Design of threaded fasteners

I/C: KALLURI VINAYAK

Introduction • Design of machine elements needs geometry and joints; single integral parts will not do • Parts are joined by fasteners and they are conveniently classified as permanent, semi-permanent and nonpermanent joints • Permanent :Welded joints, adhesive bonding Semi-permanent : Riveted joints Non-permanent fasteners: Threaded/ non threaded joints Non threaded joints- keys, pins, Cotter and Knuckle joints Threaded joints- screws, bolts and nuts, studs etc.

Thread Standards and Definitions

Lead and multi-threaded screws

• Lead is the distance moved by nut parallel to the screw axis when the nut is given one turn. l = p * multiplicity of threading • Single threaded, l=p • Double threaded, l=2p • Triple-threaded, l=3p

Single & Double thread screws

The Metric Thread: nominal major diameter of 12 mm

M12× ×1.75 mm metric designation

pitch of 1.75 mm

d = major diameter dr = minor diameter = d - 1.226 869p dp = pitch diameter = d - 0.649 519p p = pitch

basic ISO 68 profile with 60◦ symmetric threads

H =

3 p 2

(a)Square (b) Acme threads

Power screw or Screw jack mechanism
A device used in machinery to change angular motion into linear motion, and, usually, to transmit power
Find use in machines such as universal tensile testing machines, lead screws of lathes and other machine tools, automotive jacks, vises, linear actuators, adjustable floor posts and micrometers etc

The Mechanics of Power Screws square-threaded power screw single thread Mean diameter ‘dm’ pitch ‘p’ lead angle ‘λ’ helix angle ‘ψ’ loaded by the axial compressive force ‘F’

Helix angle: Angle that thread makes with plane perpendicular to thread axis Lead angle : Angle between the helix and a plane of rotation

FBD of one thread, (a) lifting (or raising) and (b) lowering A single thread of the screw is unrolled or developed for exactly a single turn. Then one edge of the thread will form the hypotenuse of a right triangle whose base is the circumference of the mean-thread-diameter circle and whose height is the lead

tanλ=(l / πdm)

Raising:

Lowering:

∑F = P − N sinλ − fNcosλ = 0 ∑F = −P − N sinλ + fNcosλ = 0 F = F − fN sinλ − N cosλ = 0 ∑ F = F + fN sin λ − N cos λ = 0 ∑V H

R

H

V

L

Raising: F (sin λ + f cos λ ) PR = cos λ − f sin λ

Lowering:

Fd m TR = 2

F ( f cos λ − sin λ ) PL = cos λ + f sin λ

 + f  F  l   π d m   PR =  1 −  f .l  π d m 

 l + π fd m   π d m − fl

  

  F  f −  l  d π m    PL =  1 +  f .l  π d m 

Fd m  πfd m − l    TL = 2  πd m + fl 

Self locking of power screws • TL gives the torque required to overcome the friction in order to lower the load • In certain instances, the load may itself lower by causing the screw to spin • In such cases, TL is either zero or negative. • Whenever, the load does NOT lower by itself unless a positive TL is applied, the screw is said to be self-locking

Self-locking of power screws The condition for self-locking is

TL ≥ 0 ⇒ πfd m > l Divide both sides by πdm. Since l/ πdm =tanλ

f > tan λ The screw is self locking whenever the coefficient of friction is greater than the tangent of the lead angle.

Accounting for collar friction • Normally a collar is employed to enable the power screw system to have sufficient bearing area hold the component being raised • Since the collar slides against the component being raised, additional torque needs to be applied to raise the load, this is called as collar friction torque Tc • To estimate the Tc, whenever the collar is not too big, it is enough to use a mean diameter, dc, at which the collar friction force is concentrated

Ff c d c Tc = 2 Total torque required to rise the load; TR’ = TR + Tc Total torque required to rise the load; TL’= TL + Tc

Power screw ‘s raising efficiency • It is the ratio of raising torque without friction to the raising torque with friction • Can be defined both with and without collar friction Fl To Fl Q To = = e= 2π TR 2π T R Use Tables 8-5 and 8-6 for values of coefficient of f and fc.

Table 8–5 Coefficients of Friction ‘f’ for Threaded Pairs

Table 8–6 Thrust-Collar Friction Coefficients

Raising torque for ACME screws

• A simple approximate equation is

Fd m  l + πfd m sec α    TR = 2  πd m − fl sec α  The effect of the thread angle in ACME thread is to increase the friction force between the screw and the nut due to the wedging action of the thread For power screw application, though the ACME thread is not suitable due to higher frictional force resulting from wedging action, is invariably used because it is easier to manufacture than the square threads.

Body stresses in power screws Bearing pressure

Critical element at which the von-Mises stress is evaluated

T F

Body stresses in power screws • Bending stress, σx resultant is von-mises stress at top of the • Torsional shear stress, τxz root plane • Axial compressive stress, σy • Transverse shear (no contribution to von-Mises stress because it is maximum where bending stress is zero and is zero where bending stress is maximum; hence needs to be only independently checked for) • Bearing pressure (no contribution to von-Mises stress because it is distributed over the thread and is maximum at the middle of thread and is zero at the root of the thread)

Body stresses in the screw: those that need only to be independently checked (no need to take into account in von-Mises stress)

σ

B

=

F 2F = πd m nt p 2 πd m nt p

Must be less than the safe bearing pressure given in Table 8-4. Causes too much wear and sometimes crushing.

The engaged threads cannot share the load equally. Some experiments show that the first engaged thread carries a maximum of 0.38 of the load. In estimating thread stresses by the equations above, substituting 0.38F for F and setting nt to 1 will give the largest level of stresses in the thread-nut combination.

σ

B

=

Table 8–4 Screw Bearing Pressure

2F 0 . 76 F = πd m nt p πd m p

F 3V 3 3F 1.14F τ= = = = 2 A 2 πdr nt p 2 πdr nt p πdr p

It is at the centre of the root area. Must be less than the shear yield strength of material.

Body stresses in the screw threads: those to be taken into account for estimation of the von-Mises stress at the critical element

Power screws are operated normally at low speeds and hence static design is enough.

16TR τ xz = 3 πd r

or

16TL τ xz = 3 πd r

F − 4F σy = = 2 A πd r

M 6F 2.28 F = = σ x = σb = I c πd r nt p πd r p 3

Resultant von-Mises stress

[

1 p p  p Q M = F × ; I = (πd r nt )  and c = 12 4 4 2

(

1 (σ x − σ y )2 + (σ y − σ z )2 + (σ z − σ x )2 + 6 τ xy2 + τ yz2 + τ zx2 σ '= 2

)]

1

2

Problem A power screw has triple thread of major diameter 25 mm, minor diameter 21.5 mm, pitch diameter 23 mm and pitch of 3 mm. A vertical load on the screw reaches a maximum of 6 kN. The coefficient of friction is 0.06 for threads and 0.03 for collar. The friction diameter of the collar is 30 mm. Find the following: (a) total torque required to raise the load, (b) total torque required to lower the load, (c) efficiency, (d) bending stress, axial normal stress, torsional shear stress and the resultant von-Mises at the root for one thread (by assuming the first engaged thread carries a maximum of 0.38 of the load). (e) bearing and transverse shear stress

+

Force, F = 6 kN d = major diameter = 25 mm; dr = minor diameter = 21.5 mm; dp = pitch diameter= 23 mm p = pitch = 3 mm; For triple threads, l= 3p = 9 mm f= 0.06; fc = 0.03; dc =30 mm mean diameter, dm= (d+dr)/2 = 23.25 mm

(a) Total torque required to raise the load, Fd m  l + πfd m  Ff c d c   + TR’ = TR + Tc = 2  πd m − fl  2 6 x10 3 x 23.25  9 + π (0.06) (23.25)  6 x10 3 (0.03)30  +  = 2 2  π (23.25) − (0.06)(9) 

= 12.874 + 2.7 = 15.574 N-m

(b) Total torque required to lower the load TL’= TL + Tc =

Fd m  πfd m − l  Ff c d c   + 2  πd m + fl  2

6 x103 x 23.25  π (0.06) (23.25) − 9  6 x103 (0.03)30   + = 2 2  π (23.25) + (0.06)(9)  = - 6.071 + 2.7 = - 3.371 N - m To Fl 6 x103 x9 (c) Efficiency, e = = = = 66.75% without collar friction 3 TR 2πTR 2π (12.874 x10 ) To Fl 6 x103 x9 Efficiency, e = = = = 55.18% with collar friction 3 TR 2πTR 2π (15.574 x10 )

6 x0.38 x6 x103 d ) Bending stress, σ x = σ b = = 67.5 MPa −3 −3 π (21.5 x10 )(1)(3x10 ) − 4F − 4 x 6 x10 3 Axial normal stress, σ z = = 2 πd r π 21 .5 x10 − 3

(

)

2

= − 16 .52 MPa

16TR 16 x(15.574) Torsional shear stress,τ xz = = = 7.98 MPa 3 −3 3 πd r π (21.5 x10 )

[

(

1 2 2 2 (σ x − σ y ) + (σ y − σ z ) + (σ z − σ x ) + 6 τ xy2 + τ yz2 + τ zx2 σ '= 2 1 1 2 2 2 2 2 (67.5 − (−16.52) ) + (16.52) + (− 67.5) + 6(7.98) = 2 = 78.39 MPa < yeild strength of the material

[

]

)]

1

2

Bearing Pressure σ

B

=

πd

m

F nt p

2F = πd m nt p 2

2 × 0 . 38 × 6 × 10 = π × 23 . 25 × 1 × 3 = 20 . 81 MPa < safe bearing pressure given in Table 8 - 4. 3

Transverse Shear stress 3V 1.14 F 1.14 × 6 ×10 τ= = = = 33.76 MPa 2 A πd r p π × 21.5 × 3 3

Must be less than shear yield strength of the material

Bolts/Screws • Bolt is the ubiquitous machine element; there is virtually no machine without at least one bolt/screw • The purpose of a bolt is to clamp two or more parts together • The clamping load stretches or elongates the bolt • The load is obtained by twisting the nut until the bolt has elongated almost to the elastic limit; this load acts as preload • While tightening the nut, if possible, hold the bolt head stationary and twist the nut, this protects the shank (shank will not feel the friction torque)

Bolts/Screws • The ideal bolt length is one in which only one or two threads project from the nut after it is tightened • Locations of stress concentration in a bolt – At the fillet – At the run-out – At the thread-root fillet in the plane of the nut

1

3

2

• The washer face and washers are used to distribute the load under the bolt head and nut face. It also prevent fatigue failure of bolt that may result when the burs on the imperfectly drilled bolt holes cut into the bolt head

Nomenclature of bolt

(See Table A-29; page-1053)

or 0.4 mm

d, Nominal diameter or major diameter

L ≤ 125 d ≤ 48  2d + 6  Threaded length, LT =  2d + 12 __ __ 125 < L ≤ 200 2d + 25 L > 200 

Table A-29; page-1053

CAP screws may have hexagonal head also similar to a bolt but with a thinner head (not shown here)

Fillister

Flat

Socket; may be hexagonal or square

See Table A-30, page-1054 for dimensions.

Machine screws:


notice that there is no shank
driven all the way up to the head into the part
sometimes having a hole for screw head to seat in the part

Hexagonal nut

Jam nuts: they are thinner (a) end view, general; (b) washer-faced regular nut; (c) regular nut chamfered on both sides; (d) jam nut with washer face; (e) jam nut chamfered on both sides.

See Table A-31, page-1055 for dimensions.

Note: that the dimension H includes the washer face thickness which is normally 0.4 mm.

Table A–31; Dimensions of Hexagonal Nuts

A bolt may be used in two different cases (1) with nut, and (2) without nut IMPORTANT: The grip length needs to be differently estimated in the two cases.

Case 2

Case 1

Design methodologies for the two cases described in the last slide. Given: fastener diameter d and pitch p or number of threads.

Grip length, l = ld + lt

Fastener Length, L ≥ l + H

-Roundup Table A-17

Grip length, l ' = h + min t 2 , d  2  2

Length of unthreaded portion in grip, ld = L − LT Fastener Length, L ≥ h + 1.5d Length of threaded portion in grip, lt = l − ld d ≤ 48 L ≤ 125  2d + 6  Threaded length, LT =  2d + 12 __ __ 125 < L ≤ 200 2d + 25 L > 200 

Length of useful unthreaded portion, ld = L − LT

Length of the useful threaded portion, lt = l '−ld

Bolt Stiffness Scenario of springs in series

⇒

At E kT = lt Ad =

πd 4

1 1 1 = + k k1 k 2

⇒

Ad E ⇒ kd = ld

k1 k 2 k= k1 + k 2

⇒

k d kt kb = k d + kt

Ad At E ⇒ kb = Ad lt + At ld

2

; area of the shank (unthreaded) portion

At = Thread tensile stress area from Table 8-1

Table 8–1 (p:412) Diameters and Areas of Metric Threads.

dr = minor diameter = d - 1.226 869p dp = pitch diameter = d - 0.649 519p

The mean of dr and dp was used to compute the tensile-stress area.

Joints- Member Stiffness Scenario of springs in series

⇒

1 1 1 1 1 1 = + + + + .... + k m k1 k 2 k 3 k 4 ki

When one of the members is a soft gasket, km is simply the gasket stiffness.


If there is no gasket, the stiffness of the members is rather difficult to obtain, except by experimentation, because the compression spreads out between the bolt head and the nut and hence the area is not uniform.
Ito’s ultrasonic technique method shows that the pressure stays high out to about 1.5 bolt radii and falls off farther away from the bolt and suggested to use pressure cone method for stiffness calculation.

Joints- Member Stiffness

Figure 8–15 Compression of a member with the equivalent elastic properties represented by a frustum of a hollow cone. Here, l represents the grip length.

The Theory of the Frustum of a Hollow Cone

P.dx dδ = EA

2 2  D d   A = π (r − ri ) = π  x tan α +  −    2   2    2 o

2

D + d  D−d   = π  x tan α +   x tan α + 2  2  

t

P dx δ= π .E ∫0 [x tanα + (D + d ) 2][x tanα + (D − d ) 2]

( P 2t tan α + D − d )(D + d ) δ= ln π .Ed tan α (2t tan α + D + d )(D − d ) π .Ed tan α k= = δ ln (2t tan α + D − d )(D + d ) (2t tan α + D + d )(D − d ) P

Resultant member stiffness:

If for ith member,

π .Ei d tan α ki =  (2ti tan α + Di − d )(Di + d )  ln    (2ti tan α + Di + d )(Di − d ) 

i=1, 2, …n

1 1 1 1 1 1 = + + + + .... + k m k1 k 2 k3 k 4 kn

When all members are made of same material then we will use only two identical frusta arranged back to back, and hence

1 1 1 1 1 2 = + = + = k m k1 k 2 k k k k ⇒ km = 2

⇒

km = Ed

π . tan α  (2t tan α + D − d )(D + d )  2 ln   ( )( ) 2 tan α t + D + d D − d  

Resultant member stiffness:

km = Ed

π . tan α  (2t tan α + D − d )(D + d )  2 ln    (2t tan α + D + d )(D − d ) 

Ito suggested an angle α = 45◦ to use, but Little reports that this overestimates the clamping stiffness. When loading is restricted to a washerface annulus (hardened steel, cast iron, or aluminum), the proper apex angle is smaller. Osgood reports a range of 25◦ ≤ α ≤ 33◦ for most combinations.

If α is 30o and D is taken as the waster face diameter dw. The washer face diameter is roughly 1.5 the major diameter of the bolt.

π .Ed tan α km = ( l tan α + d w − d )(d w + d ) 2 ln (l tan α + d w + d )(d w − d )

⇒

0.5774 π Ed km =  0.5774 l + 0.5d  2 ln 5  0 . 5774 2 . 5 l + d  

Wileman et al. FEM results for the same situation: when all members were made of same material and hence having same Young’s modulus.

km = A exp( Bd / l ) Ed

Use table 8-7 to read A and B or use graph.

A, B are constants; l is the grip length; d is the major diameter of bolt.

Bolt Strength
According to SAE or ASTM, the bolt strength is specified by its minimum tensile strength and its minimum proof strength
The proof load is the maximum load (force) that a bolt can withstand without acquiring a permanent set (i.e. first measurable deviation from elastic behavior, which is approximately equal to 0.0001 in )
The proof strength is the quotient of the proof load and the tensile-stress area
Use Table 8–11 (page:435) for proof strength of steel
For other materials, take Sp= 0.85 Sy

Table 8–11 (page:435)

For other materials, take Sp= 0.85 Sy

Estimation of Preload, Fi For both static and fatigue loading 0.75 Fp for nonpermanent connections, reused fasteners Fi =  for permanent connections  0.90 Fp

where Fp is the proof load, obtained from the equation, Fp = At Sp Here At is the tensile stress area obtained from Table 8-1 Sp is the proof strength obtained from Table 8–11

Tension joints- The External load

Tension joints- The External load

km Pb Pm kb δ = = ⇒ Pm =Pb or Pb =Pm kb kb km km and P= Pb +Pm

kb P Fb = Pb + Fi = + Fi = CP + Fi kb + k m

Fm l+H From table A-17, L=60 mm

LT=2d+6=2x12+6=30 mm ld=L-LT=60-30=30 mm lt=l-ld=40-30=10 mm Ad=(π/4)x(122)=113.09 mm2 From Table 8.1, At=84.3 mm2 E=207 GPa, Ad At E kb = Ad lt + At ld

Kb=(113.09x84.3x207x103)/(113.09x10+84.3x30) =539.2x103N/mm

Steel cylinder head Table 8-8, A=0.78715, B=0.62873 km=Ed A exp(Bd/l)=2361.15x103 N/mm kst=2km=4722.31x103 N/mm Cast iron cylinder, E=100 GPa Table 8-8, A=0.77871, B=0.61616 km=Ed A exp(Bd/l)=1124.125x103 N/mm kci=2km=2248.25x103 N/mm Km=kstkci/(kst+kci)=1523.11x103 N/mm

C=kb/(kb+km)=0.261 Table 8-11, Sp=600 MPa Fi=0.75xAtxSp=37935 N n=(AtxSp-Fi)/(CP)=4.569

Fatigue Loading of Tension Joints • For a general case with constant preload and an external load on per bolt is fluctuating between Pmin to Pmax.

Fb min = CPmin + Fi Fb max = CPmax + Fi

( Fb max − Fb min ) σa = 2 At

( Fb max + Fb min ) σm = 2 At

C (Pmax − Pmin ) ⇒ σa = 2 At

C (Pmax + Pmin ) Fi ⇒ σm = + 2 At At

Substitute the above midrange and amplitude stresses in Goodman, Gerber and ASME elliptic failure equations to solve for fatigue loading.

Fatigue Loading of Tension Joints The statistics indicate that the bolts under dynamic loading failed 65% in the thread at the nut face, 20% at the end of the thread (at the runout point) and 15% under the head.

Table 8–17 Fatigue Stress- Concentration Factors Kf for Threaded Elements

Table 8–17 Fully Corrected Endurance Strengths for Bolts and Screws with Rolled Thread

Fatigue Loading of Tension Joints In the common bolted joints of pressure cylinders, the external load varies from a lower extreme of P=0 to the upper extreme of P itself. In this case

1 1  CP Fi Fi  CP + −  = σ a = (σ max −σ min ) =  2 2  At At At  2 At

σm

1 = (σ 2

max

+σ

min

1 )= 2

The equation of the load hence is,

 CP Fi F i  CP Fi   = + + + At At  2 At At  At

σ a = σ m −σ i

Gerber Goodman Sa Sm Sa  Sm + = 1 +  Se S ut S e  S ut

Goodman

S e (S ut − σ i ) Sa = S ut + S e

Sm = S a + σ i

ASME elliptic 2

  = 1 

2

2

 Sm   Sa   =1   +  S   Se   p 

Gerber

[

1 Sut S 2ut + 4Se (Se +σi ) − S 2ut − 2σi Se Sa = 2Se

]

Sm = S a + σ i

Factor of safety ASME elliptic Se 2 2 2 = S S + S − σ p e i − σ i Se Sa Sa p 2 2 S p +S e nf =

[

Sm = S a + σ

]

i

σa

Problem: Fig illustrates the connection of a cylinder head to a pressure vessel using 10 bolts and a confined gasket seal. The effective sealing diameter is 150 mm. A=100, B=200, C=300, D=20, E=20 (all in mm). The cylinder is subjected to cyclic pressure between 0 to 6 MPa. ISO class 8.8 bolts with a diameter of 12 mm have been selected. Determine the fatigue factor of safety using Goodman criteria

Values of Fi, At, C & P are from solution of previous problem i.e. P=10602 N; C= 0.261; Fi=37935 N & At=84.3 mm2

σi=Fi/At=37935/84.3=450 MPa σa=CP/(2At)=0.261x10602/(2x84.3)=16.412 MPa From Table 8-17, Se=129 MPa From Table 8-11, Sut=830 MPa Modified Goodman criteria Sa=Se(Sut-σi)/(Sut+Se)=51.115 MPa nf=Sa/σa=51.115/16.412=3.114

Bolted and Riveted Joints in Shear (a) shear loading; (b) bending of rivet; I/c is the section modulus for the weakest member

(c) Shear of rivet; A is the cross-sectional area of all the rivets

(d) tensile failure of members; A is the net area of the plate

(e) bearing of rivet on members or bearing of members on rivet; projected area for a single rivet is A = td. Here, t is the thickness of the thinnest plate

Modes of failure in shear loading of a bolted or riveted connection

(f ) shear tear-out & (g) tensile tear-out. These failures are avoided by spacing the rivets at least 1.5 diameters away from the edge

Shear Joints with Eccentric Loading

Free-body diagram of beam Beam bolted at both ends with distributed load

enlarged view of bolt group centered at O showing primary and secondary resultant shear forces.

Primary and secondary shear forces on each bolt: Primary shear or direct load

V F '= n Same for all bolts. Secondary shear or moment load:

FA// rA + FB// rB + FC// rC + ... = M The force taken by each bolt depends upon its radial distance from the centroid; that is, the bolt farthest from the centroid takes the greatest load, while the nearest bolt takes the smallest

F /r = F /r = F /r

// // // A A B B C C Hence from the above two equations:

Mrn F = 2 2 2 rA + rB + rC + ... // n

Next find the vectorial sum of the primary and secondary shear forces to know the total load.

Centroid of pattern of bolts:

n

A x + A2 x 2 + A3 x3 + A4 x 4 + A5 x5 x= 1 1 = A1 + A2 + A3 + A4 + A5

∑Ax i

i

1

n

∑A

i

1

n

A y + A2 y 2 + A3 y3 + A4 y 4 + A5 y5 = y= 1 1 A1 + A2 + A3 + A4 + A5

∑Ay i

1

n

∑A

i

1

i

Problem: Find the total shear load on each of the three bolts for the connection shown in the figure and compute the significant bolt shear stress. Sol: F1' = F2' = F3' = 4 kN F1'' = 0

2

12 × 200 × 32 '' '' F2 = F3 = = 37.5 kN 2 2 32 + 32

1

F2 = F3 =

(F ) + (F ) ' 2 2

'' 2 2

3

= 37.72 kN

For M 12 bolts, from table 8 − 1, Ar = 76.3mm 2 37.72 ×1000 Shear stress in bolts 2,3 = = 494.32MPa 76.3 4 ×1000 Shear stress in bolt 1 = = 52.42MPa 76.3

Problem: A vertical channel 152x76(see Table A–7) has a cantilever beam bolted to it as shown. The channel is hotrolled AISI 1006 steel. The bar is hot-rolled AISI 1015 steel. The shoulder bolts are M12x1.75 ISO 5.8. For a design factor of 2.8, find the safe force F that can be applied to the cantilever.

26

Bolts: M12x1.75 ISO 5.8 From Table 8-11, Sp=380 MPa, Sut=520 Mpa, Sy=420 MPa Channel: HR AISI 1006 steel From Table A-7, t=6.4 mm, Table A-20, Sy=170 MPa Cantilever beam: AISI 1015 steel From Table A-20, Sy=190 MPa M=201 F F’b=F’A=F’B=F/3 F’’A=F’’B=2.01F Fb=F’b+F’’b=2.343 F Bolt shear: From Table 8-1, Ar=76.3 mm2 Induced shear stress due to Fb = allowable shear stress Fb/Ar=0.577Sp/nd F=2.55 kN=F1

Bearing on bolt: Ab=td=76.8 mm2 Fb/Ab=Sp/nd F=4.447 kN=F2 Bearing on channel : Fb/Ab=Sy/nd F=1.99 kN=F3 Bending of Cantilever: I=1.23x105 mm4 MC/I=Sy/nd F=2.21 kN=F4

Safe load = min{F1, F2,F3,F4}=1.99kN