Question for GATE Past Year Questions: Brakes, Clutches & Ropes
Try yourself:A wire rope is designated as 6 x 19 standard hoisting. The numbers 6 × 19 represent
[2003]
Question for GATE Past Year Questions: Brakes, Clutches & Ropes
Try yourself:A clutch has outer and inner diameters 100 mm and 40 mm respectively. Assuming a uniform pressure of 2 MPa and coefficient of friction of liner material 0.4, the torque carrying capacity of the clutch is
[2008]
Explanation
T = ∫ r . m. p. 2πr . dr
= ∫ r. 0.4. 2 . 2πr. dr
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Question for GATE Past Year Questions: Brakes, Clutches & Ropes
Try yourself:A disk clutch is required to transmit 5 kW at 2000 rpm. The disk has a friction lining with coefficient of friction equal to 0.25. Bore radius of friction lining is equal to 25 mm. Assume uniform contact pressure of 1 MPa. The value of outside radius of the friction lining is
[2006]
Explanation
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Question for GATE Past Year Questions: Brakes, Clutches & Ropes
Try yourself:Axial operation claw clutches having selflocking tooth profile
[1987]
Question for GATE Past Year Questions: Brakes, Clutches & Ropes
Try yourself:For the brake shown in the figure, which one of the following is TRUE?
[2016,Set-2]
Explanation
Taking moment about hinge in clockwise
Nb – Fl – μNc = 0
F = N(b - μc)/l
Self energizing if b > mc so for clockwise rotation of drum the brake is self energizing.
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Question for GATE Past Year Questions: Brakes, Clutches & Ropes
Try yourself: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
[2012]
Explanation
As the drum is rotating in anti clock wise direction, T1 will be right side & T2 will be clock side.
T1/T2 = eμθ = e0.25 x π = 2.19
⇒ T2 = 182.375 N
Braking torque = (T1 – T2)r
= 54.4 N – m
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Question for GATE Past Year Questions: Brakes, Clutches & Ropes
Try yourself: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
[2010]
Explanation
Given : b = 80 m m = 0.08 m
D = 250 mm = 0.25 m
R = D/2
μ = 0.25
θ = 270° = 270 x π/180 = 4.712 radians
Ft = 1000 N-m
As p1/p2 = eμθ
⇒ p1 = eμθ· p2 = e0.25 x 4.712· p2 = 3.25p2
Now Ft = (p1 - p2) R
∴ 1000 = 0.125
Solving, we get maximum tension in the belt,
p1 = 11.56 kN
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Question for GATE Past Year Questions: Brakes, Clutches & Ropes
Try yourself:A block-brake shown below has a face width of 300 mm and a mean coefficient of friction of 0.25. For an activating force of 400 N, the braking torque in Nm is
[2007]
Explanation
Taking moment about A,
RN × 200 = 400 × 600
⇒ RN = 1200 N
∴ Braking torque
TB = μ × RN x D/2 = 45Nm
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Question for GATE Past Year Questions: Brakes, Clutches & Ropes
Try yourself:In a band brake the ratio of tight side band tension to the tension on the slack side is 3. If the angle of overlap of band on the drum is 180°, the coefficient of friction required between drum and the band is
[2003]
Explanation
Given,
Tension in tight side/Tension is slack side = T1/T2 = 3
T1/T2 = eμθ
where, θ = angle of overlap = 180 = π radians
3 = eμθ
ln(3) = μπ
∴ μ = 1/π In(3) = 0.35.
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