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.

torque to raise the load = 16.411 Nm

torque to lower the load = 8.40 Nm

overall efficiency = 0.24

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 = 0.06

To find:

torque to "raise" the load

torque to and "lower"

overall efficiency

Solution:

Compute torque to raise the load:

where

is the torque

F is the load

is diameter of thread

is diameter of collar

L is the thread pitch distance

µ is coefficient of friction for thread

µ  is coefficient of friction for collar

Putting the values in above formula:

= 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

= 16.411 Nm

Compute torque to lower the load:

     = 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

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

                             = 25/ 103.06108

overall efficiency = 0.24

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