Micromechanics- Multimodal, Single Fibre and Bundle Testing

Question 1

Are short or long fibres stronger and why?

Answer 1

Shorter fibres are stronger. There is a smaller chance of having catastrophic flaw in a shorter length

Question 2

Effect of doing Weibull analysis on fewer data points

Answer 2

As number of data points N reduces the averages for m and σ* become less accurate and error bars increase. The graph of double log things vs ln(σ) fits the regression line poorer for less data

Question 3

What happens when there is more than 1 flaw type?

Answer 3

Multimodal distributions are needed

Question 4

Bimodal case

Answer 4

A combined analysis fits the data. Long equation on page 11 lecture 2 used. Need two different m and σ* values

Question 5

Problem of combining distributions

Answer 5

When compared with the original strength data there is poor fit at the low (more important) strengths

Question 6

Solution to combining distributions problem

Answer 6

For bimodal, there are two lines of best fit used on the double log vs ln(σ) graph which intersect. Split the two regions (of data that follow one line or the other) and treat them as separate distributions doing two analyses. The combined fit then is better at the lower strengths. In reality more complex analyses are required

Question 7

How does two different fibre lengths affect arithmetic mean?

Answer 7

Get two different arithmetic means one for each fibre length.
σbar1=(σ0/L1^1/m)Γ(1+1/m), same for 2
So relation between two arithmetic means is
σbar1/σbar2=(L2/L1)^1/m (databook)
Assumes m doesn’t vary with length

Question 8

Using relation between two arithmetic means for a graph

Answer 8

σbar1/σbar2=(L2/L1)^1/m start
In principle ln(σbar)=-1/m ln(L)+C
Plot graph ln(σ) vs ln(L) and gradient is -1/m
Still assumes m doesn’t vary with L
So if have mean strength at one L then can use first equation to find mean strength at other L

Question 9

How to test a single fibre

Answer 9

Test in tension. Fibres selected from two (bundle). Individually fix to a card of paper frame. Load frame in tensometer. Cut frame along horizontal dotted lines (see diagram page 16 lecture 2). Apply load. Practically very challenging

Question 10

What happens in fibre bundles if one fibre fails?

Answer 10

If a single fibre fails, the load it was bearing is redistributed through load sharing to other fibres and the matrix. Ignoring matrix gives lower limit. Subsequent failure possibilities are that adjacent fibres near initial failure if redistributed load raises above their strength, other wise fibres well away from initial failure site

Question 11

What is a bundle?

Answer 11

Multiple fibres collected together. Aka a tow

Question 12

Force applied to bundle formula

Answer 12

F=σAN
N is number of fibres that survive at that load
A is individual CSA

Question 13

Can you use bundles to determine m?

Answer 13

Yes. Plot scaled applied load against scaled bundle stresses. Slope of line joining origin to maximum of curve is exp(-1/m). So can use to find m. But gauge length is also an issue. Compatible results with single filaments of the same gauge length.

Question 14

Bundle strength formula

Answer 14

σsubB=Fmax/N0A
Nsub0 is original number of fibres
Eventually get to relation of σB/σbar in databook

Question 15

Bundle strength compared to mean fibre strength

Answer 15

Bundle strength lower than mean fibre strength (fails at weakest part) but at higher m values bundle strength tends towards mean fibre strength. Unlikely to get m above 30

Question 16

How are fibres arranged in bundles?

Answer 16

All parallel in bundles (assuming limited bending in weaving, etc). Similarly fibres in unidirectional laminae are all parallel.
Two ideal fibre packing arrangements in bundles are square or hexagonal. Real géométries tend to be non-ideal

Question 17

Terms when considering fibre packing in bundles

Answer 17

2R is distance between centres of adjacent fibres.
2r is diameter of 1 fibre
h is the fibre spacing (0 if fibres touching nearest neighbours)

Question 18

Fibre volume fraction in square array

Answer 18

CSA of 1 fibre is Af=πr2
CSA of uni cell (share from 4 centres) is A=2Rx2R=4R^2
So fibre volume fraction fsubf=(π/4)(r/R)^2
Theoretical maximum 0.785 but in practice max is 70%

Question 19

How does h for square array vary with ff?

Answer 19

h=2(R-r)
Rearrange ff equation to make R subject and sub in
Get
2((π/4ff)^1/2 -1)r

Question 20

Fibre volume fraction for hexagonal array

Answer 20

Af=πr2
CSA unit cell (rhombus) A=2rt(3)R^2
So ff=(π/2rt(3))(r/R)^2
Theoretical max is 0.907 but real max is about 70%

Question 21

How does h vary with ff for hexagonal array?

Answer 21

Start h=2(R-r)
Rearrange ff to make R subject and sub in
Get
2((π/2rt(3)ff)^1/2 -1)r

Question 22

Graph of fibre spacing vs fibre volume fraction

Answer 22

h/r vs ff.
Exponential decrease type curve but meets x axis at theoretical maxima ff for hexagonal or square.
Hexagonal curve above square by a bit