Machine Design - Bolts

1

University of WaterlooDepartment of Mechanical Engineering

ME 322 - Mechanical Design 1

Partial notes – Part 3(Bolted Connections)

G. Glinka

Fall 2007

2

Design of Design of BoltsBolts and and BoltedBolted ConnectionsConnections

Bolts are very important engineering members and their principal use is to clamp parts together with enough force to seal pressure within a vessel (or pipe) or to prevent motion between parts.

Bolts and other threaded connectors have long been standardized machine components ; therefore the task of the designer is not to design bolts, but to select them, and to ensure that the bolts and the parts that they connect have adequate strength.

Thread standards and DefinitionsThe terminology of screw threads, illustrated below is explained as follows.

Sharp Vee threads are shown for clarity; the crests and roots are actually flatted or rounded during the forming operation.

3

* The pitch is the distance between adjacent thread forms measured parallel to the thread axis

* The major diameter d is the largest diameter of a screw thread

* The minor diameter d r or d 1 is the smallest diameter of a screw thread

* The lead l, not shown, is the distance the nut moves parallel to the screw axis when the nut is given one turn. For a single thread, the lead is the same as the pitch

Basic thread profile for metric M and MJ threads. D (d) = basic major diameter of internal (external) thread; Dr (dr) = basic minor diameter of internal (external) thread; Dp (dp) = basic pitch diameter of internal (external) thread;

ρ = pitch; 1/ 20.5(3) .H ρ=

The American National (United) thread standard has been approved in the U.K. for use on all standard threaded products (shown below).

Internal threads

External threads

4

A thread can be likened to a piece of string wound in a tight helix around a cylinder - or around a conical frustum in the case of pipe thread designed to eliminate leakage. When a nut on a screw is rotated by one turn, it travels along the screw a distance known as the lead L. Developing one turn of the thread at the mean diameter dm ( the average of major and minor diameters ) gives the lead angle (or helix angle ) λas tanλ = L /π·dm. Power screws may employ multiple threads, or starts, so L = p ∗ number of starts as illustrated. Fasteners on the other hand are almost invariably single start (L = p). They are also right handed to avoid confusion in tightening, though LH screws appear in turnbuckles and in certain bicycle parts where the prevailing torque would tend to loosen RH fasteners.

From: Douglas Wright, The University of Western Australia

5

Two major United thread series are in common use : UNUN and UNRUNR. The difference between these is that a root radius must be used in the UNRUNR series. Because of decreased thread stress-concentration factors, UNRUNR series have improved fatigue strength. United threads are specified by stating the nominal major diameter, the number of threads per inch, and the thread series:

''5 18 ;8

UNRF−''5 11 ;

8UNRC−

0.625'' 18UNRF−

or

F − fine threadC − coarse thread

a) Round head b) Flat head

c) Fillister head d) Oval head

e) Truss head f) Binding head

g) Hex head (trimmed) h) Hex head (upset)

6

7

Metric threads are specified by writing the diameter and pitch in millimeters, in that order

12 1.75M × M - metricThe symbol above means : metric thread having a nominal major diameter of 12 mm and a pitch of 1.75 mm

Tensile stress area At (or effective area)

A great many tensile tests of threaded rods have shown that an unthreaded rod having a diameter equal to the mean of the pitch diameter ‘dp’ and minor diameter will have the same tensile strength as the threaded rod. The area of this unthreaded rod iscalled the tensile-stress area At of the threaded rod. This is the area used for strength calculations

Run-out, end of threadMajor diameter( about the same as nominal diameter)

Helix angle

Diameter corresponding to tensile stress area, At(Area of an unthreaded rod having the same strength)

Root, or minor diameter Pitch, (= reciprocal of number of threads per inch, also same as lead for single thread)

8

Basic dimensions of bolt heads and nutsBasic dimensions of bolt heads and nutsL

a) end view; general

Hexagonal nuts:

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

L – bolt length

9

10

Table A- 31; Dimensions of Hexagonal Nuts

11

12

Recommended procedure for Recommended procedure for the the determination of determination of bolted joint dimensions bolted joint dimensions (bolt and nut joint)(bolt and nut joint)

Given:

1. Grip or thickness of connected members, l = LG

2. Bolt diameter d and pitch p or number of threads per inch

1. Nut height H from Table A-31

2. Washer thickness t from Table A -32 or A -33.

3. Thread length LT from Table 8-7

4. Bolt length L from the relation L > l+H (Table 8-7 for preferred length)

5. Length of the unthreaded portion from the relation ld = L - LT

6. Length of the threaded portion within the grip from the equation LT = L- ld7. Area of the unthreaded portion from the equation Ad = πd2/4

8. Area of threaded portion At is the same as the tensile-stress area.

9. Bolt stiffness kb from Eq. (8-17)

Find:

l = LG

lt

LT

ldt

13

Find:1. Washer thickness t from Table A-32 or A-33.

2. Thread length LT; same as for bolts; use Table 8-7

3. Cap screw length from expression: L>h+1.5d

4. Length of unthreaded portion from: ld= L - LT

5. Length of useful threaded portion from: l = l’ - ld

6. Area of unthreaded portion: Ad= πd 2/ 4

7. Area of threaded portion: AT = At (from Table 8 -1, 8 -2)

8. Cap screw (fastener) stiffness kb from: Eq. (8-17)

Effective grip

Recommended procedure for Recommended procedure for the the determination of determination of bolted joint dimensions bolted joint dimensions (cap srew joint)(cap srew joint)

t t1 t2

l

l’LT

Lld

h

d

22

'

2

;2'

;2

G

th t dl L

dh t d

⎧ + <⎪⎪= = ⎨⎪ + ≥⎪⎩

14

Bolt StrengthBolt Strength

The bolt strength is the key factor in the design or analysis of bolted connections. In the specification standard for bolts, the strength is specified by stating the minimum proof strength, or minimum proof load, and the minimum tensile strength.

The proof load is the maximum load (force) that a bolt can withstand without acquiring a permanent set (corresponding to 0.0001 permanent strain in the fastener)

The proof strength is the quotient of the proof load and the tensile-stress area.

The proof strength thus correspond roughly to the yield-point strength and is about 90% of the 0.2 percent offset yield strength.

The tensile strength is the stress at which failure of the joint may occur (analogous to the ultimate strength).

The bolts strength parameters are given in various specification codes.

Note! All specification-grade bolts made in the US bear a manufacturer’s mark or logo, in addition to the grade marking, on the bolt head. This is confirmation that the bolt meets or exceeds specifications. If any mark is missing the bolt may be imported or home made.

The prooof loadThe proof strengthTensile stress area

=

15

16

Basic bolting arrangementsBasic bolting arrangements

Our discussion concerns that class of fasteners known as bolts, which includes the subcategories cap screws and studs. These names are associated with the three basic clamping arrangements, shown below.

In the majority of cases, the bolt passes through the hole in the part being clamped and mates with a nut (Fig. a).Occasionally, the bolts mates with threads with one of the member, rather than with a nut. The unthreaded shank of a stud or bolt helps to locate the part being assembled.

Basic bolted arrangements. a) Bolt and nut. b) Cap screw or tap bolt. c) Stud.

a) b) c)

P PP

P P P

17

Joint Analysis Joint Analysis

Tension Connections Tension Connections –– Preloading TheoryPreloading Theory

Stiffeness of a fastener

Sections through two typical tension-loaded joints are illustrated in Figures below.

Notice the clearance space provided by the bolts holes, and notice how the bolt threads extend into the body of the connection.

Twisting the nut in Fig. A streches the bolts to produce the clamping force. This clamping force is called the pre-tension or the bolt preload. Since the members are being clamped together, the clamping force whichproduces tension in the bolt induces compression in the members.

a) b)

p

l l’

18

We have noted that the elastic behaviour of connected members has a significant effect upon the distribution of load between their connectors. In an elastic bolt and nut connected by a number of thread turns, the first 'thread' takes a disproportionate fraction of the total tensile load (see the sketch above). It is clear that there is little benefit from an engagement length exceeding half-a-dozen threads. Special nuts which alleviate non-uniform load distribution are available. The variation of normal stress σ over a cross-section adjacent to the nut face ( as sketched ) is dominated bystress concentration induced by the thread root, so it is hardlysurprising that bolt fractures usually occur in the exposed threads close to the nut face.

Nut

BoltP

Stress concentration and load distribution along the Stress concentration and load distribution along the threaded part of the boltthreaded part of the bolt

From: Douglas Wright, The University of Western Australia

19

The difference between the solution A and B can be illustrated by the locations of the points of external load application.

It is generally assumed that in the case of a bolt with a nut (Fig. a) that the forces are applied under the head of the bolt and of the nut.

In the case of a cup screw the threaded part of the joint (Fig. b) is, in reality, the nut, and the load can be assumed to act where the threads begin.

P

PP

P

P

P

Extremes of external loading application, (a) and (b). The actual location is somewhere between the two. (a) Forces applied under the head of the bolt and the nut. This is the worst case. (b) Opposite extreme is application of the forces at the joint interface. (c) Force application in a tap bolt.

(a) (b) (c)

20

Modeling of the bolted region in gripModeling of the bolted region in grip

It is assumed that the two cone (frustum) region in grip is resisting the gripping force from the bolt!

,3 ,4 ,5

1 1 1 1

m m m mk k k k= + +

t

t2

P

P

d 3

4

5

1

2

The resultant stiffness of all members in grip:

The resultant stiffness of all members in grip:

,1 ,2

1 1 1

b b bk k k= +

21

The Bolt StiffenessThe Bolt StiffenessThe spring constant or the stiffness constant of an elastic member such as a bolt, is the ratio between the force appliedto the member and the deflection produced be the force.

,1 ,21 2; ;b b

b bF Fk kδ δ= =

,1 ,2

1 1 1b b bk k k

+=

Calculating bolt stiffness. (a) Defining stiffness. (b) Bars in series are springs in series.

bFk δ=

a)

b)

kb,1 kb,2

kb,2kb,1

Fb

Fb

FbFbFb Fb

Fb ( )bSlope k F kδ=

0

22

Bolt stiffness kb!

d = nominal diameter, A t = stress area, σ - average stress F/A

;b dd

d

F lA E

δ =

;b TT

t

F lA E

δ =

1 ;d

d d

lk A E

=

1 ;T

T t

lk A E

=

1 1 1 d tb

b d T d T t d

A A Ekk k k A l A l

= + ⇒ =+

The stiffness of the portion of a bolt or screw within clamped zone will generally consists of two parts, that of the unthreaded portion and that of the threaded portion. Thus the stiffness constant of the bolt is equivalent to the stiffness of the spring in series.

l

σ

2

4 bFd

σπ

=

4 b

t

FA

σ =

0

Fb

Fb FbFb

Fb

23

Stiffness of clamped members Stiffness of clamped members –– the the modelmodel

Compression of a member in the clamp assumed to be confined to the frustum of a hollow cone!

di = 2r0

Joint stiffness model (Cone – frustum model)

d = 2ri

di = 2r0

24

dw – washer diameter

25

Analysis of bolted joints under external load Analysis of bolted joints under external load To start the analysis let examine the bolt holding two parts of a joint together, as shown in the Figure below.

It helps to consider the bolt as a tension spring, with stiffness kb and the joint members it clamps as compression spring with stiffness km. The load vs. deflection curves of the bolt and the joint members can be plotted as shown above. When the joint is assembled, the bolt is put into an initial tension Fi (preload) and the joint will be under equal butopposite compression.

When we tighten a bolt, ( a) we apply torque to the nut, ( b) the nut turns, ( c) the bolt stretches, ( d) creating preload.'

26

The The staticsstatics of a bolted joint in gripof a bolted joint in grip

P

P

It is assumed that the gripping force Fb and the load P are transferred by the material volume represented by the two cone shaped frusta (shaded).

27

Fi

Fi

Fi

Fi

Fi

Fi

Free body diagrams

Static equilibrium of preStatic equilibrium of pre--loodedlooded (pre(pre--torquedtorqued) bolted joint ) bolted joint (no external load, i.e. P=0)(no external load, i.e. P=0)

Fi -initial pre-load

28

Fb

Fb

Fb

Fm

Fm

Fb

Static equilibrium of preStatic equilibrium of pre--loodedlooded (pre(pre--torquedtorqued) bolted joint ) bolted joint (with applied external load, (with applied external load, PP))

Fi -initial pre-load

Free body diagrams P P

δ

Fi

Fi

P P

Fb

Fb

P P

29

Equilibrium of forcesEquilibrium of forcesThe task of the analyst is to determine the tensile load on the bolt and the compressive load on the members in grip!

0

0

m b

m mi

b i b

mi i b b

mb

mb

P F FF F kF F k

P F k F kP k k

Pk k

δδ

δ δδ δ

δ

+ − == −= +

+ − − − == +

= +The force on the bolt:

The resultant forces Fb and Fm on the bolt and members in grip respectively are linear functions of the load but the increase (change) of the load from 0 to P changes the bolt and members load by a fraction of the applied load P because:

bb i b i mb

kF F k F Pk kδ+ += = +

mm mi i mb

kF F k F Pk kδ= − = − +

The force on the members in grip:

1b mm mb b

k kk k k k< <+ +

30

Fi

0 P

kb·δ

km·δ

P

Fb

Fm

bb

b mi

kFk k

F P++

=

mm

b mi

kFk k

F P−+

=

Variation of the force Variation of the force FFbb and and FFmm as a function of the as a function of the external load external load PP

Because the stiffness kb is less than the stiffness km therefore the increase of the bolt load Fb is smaller than the decrease of the load Fm on members in grip.

31

Fi

0

FbFm

kb·δo

km·δo

Po

PPo

45o

Separation of the joint and the separation load Separation of the joint and the separation load PPoo

00 0 0 0

0

00 0

0

;

:

1

; :1

b mb m

i m

ib i

bb m

b m

b m i bi

m b m

PP k kk k

and at separation F kP FP k F P or

kk kk k

k k F kP F where Ck C k k

δ δ δ

δ

= + ⇒ =+

=

= + ⇒ =+ −

+

+= = =

− +

32

Analysis of bolted joints under external load Analysis of bolted joints under external load (pre(pre--torquedtorqued, no external load, i.e. P=0), no external load, i.e. P=0)

The same analysis can be carried out in terms of the displacement δ !

0b0m

FmFb

δm

Contraction δmExtension δb

FiFi

Com

pres

sion

Tens

ion

33

NomenclatureNomenclature::Fi – preload

P – external tensile load

Pb – portion of P taken by the bolt

Pm – portion of P taken be members

Fb=Pb+Fi – resultant bolt load

Fm=Pm-Fi – resultant load in member

The load P is tension, and it causes the connection to stretch, through some distance δ which can be related to the load

δ =Pb/kb and δ =Pm/km

δ

Fi +kbδ = Fb

PP

Fi - kmδ

34

δ – extension of bolt = reduction in contraction of members in grip

P + Fi – kmδ = Fi + kbδ → P = kmδ + kbδ = Pm + Pb

δ = P⁄ (kb +km)

Fb = Fi + P·kb/(kb + km)

Fm = Fi - P·km/(kb + km)

Analysis of bolted joints under external load Analysis of bolted joints under external load (under external load, P)(under external load, P)

0b0m

Fm

Fb

δm

Contraction δmExtension δb

FiFi

Com

pres

sion

Tens

ion

Fb=Fi + kbδ

Fm=Fi - kmδ

δ

Pb

Pm

35

The ratios C and (1-C) are the coefficients of load P and they describe the proportion of the external load taken by the bolt (Pb=CP) and by the member (Pm=(1-C)P), respectively.

Since C is always less than 1 it means that only fraction of the load is taken by the bolt!

36

Fatigue loading of bolted joints Fatigue loading of bolted joints

It has been observed that the distribution of typical bolt failures is about 15% under the head, 20% in the end of the thread, and 65% in the thread at the nut face. This coincides with the section of the highest stress concentration.

The stresses used in the calculations are the average stresses σ and therefore all necessary corrections such as correction for the surface, size, load, stress concentration etc. should be accounted for.

But first, loading on the bolt has to be determined!

σ

σ

0l

37

0 m0 bF iF b

F m

δ bδ m

δ

F b,m

P b

F b,m

ax

F b,m

in=F

i

F b=

F i+k

bδ

F b,a=

(Fb,

max

-Fi)/

2

F b,a

F m=

F i-k

mδ

P=P m

ax

P

0

Due

to c

yclic

lo

adin

g: 0

-P m

axt

Load

ing

hist

ory

Cyc

lic lo

ad a

nd c

yclic

str

ess

in th

e bo

ltC

yclic

load

and

cyc

lic s

tres

s in

the

bolt

38

Relation between stress amplitude σa and the mean stress σm

σa=CP/2At; σm=σa+Fi/At

Because the term (Fi/At) is constant then the relation σa-σm can be represented by straight line with the slope 1 beginning at σm=Fi/At.

1

im a

t

a m

e uts

FA

S S

σ σ

σ σ

= +

+ =

39

Sm

The σm - σa relation and the Goodman diagram can also be shown in the form of the diagram below.

Safety factor against fatigue: n = Sa /σa

After solving for the factor of safety guarding against fatigue failure, one should also check the possibility of yielding.

max 2

y y y

im aa

t

y t y

i it t

S S Sn F

AS A S

n F F CPCPA A

σ σ σ σ= = =

+ +

= =++

C – failure point, B - safe

40

Fatigue Strength of BoltsFatigue Strength of Bolts

The fatigue endurance of a bolt can be estimated as in cases described in the section on Fatigue.

Se = ka ·kb·kc·kd·ke·S’eHowever, determination of the fatigue factor Kf for bolts is somewhat complex. Therefore, emprically determined fatigue notch factors are recommended to be used in the equation above. In addition, the fatigue notche factors Kf for bolts, given below, are also corrected for the surface finish. Therefore the fatigue limit of bolts should be determined as:

Se = 1 ·kb·kc·kd·1/Kf’·S’e

Table A-31; Fatigue notch factors Kf’ for threaded elements

Note! The fatigue notch factors Kf’ given in Table A-31 are not exactly as the fatigue notch factors Kf discussed in the section on Fatigue because the Kf’ factors include both the effect of the stress concentration factor Ktand the surface finish factor ka.

i.e. Kf’ = ka/Kf

41

In the case of SAE and ISO rolled threads the amount of cold work and strain hardening is known to the designer; therefore full corrected axial fatigue limit (endurance limit) Se is reported.

It means that the designer does not need to make any corrections to the fatigue limits given in the Table below!

42

ExampleExample::

The section of the scaled joint shown in the figure is loaded by a repeated force P= 6 kip.

The members have modulus of elasticity E= 16 Mpsi:

All bolts have been carefully preloaded to the load level Fi=25 kip each:

a) If hardened-steel washers 0.134 in thick are to be used under the heal and nut, what length of bolts should be used?

b) Find stiffness kb ,km and C

c) Using the modified Goodman line, find the factor of safety guarding against fatigue failure

d) Find the load factor guarding against over-proof loading.

43

Bolted and riveted joints loaded in shearBolted and riveted joints loaded in shear

Riveted and bolted joints loaded in shear are treated exactly alike in design and analysis.

When the forces on a joint tend to slide the members over each other , the joint is loaded in shear. The Figure below shows joints of this type.

Bolted/riveted jojnts in shear. Bolt locations are designated by a circle ‘●’.

a) joint ( butt splice) under axial loading

b) gusset plate – axial and eccentric loading

c) bracket – eccentric loading

44

Modes of failure Modes of failure –– for bolted joints in shearfor bolted joints in shear

There are five basic modes of failure of a bolted/riveted joint subjected to shear loading :

1. failure by shearing of the bolt (rivet)

2. failure by crushing of the bolt (bolt in bearing)

3. failure by crushing of the plate at a bolt location (member in bearing)

4a. failure of a member in tension

4b. failure of a member in combinedtension and bending

5. failure of a member in shear tear-out

45

Bolted joints under axial shear loadBolted joints under axial shear loadIf the load application is through the centroid of the bolt pattern, the loading is said to be axial.

2 2

4/ 4

F F FA d d

τπ π

= = =

Allowable force based on the distortion strain energy hypothesis and the bolt material yield strength:

2

2

3 3

4 3

ys ys

ys

S S

F Sd

τ τ

π

≤ ⇒ ≤

≤

2

4 3ysS d

Fπ

≤

1. Bolt in shear (bolt failure due to shear); shear stress in the bolt:

46

Bolted and riveted joints under eccentric shear loadBolted and riveted joints under eccentric shear load

A joint is eccentrically loaded when the line of action of the load not pass through the bolt pattern centroid. Therefore one has to know the coordinates of the bolt pattern centroid G (x, y)

Coordinates of the bolt pattern centroid GLet (figure below) A1, A2…..A5 be the respective cross-sectional areas of a group of five bolts. The location of each bolt is characterized by the coordinates of each bolt, i.e. A1 (x1,y1) , A2 (x2,y2), A3 (x3,y3), A4 (x4,y4), A5 (x5,y5).Using rules of statics we can calculate the coordinates of the centroid G.

5 51 1 2 2 3 3 4 4 1

51 2 3 4

1

n

i i

n

i

A xA x A x A x A x A xxA A A A A A

+ + + += =

+ + + +

∑

∑

5 51 1 2 2 3 3 4 4 1

51 2 3 4

1

n

i i

n

i

A yA y A y A y A y A yyA A A A A A

+ + + += =

+ + + +

∑

∑

A5

A4

A3A2

A1

G

y

x0

y

x

47

Shear forces acting upon boltsShear forces acting upon bolts

The location of the centroid of the group of bolts is needed to be known in order to determine the shear forces which act upon each bolt.

When an eccentric joint is a bearing joint, i.e., motion is prevented by interference between bolts and plates, the load can be transferred to the centroid of the bolt group as a force plus moment as illustrated below

48

Loads on BoltsLoads on Bolts under under eccentric loadingeccentric loading -- ExampleExample

49

2A

50

ExampleExampleThe figure below shows a welded fitting which has been tentatively designed to be bolted to the channel so as to transfer the 2500 lb load into the channel. The channel is made of hot-rolled law-carbon steel having a minimum yield strength 46ksi; the two fitting plates are of hot-rolled stock having a minimum Sys of 45.5ksi. The fitting is to be bolted using six standard SAE grade-2 bolts.

Check the strength of the design by computing the factor of safety for all possible modes of failure!