1. SCREW J3010/5/1
UNIT 5
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.

2. SCREW J3010/5/2
INPUT
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.
We use The principle of the
inclined plane in the screw
thread concept/applications.
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.

3. SCREW J3010/5/3
Fig. 5.1 Fig. 5.2
The angle α,
p
Tan α =
πD
(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 α =
πD
2p
Tan α =
πD

4. SCREW J3010/5/4
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 β,
Fig. 5.3., then normal force = W sec β .
The friction force is therefore increased in the ratio sec β : 1, so that the V-thread is
equivalent to a square thread having a coefficient of friction µv sec β
Fig. 5.3

5. SCREW J3010/5/5
INPUT
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).
Motion
R R
α +φ
P
W
α α φ P
W
Fig. 5.4
From the triangle of forces:
P
Tan (α + φ) =
W
P = W tan (α + φ)

6. SCREW J3010/5/6
Torque
The torque T required to rotate the screw against the load is:
1
T = P( D)
2
+O
1
T = PD
2
1
T = WD tan (α + φ)
2 1 P
D
2
Fig. 5.5: Cross-section of Screw
Efficiency
The efficiency of the screw is equal to:
Forward efficiency :
ηforward = force P required without friction (φ = 0)
force P required with friction
w tan α
η =
w tan(α + φ )
tan α
η=
tan(α + φ )
or,
η = work done on load W in 1 revolution
work done by P in 1 revolution
w
η=
P (πD)
w 
η= x
P πD

7. SCREW J3010/5/7
But,
w 1 
= and tan α =
P tan(α + φ ) πD
tan α
η=
tan(α + φ )
Maximum Efficiency
1 − sin φ
ηmak =
1 + sin φ
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
p
(a) tan α =
πD
p = tan 10˚ (3.14)(72.5)
= 40.1 mm

8. SCREW J3010/5/8
(b) μ = tan α
α = tan-1 0.3
α = 16.7˚
tan α
Efficiency for raise load, η=
tan(α + φ )
tan 10 0
η=
tan(10 0 + 16.7 0 )
0.1763
η=
0.5027
η = 0.35 x 100%
η = 35%
1
(c) Torque, T = WD tan (α + φ)
2
1
T = (1000)(72.5) tan (10˚ + 16.7˚)
2
1
T = (1000)(72.5)(0.5027)
2
T = 18.22 x 103 kN mm
T = 18.22 Nm

9. SCREW J3010/5/9
5.4 LOAD BEING LOWERED
(i) When α > φ
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.
R
P
P
Motion
α α R α-φ W
W
Fig. 5.6
From the triangle of forces:
P
Tan (α - φ) =
W
P = W tan (α - φ)
Efficiency
Downward efficiency :
ηdownward = work done by P in 1 revolution
work done on load W in 1 revolution
P(πD)
η=
w
tan(α − φ )
η=
tan α

10. SCREW J3010/5/10
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, α = φ
Given :
μ = 0.08 μ = tan φ φ = 4.57˚
p
Tan α = then, α = φ
πD
p
Tan 4.57˚ =
(3.14)(8)
p = 20 mm
p
If the pitch is 15 mm, Tan α =
πD
15
Tan α =
(3.14)(8)
Tan α = 5.97
α = 80.49˚
1
Torque, T = WD tan (α - φ)
2
1
T = (3000)(8)tan (80.49˚- 4.57˚)
2
1
T = (3000)(8)(tan 75˚ 92’)
2
T = 2.4 Nm

11. SCREW J3010/5/11
(ii) When φ > α
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.
R
Motion
α φ-α
R φ-α W
P
α α P
W
Fig. 5.7
From the triangle of forces:
P
Tan (φ - α ) =
W
P = W tan ( φ - α )
Efficiency
tan(φ − α )
ηterbalik =
tan α

12. SCREW J3010/5/12
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 φ Then, φ = 8.32˚
Length of lever, L = 70 cm Load W = 2 t = 2000 kg
Force required to lower the load,
P = W tan (φ – α)
= 2000 tan (8. 32˚-3.39˚)
= 2000 tan 4. 53˚
= 2000 x 0.0854
= 170.8 kg

13. SCREW J3010/5/13
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.

14. SCREW J3010/5/14
Feedback to Activity 5A
Have you tried the questions????? If “YES”, check your answers now
5.1 161 kg, 3.47 Nm
5.2 191 W
5.3 14.07 kg
5.4 4450 Nmm

15. SCREW J3010/5/15
INPUT
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 ( α - φ )
When the load overhauls, P = 0
Therefore,
tan ( α - φ ) = 0
α-φ =0
α=φ
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 α
η=
tan(α + φ )
but α = φ
tan α
η=
tan 2α

16. SCREW J3010/5/16
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:
1
W = 6000 N D mean = 45 + x 10 = 50 mm μ = 0.05 p = 10
2
mm
Then,
p
tan φ = μ tan α =
πD
10
tan φ = 0.05 tan α =
(3.14)(50)
φ = 2.86˚ tan α = 0.0637
α = 3.64˚
1
Torque required, T = WD tan (α + φ)
2
1
= (6000)(50) tan (3.64˚ + 2.86˚)
2
= 3000(50) tan (6.50˚)
= 3000(50) (0.114)
= 17.1 Nm
tan α
Efficiency, η=
tan(α + φ )
tan 3.64 0
η=
tan(3.64 0 + 2.86 0 )
0.0636
η=
0.114
η = 0.558

17. SCREW J3010/5/17
η = 56% > 50% (overhauling)
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.

18. SCREW J3010/5/18
Feedback To Activity 5B
Have you tried the questions????? If “YES”, check your answers now
5.5 10 gN
1
5.6 7 gN
2
5.7 12.96 m/min, 60.5 W, 0.11
2
5.8 50, 16 gN.
3

19. SCREW J3010/5/19
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,
µ = 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.

20. 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 %
CONGRATULATIONS!!!!
…..May success be with
4. 41.5 %
you always….