Question

A single-threaded 25-mm power screw hasa pitch of 5 mm. The frictional diameter of the collar is 45 mm. The max load onvertical direction of the screw is5kN. The collar has a coefficients of friction of0.06, and he threads hasa coefficients of friction of0.09. Find the overall efficiency and the torque to "raise" and "lower" the load.

Answer 1

torque to raise the load = 16.411 Nm

torque to lower the load = 8.40 Nm

overall efficiency = 0.24

Step-by-step explanation:

Given:

max load on vertical direction of the screw = Force = F = 5kN

frictional diameter of the collar = 45 mm

Diameter = 25 mm

length of pitch = 5 mm

coefficient of friction for thread µ = 0.09

coefficient of friction for collar µ
`_(c)` = 0.06

To find:

torque to "raise" the load

torque to and "lower"

overall efficiency

Solution:

Compute torque to raise the load:

`T_(R) = ( Fd_(m))/(2) ((L+(\pi ud_(m)) )/(\pi d_(m)-uL )) +(Fu_(c) d_(c) )/(2)`

where

`T_(R)` is the torque

F is the load

`d_(m)` is diameter of thread

`d_(c)` is diameter of collar

L is the thread pitch distance

µ is coefficient of friction for thread

µ
`_(c)` is coefficient of friction for collar

Putting the values in above formula:

`T_(R)` = 5(25) / 2 [5+ (π(0.09)(25) / π(25)-0.09(5)] + 5(0.06)(45) / 2

= 125/2 [5 + (3.14)(0.09)(25)/ 3.14(25)-0.45] + 13.5/2

= 62.5 [(5 + 7.065) / 78.5 - 0.45] + 6.75

= 62.5 [12.065 / 78.05 ] + 6.75

= 62.5 (0.15458) + 6.75

= 9.66125 + 6.75

= 16.41125

`T_(R)` = 16.411 Nm

Compute torque to lower the load:

`T_(L) = ( Fd_(m))/(2) (((\pi ud_(m)) - L )/(\pi d_(m)-uL )) +(Fu_(c) d_(c) )/(2)`

= 5(25) / 2 [ (π(0.09)(25) - L / π(25)-0.09(5) ] + 5(0.06)(45) / 2

= 125/2 [ ((3.14)(0.09)(25) - 5) / 3.14(25)-0.45 ] + 13.5/2

= 62.5 [ (7.065 - 5) / 78.5 - 0.45 ] + 6.75

= 62.5 [ 2.065 / 78.05 ] + 6.75

= 62.5 (0.026457) + 6.75

= 1.6535625 + 6.75

= 8.40 Nm

Since the torque required to lower the the load is positive indicating that an effort is applied to lower the load, Hence the thread is self locking.

Compute overall efficiency:

overall efficiency = F(L) / 2π
`T_(R)`

= 5(5) / 2(3.14)( 16.411)

= 25/ 103.06108

overall efficiency = 0.24