1. A spur gear has a module of 3 mm, number of teeth 16, a face width of 36 mm and a pressure angle of 20°. It is transmitting a power of 3 kW at 20 rev/s. Taking a velocity factor of 1.5 and a form factor of 0.3, the stress in the gear tooth is about

1. 1. 32 MPa

   
   
   

   2. 46 MPa

   
   
   

   3. 58 MPa

   
   
   

   4. 70 MPa

   
   
   

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1. View Hint View Answer Discuss in Forum

   Lewis equation is given as
   F = σ_d C_v bπ ym
   F can be calculated as P / ωr ...[ω = 20 × 2π rad /sec]
   D = mZ = 3 × 16 = 48mm
   C_v = 1.5, b = 36, y = 0.3, m = 3
   

   ∴ F =	3kW
   	40π 24

   
   

   →	3 × 103 × 103	σd × 1.5 × 36 × π × 0.3 × 3
   	960 π

   
   ⇒ ω_d65 ≅ 65
   ⇒ nearest stress in gear tooth taking in account some F.O.S., σ = 58

   Correct Option: B

   Lewis equation is given as
   F = σ_d C_v bπ ym
   F can be calculated as P / ωr ...[ω = 20 × 2π rad /sec]
   D = mZ = 3 × 16 = 48mm
   C_v = 1.5, b = 36, y = 0.3, m = 3
   

   ∴ F =	3kW
   	40π 24

   
   

   →	3 × 103 × 103	σd × 1.5 × 36 × π × 0.3 × 3
   	960 π

   
   ⇒ ω_d65 ≅ 65
   ⇒ nearest stress in gear tooth taking in account some F.O.S., σ = 58

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2. The resultant force on the contacting gear tooth is N is

1. 1. 77.23

   
   
   

   2. 212.20

   
   
   

   3. 2258.1

   
   
   

   4. 289.43

   
   
   

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1. View Hint View Answer Discuss in Forum

   P = Tω
   ∴ 20 × 10³ =T.2π × 30
   ⇒ T =106. 1 Nm
   Now, T = F_t. r
   

   Here, r =	mTp	=	5 × 20	= 50 nm = .05 m
   	2		2

   
   

   Ft =	106.1	=	106.1	= 2122N
   	r		.05

   
   ∴ Resultant force,
   

   F =	Ft	=	2122	= 2258 N.
   	cos θ		cos 20°

   
   

   Correct Option: C

   P = Tω
   ∴ 20 × 10³ =T.2π × 30
   ⇒ T =106. 1 Nm
   Now, T = F_t. r
   

   Here, r =	mTp	=	5 × 20	= 50 nm = .05 m
   	2		2

   
   

   Ft =	106.1	=	106.1	= 2122N
   	r		.05

   
   ∴ Resultant force,
   

   F =	Ft	=	2122	= 2258 N.
   	cos θ		cos 20°

   
   

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3. The contact ratio of the contacting tooth is

1. 1. 1.21

   
   
   

   2. 1.25

   
   
   

   3. 1.29

   
   
   

   4. 1.33

   
   
   

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1. View Hint View Answer Discuss in Forum

   Arc of contact (AOC) =	19	= 20.21 mm
   	cos 20°

   
   

   Circular pitch =	πd	= π m = π × 5 = 5πmm
   	z

   
   

   Contact ratio =		LOA	=	20.21	= 1.29
   		Circular pitch		5.π

   
   

   Correct Option: C

   Arc of contact (AOC) =	19	= 20.21 mm
   	cos 20°

   
   

   Circular pitch =	πd	= π m = π × 5 = 5πmm
   	z

   
   

   Contact ratio =		LOA	=	20.21	= 1.29
   		Circular pitch		5.π

   
   

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4. The center distance for the above gear set in mm is

1. 1. 140

   
   
   

   2. 150

   
   
   

   3. 160

   
   
   

   4. 170

   
   
   

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1. View Hint View Answer Discuss in Forum

   Centre distance
   

   =		m	(Tp + Tg) =	5 (20 + 40)	= 150 mm
   		2		2

   
   

   Correct Option: B

   Centre distance
   

   =		m	(Tp + Tg) =	5 (20 + 40)	= 150 mm
   		2		2

   
   

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5. 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

1. 1. 242.0

   
   
   

   2. 166.5

   
   
   

   3. 121.0

   
   
   

   4. 74.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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