ken youssefisjsu 1 various fasteners non-permanent fasteners (threaded) machine screws (cap...

Ken Youssefi SJSU 1

Various FastenersNon-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 joint material, the force to be transmitted, whether detachable fastener is desired, fastener cost, cost of assembly, and weight.

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Threaded FastenersSome 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 TerminologyUNF (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.

1/2 - 20UNF – 2A or (1/2 – UNF)

Fit classThread series

Threads per inchMajor diameter

M12 x 1.75

Metric standards

Major diameter, mmPitch, p in 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


h

Stresses in Threads

Axial load

σt = P / At

Tensile stress area

Torsion

τ = 16 T / πdr3

Root diameter

Bearing stress

σt = 4P / [π (d2 – dr2)] (h/p)

Number of threads in contact

Use a safety factor of 2 with these equations

Stripping stress

τ (screw) = P / πdr (.8h) , τ (nut) = P / πd (.88h)


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 engagementA longer thread engagement is needed if a screw is threaded into a tapped (blind) hole.

Same material (bolt and member) L(tapped hole length) ≥ d

Steel bolt in cast iron, bronze, or brass L(tapped hole length) ≥ 1.5d

Steel bolt in aluminum L(tapped hole length) ≥ 2d


Bolted Joints in Tension – Effect of StiffnessBolt carries all of the load

member carries all of the load


Bolt Force

If there is no separation then, the deflection of the bolt and member has to be the same

No separation

Bolt force

Stiffness ratio


Bolt StiffnessThe portion of the bolt in the clamping zone (grip) consists of unthreaded and threaded sections.

Springs in series

Bolt stiffness, or use


Member Stiffness – Shigley methodStiffness 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

Member stiffnessusing


Member Stiffness – Juvinal method

where Ac is the effective area of clamped

members

km = Ac E / grip length

The stiffness of clamped members


Member StiffnessGasketed joints

Unconfined gasket

Confined seals allow the hard faces of the members to contact, joint behaves as unsealed one, same member stiffness km as before.

Gasket material

Copper E = 17.5x106 psi

Cork E = 12.5x103 psi

Plain rubber E = 1000 psiTeflon E = 35x103 psi

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

Confined gasket Confined O-ring


Initial Tension (preload)

Recommendation for both static and fatigue loading

Fi = Ki At Sp

Proof strength

Tensile stress area

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.


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.


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.


Design of Bolted Joints in Tension under Static loads

Bolt force

Bolt stress

Bolt stress has to be less than the proof strength

Joint separation

,

Safety factor against separation

Safety factor for static load



Design stepsStatic 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) Sp , Sy , Su , Se

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



Design stepsConsidering stiffness (joint not separating)

1. Select bolt grade ( grade 5 and 8 are common) Sp , Sy , Su , Se

2. Determine the maximum load per bolt, P.

4. Assume bolt diameter, d.

3. Select a safety factor, n = 2.

6. Determine the preload, Fi = 0.9At Sp

8. Specify torque, T = 0.2Fi d

5. Look up At , and calculate bolt and member stiffness to find C.

7. Solve for safety factor n, check against the value selected in step 3, iterate if until the desired safety factor is reached.


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


Design of Bolted Joints in Tension under Fatigue loading

Pmax = maximum applied load to the joint

Pmin = minimum applied load to the joint

Fa = alternating load = (Fmax – Fmin)/2

Fm = mean load = (Fmax + Fmin)/2

σa = alternating stress = Fa /At

σm = mean stress = Fa /At

(Fbolt)max = maximum load applied to the bolt = C Pmax + Fi

(Fbolt)min = minimum load applied to the bolt = C Pmin + Fi

Use Goodman line as design criteria

σa σm

Se Sutnf

+ =1

Endurance limitUltimate strength

Fatigue safety factor



nf=

AtSu(Pmax – Pmin) + Se(Pmax + Pmin) + 2(Se Fi )/C

2(Se Su )/C

Design equation

Common case, Pmin = 0, so (Fbolt)min = Fmin = Fi

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.

Endurance Strength, Se



Fatigue failure criteria

Se = endurance limit

Sut = ultimate strength in tension , Sp = proof strength

ASME-elliptic

+ = 1nf

σm

Sp

2σa

Se

2nf

Goodman

σa σm

Se Sut

+ =1nf

nf = safety factor guarding against

fatigue failure.

Yielding

σa σm+=ny

Sp ny = safety factor guarding against

yielding.



Design steps

1. Select bolt grade ( grade 5 and 8 are common) Sp, Sy, Su, Se 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

6. Decide on preload Fi. Use .6SpAt ≤ Fi ≤ .9SpAt as guideline, (Fi = .75 SpAt is common). Unless specified otherwise by seal manufacturer.

5. Choose a bolt diameter, d, look the tensile stress area, At and calculate the stiffness ratio C.

7. Use the design equation and calculate the safety factor, nf , iterate

until the calculated safety factor matches the chosen one in step 4.


Design Example – Fatigue LoadingConsider 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 Di

2/4) = [(2)(π)(250)2/4] / 14 = 7013 NPmin = 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)


Design Example – Fatigue Loading6. Calculate stiffness ratio, C.

Bolt stiffness,

kb = Abolt E / grip length = πd2E / 4g = π122x207x103 / 4x20

kb = 1.17x106 N / mm

Ac = d2 + .68dg + .065g2 = 122 + .68x12x20 +.065x202 = 333.2

Stiffness ratio,

C = kb / (kb + km) = .46

Member stiffness,1/km = 1/kAl + 1/kcast k = A E / grip length

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


Design Example – Fatigue Loading

Fi = .75 SpAt = .75 x 650 x 84.3 = 41,100 N

7. Select preload

nf

=At

8. Calculate safety factor

Su(Pmax – Pmin) + Se(Pmax + Pmin) + 2(Se Fi )/C

2(Se Su )/C

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)


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 12 mm# of bolts 24

Bolt grade 10.9 metric


Bolted Joints in Shear

Primary shear – same for all bolts

F‘ = V / # of bolts

Secondary shear – for the nth bolt

FnMrn = ״

rA2 + rB

2 + rC2 + …..



τ = F / As = 21/144 = 146 MPa

Use grade 4.8, Ssy = 155 MPa

V = 16 kN , M = 16(425) = 6800 N-m

FC = FD = 14.8 kN

FA = FB = 21.0 kN

What grade of bolt should be used?



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

ken youssefisjsu 1 various fasteners non-permanent fasteners (threaded) machine screws (cap...

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