MACHINE DESIGN
ENRG-410
Lecture No. 7 & 8
Sajid Ali, PhD
Assistant Professor, College of Engineering
1
Types of Screw Threads
There are two distinct uses for screw threads, and they usually demand
different behavior from the threads
Power Screw transforms rotary motion into
substantial linear motion (or vice versa in certain
applications).
Examples:
➢Lead screw in Lathe
➢The screw in a car lifting jack
Threaded Fastener similar to a nut and bolt which joins a number of
components together again by transforming rotary motion into linear motion,
though in this case the translation is small.
2
8-1: Thread Standards and Definitions
Terminology of screw threads. Sharp Vee Threads shown for clarity; the crests and roots
are actually flattened or rounded during the forming operation
All threads are made according to the right-hand rule unless otherwise noted.
3
Threads
A multiple-threaded product is one having two or more threads cut beside each other
(Imagine two or more strings wound side by side around a pencil).
(a) Single (STANDARD)-, (b) Double-, and (c) Triple threaded screws.
Lead (L) is the distance the nut moves parallel to the screw axis when the nut is
given one turn.
L=n p
4
Thread Systems
A thread 'system' is a set of basic thread proportions which is scaled to different screw
sizes to define the thread geometry.
• ISO Metric thread system.
• The American National (Unified) Thread standards
• Square and ACME
(a)
(b)
(c)
Thread profiles. a) Square (b) ACME; (c) UN, ISO
5
ISO Metric threads
Major
Dia
H=
3
p
2
Pitch
Dia
Minor
Dia
Figure 8-2: Thread Profile for Metric (M, MJ). MJ has rounded fillet at the root.
6
ISO Metric threads
1) ISO Metric thread system:
Table 8-1
Major diameter (mm),
2a= 60°
Standard thread is RH
Specifications:
M12 x1.75 or MJ12 x 1.75
M = Basic Metric, J = round root;
12 = nominal major diameter (mm);
1.75 = pitch (mm)
7
ISO Metric threads
Tensile-stress area
(At) of the threaded
rod: An unthreaded
rod having a
diameter equal to
the mean of the
pith diameter and
minor diameter
have same tensile
strength as the
threaded rod.
  d p + dr
At = 
4 2



2
8
The American National (Unified) Thread standards
2) The American National (Unified) Thread standards is used
mainly in the US:
• Table-8-2 (Size designation) use d
• UN=regular thread,
• UNR=round root (use root radius)
Specifications:
5/8”-18 UN, UNC, UNF, UNR, UNRC, UNRF
5/8”= d
18 = N (thread size i.e. no. of threads per inch)
UN = Unified, F=fine, C=Coarse, R =Round Root
9
The American National (Unified) Thread standards
10
The American National (Unified) Thread standards
3.
•
•
Square (a) and The ACME Threads (b)
Used mainly in power screws.
Table 8.3 gives preferred pitches for ACME threads
11
Example-1
A power screw is 23 mm in diameter and has a thread pitch of 7 mm.
(a) Find the thread depth, the thread width, the mean and root diameters, and the lead,
provided square threads are used.
(b) Repeat part (a) for Acme threads.
Given:
Diameter of the power screw, d = 23 mm
Thread pitch, p = 7 mm
(a) Square Threads
12
Example-1
A power screw is 23 mm in diameter and has a thread pitch of 7 mm.
(a) Find the thread depth, the thread width, the mean and root diameters, and the lead,
provided square threads are used.
(b) Repeat part (a) for Acme threads.
Given:
Diameter of the power screw, d = 23 mm
Thread pitch, p = 7 mm
(b) ACME Threads
13
8-2: The Mechanics of Power Screws
A power screw is a device used in machinery to change the angular motion into linear
motion, and usually, to transmit power.
Applications:
❑ Lead screws of lathes
❑ Screws for vises, presses and jacks
Weight supported by three screw jacks. In each screw jack, only the
shaded member rotates.
14
8-2: The Mechanics of Power Screws
In Figure a square threaded power screw with
single thread having a mean diameter dm, a pitch p,
a lead angle λ, and a helix angle ψ is loaded by the
axial compressive force F.
We wish to find an expression for the torque
required to raise this load, and another expression
for the torque required to lower the load.
Portion of a power screw
(Square)
15
8-2: The Mechanics of Power Screws
(b) Inclined Plane Wrapped Round
a Base Cylinder to Form a Thread
16
8-2: The Mechanics of Power Screws
PR
PL
Force Diagrams (a) Lifting the load; (b) Lowering the load
Imagine that a single thread of the screw is unrolled or developed for exactly a single
turn. Then on edge of the thread will form the hypotenuse of a right triangle whose
base is the circumference of the mean-thread- circle and whose height is the lead. The
angle λ is the lead angle of the thread . For raising the load a force PR acts to the right
and to lower the load, PL acts to the left.
17
8-2: The Mechanics of Power Screws
PL
PR
Force Diagrams (a) Lifting the load; (b) Lowering the load
For raising the load
For lowering the load
F
F
H
= PR − N sin  − f N cos  = 0
V
= F − N cos  + f N sin  = 0
F
F
H
= − PL − N sin  + f N cos  = 0
V
= F − N cos  − f N sin  = 0
(a)
(b)
18
8-2: The Mechanics of Power Screws
PL
PR
Force Diagrams (a) Lifting the load; (b) Lowering the load
Eliminating N from the previous equations and solving for P gives
For raising the load
For lowering the load
F ( sin  + f cos  )
PR =
cos  − f sin 
PL =
F ( f cos  − sin  )
cos  + f sin 
(c)
(d)
19
8-2: The Mechanics of Power Screws
PL
PR
Force Diagrams (a) Lifting the load; (b) Lowering the load
Next, divide the numerator and the denominator of these equations by cos λ and use
the relation tan  = l  d m
For raising the load
For lowering the load
F ( l  d m ) + f 
PR =
1 − ( f l  dm )
LPR =
F  f − ( l  d m ) 
1 + ( f l  dm )
(e)
(f)
20
8-2: The Mechanics of Power Screws
PL
PR
Force Diagrams (a) Lifting the load; (b) Lowering the load
The torque is the product of the force P and the mean radius
Torque required for raising the load to
overcome thread friction and to raise the
load
TR =
Torque required for lowering the load to
overcome part of the thread friction in
lowering the load
Fd m  f  d m − l 
TL =

 (8-2)
2   dm + f l 
Fd m  l + f  d m 

 (8-1)
2   dm − f l 
21
Self Locking Condition
❑ If the lead is large or the friction is low, the load will lower itself by causing the
screw to spin without any external effort. In such cases the torque
from Eq. (8-
2) will be negative or zero.
❑ When a positive torque is obtained from this equation, the screw is said to be self
locking
Condition for Self Locking:
 fd m  l
Dividing both sides of the above inequality by  d mand recognizing that, we get
l  d m = tan 
f  tan 
(8-3)
22
Efficiency
❑ If we let f = 0 in Eq. (8-1), we obtain
T0 =
Fl
2
(g)
which, is the torque required to raise the
load.
❑ The efficiency is therefore
T0
Fl
efficiency = e =
=
(8-4)
TR 2TR
23
Power Screw- ACME Thread
F is parallel to screw axis i.e. makes angle α= 14.5° with
thread surface ignoring the small effect of l, the resultant
normal force N is F/cos α . The frictional force = f N is
increased and thus friction terms in Eq. (8.1) are modified
accordingly:
Torque required to raise load F
d m  l + πfd m sec  
TR = F


2  πd m − fl sec  
(8-5)
ACME thread is not as efficient as square thread because of additional friction due to
wedging action but it is often preferred because it is easier to machine.
24
Power Screw with Collar
When the screw is loaded axially, a thrust or
collar bearing must be employed between the
rotating and stationary members in order to carry
the axial component
If f c is the coefficient of collar friction,
the torque required is
d m  l + πfd m sec  
TR = F

 + Tc
2  πd m − fl sec  
Ff c d c
(8-6)
Tc =
2
fc= collar friction coefficient
dc = collar mean diameter
25
Power Screws-friction coefficients
Table 8-5 Coefficients of friction f for Threaded Pairs
Table 8-6 Thrust Collar friction coefficient, fc
26
Recap (Power Screw)
Torque for raising the load
Torque for lowering the load
Square Threads
Square Threads
Fd m  l + f  d m 
TR =


2   dm − f l 
ACME Threads
d m  l + πfd m sec  
TR = F


2  πd m − fl sec  
ACME Threads with Collar
TR = F
Tc =
dm
2
Ff c d c
2
 l + πfd m sec  

 + Tc
 πd m − fl sec  
TL =
Fd m  f  d m − l 


2   dm + f l 
Condition for Self
Locking:
 fd m  l
or,
f  tan 
T0
Fl
efficiency = e =
=
TR 2TR
27
Problem # 8.8 (modified)
28
Problem # 8.8 (modified)
Problem # 8.8 (modified)
Given:
• 5/8”-6 Acme thread
i.e. d=5/8” and N=6
• f=fc= 0.15
• dc=7/16 in
• P = 6 lb
• Larm=2-3/4 in
Required:
F, Efficiency, Self-Lock?
Larm
P
F
29
Problem # 8.8 (modified)
Lever torque for power screw with collar
TR −total = F
Tc =
dm
2
 l + πfdm sec

 πdm − fl sec

 + Tc

Ff c d c
2
30
Problem # 8.8 (modified)
Efficiency
Efficiency = e =
Self-lock
Fl
161 0.1667
=
= 0.26
2 TRR
2    16.5




fd mm 
 ll
fd
m
fd mm =
=


0.15

0.5417
=
0.255
 0.15
0 .1 5 
 0.5417
0 .5 4 1 7 =
= 0.255
0 .2 5 5
fd
m
= 0.1667
0.1667
0 .1 6 6 7
ll =
which is clear that it is self lock
31
Example
A single-threaded 20 mm power screw is 20 mm in diameter with a pitch of 5 mm. A
vertical load on the screw reaches a maximum of 3 kN. The coefficients of friction are
0.06 for the collar and 0.09 for the threads. The frictional diameter of the collar is 45
mm. Find the overall efficiency and the torque to "raise" and "lower" the load.
Given
32
Self Locking Condition
Solution
33
Power Screws Stress Analysis
The stresses in a power screw may be divided into:
A-Stresses in the Screw Body
B-Stresses in the Thread
A-Stresses in the Screw Body
1-Shearing stress in screw body.
F
(8-7)
T
2-Axial stress in screw body
(8-8)
F
35
B-Stresses in the Thread
3- Thread Bearing Stress
Compressive or crushing stresses in
threads
F
(8-10)
where nt is the number of engaged
threads.
F
36
B-Stresses in the Thread
4-Thread bending stress
The bending stress at the root of the thread is
F
(8-11)
πdr
p/2
37
37
B-Stresses in the Thread
5- Transverse shear stress at the center of the thread root
F
(8-12)
Notice that the transverse shear stress at
the top of the root is zero
πdr
p/2
38
Power Screws Stress Analysis
The state of stress at top of root “plane” is:
The tri-axial stress state may be represented
by Von-Mises equivalent Stress.
39
Power Screws Stress Analysis
❑ The engaged threads cannot share the load equally. Some experiments show that
➢ The first engaged thread carries 0.38 of the load
➢ The second engaged thread carries 0.25 of the load
➢ The third engaged thread carries 0.18 of the load
➢ The seventh engaged thread is free of load
❑ In estimating thread stresses by the equations above, substituting nt to 1 will give the
largest level of stresses in the thread-nut combination
40
Example 8-1
41
Example 8-1
42
Example 8-1
43
Example 8-1
44
Example 8-1
45
Threaded Fasteners
BOLTS:
To clamp two or more members
together.
Parts:
(1) Head (Square or Hexagonal)
(2) Washer (dw=1.5d)
(3) Threaded part
(4) Unthreaded part
Dimensions of square and hexagonal bolts are given in TABLE A-27
Threaded Length:
English Threads
Metric Threads (Dimensions are in mm)
d
46
Nuts
(a) End view, general; (b) Washer-faced regular nut; (c) Regular nut
chamfered on both sides; (d) Jam nut with washer face; (e) Jam nut
chamfered on both sides.
Table A-31 gives dimensions of Hexagonal nuts
Good Practice:
1.Never re-use nuts;
2.Tightening should be done such that 1 or 2 threads come out of the nut;
3.Washers should always be used under bolt head to prevent burr stress
concentration.
47
Screws
Used for clamping members same as bolt except that 1 member should be threaded.
(a) fillister head; (b) flat head; (c) hexagonal socket head
48
Grip
The grip l of a connection is the total thickness of the clamped material.
Effective grip l′ for a screw is
is given in Table 8–7.
49
Table 8–7: Suggested Procedure for Finding Fastener Stiffness
Grip
50
Table 8–7: Suggested Procedure for Finding Fastener Stiffness
Fastener Length
51
Table 8–7: Suggested Procedure for Finding Fastener Stiffness
Washer thickness from Table
A-30 or A-31
52
Fastened Joint Preload
Preload or Clamping Force or Pretension
Clamping load stretches or elongates the bolt; The load is obtained by twisting the nut
until the bolt has elongated almost to the elastic limit ➔ Preload or Clamping force
or Pretension ➔ present even with no external load.
In bolted joint:
➢The members are under compression and
➢The bolt under tension
kb
km
53
Fastener Stiffness
kb = Equivalent spring constant of bolt composed of:
1
1
1
kt = 1
Threaded
and
k
=
Unthreaded
parts
=
+
Acting as springs in series
1 1 1 1 1d
= 1= + 1+ k1b k d kt
k
k
k
1 b 1kbk d 1=k dk t +ktk
=
+kb k k dk kbt = k d kt
kb kb k=kd =dkt t d t
k d + kt
b
k
k
d
t
k +k
k dkkbdt =k dkt++ktk
Ad E
At E
kb =
d
t
AdkEAd E k dAt=EAt E kt =
k
+
k d =dk d = t AktE=kt = A
ld E
lt
d
t
l
l
Ad kEd d= ldl At Ekt t = lt A
l
d At E
kd =
k
=
d
t
=
t AEA k
Ad A
t d tlEb
l
kb =kdb =
Ad lt + At ld
t
A
A
E
d
t
A l +d lA
l
AkdbdA=ttA
EA t lt+d+AtAldl
kb =
d t
t d
Ad lt + At ld
kd
kb
kt
54
Member Stiffness
There may be more than two members included in
the grip of the fastener. All together act as springs
under compression in series: (Eq. 8-18)
1
1
1
1
1
= +
+
+ ....
k m k1 k 2 k3
ki
If one of the members is a soft gasket, its stiffness
relative to the other members is usually so small
that for all practical purposes the others can be
neglected and only the gasket stiffness is used.
If there is no gasket, the stiffness of the members
is rather difficult to obtain, except
experimentation., because compression spreads
out and hence the area is not uniform
55
Member Stiffness
Rutscher’s pressure cone method with fixed cone angle
Compression stress distribution from experimental data
Pdx
d =
EA
Integrating from 0 to l
P
(2t tan  + D − d )( D + d )
 =
ln
Ed tan  (2t tan  + D + d )( D − d )
P
Ed tan 
k = =
 ln (2t tan  + D − d )( D + d )
( 2t tan  + D + d )( D − d )
56
Member Stiffness
For Members made of Aluminum, hardened steel and cast iron 25 <α< 33°
For α = 30°
If Members have same E with symmetrical frusta (l= 2t) they act as 2 identical springs
km = k/2. For α = 30° and D = dw = 1.5d
Alternatively
57
Member Stiffness
Finite element
analysis results:
km
= A exp( Bd / l )
Ed
➢For standard washer
faces and members of
same material
➢A and B from Table 8.8
58
EXAMPLE 8–2
59
EXAMPLE 8–2
60
8-11 Fatigue Loading of Tension Joints
Fmax = Fb
Fmin = Fi
Fmax = Fb
Force
Generally, bolted joints are subject to 0-Fmax,
(e.g pressure vessels, flanges, pipes, …)
Fmin = Fi
Fb − Fi
Fa = Fb − Fi
Fa = 2
2
Fb + Fi
Fm = Fb + Fi
Fm = 2
2
Fb − Fi (CP + Fi ) − Fi CP
 a = F − F = (CP + F ) − F = CP
i
2 Ait
 a = 2b At i =
= 2 At
At F (CP +2 A
2 At F
t )+ F
F2b +
F
CP
i
i
i
 m = F + F = (CP + F ) + F = CP + Fi =  a +  i
i
2 Ait
 m = b2 At i =
= 2 At + Ati =  a +  i
2 At
2 At
2 At At
61
8-11 Fatigue Loading of Tension Joints
Strength components Sa and Sm depend on the
Failure Criteria.
62
8-11 Fatigue Loading of Tension Joints
Strength Components (Sa & Sm)
Goodman:
Sa =
Se ( Sut −  i )
Sut + Se
Sm = Sa +  i
ASME Elliptic:
(
Se
2
2
2
Sa = 2
S
S
+
S
−

−  i Se
p
p
e
i
2
S p + Se
)
S m = Sa +  i
Gerber:
Sa =
1 
Sut Sut2 + 2 Se ( Se +  i ) − Sut2 − 2 i S 

2 Se 
S m = Sa +  i
63
8-11 Fatigue Loading of Tension Joints
Stress Concentration and Endurance Strength
Use Kf for both Sa and Sm
Including Kf
64
8-11 Fatigue Loading of Tension Joints
Fatigue Factor of Safety (nf )
Goodman
If no preload, then C = 1, and Fi = 0,
Preload is beneficial for resisting fatigue
Given by:
Factor of Safety against yielding (ny)
65
EXAMPLE 8–5
66
EXAMPLE 8–5
The joint is composed of three frusta; the upper two frusta
are steel and the lower one is cast iron.
67
EXAMPLE 8–5
68
EXAMPLE 8–5
Fatigue Factor of Safety (nf )
Factor of Safety against first cycle yielding (ny)
69
EXAMPLE 8–5
Gerber
Proof Strength
Line
Load Line
Modified
Goodman
70
EXAMPLE 8–5
Point A
Point B
Point C
71
EXAMPLE 8–5
Point D
(1)
Also
(2)
72
EXAMPLE 8–5
Point E