Design of power screws - Module 6 Power Screws Lesson 2 Notes | EduRev

: Design of power screws - Module 6 Power Screws Lesson 2 Notes | EduRev

 Page 1


 
 
 
 
 
 
 
 
 
Module  
6 
Power Screws 
Version 2 ME, IIT Kharagpur 
Page 2


 
 
 
 
 
 
 
 
 
Module  
6 
Power Screws 
Version 2 ME, IIT Kharagpur 
 
 
 
 
 
 
Lesson  
  2 
Design of power screws 
Version 2 ME, IIT Kharagpur 
Page 3


 
 
 
 
 
 
 
 
 
Module  
6 
Power Screws 
Version 2 ME, IIT Kharagpur 
 
 
 
 
 
 
Lesson  
  2 
Design of power screws 
Version 2 ME, IIT Kharagpur 
 
Instructional Objectives 
 
At the end of this lesson, the students should have the knowledge of 
 
• Stresses in power screw. 
• Design procedure of a power screw. 
 
6.2.1 Stresses in power screws 
 
Design of a power screw must be based on the stresses developed in the 
constituent  parts. A power screw is subjected to an axial load and a turning 
moment. The following stresses would be developed due to the loading: 
 
a) Compressive stress is developed in a power screw due to axial load. 
Depending on the slenderness ratio it may be necessary to analyze for 
buckling. The compressive stress s
c
 is given by s
c
 = 
2
c
P
d p
 where d
c
 is the 
core diameter and if slenderness ratio ? is more than 100 or so buckling 
criterion must be used. ? is defined as ? = 
L
k
 where I=Ak
2
 and L is the 
length of the screw. Buckling analysis yields a critical load P
c
 and if  
 
both 
ends are assumed to be hinged critical load is given by P
C
 = 
2
2
EI
L
p . In 
general the equation may be written as P
C
 = 
2
2
EI
n
L
p where n is a constant 
that depends on end conditions. 
b) Torsional shear stress is developed in the screw due to the turning 
moment and this is given by t =
3
c
16T
d p
 where T is the torque applied. 
Version 2 ME, IIT Kharagpur 
Page 4


 
 
 
 
 
 
 
 
 
Module  
6 
Power Screws 
Version 2 ME, IIT Kharagpur 
 
 
 
 
 
 
Lesson  
  2 
Design of power screws 
Version 2 ME, IIT Kharagpur 
 
Instructional Objectives 
 
At the end of this lesson, the students should have the knowledge of 
 
• Stresses in power screw. 
• Design procedure of a power screw. 
 
6.2.1 Stresses in power screws 
 
Design of a power screw must be based on the stresses developed in the 
constituent  parts. A power screw is subjected to an axial load and a turning 
moment. The following stresses would be developed due to the loading: 
 
a) Compressive stress is developed in a power screw due to axial load. 
Depending on the slenderness ratio it may be necessary to analyze for 
buckling. The compressive stress s
c
 is given by s
c
 = 
2
c
P
d p
 where d
c
 is the 
core diameter and if slenderness ratio ? is more than 100 or so buckling 
criterion must be used. ? is defined as ? = 
L
k
 where I=Ak
2
 and L is the 
length of the screw. Buckling analysis yields a critical load P
c
 and if  
 
both 
ends are assumed to be hinged critical load is given by P
C
 = 
2
2
EI
L
p . In 
general the equation may be written as P
C
 = 
2
2
EI
n
L
p where n is a constant 
that depends on end conditions. 
b) Torsional shear stress is developed in the screw due to the turning 
moment and this is given by t =
3
c
16T
d p
 where T is the torque applied. 
Version 2 ME, IIT Kharagpur 
c)  Bending stresses are developed in the screw thread and this is illustrated 
in figure-6.2.1.1. The bending moment M=
'
Fh
2
 and the bending stress on 
a single thread is given by s
b
=
My
I
. Here y =
t
2
, I = 
3
m
dt
12
p
and F
' 
is the load 
on a single thread. Figure-6.2.1.2 shows a developed thread and figure-
6.2.1.3 shows a nut and screw assembly. This gives the bending stress at 
the thread root to be s
b
 =
'
2
m
3Fh
dt p
 . This is clearly the most probable place 
for failure. 
Assuming that the load is equally shared by the nut threads 
d) Bearing stress s
br
 at the threads is given by s
br
=
'
m
F/n
dh
'
p
 
e) Again on similar assumption shear stress t at the root diameter is given by  
      t= 
'
c
F/n
dt
'
p
 
Here n
/
 is the number of threads in the nut. Since the screw is subjected to 
torsional shear stress in addition to direct or transverse stress combined 
effect of bending, torsion and tension or compression should be considered in 
the design criterion. 
 
6.2.2 Design procedure of a Screw Jack  
 
A typical screw jack is shown in figure-6.2.2.1 . It is probably more 
informative to consider the design of a jack for a given load and lift. We 
consider a reasonable value of the load to be 100KN and lifting height to be 
500mm. The design will be considered in the following steps: 
 
 
 
Version 2 ME, IIT Kharagpur 
Page 5


 
 
 
 
 
 
 
 
 
Module  
6 
Power Screws 
Version 2 ME, IIT Kharagpur 
 
 
 
 
 
 
Lesson  
  2 
Design of power screws 
Version 2 ME, IIT Kharagpur 
 
Instructional Objectives 
 
At the end of this lesson, the students should have the knowledge of 
 
• Stresses in power screw. 
• Design procedure of a power screw. 
 
6.2.1 Stresses in power screws 
 
Design of a power screw must be based on the stresses developed in the 
constituent  parts. A power screw is subjected to an axial load and a turning 
moment. The following stresses would be developed due to the loading: 
 
a) Compressive stress is developed in a power screw due to axial load. 
Depending on the slenderness ratio it may be necessary to analyze for 
buckling. The compressive stress s
c
 is given by s
c
 = 
2
c
P
d p
 where d
c
 is the 
core diameter and if slenderness ratio ? is more than 100 or so buckling 
criterion must be used. ? is defined as ? = 
L
k
 where I=Ak
2
 and L is the 
length of the screw. Buckling analysis yields a critical load P
c
 and if  
 
both 
ends are assumed to be hinged critical load is given by P
C
 = 
2
2
EI
L
p . In 
general the equation may be written as P
C
 = 
2
2
EI
n
L
p where n is a constant 
that depends on end conditions. 
b) Torsional shear stress is developed in the screw due to the turning 
moment and this is given by t =
3
c
16T
d p
 where T is the torque applied. 
Version 2 ME, IIT Kharagpur 
c)  Bending stresses are developed in the screw thread and this is illustrated 
in figure-6.2.1.1. The bending moment M=
'
Fh
2
 and the bending stress on 
a single thread is given by s
b
=
My
I
. Here y =
t
2
, I = 
3
m
dt
12
p
and F
' 
is the load 
on a single thread. Figure-6.2.1.2 shows a developed thread and figure-
6.2.1.3 shows a nut and screw assembly. This gives the bending stress at 
the thread root to be s
b
 =
'
2
m
3Fh
dt p
 . This is clearly the most probable place 
for failure. 
Assuming that the load is equally shared by the nut threads 
d) Bearing stress s
br
 at the threads is given by s
br
=
'
m
F/n
dh
'
p
 
e) Again on similar assumption shear stress t at the root diameter is given by  
      t= 
'
c
F/n
dt
'
p
 
Here n
/
 is the number of threads in the nut. Since the screw is subjected to 
torsional shear stress in addition to direct or transverse stress combined 
effect of bending, torsion and tension or compression should be considered in 
the design criterion. 
 
6.2.2 Design procedure of a Screw Jack  
 
A typical screw jack is shown in figure-6.2.2.1 . It is probably more 
informative to consider the design of a jack for a given load and lift. We 
consider a reasonable value of the load to be 100KN and lifting height to be 
500mm. The design will be considered in the following steps: 
 
 
 
Version 2 ME, IIT Kharagpur 
1. Design of the screw  
A typical screw for this purpose is shown in figure-6.2.2.2. 
Let us consider a mild steel screw for which the tensile and shear strengths 
may be taken to be approximately 448MPa and 224 MPa respectively. Mild 
steel being a ductile material we may take the compressive yield strength to 
be also close to 448MPa. Taking a very high factor of safety of 10 due to the 
nature of the application and considering the axial compression the core 
diameter of the screw d
c
 is given by 
3
c
6
100x10
d
448x10
410
=
?? p
??
??
 which gives d
c
 ˜ 54 
mm. 
From the chart of normal series square threads in table- 6.1.1.1  the nearest 
standard nominal diameter of 70 mm is chosen, with pitch p = 10 mm. 
Therefore, core diameter d
c
 = 60 mm , Major diameter d
maj
 = 70mm , Mean 
diameter d
m
 = 65 mm , Nominal diameter d
n
 = 70mm. 
The torque required to raise the load is given by  
T = 
mm
m
Fd d
2d
?? +µp
??
p-µ
??
l
l
 
Where l = np, n being the number of starts. Here we have a single start screw 
and hence l = p =10mm, d
m 
= 65mm, F = 100X10
3
N 
Taking a safe value of µ for this purpose to be 0.26 and substituting the 
values we get  
T = 1027 Nm. 
 
Check for combined stress  
 
 The screw is subjected to a direct compressive stress s
c
 and a torsional 
shear stress t. The stresses are given by  
s
c
 = 
3
22
c
4F 4x100x10
35.3MPa
d x(0.06)
==
pp
 
Version 2 ME, IIT Kharagpur 
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!