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Failure modes of Bolted Joints
Citation preview
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 mmLaminate thickness (t) = 1.25 mmJoint width, or bolt spacing (w) = 25 mmEdge distance (e) = 25 mmApplied load (F) = 5000 N
1. Bearing (compression)
Compressive stress is x
c = 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
tons
bearingtensionsplittingsheartension+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/mmBolt diameter (d) = 6 mmBolt spacing (p) = 6 bolt diameters = 36 mmEdge 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)x
T + 2 e xy]
= 18248 NThis 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