Micromechanics- Composite Strength 1 Flashcards

1
Q

What does strength refer to?

A

Failure of materials

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2
Q

What are the two failure cases?

A

Matrix fails first
Fibres fail first

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3
Q

Assumptions fibres and matrix for long fibre composites

A

Equal strain
Perfectly elastic behaviour
Perfect interfacial bonding
Brittle failure

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4
Q

Equations for stress and YM for uniaxial loading of long fibre composite

A

σ1=ffσf+(1-ff)σm
E1=ffEf+(1-ff)Em

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5
Q

Describe the stress-strain graph for when matrix fails first

A

There are 4 diagonal lines from the origin. The top tracks the stress in the fibres (σf) and finishes at (εfu, σfu). The bottom tracks the stress in the matrix (σm) and is shallower and shorter finishing at (εmu, σmu). Second highest is for stress in composite with no failed components so ffσf+(1-ff)σm. Second lowest is stress in composite only by fibres so ffσf. Stress in composite (σ1) follows second highest line until εmu then moves horizontally to follow second lowest line until failure. Stress in fibres at εmu is σfmu (on top line)

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6
Q

When matrix fails first, what is true for strains and what is the total stress on fibres when matrix fails?

A

εmu<εfu (where u means ultimate so a failure strain or stress).
The stress on fibres when matrix fails (εmu) is σfmu so:
ffEfεmu=ffσfmu
Stress from failed matrix transferred to fibres is:
(1-ff)Emεmu=(1-ff)σmu
So when matrix fails total stress on fibres is:
ffσfmu+(1-ff)σmu
This is the value at the start of the horizontal region of the graph

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7
Q

When matrix fails, what happens if the failure stress of the fibres is higher than the current stress on the fibres?

A

So ffσfu > ffσfmu+(1-ff)σmu
Further matrix failure events (cracking) are likely
Once matrix starts to crack, little further increase in composite stress until all load transferred to fibres (horizontal region)
Typical case for ceramic matrix composites

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8
Q

The strength of the composite when matrix fails first depending on whether all load is transferred to fibres

A

If final failure before all load is transferred, composite strength is:
σ1u=ffσfmu+(1-ff)σmu
If final failure after all load transferred:
σ1u=ffσfu

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9
Q

For matrix failing first describe the graph of composite strength against fibre volume fraction

A

Steeper line from origin for fibre controlled.
Shallower line from some y-intercept for matrix controlled.
They cross at the minimum ff required for strengthening.
Composite strength follows matrix controlled until cross then fibre controlled

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10
Q

Formula for minimum fibre volume fraction required for strengthening when matrix fails first

A

ff’=σmu/(σfu+σmu+σfmu)
In databook
This is the min ff required for full transfer of load from the failed matrix to the fibres.
Lower than this means fibres don’t improve strength at all

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11
Q

Describe the stress strain graph for when fibres fail first

A

Has 3 diagonal lines from origin. Top line is fibre stress (σf) and ends first at (εfu, σfu). Bottom line is matrix stress and ends later at (εmu, σmu). σfu is still much higher than σmu. Middle line is for composite stress with no failed components so ffσf+(1-ff)σm. Composite stress follows this line until εfu then goes horizontal until εmu (doesn’t join another line). Matrix stress when fibres fail is σmfu.

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12
Q

When fibres fail first, what is true for strains and what is the total stress on matrix when fibres fail?

A

εmu>εfu
Stress on matrix when fibres fail is σfmu so:
fmEmεfu=(1-ff)σmfu
Stress from failed fibres transferred to matrix is:
ffEfεfu=ffσfu
So when fibres fail, total stress on matrix is:
ffσfu+(1-ff)σmfu

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13
Q

When fibres fail first, what happens if failure stress on matrix is higher than the current stress in the matrix?

A

So (1-ff)σmu > ffσfu+(1-ff)σmfu
Multiple fibre failure events are likely.
Once fibres start to fail, they break into progressively shorter lengths.
Typical case for polymer matrix composites

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14
Q

The strength of the composite when fibres fail first depending on length of fibre fragments

A

If fibre fragments too short for load transfer, composite strength is:
σ1u=(1-ff)σmu
If fibres still carrying some load:
σ1u=ffσfu+(1-ff)σmfu

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15
Q

For fibres failing first, describe the graph for composite strength vs fibre volume fraction

A

Two diagonal lines. Fibre controlled is positive gradient with lower y-intercept. Matrix controlled is negative gradient with higher y-intercept. They cross at the minimum ff for strengthening. Composite strength follows matrix line (down) until cross then fibre line (up).

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16
Q

Formula for minimum ff for strengthening when fibres fail first

A

ff’=(σmu-σmfu)/(σfu+σmu+σmfu)
In databook
This is the minimum ff required for full transfer of load from failed fibres to matrix. Lower than this fraction means fibres don’t improve strength at all.

17
Q

Critical ff for strengthening

A

For fibres failing first. In the composite strength vs ff graph, ff’ is at a min which is below a purely unreinforced matrix. Then need a bit more ff to get back up to this level. This is the critical fibre volume fraction (ff^crit) which is when the composite will be stronger than just unreinforced matrix. Generally a very small effect.
ff^crit=(σmu-σmfu)/(σfu-σmfu)

18
Q

Potential deviation from model behaviour

A

Yielding matrices (thermoplastics, metals).
Both matrix and fibres still carry some load after fracture.
Fibre failure isn’t defined by a single strength value (e.g a Weibull statistics).
Assumed fibres fail in isolation and don’t interact with neighbouring fibres (not true)

19
Q

What happens if fibres and matrix fail with the same failure strain?

A

Situation much more complicated. More stress is carried by fibres than matrix. Therefore fibres failing would redistribute more stress than when the matrix fails so fibres fail first case would probably dominate. Unlikely scenario due to requirements for fibre and matrix properties to be different

20
Q

What is true about composite strength transverse to fibre direction?

A

σ2u is often less than the unreinforced matrix strength σmu. A significant consequence of this is that in a laminate, transverse plies will often fail before axial plies

21
Q

Reasons for reduced strength in transverse loading (weak or strong interfacial bonding)

A

Weak interfacial bonding leads to fibre failure by matrix cracks linking up and propagating through the matrix.
Strong interfacial bonding leads to high local stresses accumulating in the matrix near the interface.
The fibres make little contribution to transverse strength, especially fibres with intrinsically poor transverse properties (e.g carbon)

22
Q

Formula for transverse composite strength and conditions for it

A

For a square array of fibres valid for ff between 0 and 0.6
σ2u=σmu(1-2(ff/π)^1/2)
In databook

23
Q

Graphs of composite transverse strength vs ff for different composites

A

Highest at 0 ff. Steep decline getting shallower until nearly linear.
C in epoxy is bit higher than glass fibres in unsaturated polyester but both pretty low (start at 80 and 60Mpa). SiC in glass is much higher (starts at 10000MPa)

24
Q

Critical fibre length for fibre failure in short fibre composites

A

Lc=σfu x r/2τic
From previous lecture

25
Q

Fibre stress vs fibre length graph for short fibre composites

A

For 2L<2Lc is a triangle below σfu
For 2L=2Lc is a triangle with top at σfu
For 2L>2Lc is trapezium with top at σfu
Although shear lag theory shows non-linear stress build-up is more realistic

26
Q

For L>Lc what is average stress along fibre length and what is composites strength in an aligned short fibre composite?

A

σf bar=σf(1-Lc/2L)
σcu=ff(1-Lc/2L)σfubar+(1-ff)σmfu

27
Q

For L

A

σf bar=Lσfu/2Lc
σcu=ff(L/2Lc)σfubar+(1-ff)σmfu

28
Q

Formula for aligned short fibre composite strength with a mixture of sub and super critical length fibres

A

Very long with sums in databook

29
Q

Failure mechanisms in short fibre composites with a typical mixture of sub and super critical length fibres

A

Fibre failure, matrix failure, composite failure.
Order depends on type of composite

30
Q

What do super-critical length fibres provide?

A

An energy absorption mechanism as they fracture into sub-critical length fibres