The American University in Cairo MENG 356

MECHANICAL DESIGN I

Dr. Hesham Hegazi Lecture # 23: Bolted Connections

Fall 2003

Bolted Joints in Tension

The stiffness of any component is given by:

l

EAFk ==δ

The stiffness of the bolt is given by:

= 24

dlEkb

π

There are two components in the grip of the bolt which act as two compression springs in series. Their combined stiffness (kc) is given by:

21

111kkkc

+=

where k1 and k2 are the stiffness of the two parts. It is difficult to predict the area of the two components which is compressed by the bolt-head and the nut. As shown in Figure it is assumed that an annular area with (3d) and (d) as outer and inner diameters, respectively, is under the grip of the bolt.

( )[ ] 222 234

dddA ππ =−=

td

lEAkk

2

212π===

The American University in Cairo MENG 356

MECHANICAL DESIGN I

Dr. Hesham Hegazi Lecture # 23: Bolted Connections

Fall 2003 when the nut is initially tightened, the bolt is subjected to an initial tension, which is called as pre-load (Fi). Under the action of pre-load, the bolt is elongated by an amount δb and the two parts are compressed by an amount δc; thus

c

ic

b

ib k

FkF == δδ

The load deflection diagram is shown in figure. Line OA indicates the elongation of the bolt, while line CA shows the compression of the parts. The slope of line CA is negative because of the compressive force. When the parts are clamped and are put in service, they are subjected to an external force F. The effect of the external force is as follows:

a) The bolt is further elongated by ∆δ and the bolt load is increased by ∆F (Line AB). b) The compression of the two parts is relieved by the same amount ∆δ and there is a

corresponding reduction in load. The reduction in load is (F - ∆F). This deformation is shown by the line AD.

( )

δδ ∆∆−=

∆∆= FFkFk cb

dividing the two equations, we have

���

����

�

+=∆

cb

bkk

kFF

and the resultant load on the bolt Fb = Fi + ∆F As the external load further continues on line OA, the limitation is point M where the compression of the two parts becomes zero. At this point, the joint will begin to open, since the parts can no longer expand to maintain the tight joint.

The American University in Cairo MENG 356

MECHANICAL DESIGN I

Dr. Hesham Hegazi Lecture # 23: Bolted Connections

Fall 2003

cb

b

b

i FFδδδ +

= max,

or ���

����

� +=b

cbib FF

δδδ

max,

Then ���

����

� +=c

cbib k

kkFF max,

- Eccentrically Loaded Bolted Joints in Shear

The American University in Cairo MENG 356

MECHANICAL DESIGN I

Dr. Hesham Hegazi Lecture # 23: Bolted Connections

Fall 2003

)./('4

'3

'2

'1 boltsofNoFFFFF ====

4''

43''

32''

21''

1 rFrFrFrFFe +++=

Where ( )24

23

22

21

1''1 rrrr

FerF+++

=

( )24

23

22

21

2''2 rrrr

FerF+++

=

( )24

23

22

21

3''3 rrrr

FerF+++

=

( )24

23

22

21

4''4 rrrr

FerF+++

=

The American University in Cairo MENG 356

MECHANICAL DESIGN I

Dr. Hesham Hegazi Lecture # 23: Bolted Connections

Fall 2003

- Bolted Joint With Combined Stresses

1- Direct Shear

).(

'2

'1 BoltsofNo

FFF ==

2- Tension The moment (Fe) tends to cause the bracket to rotate about edge C, the stretching of the bolt is proportional to its distance from the edge C.

Fe = 2(F1l1 + F2l2)

2

2

1

1llδδ = or F = k δ then

22

2

11

1lk

Flk

F =

substitute in the previous equation

)(2)(2 2

221

222

221

11 ll

FelFll

FelF+

=+

=

Then the bolts are designed according to the maximum shear stress theory.