Question

A single square-thread screw has an input power of 3 kW at a speed of 1 rev/s. The screw has a diameter of 40 mm and a pitch of 8 mm. The frictional coefficients are 0.18 for the threads and 0.09 for the collar, with a collar friction radius of 50 mm. Find the axial resisting load F and the combined efficiency of the screw and collar if the screw is raising the load.

Answer 1

Load force = 31.24kN

Efficiency = 16.67%

Step-by-step explanation:

It was stated that

Input Power = P,in = 3kW = 3,000W

Speed, S = 1rev/s

Pitch, p = 8mm

Thread frictional coefficient = μt = 0.18

Collar frictional coefficient = μc = 0.09

Friction radius of collar, Rc = 50mm

Step 1, calculate the torque while the load is being lifted in terms of 'F'.

This is determined by using the formula

T = ½FDm[1 + πDmμt]/[πDm - μtp]

By substituton.

T = ½F(40-4)[1 + π(40-4)0.18]/[π(40-4) - 0.18 * 8]

T = 18F(1 + 6.48π)/(36π - 1.44)

T = 3.44F.Nmm

T = 3.44 * 10^-3F Nm

Step 2 is to calculate the torque due to friction from the collar

T = Fμc * Rc

T = F * 0.09 * 50

T = 4.5F. Nmm

T = 4.5 * 10^-3F Nm

Then, we calculate the axial resisting load 'F' by using the the following power input relation.

P,in = Tw

P,in = (T1 + T2) * 2πN

Substitute each value

3,000 = (3.44 + 4.5) * 10^-3 * F * 10^-3 * 2 * π * 2

F = 3000/((3.44 + 4.5) * 10^-3 * 10^-3 * 2 * π * 2

F = 31,247.69N

F = 31.24kN

Hence, the axial resisting load is

F = 31.24kN

Calculating Efficiency

Efficiency = Fp/2πP

Efficiency = 2Fp/P,in

Substitute each value

Efficiency = 2 * 31,247.69 * 8 * 10^-3/3000

Efficiency = 0.166654346666666

So approximately efficiency = 16.67%

Answer 2

Axial Resisting Load, F = 31.24kN

Efficiency = 16.67%

Step-by-step explanation:

Given

Input Power = P,in = 3kW = 3,000W

Speed, S = 1rev/s

Pitch, p = 8mm

Thread frictional coefficient = μt = 0.18

Collar frictional coefficient = μc = 0.09

Friction radius of collar, Rc = 50mm

First, we calculate the torque while the load is being lifted in terms of 'F'.

This is calculated by

T = ½FDm[1 + πDmμt]/[πDm - μtp]

By substituton.

T = ½F(40-4)[1 + π(40-4)0.18]/[π(40-4) - 0.18 * 8]

T = 18F(1 + 6.48π)/(36π - 1.44)

T = 3.44F.Nmm

T = 3.44 * 10^-3F Nm

Then we calculate the torque due to friction from the collar

T = Fμc * Rc

T = F * 0.09 * 50

T = 4.5F. Nmm

T = 4.5 * 10^-3F Nm

Then, we calculate the axial resisting load 'F' by using the the following power input relation.

P,in = Tw

P,in = (T1 + T2) * 2πN

Substitute each value

3,000 = (3.44 + 4.5) * 10^-3 * F * 10^-3 * 2 * π * 2

F = 3000/((3.44 + 4.5) * 10^-3 * 10^-3 * 2 * π * 2

F = 31,247.69N

F = 31.24kN

Hence, the axial resisting load is

F = 31.24kN

Calculating Efficiency

Efficiency = Fp/2πP

Efficiency = 2Fp/P,in

Substitute each value

Efficiency = 2 * 31,247.69 * 8 * 10^-3/3000

Efficiency = 0.166654346666666

Efficiency = 16.67%