Micromechanics- Composite Strength 2

Question 1

For a crack tip in an isotropic system, where are the stresses?

Answer 1

They are located at the crack tip in the axial plane.
Run ahead of the crack tip in the transverse plane

Question 2

What does a transverse crack lead to?

Answer 2

Can cause debonding at the fibre/matrix interface allowing it to circumvent the fibre, leaving it intact. This is crack deflection and is an important toughening mechanism in composites

Question 3

What is the peak stress in a fibre neighbouring a failed fibre a function of?

Answer 3

The interfacial shear stress or the matrix stress

Question 4

Graphs of relative stress concentration (transmitted to neighbouring fibre) vs relatively sliding stress of relative far field matrix stress

Answer 4

Δ σ1/σf vs τ/σf: concave curves up from origin
Δ σ1/σf vs σm/σYm (matrix stress over yield stress): starts roughly horizontal at a y-intercept, curves up a bit around 0.5 and then settles to higher horizontal near 1

Question 5

What influences neighbouring stress?

Answer 5

Higher for higher ratio of fibre/matrix stiffness.
Tends to zero as shear stress tends to zero.
Increased when the distance is less.
Increased with a yielding matrix

Question 6

What do shear forces present in composites dictate?

Answer 6

That failure will occur on the orthogonal planes

Question 7

The three pairs of shear forces and how important they are

Answer 7

1- typically great resistance to fibre fracture so forces of the type τ12=τ13 are unlikely to occur.
2- in thin lamina in the 1-2 plane, stresses in the 3 direction do not occur so τ23=τ32 are less important.
3- the remaining τ21=τ31 shear forces are most important (fibres sliding along each other, not across)
See slide 10 for diagrams

Question 8

Important shear forces τ21 and τ31

Answer 8

Similar to tensile failure, stress concentrations in the matrix cause failure. Local matrix deformation (without cracking) can occur. No simple expression to predict τ21u but empirical observations suggest it is similar to τmu for ff<0.6. Higher ff impose sever restrictions on the matrix

Question 9

Shear stress concentration factor vs ff

Answer 9

Very flat until after 0.6 where it shoots up to nearly vertical

Question 10

How does compressive failure work?

Answer 10

Under axial loading in compression, fibres will tend to buckle. Lateral constraints (like matrix and other fibres) require the buckling to occur in phase. Extensive buckling leads to collapse and failure of composite but localised buckling more common. Local buckling leads to kink-bands in fibres where plastic deformation of the matrix is required, leading to plastic micro-buckling.

Question 11

How does the type of fibre influence compressive failure?

Answer 11

Brittle fibres like carbon fracture when they form kink-bands.
Ductile fibres like Kevlar don’t

Question 12

Euler buckling stress

Answer 12

The stress at which buckling occurs
σb=(π^2 E/16)(d/L)^2
Where E is YM
L is full fibre length
d is fibre diameter
Means buckling favoured for high aspect ratios (σb decreases)

Question 13

Compressive strengths of carbon, glass and Kevlar

Answer 13

Carbon and glass have similar tensile and compressive strength.
Kevlar has compressive strength about 20% of its tensile strength. In tension, covalent bonding along polymer backbone controls properties. In compression, van der Waals interactions are dominant but ultimately weaker

Question 14

Formula for compressive failure stress of a composite

Answer 14

σcu=τ(ym)/Δφ
Where τ sub ym is the shear yield stress of the matrix and Δφ is the average fibre misalignment angle in radians. Data suggest that Δφ is roughly 3°.

Question 15

Compressive failure stress vs matrix yield stress

Answer 15

When both on log scales is linear graph with positive gradient.

Question 16

What are the two failure criteria for off-axis loads?

Answer 16

Maximum stress criterion.
Tsai-Hill criterion

Question 17

For maximum stress criterion, when will failure occur?

Answer 17

When any one of the following conditions is satisfied.
σ1>=σ1u (axial failure)
σ2>=σ2u (transverse failure)
τ12>=τ12u (shear failure)
Use stress transformation matrix to convert off-axis stresses to principal axes stresses.

Question 18

Formula for magnitudes of stresses required to cause failure for uniaxial tension (maximum stress criterion)

Answer 18

In databook
σ’1u=σ1u/cos^2(φ)
σ’1u=σ2u/sin^2(φ)
σ’1u=τ12u/cos(φ)sin(φ)
Where φ is angle between uniaxial stress and the 1 direction
Note that shear strength is independent of the sign of τ12

Question 19

Composite strength vs loading angle for different types of failure and maximum stress criterion

Answer 19

Loading angle from 0 to 90. Axial failure from some y-intercept curves upwards. Shear failure a really wide and deep u shape. Transverse failure a 1/x shape not as close to the y-axis as shear failure but later crosses to go below it. Overall composite follows axial failure of a small amount until meets shear failure and follows it down, then transverse failure at roughly 25° and stays low.

Question 20

How does Thai-Hill criterion work?

Answer 20

Modified von Mises criterion. Replaces yield stresses with measured failure stresses. Accounts for transversely isotropic system (σ2u=σ3u).
(σ1/σ1u)^2+(σ2/σ2u)^2-(σ1σ2/σ1u^2)+(τ12/τ12u)^2=1
Provides a failure envelope. If the stress state (σ1, σ2, τ12) lies outside the envelope (left side of expression >=1), failure is predicted. Mode of failure no identified but can be inferred.

Question 21

How to work out uniaxial stress required for failure by Tsai-Hill criterion

Answer 21

Long formula for σ’1 in databook. φ is the angle between the stress and the 1 direcrion

Question 22

How does composite strength vs loading angle graph using Tsai-Hill compare with maximum stress criterion?

Answer 22

Very similar lines. Tsai-Hill doesn’t go horizontal at start it curves down straight away and generally stays lower but gets close to other line as angle increases. Experimental data tends to agree with Tsai-Hill, in particular the value at 30° which failed in both shear and transverse modes (for carbon in epoxy), where max stress criterion predicts only shear failure. Although practical difficulties make obtaining the failure stresses very challenging.

Question 23

How to calculate the loading angle at which failure transitions from axial to shear or from shear to transverse

Answer 23

Axial to shear equates σ1u/cos^2(φ)=τ12u/cosφsinφ.
Shear to transverse equates σ2u/sin^2(φ)=τ12u/cosφsinφ.
Given all values you need to solve for φ.
Normally in region of 4° and 25° respectively.
Can then use Tsai-Hill criterion equation to calculate failure stress at these angles (much higher for axial to shear)