Machine Design Miscellaneous
- A force of 400 N is applied to the brake drum of 0.5 m diameter in a band brake system as shown in figure, where the wrapping angle is 180°. If the coefficient of friction between the drum and the band is 0.25, the braking torque applied, in Nm is
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As the drum is rotating in anti clock wise direction, T1 will be right side & T2 will be clock side.T1 = eμθ = e0.25 × π = 2.19 T2
⇒ T2 = = 182.375 N
Braking torque = (T1 – T2 )r
= 54.4 N – mCorrect Option: B
As the drum is rotating in anti clock wise direction, T1 will be right side & T2 will be clock side.T1 = eμθ = e0.25 × π = 2.19 T2
⇒ T2 = = 182.375 N
Braking torque = (T1 – T2 )r
= 54.4 N – m
- A band brake having bandwidth of 80 mm, drum diameter of 250 mm, coefficient of friction of 0.25 and angle of wrap of 270 degrees is required to exert a friction torque of 1000 N m. The maximum tension (in kN) developed in the band is
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Given: b = 80 mm = 0.08 m D = 250 mm = 0.25
R = D = 0.25 2
μ = 0.25θ = 270° = 270 × π = 4.712 radians 180
Ft = 100 N-mAs p1 = eμθ p2
⇒ p1 = eμθ ⋅ p2 = e0.25 × 4.712 ⋅ p2 = 3.25p2
Now, Ft = (p1 − p2)R∴ 1000 = p1 − p1 0.125 3.25
Solving, we get maximum tension in the belt,
p1 = 11.56 kNCorrect Option: D
Given: b = 80 mm = 0.08 m D = 250 mm = 0.25
R = D = 0.25 2
μ = 0.25θ = 270° = 270 × π = 4.712 radians 180
Ft = 100 N-mAs p1 = eμθ p2
⇒ p1 = eμθ ⋅ p2 = e0.25 × 4.712 ⋅ p2 = 3.25p2
Now, Ft = (p1 − p2)R∴ 1000 = p1 − p1 0.125 3.25
Solving, we get maximum tension in the belt,
p1 = 11.56 kN
- In spur gears, the circle on which the involute is generated is called the
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NA
Correct Option: C
NA
- Interference in a pair of gears is avoided, if the addendum circles of both the gears intersect common tangent to the base circles with in the points of tangency.
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NA
Correct Option: A
NA
- For a pinion of 15 teeth, under cutting ______ (increases/decreases) with _______(increase/ decrease) of pressure angle.
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Interference decreases with increase in pressure
Correct Option: A
Interference decreases with increase in pressure