Ken YoussefiSJSU 1 Various Fasteners Non-permanent fasteners (threaded) Machine screws (cap screws)Bolt and NutStud 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 YoussefiSJSU 2 Threaded Fasteners Some common screw and bolt head type Tamper resistant screw heads
Ken YoussefiSJSU 3 Threaded Fasteners – Nut and Washer
Ken YoussefiSJSU 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 YoussefiSJSU 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. 1/2 - 20UNF – 2A or (1/2 – UNF) Fit class Thread series Threads per inchMajor diameter M12 x 1.75 Metric standards Major diameter, mm Pitch, p in mm
Ken YoussefiSJSU 6 Basic dimensions of Unified threads Use tensile stress area, A t, for all stress calculations The mean of the pitch diameter and the minor diameter is used to calculate the tensile stress area.
Ken YoussefiSJSU 10 Threaded Fastener Materials - Metric
Ken YoussefiSJSU 11 h Stresses in Threads Axial load σ t = P / A t Tensile stress area Torsion τ = 16 T / πd r 3 Root diameter Bearing stress σ t = 4P / [π (d 2 – d r 2 )] (h/p) Number of threads in contact Use a safety factor of 2 with these equations Stripping stress τ (screw) = P / πd r (.8h), τ (nut) = P / πd (.88h)
Ken YoussefiSJSU 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 brassL(tapped hole length) ≥ 1.5d Steel bolt in aluminumL(tapped hole length) ≥ 2d
Ken YoussefiSJSU 13 Bolted Joints in Tension – Effect of Stiffness Bolt carries all of the load member carries all of the load
Ken YoussefiSJSU 14 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
Ken YoussefiSJSU 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 YoussefiSJSU 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 YoussefiSJSU 17 Member Stiffness – Juvinal method where A c is the effective area of clamped members k m = A c E / grip length The stiffness of clamped members
Ken YoussefiSJSU 18 Member Stiffness Gasketed joints Unconfined gasket Confined seals allow the hard faces of the members to contact, joint behaves as unsealed one, same member stiffness k m as before. Gasket material Copper E = 17.5x10 6 psi Cork E = 12.5x10 3 psi Plain rubber E = 1000 psi Teflon E = 35x10 3 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, k m = k g = A g E g / t g Confined gasket Confined O-ring
Ken YoussefiSJSU 19 Initial Tension (preload) Recommendation for both static and fatigue loading F i = K i A t S p Proof strength Tensile stress area Constant, 0.75 to 0.9 Bolt should not be reused If tightened to 90% of the proof load (F p = A t S p ), 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 YoussefiSJSU 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 YoussefiSJSU 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 F i d For critical applications a torque wrench should be used to apply the proper preload.
Ken YoussefiSJSU 22 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
Ken YoussefiSJSU 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. F m = 0, F b = P 1.Select bolt grade ( grade 5 and 8 are common) S p, S y, S u, S e 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. S p = P (design) / A t 4.Select bolt size, d, from the table 5.Use preload, F i = 0.9A t S p 6.Calculate torque, T = 0.2F i d
Ken YoussefiSJSU 24 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) S p, S y, S u, S e 2.Determine the maximum load per bolt, P. 4.Assume bolt diameter, d. 3.Select a safety factor, n = 2. 6.Determine the preload, F i = 0.9A t S p 8.Specify torque, T = 0.2F i d 5.Look up A t, 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.
Ken YoussefiSJSU 25 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. A t = P (design) / S p = 23.68 mm 2 4.Select bolt size, d, from the table, d = 7 mm, (A t = 28.9 mm 2 ) 5.Use preload, F i = 0.9A t S p = 0.9 (28.9)(380) = 9883.8 N 6.Calculate torque, T = 0.2F i d =.2 (9883.8)(7) = 13.84 N-m
Ken YoussefiSJSU 26 Design of Bolted Joints in Tension under Fatigue loading P max = maximum applied load to the joint P min = minimum applied load to the joint F a = alternating load = (F max – F min )/2 F m = mean load = (F max + F min )/2 σ a = alternating stress = F a /A t σ m = mean stress = F a /A t (F bolt ) max = maximum load applied to the bolt = C P max + F i (F bolt ) min = minimum load applied to the bolt = C P min + F i Use Goodman line as design criteria σaσa σmσm SeSe S ut nfnf + = 1 Endurance limit Ultimate strength Fatigue safety factor
Ken YoussefiSJSU 27 Design of Bolted Joints in Tension under Fatigue loading nfnf = AtAt S u (P max – P min ) + S e (P max + P min ) + 2(S e F i )/C 2(S e S u )/C Design equation Common case, P min = 0, so (F bolt ) min = F min = F i Suggested values for endurance limit for common bolts with rolled threads. For cut threads use K f = 3.8 for grade 4 and higher. multiply the alternating component of stress by K f. Endurance Strength, S e
Ken YoussefiSJSU 28 Design of Bolted Joints in Tension under Fatigue loading Fatigue failure criteria S e = endurance limit S ut = ultimate strength in tension, S p = proof strength ASME-elliptic + = 1 nfnf σmσm SpSp 2 σaσa SeSe 2 nfnf Goodman σaσa σmσm SeSe S ut + = 1 nfnf n f = safety factor guarding against fatigue failure. Yielding σaσa σmσm + = nyny SpSp n y = safety factor guarding against yielding.
Ken YoussefiSJSU 29 Design of Bolted Joints in Tension under Fatigue loading Design steps 1.Select bolt grade ( grade 5 and 8 are common) S p, S y, S u, S e 2.Select number of bolts. If circular pattern, use the restriction 3 ≤ (π D b )/N d ≤ 6 to allow access to bolt head for tightening. D b 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, P max and P min. 4.Choose a safety factor, n f = 1.5 to 2.5 6.Decide on preload F i. Use.6S p A t ≤ F i ≤.9S p A t as guideline, (F i =.75 S p A t is common). Unless specified otherwise by seal manufacturer. 5.Choose a bolt diameter, d, look the tensile stress area, A t and calculate the stiffness ratio C. 7.Use the design equation and calculate the safety factor, n f, iterate until the calculated safety factor matches the chosen one in step 4.
Ken YoussefiSJSU 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. 1.Select 14 grade 9.8 bolt. S p = 650, S y = 720, S u = 900, and S e = 140 MPa 2.Choose a safety factor, n f = 1.5 3.Determine the maximum and minimum applied load per bolt, P max and P min. P max = (pressure)(Area) / N= (2.0)(π D i 2 /4) = [(2)(π)(250) 2 /4] / 14 = 7013 N P min = 0 4.Choose a bolt diameter, d = 12 mm, look up A t = 84.3 mm 2 5.Check bolt spacing, 3 ≤ (π D b )/N d ≤ 6, 3 ≤ (π 350)/12x14 = 6.5 ≤ 6 (okay)
Ken YoussefiSJSU 31 Design Example – Fatigue Loading 6.Calculate stiffness ratio, C. Bolt stiffness, k b = A bolt E / grip length = πd 2 E / 4g = π12 2 x207x10 3 / 4x20 k b = 1.17x10 6 N / mm A c = d 2 +.68dg +.065g 2 = 12 2 +.68x12x20 +.065x20 2 = 333.2 Stiffness ratio, C = k b / (k b + k m ) =.46 Member stiffness, 1/k m = 1/k Al + 1/k cast k = A E / grip length k Al = A c E Al / g = 333.2x70,000/10 = 2.332x10 6 k cast = A c E cast / g = 333.2x100,000/10 = 3.332x10 6 k m = 1.37x10 6 N / mm
Ken YoussefiSJSU 32 Design Example – Fatigue Loading F i =.75 S p A t =.75 x 650 x 84.3 = 41,100 N 7.Select preload nfnf = AtAt 8.Calculate safety factor S u (P max – P min ) + S e (P max + P min ) + 2(S e F i )/C 2(S e S u )/C n f = 1.1 < 1.5, select larger diameter or higher strength bolt 9.Select grade 10.9 bolt, S p = 830, S y = 940, S u = 1040, and S e = 162 MPa F i =.75 S p A t =.75 x 830 x 84.3 = 52,477 N n f = 1.36 < 1.5, use more bolts, select 24 bolts Check bolt spacing, 3 ≤ (π 350)/12x24 = 3.8 ≤ 6 (okay)
Ken YoussefiSJSU 33 Design Example – Fatigue Loading Determine the maximum applied load per bolt, P max. P max = (pressure)(Area) / N= (2.0)(π D i 2 /4) = [(2)(π)(250) 2 /4] / 24 = 4090 N n f = 1.47 ≈ 1.5 Specification Bolt diameter12 mm # of bolts24 Bolt grade10.9 metric
Ken YoussefiSJSU 34 Bolted Joints in Shear Primary shear – same for all bolts F ‘ = V / # of bolts Secondary shear – for the nth bolt Fn״ =Fn״ = Mr n r A 2 + r B 2 + r C 2 + …..
Ken YoussefiSJSU 35 Bolted Joints in Shear τ = F / A s = 21/144 = 146 MPa Use grade 4.8, S sy = 155 MPa V = 16 kN, M = 16(425) = 6800 N-m F C = F D = 14.8 kN F A = F B = 21.0 kN What grade of bolt should be used?
Ken YoussefiSJSU 36 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