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3
Threaded Fasteners - Introduction
Budynas, p. 396
“… the curiosity of any person
interested in mechanical
engineering naturally results in the
acquisition of a good background
knowledge of fastening methods.”
“Contrary to first impressions, the
subject is one of the most
interesting in the entire field of
mechanical design.”
5
Metric ‘M’ and ‘MJ’ Thread Profile
d – major diameter
dr – minor (or root) diameter
dp – pitch diameter
p – pitch
H = p2
3
10
Mechanics of Power Screws
Objectives:
– Determine relation between applied torque
and axial load
– Consider both raising and lowering the load
– Compute stresses at the thread root
12
Free-Body Diagrams
• Clear diagram of each body being analyzed
• Label coordinate system
• Isolate the body from it’s supports
• Replace supports with reaction forces
• Label all applied forces and moments
• Apply equilibrium equations
• Solve for unknown forces and moments
14
Analysis of Thread Forces
pnl
l
where
n = 1 single thread
n = 2 double thread
n = 3 triple thread
15
Free Body Diagrams
Raising the Load Lowering the Load
PR, PL – net force due to applied torque
f - friction coefficient
N - normal force
18
Torque Requirements
sincos
sincos
sincos
cossin
f
fFP
f
fFP
L
R
Eliminating N
and substituting
2
2
tan
mLL
mRR
m
dPT
dPT
d
l
20
Effect of Friction – Raising the Load
fld
dflFdT
m
mmR
2
20
FlT
RT
Te 0
if f = 0
and the efficiency, e, is given by
21
Effect of Friction (cont.)
Lowering the load:
Note, if
f dm > l
then
TL > 0
and an applied torque is needed to lower the load. For TL 0, the screw is “self-
locking.” Since l / dm = tan , the screw is self-locking if
f > tan
fld
ldfFdT
m
mmL
2
27
Stresses in Threads
Bearing Stress
Bending Stress
Shear Stress
pnd
F
trb
6
pnd
F
tr
3
pnd
F
tmB
2
27
28
Distribution of Load in Engaged Threads
Experimental result:
First thread carries 38% of the load
Second thread carries 25% of the load
Third thread carries 18% of the load...
Seventh thread is stress free
30
Mechanics of Bolted Joints(Sections 8-4 – 8-10)
Tensile load applied to joint
Bolt tension = preload due to
tightening of nut + stress due
to applied load
Member stress = compressive
stress due to tightening of nut
+ stress due to applied load
First, need to standard bolt dimensions => Section 8-3
32
Hexagon-head Thread Length
inLind
inLindLT
6,2
12
6,4
12
200,252
200125,122
48,125,62
Lmmd
mmLmmd
mmdmmLmmd
LTLT
Note: Ideal bolt length – one or two threads project from the nut
38
Hexagonal Nuts
a) end view
b) washer faced regular nut
c) regular nut chamfered on both faces
d) jam nut with washer face
e) jam nut chamfered on both faces
41
Mechanics of Bolted Joints(Sections 8-4 – 8-10)
Tensile load applied to joint
Bolt tension = preload due to
tightening of nut + stress due
to applied load
Member stress = compressive
stress due to tightening of nut
+ stress due to applied load
Grip length, l
44
Bolt Stiffness
d
dd
l
EAk
t
tt
l
EAk
dttd
td
td
tdb
lAlA
EAA
kk
kkk
Unthreaded portion
Threaded portion
Effective bolt stiffness (springs in series)
48
Member Stiffness(Section 8-5)
For multiple members, treat as springs in series
If one of the members is a gasket with low stiffness
im kkkkk
1...
1111
321
gasketm kk
49
Member Stiffness (no gasket)
Pressure cone method
Frustum - portion of a cone which lies
between two parallel planes
- half-apex angle
50
Member Stiffness (cont.)
Frustum stiffness – two members with same material and thickness and taking = 30º, l = 2t (see text for derivation)
dl
dl
dEkm
5.25774.0
5.05774.05ln2
5774.0
55
b) Both members are steel
c) Repeat using Eq. 8-23
Example 8-2 parts (b & c)
dl
dl
dEkm
5.25774.0
5.05774.05ln2
5774.0
l
BdA
dE
km exp
57
Bolt Strength(Section 8-6 )
Sy – yield strength
Sut – ultimate tensile strength
Sp - The proof strength is the stress
above which permanent set is
induced
Permanent set – first measurable
deviation from linear elastic behavior
58
Variability of Strength
Statistical measures:
Mean strength = 145.1 ksi
Standard deviation = 10.3 ksi
Minimum strength = 120 ksi
59
SAE Specifications for Steel Bolts
(Table 8-9)
Note: Head Markings
Also note: grade of
nut must match the
bolt grade
62
Tension Joints – The External Load
Preload: Fi
Load applied to joint: P
Portion of P taken by bolt: Pb
Portion of P taken by members: Pm
P = Pb + Pm
Resultant bolt load: Fb = Pb + Fi
Resultant member load: Fm = Pm - Fi
Fraction of external load P carried by bolt: C
Fraction of external load P carried by members: 1 - C
65
Example Bolt and Member Stiffnesses(1/2 in – 13 NC steel bolts)
C – fraction of load carried by bolt (< 20%)
(1-C) - fraction of load carried by members
66
Relation between Bolt Torque and Bolt Tension
The torque required to raise a fastener and for overcoming collar
friction were derived in section 8-2.
sec
sec
2 lfd
dflFdT
m
mmR
2
CCC
dfFT
67
Relation between Bolt Torque and Bolt Tension
(cont.)
Setting the mean collar diameter to:
leads to
where
and
dc d 1.5d 2
T KFid
K dm
2d
tan f sec
1 f tan sec
0.625 fc
tan l
dm,
73
Tension Joint with Preload
b CP
AtFi
At
n SpAt Fi
CP
PnP
The total tensile stress in bolt
Introducing a load factor, n, where
and solving for n gives
where n is the load factor associated with bolt strength
74
Tension Joint with Preload (cont.)
We also need to check for joint separation where Fm = 0
Introducing a load factor, n0, where
gives
where n0 is the load factor associated with joint separation
PnP 0
n0 Fi
P 1C
75
Recommended Preload
sconnectionpermanentforF
fastenersreusedsconnectionntnonpermaneforFF
p
p
i ,90.0
,,75.0
ptp SAF
Where the proof load is given by
80
Example 8-4 Summary
0.0
10.0
20.0
30.0
40.0
50.0
0.0 10.0 20.0 30.0 40.0 50.0
Bo
lt L
oa
d (
ksi)
External Load (kip)
For P < 22.8 kip
Bolt load = 14.4 + 0.368 * P
Plates separate when
P=3.80 (6 kip) = 22.8 kip
For P > 22.8 kip
Bolt load = P
81
Finite Element Validation of Example 8-4
Bolt
Bottom Plate
Top plate
Assembly
Contact
Surfaces
Element type - Axisymmetric
Three parts: bolt, top plate and bottom plate
(all axisymmetric)
82
Finite Element Validation of Example 8-4 (cont.)
Three Steps
Step 1 - Apply tensile stress to bolt to stretch bolt beyond plate thickness
(undeformed bolt length smaller than plates)
Step 2 - Define and activate contact surfaces
Release tensile stress to squeeze members
(simulates tightening of bolt)
Step 3 - Apply external tensile load on joint
Measure bolt load as function of external load
83
Finite Element Results - Animation
http://personal.egr.uri.edu/taggart/courses/mce301/ex84/ex84.html
84
Finite Element Results
0.0
10.0
20.0
30.0
40.0
50.0
0.0 10.0 20.0 30.0 40.0 50.0
Bo
lt L
oa
d (
ksi)
External Load (kip)
FEA
Theory
86
Gasketed Joints
The bolt tension determines the pressure acting on the gasket. For N bolts, the
pressure is
Substituting
Gives
For uniform pressure on the gaskets, the recommended distance between the
bolts should be between 3 and 6 times the bolt diameter
where Db is the bolt circle diameter
g
iA
NCnPFp 1
NA
Fp
g
m
im FnPCF 1
63 dN
Db
90
Tensile Failure of Members
A
F
where A is the net section area.
May need to include stress
concentration effects – to be
discussed later in course
92
Tear-out Failures
Shear Tear-out Failure Tensile Tear-out Failure
Tear-out failure can usually be avoided by placing the bolt or
rivet far enough from the edge
at
F
2
95
Joint Loads
Direct load (primary shear)
Moment load (secondary shear)
n
VF 1'
DDCCBBAA rFrFrFrFM """"1