Various Fasteners

Various Fasteners
Non-permanent fasteners (threaded)
Machine screws (cap screws)
Bolt and Nut
Stud and Nut
Permanent fasteners
•
Welding
•
Bonding (adhesive, brazing, and soldering)
•
Rivets
Which method to select depends on the type of joint, the force to be
transmitted, whether detachable fastener is desired, fastener cost, cost
of assembly, and weight.
Ken Youssefi
SJSU
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Threaded Fasteners
Some common screw and bolt head type
Tamper resistant screw heads
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Threaded Fasteners – Nut and Washer
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Thread Standards and Terminology
The American National (Unified, UN) standard thread
UNC (Coarse thread) – has fewest threads per inch than other series, good for
frequent assembly and disassembly, use where vibration is not a problem,
reduces the likelihood of cross-threading.
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Thread Standards and Terminology
UNF (Fine thread) – has more threads per inch than UNC, use where higher
bolt strength is needed, has less tendency to loosen under vibration (smaller
lead angle).
UNEF (Extra Fine thread) – has more threads than other series, use for
precision applications or for thin-wall applications.
UNRF (Fine thread) – has rounded root contour to reduce stress concentration
and enhance resistance to fatigue failure.
Thread Classes – specifies ranges of dimensional tolerance and allowance.
Class 1A, 2A, and 3A apply to external threads and 1B, 2B, and 3B
apply to internal threads. The higher the class the smaller the tolerance.
Thread series
Fit class
Metric standards
1/2 - 20UNF – 2A or (1/2 – UNF)
Major diameter
M12 x 1.75
Threads per inch
Pitch, p in mm
Major diameter, mm
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Basic dimensions
of Unified threads
Use tensile stress
area, At , for all
stress calculations
The mean of the pitch
diameter and the
minor diameter is
used to calculate the
tensile stress area.
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Power Screw Thread Standards
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Threaded Fastener Materials
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Threaded
Fastener
Materials
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Threaded Fastener Materials - Metric
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Stresses in Threads
Axial load
σt = P / At
Torsion
Tensile stress area
Root diameter
τ = 16 T / πdr3
h
Bearing stress
σt = 4P / [π (d2 – dr2)] (h/p)
Stripping stress
τ (screw) = P / πdr (.8h) ,
Number of threads
in contact
τ (nut) = P / πd (.88h)
Use a safety factor of 2 with these equations.
Ken Youssefi
SJSU
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Minimum Nut Height
If the nut is long enough, the load require to strip the threads will be larger
than the load needed to fail the screw in tension.
For unified and metric threads, a nut height of at least 0.5d will
have a strip strength higher than the screw’s tensile strength.
Minimum tapped-hole engagement
A longer thread engagement is needed if a screw is threaded into a
tapped (blind) hole.
Same material (screw and member)
L(tapped hole length) ≥ d
Steel screw in cast iron, bronze, or brass
L(tapped hole length) ≥ 1.5d
Steel screw in aluminum
L(tapped hole length) ≥ 2d
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SJSU
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Bolted Joints in Tension – Effect of Stiffness
Bolt carries all of the load
member carries all of the load
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Bolt Force
If there is no separation then, the deflection of the bolt and member has to be the same
Bolt force
No separation
Stiffness ratio
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Bolt Stiffness
The portion of the bolt in the clamping zone (grip)
consists of unthreaded and threaded sections.
Springs in series
Bolt stiffness, or use
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Member Stiffness – Shigley method
Stiffness of a joint made of different members
Distribution of the pressure through
the member resembles a cone.
If the members in the joint are made of the same
material and have the same thickness then,
and
using
Member stiffness
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SJSU
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Member Stiffness – Juvinal method
The stiffness of clamped members
km = Ac E / grip length
where Ac is the effective area of clamped
members
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Member Stiffness
Gasketed joints
Unconfined gasket
Confined gasket
Confined O-ring
Confined seals allow the hard faces of the members to contact, joint
behaves as unsealed one, same member stiffness km as before.
Unconfined gasket should be considered as a member. If the gasket is made
of a soft material (low E), the gasket stiffness will dominate the total
member stiffness,
km = kg = AgEg / tg
Gasket material
Copper E = 17.5x106 psi
Plain rubber E = 1000 psi
Teflon E = 35x103 psi
Cork E = 12.5x103 psi
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SJSU
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Initial Tension (preload)
Recommendation for both static and fatigue loading
Tensile stress area
Fi = Ki At Sp
Proof strength
Constant, 0.75 to 0.9
Bolt should not be reused If tightened to 90% of the proof load (Fp = At
Sp), yielding may have occurred.
For static loads and permanent connection, tighten to 90% of the proof load.
For fatigue loading and non-permanent connections (reused fastener)
tighten to 75% of the poof load.
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SJSU
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Why High Preload?
•
External loads (tensile) tend to separate members,
bolt force cannot increase much unless the
members separate, the higher the preload the less
likely the members are to separate.
•
For external loads tending to shear the bolt, the
higher the preload the greater the friction force
resisting the relative motion in shear.
•
Higher preload reduces the dynamic load on the bolt because the
effective area of the clamped members is larger.
•
Higher preload results in maximum protection against overloads,
which can cause joint separation, and provides protection against
thread loosening.
Ken Youssefi
SJSU
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Bolt Tightening - Torque
Tightening torque related to preload and bolt diameter.
The constant value, .2, remains approximately the same
regardless of the bolt size.
T = 0.2 Fi d
For critical applications a torque wrench should be used to apply the
proper preload.
Ken Youssefi
SJSU
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Design of Bolted Joints in Tension under
Static loads
Bolt force
Bolt stress
Bolt stress has to be less than the
proof strength
Safety factor for
static load
Joint separation
,
Safety factor against
separation
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SJSU
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Design of Bolted Joints in Tension under Static loads
Design steps
Static load with no seal and considering separation as the worst case,
bolt carries all of the external load, stiffness not considered.
Fm = 0, Fb = P
1.
Select bolt grade ( grade 5 and 8 are common)
2.
Determine the maximum load per bolt, select a safety factor (n = 2)
and calculate the design load. P (design) = n P
3.
Calculate the required tensile stress area. Sp = P (design) / At
4.
Select bolt size, d, from the table
5.
Use preload, Fi = 0.9At Sp
6.
Calculate torque, T = 0.2Fi d
Ken Youssefi
SJSU
Sp , Sy , Su , Se
23
Design of Bolted Joints in Tension under
Static loads
Design steps
Considering stiffness (joint not separating)
1.
Select bolt grade ( grade 5 and 8 are common)
2.
Determine the maximum load per bolt, P.
3.
Select a safety factor, n = 2.
4.
Assume bolt diameter, d.
5.
Look up At , and calculate stiffness ratio C .
6.
Determine the preload, Fi = 0.9At Sp
7.
Solve for safety factor n, check against the value selected in step 3,
iterate if until the desired safety factor is reached.
8.
Specify torque, T = 0.2Fi d
Ken Youssefi
SJSU
Sp , Sy , Su , Se
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Design Example
A pillow block is attached by two machine screw.
You are asked to select appropriate screws and
specify the tightening torque.
1.
Select a relatively inexpensive bolt grade, 5.8
with proof strength of 380 MPa
2.
Determine the maximum load per bolt, select a safety factor (n = 2)
and calculate the design load. P (design) = 2 (9000/2) = 9000 N
3.
Calculate the required tensile stress area. At = P (design) / Sp = 23.68 mm2
4.
Select bolt size, d, from the table, d = 7 mm, (At = 28.9 mm2)
5.
Use preload, Fi = 0.9At Sp = 0.9 (28.9)(380) = 9883.8 N
6.
Calculate torque, T = 0.2Fi d = .2 (9883.8)(7) = 13.84 N-m
Ken Youssefi
SJSU
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Design of Bolted Joints in Tension under
Fatigue loading
Pmax = maximum applied load to the joint
Pmin = minimum applied load to the joint
(Fbolt)max = maximum load applied to the bolt = C Pmax + Fi
(Fbolt)min = minimum load applied to the bolt = C Pmin + Fi
Fa = alternating load = (Fmax – Fmin)/2
σa = alternating stress = Fa /At
σm = mean stress = Fa /At
Fm = mean load = (Fmax + Fmin)/2
Use Goodman line as design criteria
σa
Se
+
σm
Sut
=
Endurance limit
1
nf
Fatigue safety factor
Ultimate strength
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SJSU
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Design of Bolted Joints in Tension under
Fatigue loading
Design equation
Su(Pmax – Pmin) + Se(Pmax + Pmin) + 2(Se Fi )/C
2(Se Su )/C
=
At
nf
Common case, Pmin = 0, so (Fbolt)min = Fmin = Fi
Endurance Strength, Se
Suggested values for
endurance limit for common
bolts with rolled threads. For
cut threads use Kf = 3.8 for
grade 4 and higher. multiply
the alternating component of
stress by Kf.
Ken Youssefi
SJSU
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Design of Bolted Joints in Tension under
Fatigue loading
Fatigue failure criteria
Sut = ultimate strength in tension , Sp = proof strength
Se = endurance limit
Goodman
σa
Se
σm
+
Sut
1
= n
f
nf = safety factor guarding against
fatigue failure.
ASME-elliptic
n f σa
Se
2
+
Yielding
ny =
Ken Youssefi
n f σm
Sp
2
Sp
σa + σm
= 1
ny = safety factor guarding against
yielding.
SJSU
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Design of Bolted Joints in Tension under
Fatigue loading
Design steps
1.
Select bolt grade ( grade 5 and 8 are common)
2.
Select number of bolts. If circular pattern, use the restriction
3 ≤ (π Db)/N d ≤ 6 to allow access to bolt head for tightening. Db is
bolt pattern diameter, d is bolt diameter and N is the number of bolts.
3.
Determine the maximum and minimum applied load per bolt, Pmax and Pmin.
4.
Choose a safety factor, nf = 1.5 to 2.5
5.
Choose a bolt diameter, d, look the tensile stress area, At and calculate
the stiffness ratio C.
6.
Decide on preload Fi. Use .6SpAt ≤ Fi ≤ .9SpAt as guideline,
(Fi = .75 SpAt is common). Unless specified otherwise by seal
manufacturer.
7.
Use the design equation and calculate the safety factor, nf , iterate
until the calculated safety factor matches the chosen one in step 4.
Ken Youssefi
SJSU
Sp, Sy, Su, Se
29
Design Example – Fatigue Loading
Consider a cast iron cylinder with aluminum
cover plate with internal gage pressure that
fluctuates between 0 and 2.0 MPa. Both
members 10 mm thick. Design the bolted joint.
Specify the bolt grade, number of bolts and
bolt diameter for infinite life.
1.
Select 14 grade 9.8 bolt.
Sp = 650, Sy = 720, Su = 900, and Se = 140 MPa
2. Choose a safety factor, nf = 1.5
3.
Determine the maximum and minimum applied load per bolt, Pmax and Pmin.
Pmax = (pressure)(Area) / N= (2.0)(π Di2/4) = [(2)(π)(250)2/4] / 14 = 7013 N
Pmin = 0
4.
Choose a bolt diameter, d = 12 mm, look up At = 84.3 mm2
5.
Check bolt spacing, 3 ≤ (π Db)/N d ≤ 6 , 3 ≤ (π 350)/12x14 = 6.5 ≤ 6 (okay)
Ken Youssefi
SJSU
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Design Example – Fatigue Loading
6.
Calculate stiffness ratio, C.
Bolt stiffness,
kb = Abolt E / grip length = πd2E / 4g = π122x207x103 / 4x20
kb = 1.17x106 N / mm
Member stiffness,
1/km = 1/kAl + 1/kcast
k = A E / grip length
Ac = d2 + .68dg + .065g2 = 122 + .68x12x20 +.065x202 = 333.2
kAl = Ac EAl / g = 333.2x70,000/10 = 2.332x106
kcast = Ac Ecast / g = 333.2x100,000/10 = 3.332x106
km = 1.37x106 N / mm
Stiffness ratio,
C = kb / (kb + km) = .46
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SJSU
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Design Example – Fatigue Loading
7.
Select preload
Fi = .75 SpAt = .75 x 650 x 84.3 = 41,100 N
8.
Calculate safety factor
Su(Pmax – Pmin) + Se(Pmax + Pmin) + 2(Se Fi )/C
=
2(Se Su )/C
At
nf
nf = 1.1 < 1.5, select larger diameter or higher strength bolt
9.
Select grade 10.9 bolt,
Sp = 830, Sy = 940, Su = 1040, and Se = 162
MPa
Fi = .75 SpAt = .75 x 830 x 84.3 = 52,477 N
nf = 1.36 < 1.5, use more bolts, select 24 bolts
Check bolt spacing, 3 ≤ (π 350)/12x24 = 3.8 ≤ 6 (okay)
Ken Youssefi
SJSU
32
Design Example – Fatigue Loading
Determine the maximum applied load per bolt, Pmax.
Pmax = (pressure)(Area) / N= (2.0)(π Di2/4) = [(2)(π)(250)2/4] / 24 = 4090 N
nf = 1.47 ≈ 1.5
Specification
Bolt diameter
# of bolts
Bolt grade
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12 mm
24
10.9 metric
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Bolted Joints in Shear
Primary shear – same for all bolts
F‘ = V / # of bolts
Secondary shear – for the nth bolt
Fn‫= ״‬
Ken Youssefi
Mrn
rA2 + rB2 + rC2 + …..
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Bolted Joints in Shear
What grade of bolt
should be used?
V = 16 kN , M = 16(425) = 6800 N-m
FC = FD = 14.8 kN
FA = FB = 21.0 kN
τ = F / As = 21/144 = 146 MPa
Use grade 4.8, Ssy = 155 MPa
Ken Youssefi
SJSU
35
Bolted Joints in Shear
Consider a bracket attached to the wall
by two bolts as shown.
 Assume shear is carried by
friction
 Neglect friction and assume
shear is carried by the bolts
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SJSU
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