Micromechanics- Shear Lag Theory 1

Question 1

Stress transfer for a short, stiff well-bonded fibre

Answer 1

Tensile stress is transferred from the matrix to the fibre by means of interfacial shear stress

Question 2

What does higher strain in the matrix than the fibre lead to?

Answer 2

The matrix lagging behind in the vicinity of the fibre. Means the stress in the fibre will be high will be high in the centre of the fibre and low at the ends. Also means interfacial shear stress is 0 in the centre of the fibre and high at the ends

Question 3

Matrix annuli

Answer 3

Rings of matrix around a fibre

Question 4

Relating shear at some distance from centreline of fibre to interfacial shear stress

Answer 4

τ(ρ)=τi x r/ρ
Where ρ is distance from centreline
τi is the interfacial shear stress
r is the radius of the short fibre

Question 5

Difference between the displacement of the matrix at some radius R and that at the interface formula

Answer 5

(uR-ur)=(τi r/Gm)ln(R/r)
Where u is displacement of the matrix (from integration of differentials)
Gm is the shear modulus of the matrix
r is the radius of the short fibre
τi is the interfacial shear stress

Question 6

Assumptions for difference between displacement of matrix formula

Answer 6

R is assumed to be remote enough from the fibre that the matrix strain is uniform.
The ratio R/r is related to packing of fibres and therefore the fibre volume fraction ff.
Can assume hexagonal array for various packing efficiency equations

Question 7

Formula for stress in the short fibre along its length

Answer 7

σf=Efεc(1-cosh(nx/r)sech(ns))
Where Ef if YM of fibre
n has its own formula
s=L/r (half length over r) which is aspect ratio
εsubc is strain of composite
x is distance along fibre from the centre of the fibre

Question 8

What is the interfacial shear stress related to?

Answer 8

The change in fibre stress along the fibre

Question 9

Formula for interfacial shear stress along the fibre length

Answer 9

τi=(n/2)EfεcSinh(nx/r)Sech(ns)

Question 10

Formula for n

Answer 10

n=(2Em/(Ef(1+νm)ln(1/ff)))^1/2

Question 11

Variation of n with fibre volume fraction

Answer 11

Starts low. Curves up slowly getting faster close to ff=1. Although most short fibre composites don’t go above ff=0.4. Carbon fibre in epoxy lowest, glass fibre in unsaturated polyester next up, SiC fibre in glass fair bit higher. n goes from 0 to about 3

Question 12

Fibre stress vs length along fibre graph for carbon fibre in epoxy and SiC in glass and different aspect ratios

Answer 12

For a constant ff, e.g 0.3. Zero at ends. Maximum in the middle. Get more and more square as aspect ratio s increases. To low an aspect ratio means even centre of fibre not supporting the maximum it could. Aspect ratios need to be higher for carbon fibre in epoxy than for SiC in glass to have middle reach maximum. 50 enough for C but 10 enough for SiC.

Question 13

Why are there differences in stress in fibres along their length for C in epoxy and SiC in glass?

Answer 13

Carbon fibres stiffer than SiC so can support a higher stress. When fibre-p/matrix stiffness mismatch is significant (C in epoxy) higher aspect ratios are needed than when mismatch not as significant (SiC in glass)

Question 14

Interfacial shear stress vs length along fibre graphs for carbon fibre in epoxy and SiC in glass and different aspect ratios

Answer 14

Zero in middle. Higher aspect ratio means flatter for longer until near ends where sharp increase of decrease. Lower s means shorter flat section or diagonal and curves up or down further from ends. s makes little difference for C but big difference for SiC. Higher interfacial shear stress values for SiC as change in fibre stress along fibre is more pronounced

Question 15

What happens to fibre stress vs length along fibre graphs if ff increased?

Answer 15

For a given aspect ratio the maximum stress observed in the middle is higher if not maximum or sustains maximum for greater length

Question 16

What happens to interfacial shear stress vs length along fibre graphs if ff increased?

Answer 16

The increase or decrease at the ends is sharper and reaches greater values

Question 17

What is stress transfer aspect ratio?

Answer 17

st. The aspect ratio which exhibits maximum stress transfer.
Roughly 3/n

Question 18

Variation of st with fibre volume fraction

Answer 18

Starts high near 0. Curves down decreasing gradient then diagonal for a while and curves down increasing gradient a bit near 1. Graph highest for C in e, middle for glass fibre in PE, lowest for SiC in g

Question 19

What is true about the contribution of the fibre ends?

Answer 19

They are a greater contribution the smaller the fibres are

Question 20

What are the simplifications of the shear lag model?

Answer 20

It ignores stress transfer across the fibre ends which becomes significant at smaller aspect ratios. Assumes σf=0 at fibre ends and doesn’t provide a way of calculating the fibre end stress σe.

Question 21

What does modified shear lag theory use for the fibre end stress?

Answer 21

The simple average of the fibre stress at the centre of the fibre and the far field matrix stress.
σe=(σ(0’+σminf)/2

Question 22

Formula for fibre end stress in modified shear lag theory

Answer 22

σe=εcE’m
Where Em’ =(Ef(1-sech(ns))+Em)/2

Question 23

Modified solution defining stress in fibre along its length

Answer 23

σf’=εc(Ef-(Ef-Em’)cosh(nx/r)sech(ns)

Question 24

Modified expression for the interfacial shear stress along the fibre length

Answer 24

τi=(n/2)(Ef-Em’)εcSinh(nx/r)Sech(ns)

Question 25

Modified graphs for along fibre length

Answer 25

See slides 25-28 lecture 6