# Various Fasteners Non-permanent fasteners (threaded)

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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

Tamper resistant screw heads Ken Youssefi SJSU

Threaded Fasteners – Nut and Washer
Ken Youssefi SJSU

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

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

Ken Youssefi SJSU

Ken Youssefi SJSU

Ken Youssefi SJSU

Ken Youssefi SJSU

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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