2}=42\;\times\;10^{-3}\;m\)
P = ωT ⇒ FT × R × ω
\(15 × {10^3} = {F_t} × 42 × {10^{ - 3}} × \frac{{2\pi\;×\;960}}{{60}}\)
∴ Ft = 3552.26 N
Force transmitted,
\({F_t} = \frac{{{\sigma _b}\;\times\;b\;\times\;Y\;\times\;m}}{{{f_s}}}\)
\(3552 = {\rm{}}\frac{{{\sigma _b}\;\times\;25\;\times\;0.32\;\times\;4}}{{1.5}}\)
∴ σb = 166.5 MPa
The involute function is defined as
Answer (Detailed Solution Below)
Design of Gears Question 11 Detailed Solution
Download Solution PDFExplanation:
Involute function:
It is a function used to design profile of involute gears. This function defines tooth thickness, tooth space and other involute parameters.
Consider the involute of a circle shown in figure (a)
l = length of unwrapped
l = AB arc = rb (β + δ) = rb tan α
where rb = base radius
Thus (β + δ) = tan α
Also from figure (b)
x = rb cosβ + l sinβ = rb [cosβ + (β + δ)sinβ]
y = rb [sinβ - (β + δ)cosβ]
Also θ = α = β = tanα - δ
θ = tanα - α - δ = inv (α) - δ
where inv (α) = tanα - α
here in this question α = ϕ
∴ Inv ϕ = tan ϕ - ϕ represents the involute function
where ϕ = pressure angle
The gearbox in the automobile is placed between
Answer (Detailed Solution Below)
Design of Gears Question 12 Detailed Solution
Download Solution PDFExplanation:
Automotive gearbox:
- It is an important part of the automobile used to provide variable speed as per requirement.
- It is placed between the clutch and hooks joint
- Gearbox act as a link between Engine and Wheel.
- The gearbox consists of a driver shaft, a driven shaft, and a set of gears.
- We generally design gear on the basis of bending stress and it is weak in bending.
Parameters considered during designing of Gearbox:
- Input needs
- Output needs
- Engine power
- Maximum speed
- Load profile
- Driving style
- Road profile
- Cost
Shortcut Trick
The order of components: D G C E (Differential Gearbox Clutch Engine)
A spur gear with 20° full depth teeth is transmitting 20 kW at 200 rad/s. The pitch circle diameter of the gear is 100 mm. The magnitude of the force applied on the gear in the radial direction is
Answer (Detailed Solution Below)
Design of Gears Question 13 Detailed Solution
Download Solution PDFConcept:
The force which driving tooth exerts on the driven tooth is along a line from the pitch point to point of contact of the two teeth. This line is also the common normal at the point of contact of the two mating gears and is known as the line of action or pressure line. This is shown below.
This force along the pressure line can be resolved into two components. First is tangential component which helps in transmitting power. Other component is radial component. This is shown below.
FT = FN cos ϕ …1)
FR = FN sin ϕ …2)
\(\frac{{{F_R}}}{{{F_T}}} = \tan \phi \)
⇒ FR = FT tan ϕ …3)
Calculation:
Given data; Power (P) = 20 kW = 20 × 103 W
Angular velocity (ω) = 200 rad/s
\(P = Tω \Rightarrow T = \frac{P}{ω } = \frac{{20 \times 1000}}{{200}} = 100\;Nm\)
⇒ Torque (T) = 100 Nm
Pitch circle diameter (D) = 100 mm
Radius (R) = 50 mm
\(T = {F_T}R \Rightarrow {F_T} = \frac{{100}}{{50}} \times 1000 = 2000N\)
So, tangential force (FT) = 2000 N
But we need radial force (FR) =?
From 3)
FR = (2000) tan (20°) = 727.94 N = 0.72794 kN
Hence, magnitude of radial force (FR) = 0.72794 kN ≈ 0.73 kN
Keypoints:
Be careful about calculation of forces.
There are 3 forces i.e.
1) Normal force (FN)
2) Tangential force (FT) = FN cos ϕ
3) Radial force (FR) = FN sin ϕ
Power transmitting component is tangential force (FT).
Always see carefully which force component is asked to calculate out of 3 components and calculate accordingly.
Lewis equation is applied
Answer (Detailed Solution Below)
Design of Gears Question 14 Detailed Solution
Download Solution PDFExplanation:
The Lewis equation is applied only to the weaker of the two wheels (i.e. pinion or gear).
Lewis Equation:
Sb = mbσbY
where Sb = Beam strength of gear tooth (N), σb = Permissible bending stress
It is observed that m and b are the same for pinion as well as for gear.
- When different materials are used, the product (σb × Y) decides the weaker between pinion and gear.
- The Lewis form factor Y is always less for a pinion compared with gear.
- When the same material is used for the pinion and gear, the pinion is always weaker than the gear
The following points are to be noted in the design of spur gears:
- Lewis equation is applied to the weaker of the two mating gears.
- If the material of construction is the same, the pinion is weaker.
- If pinion and gear are of different materials, then the product (σw × Y) or (σo × Y) is the deciding factor. Lewis equation is applied on the wheel for which (σw × Y) or (σo × Y) is minimum.
20° full depth involute profile 19 tooth pinion and 37 teeth gear are in mesh. If the module is 5 mm, then the center distance between the gear pair is
Answer (Detailed Solution Below)
Design of Gears Question 15 Detailed Solution
Download Solution PDFConcept:
The center distance between the gear pair is given by:
Centre distance =\(\left( {\frac{{{D_1} + {D_2}}}{2}} \right)\),
where D1 = Pitch diameter of pinion, D2 = Pitch diameter of the gear
Module:
- The module is the ratio of the pitch diameter of gear to the no of teeth, \(m = \frac{D}{T}\)
- The module of two mating gears must be the same (m1 = m2).
Calculation:
Given:
T1 = 19, T2 = 37, m = 5
For pinion,
\(m = \frac{{{D_1}}}{{{T_1}}}\)
\( \Rightarrow 5 = \frac{{{D_1}}}{{19}} \Rightarrow {D_1} = 95\;mm\)
For gear, \(m = \frac{{{D_2}}}{{{T_2}}}\)
\( \Rightarrow 5 = \frac{{{D_2}}}{{37}} \Rightarrow {D_2} = 185\)
Center distance \( = \frac{{{D_1} + {D_2}}}{2} = \frac{{95 + 185}}{2} = 140\;mm\)