1. 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 joint material, the force to be transmitted, whether detachable fastener is desired, fastener cost, cost of assembly, and weight.
Ken Youssefi
SJSU

2. Threaded Fasteners Some common screw and bolt head type
Tamper resistant screw heads
Ken Youssefi
SJSU

3. Threaded Fasteners – Nut and Washer
Ken Youssefi
SJSU

4. 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.
Ken Youssefi
SJSU

5. 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
M12 x 1.75
Metric standards
Major diameter, mm
Pitch, p in mm
1/2 - 20UNF – 2A or (1/2 – UNF)
Major diameter
Threads per inch
Ken Youssefi
SJSU

6. 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.
Ken Youssefi
SJSU

7. Power Screw Thread Standards
Ken Youssefi
SJSU

8. Threaded Fastener Materials
Ken Youssefi
SJSU

9. Threaded Fastener Materials
Ken Youssefi
SJSU

10. Threaded Fastener Materials - Metric
Ken Youssefi
SJSU

11. Stresses in Threads σt = P / At τ = 16 T / πdr3
Axial load
Tensile stress area
σt = P / At
Torsion
τ = 16 T / πdr3
Root diameter
τ (nut) = P / πd (.88h)
Stripping stress
τ (screw) = P / πdr (.8h) , h
Bearing stress
σt = 4P / [π (d2 – dr2)] (h/p)
Number of threads in contact
Use a safety factor of 2 with these equations
Ken Youssefi
SJSU

12. 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 (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
Ken Youssefi
SJSU

13. Bolted Joints in Tension – Effect of Stiffness
Bolt carries all of the load
member carries all of the load
Ken Youssefi
SJSU

14. Bolt Force
If there is no separation then, the deflection of the bolt and member has to be the same
Stiffness ratio
Bolt force
No separation
Ken Youssefi
SJSU

15. 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
Ken Youssefi
SJSU

16. 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
Member stiffness
using
Ken Youssefi
SJSU

17. Member Stiffness – Juvinal method
The stiffness of clamped members
km = Ac E / grip length
where Ac is the effective area of clamped members
Ken Youssefi
SJSU

18. Member Stiffness Gasketed joints
Confined gasket
Confined O-ring
Unconfined gasket
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
Cork E = 12.5x103 psi
Plain rubber E = 1000 psi
Teflon E = 35x103 psi
Ken Youssefi
SJSU

19. Initial Tension (preload)
Recommendation for both static and fatigue loading
Tensile stress area
Fi = Ki At Sp
Constant, 0.75 to 0.9
Proof strength
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.
Ken Youssefi
SJSU

20. 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

21. 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

22. 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
Ken Youssefi
SJSU

23. 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
Select bolt grade ( grade 5 and 8 are common)
Sp , Sy , Su , Se
Determine the maximum load per bolt, select a safety factor (n = 2) and calculate the design load. P (design) = n P
Calculate the required tensile stress area. Sp = P (design) / At
Select bolt size, d, from the table
Use preload, Fi = 0.9At Sp
Calculate torque, T = 0.2Fi d
Ken Youssefi
SJSU

24. Design of Bolted Joints in Tension under Static loads
Design steps
Considering stiffness (joint not separating)
Select bolt grade ( grade 5 and 8 are common)
Sp , Sy , Su , Se
Determine the maximum load per bolt, P.
Select a safety factor, n = 2.
Assume bolt diameter, d.
Look up At , and calculate bolt and member stiffness to find C.
Determine the preload, Fi = 0.9At Sp
Solve for safety factor n, check against the value selected in step 3, iterate if until the desired safety factor is reached.
Specify torque, T = 0.2Fi d
Ken Youssefi
SJSU

25. Design Example
A pillow block is attached by two machine screw. You are asked to select appropriate screws and specify the tightening torque.
Select a relatively inexpensive bolt grade, 5.8 with proof strength of 380 MPa
Determine the maximum load per bolt, select a safety factor (n = 2) and calculate the design load. P (design) = 2 (9000/2) = 9000 N
Calculate the required tensile stress area. At = P (design) / Sp = mm2
Select bolt size, d, from the table, d = 7 mm, (At = 28.9 mm2)
Use preload, Fi = 0.9At Sp = 0.9 (28.9)(380) = N
Calculate torque, T = 0.2Fi d = .2 (9883.8)(7) = N-m
Ken Youssefi
SJSU

26. 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
Fm = mean load = (Fmax + Fmin)/2
σa = alternating stress = Fa /At
σm = mean stress = Fa /At
Use Goodman line as design criteria σa σm Se
Sut nf + = 1
Endurance limit
Ultimate strength
Fatigue safety factor
Ken Youssefi
SJSU

27. Design of Bolted Joints in Tension under Fatigue loading
Design equation nf = At
Su(Pmax – Pmin) + Se(Pmax + Pmin) + 2(Se Fi )/C
2(Se Su )/C
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
Ken Youssefi
SJSU

28. 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 σm Se
Sut + = 1 nf
nf = safety factor guarding against fatigue failure.
ASME-elliptic + = 1 nf σm Sp 2 σa Se
Yielding σa σm + = ny Sp
ny = safety factor guarding against yielding.
Ken Youssefi
SJSU

29. Design of Bolted Joints in Tension under Fatigue loading
Design steps
Select bolt grade ( grade 5 and 8 are common)
Sp, Sy, Su, Se
Select number of bolts. If circular pattern, use the restriction ≤ (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.
Determine the maximum and minimum applied load per bolt, Pmax and Pmin.
Choose a safety factor, nf = 1.5 to 2.5
Choose a bolt diameter, d, look the tensile stress area, At and calculate the stiffness ratio C.
Decide on preload Fi. Use .6SpAt ≤ Fi ≤ .9SpAt as guideline, (Fi = .75 SpAt is common). Unless specified otherwise by seal manufacturer.
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

30. 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.
Select 14 grade 9.8 bolt.
Sp = 650, Sy = 720, Su = 900, and Se = 140 MPa
Choose a safety factor, nf = 1.5
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
Choose a bolt diameter, d = 12 mm, look up At = 84.3 mm2
Check bolt spacing, 3 ≤ (π Db)/N d ≤ 6 , 3 ≤ (π 350)/12x14 = 6.5 ≤ 6 (okay)
Ken Youssefi
SJSU

31. Design Example – Fatigue Loading
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 = d dg g2 = x12x x202 = 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
Ken Youssefi
SJSU

32. Design Example – Fatigue Loading
Select preload
Fi = .75 SpAt = .75 x 650 x 84.3 = 41,100 N nf = At
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
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

33. 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 metric
Ken Youssefi
SJSU

34. Bolted Joints in Shear Primary shear – same for all bolts
F‘ = V / # of bolts
Secondary shear – for the nth bolt
Fn״ =
Mrn
rA2 + rB2 + rC2 + …..
Ken Youssefi
SJSU

35. Bolted Joints in Shear τ = F / As = 21/144 = 146 MPa
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

36. Bolted Joints in Shear Assume shear is carried by friction
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 Youssefi
SJSU