Mechanical efficiency of screw with a nut

MECHANICAL EFFICIENCY OF SCREW WITH A NUT

A screw when developed is an inclined plane. Threads are cut on a cylindrical body of diameter 'd', as shown in Fig.6.11(a).

The circumference of the cylinder = π d

Helix angle of the thread = tan α = p / πd

where p = Pitch of the thread, and

α = Inclination of the plane which is equal to the helix angle of the thread.

Fig.6.11(b) shows that the motion of the nut on a screw is analogous to the motion on an inclined plane.

1. When the Nut Moves Upward

When the nut moves upward on screw, it is similar to the motion of body up the plane.

Since the effort P required to move the body acts horizontally (Fig.6.11(b)), substitute θ = 90°.

Then the mechanical efficiency is given by

2. When the Nut is Lowered

Similarly when nut moves downward on screw, it is similar to motion of the body down the plane.

Since the effort P required to move the body acts horizontally (Fig.6.11(b)), substitute θ = 90°.

Then the mechanical efficiency is given by

3. Condition for Maximum Efficiency of Nut and Screw 

The efficiency of nut and screw arrangement is given by

It is the efficiency for the upward motion.

For determining condition of maximum efficiency,

This is the condition for maximum efficiency of screw jack.

Substituting the value of a in the efficiency equation, we get

This is the maximum efficiency of a nut and screw arrangement for raising the load.

Note

As discussed above, the maximum efficiency of a nut and screw arrangement for lowering a load can also be derived and its expression is same as for raising the load.

Example 6.3 

A square threaded bolt of root diameter 22.5 mm and pitch 5 mm is tightened by screwing nut whose mean diameter of bearing surface is 50 mm. If coefficient of friction for nut and bolt is 0.1 and for nut and bearing surface is 0.16, find the force required at the end of a spanner 500 mm long when the load on the bolt is 10 kN.

[A.U., Nov/Dec 2010]. 

Given data: 

Square thread; d =22.5 mm = 0.0225 m; p = 5 mm = 0.005 m; D = 50 mm or R = 25 mm = 0.025 m; μ = tan ϕ = 0.16; l = 500 mm = 0.5 m; W = 10 kN = 10 × 103 N.

Solution: 

For at the circumference of screw is given by

Example 6.4 

A 150 mm diameter valve, against a steam pressure of 2 MN/m2 is acting, is closed by means of a square threaded screw 50 mm in external diameter with 6 mm pitch. If the coefficient of friction is 0.12, find the torque required to turn the handle. 

Given data: 

D = 150 mm; psteam = 2 MN/m2 = 2 × 106 N/m2; d0 = 50 mm = 0.05 m;

p = 6 mm = 0.006 m; μ = tan ϕ = 0.12.

Solution:

We know that force required to turn the handle,

P = W tan (α + ϕ) = 35343 tan (2.33° + 6.84°) = 5705.33 N

⸫ Torque required to turn the handle,

Example 6.5

A vertical screw with single start square threads 50 mm mean diameter and 10 mm pitch is raised against a load of 5500 N by means of a hand wheel, the boss of which is threaded to act as a nut. The axial load is taken up by a thrust collar which supports the wheel boss and has a mean diameter of 65 mm. If the coefficient of friction is 0.15 for the screw and 0.18 for the collar and tangential force applied by each hand to the wheel is 140 N, find suitable diameter of the hand wheel.

Given data: 

d = 50 mm 50 × 10-3 m; p = 10 mm = 10 × 10-3 m; W = 5500 N; D = 65 mm or R = 32.5 mm = 32.5 × 10-3 m; μ = tan ϕ = 15; μ1 = 0.18; P1 = 140 N.

Solution:

Tangential force required at the circumference of the screw,

Total torque required to turn the hand wheel,

Let

D1 = Diameter of the hand wheel

We know that the torque applied to the hand wheel,

T = Tangential load on wheel × Radius of wheel

Substituting the value of T from equation (i) in (ii), we get

61.837 = 140 × D1 or D1 = 0.441 m Ans.