Design of Gears MCQ Quiz - Objective Question with Answer for Design of Gears - Download Free PDF

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.

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:

\( t \approx 6.3 \, \text{mm} \)

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.

Explanation:

Explanation:

Gear finishing processes:

1. Conventional finishing process for gears:
   1. Gear shaving
   2. Gear grinding
   3. Gear honing
   4. Gear lapping
   5. Gear burnishing
   6. Gear skiving
2. Advanced finishing processes for gears:
   1. Gear finishing by electrochemical honing.
   2. Gear finishing by electrochemical grinding.
   3. 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.

Concept:

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

Concept:

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:

1. The effect of the radial component (Wr), which induces compressive stresses, is neglected.
2. 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.
3. The effect of stress concentration is neglected.
4. It is assumed that at any time, only one pair of teeth is in contact and takes the total load.

Concept:

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:

1. The effect of the radial component (Wr), which induces compressive stresses, is neglected.
2. 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.
3. The effect of stress concentration is neglected.
4. 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.

Concept:

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:

1. The effect of the radial component (Wr), which induces compressive stresses, is neglected.
2. 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.
3. The effect of stress concentration is neglected.
4. 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

Concept:

Power transmitted by the gear (P) = Torque × Angular speed ⇒ ωT

Torque (T) = Tangential force × Radius ⇒ F_T × 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, T_p = 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

Explanation:

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 = r_b (β + δ) = r_b tan α

where r_b = base radius

Thus (β + δ) = tan α

Also from figure (b)

x = r_b cosβ + l sinβ = r_b [cosβ + (β + δ)sinβ] 

y = r_b [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

Explanation:

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)

Concept:

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.

F_T = F_N cos ϕ …1)

F_R = F_N sin ϕ …2)

\(\frac{{{F_R}}}{{{F_T}}} = \tan \phi \)

⇒ F_R = F_T 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 (F_T) = 2000 N

But we need radial force (F_R) =?

From 3)

F_R = (2000) tan (20°) = 727.94 N = 0.72794 kN

Hence, magnitude of radial force (F_R) = 0.72794 kN ≈ 0.73 kN

Keypoints:

Be careful about calculation of forces.

There are 3 forces i.e.

1) Normal force (F_N)

2) Tangential force (F_T) = F_N cos ϕ

3) Radial force (F_R) = F_N sin ϕ

Power transmitting component is tangential force (F_T).

Always see carefully which force component is asked to calculate out of 3 components and calculate accordingly.

Explanation:

The Lewis equation is applied only to the weaker of the two wheels (i.e. pinion or gear).

Lewis Equation:

Sb = mbσbY
where S_b = 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. 

Concept:

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 \( = \frac{{{D_1} + {D_2}}}{2} = \frac{{95 + 185}}{2} = 140\;mm\)