ME311 Machine Design
Lecture 9: Screws
(Chapter 16)
W Dornfeld
05Nov2009
Fairfield University
School of Engineering
Thread Geometry
Thread Pitch
Crest
Major
Pitch
n (Threads / inch ) 
Thread
Angle
1
p
Thread Height
Root
Minor
Diameters
The Pitch Diameter
is midway between
the Major and Minor
Diameters.
Hamrock
Page 707
Thread Types
Lead =
1 x Pitch
Lead =
3 x Pitch
Single-, double-, and triple-threaded screws.
Also called single-, double-, and triple-start.
Acme
Thread
Acme threads are used in C-Clamps, vices, and cartoons.
Square
Thread
Hamrock
Page 708
Details of Thread Profiles
Thread Height
ht 
0.5 p
 0.8660 p
tan( 30)
Relationships for M (metric) and UN (unified = US) screw threads.
Example:
UN: ¼-20, means 0.25in. Major diameter & 20 threads/inch.
M: M8x1.25, means 8mm Major diameter & pitch of 1.25mm
Hamrock
Page 708
Power Screws
W
Load
on nut
m
Thread
friction
Lead
rm
Looking at a square
thread screw, we unwind
one turn:
a
Lead
2p rm
Mean
thread
radius
mc
r
Mean
c
collar radius
Collar
friction
The Mean Radius is midway
between the Crest and Root Radii.
This shows an inclined
ramp with angle
1 Lead
a  tan
2p rm
Square Thread Screw Torque
m
The torque required to
raise the load W is
 m  tan a

Traise  W rm
 m c rc 
 1  m tan a

and to lower the load, we
flip two signs:
Tlower
W
Lead
rm
mc
rc
 m  tan a

 W rm
 m c rc 
 1  m tan a

Hamrock
Page 715
Power Screw Thread Angle
If the thread form is not square but has an angle b,
replace the thread friction m with the effective friction
m
me 
cos( b / 2)
b
The effect:
• Square: b = 0, b/2 = 0, 1/cos(0°) = 1.0
• Acme: b = 29°, b/2 = 14.5°, 1/cos(14.5°) = 1.033
• Unified: b = 60°, b/2 = 30°, 1/cos(30°) = 1.15
The thread angle effectively increases surface friction between 3 and 15%
Note: Instead of b/2, Hamrock uses  n  tan (cos a tan b 2)
The difference is negligible.
1
Power Screws - Overhauling
If the collar friction is small (e.g., it may have a
ball thrust bearing), too small a thread friction
may let the weight screw down on its own.
Lead
This can happen when m  tan a 
2p rm
(the numerator m  tan a goes negative).
Tlower
0
 m  tan a
 W rm
 m c rc 
 1  m tan a

m
a
2p rm
This is the same case for a weight sliding down a ramp
when the incline angle a exceeds tan-1m.
Lead
Ball Screws Have Low Friction
Recirculating balls roll between ball screw
and ball nut to minimize friction.
These almost always overhaul.
Our Scissors Jack
1015 Lb
1522 Lb Tension
Handle End with ball
thrust bearing
End with nut
Scissors Jack Analysis
Thread ID = 0.398 in.
Thread OD = 0.468 in.
Estimate dp= (0.398+0.468)/2
= 0.433 in.
Handle length = 135/25.4 = 5.31 in.
Lead = 0.10 in.
Thread angle b = 29°
Guess m = 0.20
mc = 0 due to bearing
W = 1522 Lb.
What torque is required to raise the jack?
What force is required on the handle?
C-Clamp Analysis
Thread ID = 0.391 in.
Thread OD = 0.480 in.
Handle length = 3 in.
N = 8 Threads/Inch
Thread angle b = 60°
Guess m = 0.15
mc = 0 to simplify things
W = 500 Lb.
What torque is required to cause
the 500 Lb. squeeze?
Note: If Acme, could use Eqn. 16.4
d p  d c  0.5 p  0.01  0.48  (0.5)(0.125)  0.01  0.4075 in.
But with a 60° thread angle, this is NOT an Acme.
Estimate dp= (ID+OD)/2 = (0.390+0.480)/2 = 0.436 in.
Using Dornfeld Lecture Equations
dp= 0.436 in.
N = 8 Threads/Inch
Lead = 1/N = 0.125 in.
a  tan 1
Thread angle b = 60°
m = 0.15
W = 500 Lb.
Lead
0.125
 tan 1
 tan 1 (0.09126)  5.21
2p rm
2p (0.436 / 2)
Because this is not a square thread, must use effective coefficient of
friction = m/cos(b/2) = 0.15/cos(30°) = 0.15/0.866 = 0.1732
0 
 0.436 0.1732  0.09126 
 m  tan a
Traise  W rm
 mc rc   500 

1

m
tan
a
2
1

(
0
.
1732
)(
0
.
09126
)




 (500)(0.218)
0.26446
 29.29 Lb.In.
0.9842
Using Hamrock Equations
dp= 0.436 in.
Thread angle b = 60°
N = 8 Threads/Inch
m = 0.15
Lead = 1/N = 0.125 in.
W = 500 Lb.
Lead
a  tan 1
 5.21 ; tan( a )  0.09126
2p rm
b
 n  tan 1 (cos a tan )  tan 1 (cos 5.21 tan 30)  tan 1 (0.9959  0.57735)
2
1
 n  tan (0.57496)  29.897
How close is this to b/2 = 30°?
0 
 (d p / 2)(cos  n tan a  m )
 (0.436 / 2)(cos 29.9 tan 5.21  0.15) 
Traise  W 
 mc rc   500

cos


m
tan
a
cos
29
.
9


0
.
15
tan
5
.
21



n


 (500)(0.218)
(0.866)(0.09126)  0.15
0.22903
 (109)
 29.29 Lb.In.
0.866  (0.15)(0.09126)
0.85231
[Eqn. 16.10]
The equations are equivalent. Pick whichever one suits you best.
Overhauling Revisited
• Power screws can lower all by themselves if the friction
becomes less than the tangent of the lead angle, a.
• This corresponds to the numerator in the Tlower equation going
negative, with the transition being where the numerator is Zero.
• You can use either Dornfeld or Hamrock equation, but
remember that the Dornfeld equation is Effective friction, and you
must multiply by cos(b/2) to get the actual friction.
Hamrock:
Tlower
 (d p / 2)( m  cos  n tan a )

W
 mc rc 
cos  n  m tan a


Dornfeld:
Tlower
 m  tan a

 W rm
 mc rc 
 1  m tan a

Transition when:
m  cos  n tan a
me  tan a
m  me cos( b / 2)  cos( b / 2) tan a
The equations are equivalent. Pick whichever one suits you best.
Failure Modes: Tensile Overload
When the tensile stress on a bolt exceeds the
material’s Proof Strength, the bolt will
permanently stretch.

P
At
Where At is the Tensile Stress Area for
the bolt – the equivalent area of a
section cut through the bolt.
For UN threads,
0.9743 

At  (0.7854) d c 

n


2
dc = Crest Dia (in.)
n = threads/in.
For M threads,
At  (0.7854)( d c  0.9382 p) 2
dc = Crest Dia (mm)
p = pitch (mm)
Hamrock
Page 731
Failure Modes: Thread Shear
l
Shear of Nut Threads
Ashear  pd crestl
The shear strength
of the bolt and nut
material may not
be the same.
Shear of Bolt Threads
Ashear  pd rootl
Failure Modes: Shank Shear
Ashear 
pd
2
shank
4
Ashear  2 
2
pd shank
4

2
pd shank
2
Bolts are not really intended to be used this way unless they are
Shoulder Bolts:
Typically the preload from tightening the bolt clamps the joint,
and the friction between the members holds the joint.
Bolt Preload
JH Bickford explains :
'When we tighten a bolt,
( a) we apply torque to the nut,
( b) the nut turns,
( c) the bolt stretches,
( d) creating preload.'
So the bolt is really a spring that stretches
and creates preload on the joint.
We use the Power Screw equations to determine how torque results
in preload. This can be approximated simply by:
T  KDcrest P
Where T is torque, Dcrest is the bolt crest diameter, P is the preload,
and K is a dimensionless constant. K = 0.20 for clean, dry threads
and K = 0.15 for lubricated threads.
Bolt Stiffness
Lshank
A bolt looks like two
springs in series: one
rod with the Crest
diameter and one with
the Root diameter.
Lthread
Their lengths are
increased to reflect the
head and nut.
1
4  Ls  0.4d c Lt  0.4d r



2
kb pE 
dc
d r2



Hamrock
Page 725
Bolt Stiffness Exercise
Calculate the stiffness of a 3/8-16 screw that is
4 in. long and clamps 3.5” of material. Use
Eqn. 16.23 to determine shank length.
Lshank
Lbolt
Lclamp
Lt in
16.22/23
Lthread
Lthread
Lt in
16.21
Note: Hamrock uses Lt in Eqns. 16.21 and
16.22/23, BUT THEY ARE DIFFERENT THINGS!
In 16.21 it is the Clamped thread length;
in 16.22/23 it is Total thread length.
Hamrock
Page 726
Joint Stiffness
The material clamped by
the bolt also acts like a
spring – in compression.
Effectively, only the material
in the red double conical
area matters.
There are many methods to
calculate this stiffness.
Compare these calculator
stiffness results from
tribology-abc.com with
Hamrock’s Example 16.6
Hamrock
Page 727
How Bolt Preload Works
Preload isolates the bolt from most of any external loads.
The joint stiffness factor, Cj, determines what fraction of
external loads the bolt actually sees.
Cj 
kb
kb  k j
Hamrock
Eqn. 16.17
From Norton, Chap. 14
Bolt Strength
For Metric grades, the first number x 100 = Sut in MPa. The
fraction x Sut = Sy. Ex: grade 12.9 has Sut ≈1200 MPa and
Sy ≈ 0.9x1200 = 1080 MPa.
Hamrock
Page 731
Bolt Loading
Ultimate
Generally, bolts are preloaded to:
• 75% of Proof Load for reused
connections
• 90% of Proof Load for permanent
connections
0.2%Yield
Proof
where Proof Load = Proof Strength x At.
The Proof Strength is approximately at
the elastic limit for the material.
Hamrock
Page 733
Recommended Site:
BoltScience.com