SCREW J3010/5/1

SCREW

OBJECTIVES

General Objective : To understand the concept of the working of a screw.

Specific Objectives : At the end of this unit you should be able to :

state the terms that are important for the study of screw.

match the principle of screw working to that of friction on an inclined

plane.

use these suitable concepts to solve problem involve.

calculate the answer using these concepts correctly.

UNIT 5

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5.0 INTRODUCTION

The screws, bolt, studs, nuts etc. are widely used in various machines and structures

for fastenings. These fastenings have screw threads, which are made by cutting a

continuous helical groove on cylindrical surface. If the threads are cut on the outer

surface of a solid rod, these are known as external threads. But if the threads are cut on

the internal surface of a hollow rod, these are known as internal threads.

5.1 THE SQUARE-THREADED SCREW

(a) A Single-Start Thread

Let W be the axial force against which the screw is turned, Fig. 5.1, and P the

tangential force at the mean thread radius necessary to turn the nut. The

development of a thread ia an inclined plane, Fig. 5.2, and turning the nut

against the load is equivalent to moving this load up the plane by the

horizontol force P applied at the mean radius of the thread.

We use The principle of the

inclined plane in the screw thread

concept/applications.

INPUT

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Fig. 5.1 Fig. 5.2

The angle ,

Tan = D

p

(b) A Two-Start Thread

For a two-start thread the distance moves axially by the screw in its nut in one

revolution is the lead , which is twice the pitch.

Tan = D

Tan = D

p

2

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5.2 THE VEE-THREADED SCREW

For a V-thread, the normal force between the nut and the screw is increased since the

axial component of this force must equal W. Thus, if the semi-angle of the thread is ,

Fig. 5.3., then normal force = W sec .

The friction force is therefore increased in the ratio sec : 1, so that the V-thread is

equivalent to a square thread having a coefficient of friction v sec

Fig. 5.3

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Turning the screw is equivalent to moving a mass of weight W along the inclined plane by a

horizontal force P.

5.3 RAISING LOAD

When the load is raised by the force P the motion is up the plane.(Fig.5.4).

Fig. 5.4

From the triangle of forces:

Tan ( + ) = W

P

P = W tan ( + )

Motion

W

R

P + W

R

P

INPUT

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Torque

The torque T required to rotate the screw against the load is:

T = P ( D2

1)

T = PD2

1

T = 2

1WD tan ( + )

Fig. 5.5: Cross-section of Screw

Efficiency

The efficiency of the screw is equal to:

Forward efficiency :

forward = force P required without friction ( = 0)

force P required with friction

= )tan(

tan

w

w

= )tan(

tan

or,

= work done on load W in 1 revolution

work done by P in 1 revolution

= )( DP

w

= D

xP

w

+ O

P

D2

1

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But,

)tan(

1

P

w and

D

tan

= )tan(

tan

Maximum Efficiency

mak =

sin1

sin1

Example 5.1

The helix angle of a screw tread is 10˚. If the coefficient of friction is 0.3 and the mean

diameter of the square thread is 72.5 mm, calculate:

(a) the pitch of the tread,

(b) the efficiency when raising a load of 1 kN,

(c) the torque required.

Solution 5.1

Given:

α = 10˚ μ = 0.3 Dmean = 72.5 mm

(a) tan α = D

p

p = tan 10˚ (3.14)(72.5)

= 40.1 mm

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(b) μ = tan α

α = tan-1

0.3

α = 16.7˚

Efficiency for raise load, = )tan(

tan

= )7.1610tan(

10tan00

0

= 5027.0

1763.0

= 0.35 x 100%

= 35%

(c) Torque, T = 2

1WD tan ( + )

T = 2

1(1000)(72.5) tan (10˚ + 16.7˚)

T = 2

1(1000)(72.5)(0.5027)

T = 18.22 x 103 kN mm

T = 18.22 Nm

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5.4 LOAD BEING LOWERED

(i) When >

P is applied to resist the downward movement of the load. Under

this condition the load would just about to move downwards. If P

were not applied in this direction the load would overhaul, that is

move down on its own weight.

Fig. 5.6

From the triangle of forces:

Tan ( - ) = W

P

P = W tan ( - )

Efficiency

Downward efficiency :

downward = work done by P in 1 revolution

work done on load W in 1 revolution

= w

DP )(

=

tan

)tan(

Motion

W

R

P

- W R

P

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Example 5.2

Calculate the pitch of a single-start square threaded screw of a jack which will

just allow the load to fall uniformly under its own weight. The mean diameter

of the threads is 8 cm and μ= 0.08. If the pitch is 15 mm, what is the torque

required to lower a load of 3 kN?

Solution 5.2

If the load is to fall uniformly under its own weight, =

Given :

μ = 0.08 μ = tan = 4.57˚

Tan = D

p

then, =

Tan 4.57˚ = )8)(14.3(

p

p = 20 mm

If the pitch is 15 mm, Tan = D

p

Tan = )8)(14.3(

15

Tan = 5.97

= 80.49˚

Torque, T = 2

1WD tan ( - )

T = 2

1(3000)(8)tan (80.49˚- 4.57˚)

T = 2

1(3000)(8)(tan 75˚ 92’)

T = 2.4 Nm

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(ii) When >

If the angle of friction is greater than the angle of the plane, then it can

be seen from the triangle of force, that force P must be applied to help

lower the load.

Fig. 5.7

From the triangle of forces:

Tan ( - ) = W

P

P = W tan ( - )

Efficiency

terbalik =

tan

)tan(

W

R

P

-

- W R

P

Motion

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Example 5.3

The mean diameter of a square-threaded screw jack is 5 cm. The pitch

of the thread is 1 cm. The coefficient of friction is 0.15. What force

must be applied at the end of a 70 cm long lever, which is

perpendicular to the longitudinal axis of the screw to lower a load of 2

tonnes?

Solution 5.3

Given:

Mean diameter, d = 5 cm Pitch, p = 1 cm

Coefficient of friction, μ = 0.15 = tan Then, = 8.32˚

Length of lever, L = 70 cm Load W = 2 t = 2000 kg

Force required to lower the load,

P = W tan ( – α)

= 2000 tan (8. 32˚-3.39˚)

= 2000 tan 4. 53˚

= 2000 x 0.0854

= 170.8 kg

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Activity 5A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT

INPUT…!

5.1 A nut on a single-start square thread bolt is locked tight by a torque of 6 Nm. The

thread pitch is 5 mm and the mean diameter 6 cm. Calculate :

(a) the axial load on the screw in kilogrammes

(b) the torque required to loosen the nut. μ = 0.1.

5.2 The total mass of the table of a planning machine and its attached work piece is 350

kg. The table is traversed by a single-start square thread of external diameter 45 mm

and pitch 10 mm. The pressure of the cutting is 600 N and the speed of cutting is 6

meters per minute. The coefficient of friction for the table is 0.1 and for the screw

thread is 0.08. Find the power required.

5.3 The mean radius of a screw of a square thread screw jack is 2.5 cm. The pitch of the

thread is 7.5 mm. If the coefficient of friction is 0.12, what effort applied at the end of

a lever 60 cm length is needed to raise a weight of 2000 kg?.

5.4 A screw press is used to compress books. The thread is double thread (square head)

with a pitch of 4 mm and a mean radius of 25 mm. The coefficient of friction for the

contact surfaces of the threads is 0.3. Determine the torque required for a pressure of

500 N.

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Feedback to Activity 5A

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5.1 161 kg, 3.47 Nm

5.2 191 W

5.3 14.07 kg

5.4 4450 Nmm

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5.5 OVERHAULING OF THE SCREW

When the load moves down and overcomes the thread friction by its own weight, it is

said to overhaul. When it moves down against a resisting force P we found that

P = W tan ( - )

When the load overhauls, P = 0

Therefore,

tan ( - ) = 0

- = 0

=

Hence,

when the angle of the inclined plane is equal to the angle of the friction the load will

overhaul.

The efficiency of such a screw when raising a load is given by:

= )tan(

tan

but =

=

2tan

tan

INPUT

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Example 5.4

A load of 6 kN is lifted by a jack having a single-start square thread screw of 45 mm

core diameter and pitch 10 mm. The load revolves with the screw and the coefficient

of friction at the screw thread is 0.05. Find the torque required to lift the load. Show

that the load will overhaul.

Solution 5.4

Given:

W = 6000 N D mean = 45 + 2

1 x 10 = 50 mm μ = 0.05 p = 10 mm

Then,

tan = μ tan α = D

p

tan = 0.05 tan α = )50)(14.3(

10

= 2.86˚ tan α = 0637.0

α = 3.64˚

Torque required, T = 2

1WD tan ( + )

= 2

1(6000)(50) tan (3.64˚ + 2.86˚)

= 3000(50) tan (6.50˚)

= 3000(50) (0.114)

= 17.1 Nm

Efficiency, = )tan(

tan

= )86.264.3tan(

64.3tan00

0

= 114.0

0636.0

= 558.0

= 56% > 50% (overhauling)

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Activity 5B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT

INPUT…!

5.5 The drum of a windlass is 10 cm in diameter and the effort is applied to the handle 60

cm from the axis. Find the effort necessary to support a weight of mass 120 kg.

5.6 A wheel and axle is used to raise a load of mass 50 kg. The radius of the wheel is 50

cm and while it makes seven revolutions the load rises 3.3 m. What is the smallest

force that will support the load?.

5.7 A table carrying a machine tool is traversed by a three-start screw of 6 mm pitch and

external diameter 23 mm. The mass of the table is 200 kg and the coefficient of

friction between the table and its guides is 0.1. The screw is driven by a motor at 12

rev/s. Find the speed of operation of the tool, the power required and the coefficient of

friction for the thread if the efficiency is 70 percent.

5.8 A differential pulley, the two parts of which have respectively twenty four and twenty

five teeth, is used to raise a weight of mass 500 kg. Show by sketch how the apparatus

is used, and determine its velocity ratio. Find also what effort must be exerted if the

efficiency is 60 percent.

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5.5 10 gN

5.6 72

1 gN

5.7 12.96 m/min, 60.5 W, 0.11

5.8 50, 163

2 gN.

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SELF-ASSESSMENT 5

You are approaching success. Try all the questions in this self-assessment section and check

your answers with those given in the Feedback on Self-Assessment 5 given on the next page.

If you face any problems, discuss it with your lecturer. Good luck.

1. Find the torque to raise a load of 6 kN using a screw jack with a double-start square

thread, containing two threads per centimeter and a mean diameter of 60 mm. μ =

0.12. What is the torque required to lower the load?

2. A lathe saddle of mass 30 kg is traversed by a single-start square-thread screw of 10

mm pitch and mean diameter 40 mm. If the vertical force on the cutting tool is 250 N,

find the torque at the screw required to traverse the saddle. The coefficient of friction

between saddle and lathe bed and for the screw thread is 0.15.

3. A double-start square-thread screw has a pitch of 20 mm and a mean diameter of 100

mm, = 0.03. Calculate its efficiency when raising a load.

4. A load of 2500 N is to be raised by a screw jack with mean diameter of 75 mm and

pitch of 12 mm. Find the efficiency of the screw jack, if the coefficient of friction

between the screw and nut is 0.075.

SCREW J3010/5/20

Feedback to Self-Assessment 5

Have you tried the questions????? If “YES”, check your answers now

1. 31.4 Nm , 12 Nm

2. 0.376 Nm

3. 80.5 %

4. 41.5 %

CONGRATULATIONS!!!!…..

May success be with you

always….