A machine vice whose length of the handle is 150mm and the coefficient...
Explanation: M₂=0.17 x W x (55+45)/4 or M₂=4.25W N-mm.
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A machine vice whose length of the handle is 150mm and the coefficient...
Problem Statement:
A machine vice has a handle length of 150mm, a coefficient of friction between thread and collar of 0.15 and 0.17 respectively, and a force of 125N applied at the handle. The collar has outer and inner diameters of 55mm and 45mm respectively, with a nominal diameter of 22mm and a pitch of 5mm. Find the collar torque in terms of clamping force W assuming uniform wear theory.
Solution:
Given,
Length of handle, L = 150mm
Coefficient of friction between thread and collar, μt = 0.15
Coefficient of friction between collar and vice body, μc = 0.17
Force applied at handle, F = 125N
Outer diameter of collar, D2 = 55mm
Inner diameter of collar, D1 = 45mm
Nominal diameter, d = 22mm
Pitch, p = 5mm
We know that the torque required to clamp the workpiece is given by:
T = W × R
where W is the clamping force and R is the effective radius.
Effective radius of collar, r = (D1 + D2)/4 = 0.025m
Clamping force, W = F × μc × μt
W = 125 × 0.17 × 0.15 = 3.1875N
Using uniform wear theory, the effective diameter of the collar is given by:
De = d - 0.5k
where k is the wear allowance.
Assuming a wear allowance of 0.05mm, k = 0.05mm = 0.00005m
De = 22 - 0.5 × 0.00005 = 21.999975mm = 0.021999975m
The pitch circumference of the collar is given by:
Cp = πdp
Cp = π × 0.022m × 5 = 0.03491m
The torque required to clamp the workpiece is given by:
T = W × Cp/2πDe
T = 3.1875 × 0.03491/2π × 0.021999975
T ≈ 4.25W
Therefore, the collar torque in terms of clamping force W assuming uniform wear theory is 4.25W.
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