Bolted joint failure modes

F. Matthews, in Handbook of Polymer Composites for Engineers

Lay-up 1:

[+45,0,-45,0,90,0,+45,0,-45,0]s

Lay-up 2:

[+45,-45,02,+45,90,-45,03]s

Lay-up 3:

[+45,-45,+45,-45,90,05]s

Lay-up 4:

[+45,-45,02,90,0,+45,-45,02]s

Lay-up 5:

[+45,-45,05,+45,-45,90]s

Effect of ‘blocked’ laminate stacking sequence on bearing strength

Simplified procedures for designing composite bolted joints

from CC Chamis, J Reinf Plast & Comp.,vol 9, pp614-626

Basic bolt geometry

wed

laminate thickness = t

F

y

x

1. Bearing (compression) failure

F

At failure, F = d t xc

2. Tension failure

F

At failure, F = (w-d) t xT

3. Wedge splitting(due to lateral pressure of bolt)

F

At failure, F = ½(2e - d) t yT

F/2

4. Shear out

F

At failure, F = 2 e t xy

5. Combined tension and shear

F

At failure, F = ½ t [(w - d)xT + 2 e xy]

Example failure analysis

High strength carbon/epoxy laminate.

Layup [0,±45,0,90]s - 10 plies at 0.125 mm per ply.

Fibre volume fraction: 60%

Strength values:

long. tension (xT) = 546 MPa

trans. tension (yT) = 343 MPa

long. compression (yT) = 550 MPa

in-plane shear (xy) = 267 MPa

Example failure analysis

Bolt diameter (d) = 6 mm

Laminate thickness (t) = 1.25 mm

Joint width, or bolt spacing (w) = 25 mm

Edge distance (e) = 25 mm

Applied load (F) = 5000 N

1. Bearing (compression)

Compressive stress is

xc = F / d t = 5000 / (6 x 1.25) = 667 MPa

This is greater than the compressive strength of the laminate, so bearing failure occurs.

The maximum load would be

550 x 6 x 1.25 = 4125 N

2. Tension

Tensile stress is x

T = F / (w - d) t = 5000 / (19 x 1.25) = 211 MPa

This is less than the tensile strength of the laminate, by a factor of 2.6.

And so on…each failure mode is considered separately, and a margin of safety calculated.

Geometrical aspects

It is straightforward to use a spreadsheet to examine the dependence of overall strength and failure mode on bolt geometry.

The following example takes the laminate information given above, and calulates failure loads for the 5 different modes as a function of bolt diameter:

Bolted joint failure loads

0

5000

10000

15000

20000

5 6 7 8 9 10 11 12 13 14 15

bolt diameter (mm)

new

ton

s

bearing

tension

splitting

shear

tension+shear

- shear failure load is independent of bolt diameter- bearing failure occurs for d < 12 mm- strongest joint has d between 12 and 13 mm, where several failure modes are likely (for this laminate)

Multi-bolt joints

from CC Chamis, J Reinf Plast & Comp.,vol 9, pp614-626

Example analysis of multi-bolt joint

• Connection required between composite panel and metal plate.

• Assume that all bolts share load equally.

• Bolts are ‘designed’ for the composite - we assume the metal plate is strong enough.

• High strength carbon/epoxy laminate, as defined previously.

Example analysis of multi-bolt joint

Design tensile load (P) = 400 N/mm

Bolt diameter (d) = 6 mm

Bolt spacing (p) = 6 bolt diameters = 36 mm

Edge distance (e) = 4 bolt diameters = 24 mm

Load carried per bolt

F = bolt spacing x load per unit width

= 36 mm x 400 N/mm

= 14400 N

= 14.4 kN

Number of bolts per row

1. Assuming bearing failure mode:

n = F / d t xc

= 14400 / (6 x 1.25 x 550)

= 3.5

so 4 bolts are required to avoid bearing failure.

Number of bolts per row

2. Assuming tension failure mode:

n = F / (p - d) t xT

= 14400 / (36 - 6) x 1.25 x 546)

= 0.7

so only 1 bolt is required to avoid tensile failure.

Check other failure modes for edge and centre bolts

3. Check first row centre bolt in shear-out:

Each bolt takes 14400 / 4 = 3600 N

Shear stress = 3600 / (2 e t) = 60 MPa

Compare with shear strength of laminate:

60 < 267 MPa, so OK.

Check other failure modes for edge and centre bolts

4. Check first row centre bolt in wedge splitting:

Transverse tensile stress

= 2 x 3600 / [(2e - d) t] = 137 MPa

Compare with transverse tensile strength of laminate:

137 < 343 MPa, so OK.

Check other failure modes for edge and centre bolts

5. Check corner bolt in tension/shear-out:

Force required to cause failure:

F = ½ t [(p - d)xT + 2 e xy]

= 18248 N

This is much greater than the actual load on this bolt (3600 N), so OK.

Other factors not included in preliminary design:

• Bypass load

• Friction effects

• Cyclic loading and laminate degradation

• Thermal and moisture effects

• Biaxial loads

• Flat-wise compression