Micromechanics- Intro and Brittle Materials

Question 1

What is a composite?

Answer 1

A material made from two or more constituent phases
The phases have different properties
Phases remain separate within the composite

Question 2

What is well defined and what varies for a brittle material?

Answer 2

Stiffness is well defined and related to chemical and physical structure.
Not a single strength (failure stress) value. Flaws in the structure create stress concentrations. Flaw size, shape and distribution can vary widely.

Question 3

Two types of flaws

Answer 3

Unintentional flaws- created during manufacturing, processing or handling.
Intrinsic flaws- unavoidable and the result of the material’s properties

Question 4

Weakest link theory

Answer 4

Presence of flaws leads to sudden and catastrophic failure of brittle material at the weakest part of the structure, the most serious (often but not always biggest) flaw.

Question 5

Probability vs flaw size graph

Answer 5

Like the normal distribution curve

Question 6

Failure stress relative to crack size

Answer 6

σf proportional to 1/rt(c)

Question 7

Which flaws influence strength?

Answer 7

Smaller flaws don’t initiate failure so don’t influence strength.
Only large flaws need to be considered as they control the strength

Question 8

What’s wrong with normal (Gaussian) distribution?

Answer 8

Doesn’t accurately characterise the large flaws/low strengths.

Question 9

Method used for fibre strength

Answer 9

Weibull. Estimate probability of failure at known strength then extrapolate to desired failure probability to give maximum allowed stress

Question 10

How do Weibull statistics of fibre failure work?

Answer 10

Consider fibre length L as N short lengths ΔL connected in series. When one short length fails, the fibre fails. For a given stress the risk of failure per unit length (risk function) is R(σ) assuming CSA constant. So cumulative probability of failure Pf for infinitesimal length is
Pf=R(σ)ΔL. Probability of all surviving is Ps=1-Pf which is also
(1-Pf1)(1-Pf2)…(1-PfN)

Question 11

Consequence of very small lengths in Weibull statistics

Answer 11

Each 1-Pf approximates to exp(-x)=1-x form. Combine into one exponential with sum of powers. Sum of Pf=R(σ)ΔL. Integrate

Question 12

Formula for risk function

Answer 12

R(σ)=((σ-σu)/σ0)^m
Based on experimental evidence
σ sub u is stress below which there is zero probability of failure
σ sub 0 is a scaling factor
m is the Weibull modulus
Often conservatively simplified to R(σ)=(σ/σ0)^m

Question 13

What is the Weibull modulus?

Answer 13

A unitless quantity which describes the amount of scatter in the data. Usually 2-20. High value means low scatter and evenly distributed flaws. Low means high scatter and poorly distributed flaws.

Question 14

Is the Weibull distribution symmetrical?

Answer 14

No

Question 15

Assumptions for fibres

Answer 15

Essentially homogeneous
Fail in tension
Uniform diameters
Uniform distribution of flaws per unit length
Only one flaw type
Means you get the formula for Pf in the databook

Question 16

What is σ*?

Answer 16

σ*=σ0/L^(1/m)
Is the characteristic failure stress of a set of fibres with length L

Question 17

What happens to formulae if length is a variable?

Answer 17

Multiply inside of exponential by L/L0 and σ* becomes σ0.
Where σ0 is the characteristic failure strength of a set of fibres with length L0

Question 18

What happens to probability of failure vs failure stress graph as m increases?

Answer 18

Peak gets higher and curves more steep and narrow

Question 19

Cumulative probability of failure vs failure stress graph

Answer 19

Like normal cumulative frequency graph. As m increases the middle part gets steeper. All lines of different m intersect at σ=σ*

Question 20

Relating characteristic strength to arithmetic mean strength

Answer 20

Relate σ* to σ bar. Via a gamma function Γ(z)
See formula in databook for σ bar relative for σ*
Gamma function related to the generalised factorial function
Γ(z)=(z-1)!
Can use gamma function lookup table to find it for given z
z in this case is 1+1/m

Question 21

Comparing Weibull and normal probability vs stress graphs

Answer 21

Weibull has higher peak slightly right. Comes back down on the right and crosses normal graph before reaching zero

Question 22

How to determine Weibull modulus and characteristic strength

Answer 22

Start with Pf=1-exp(-(σ/σ)^m). Rearrange and take logs twice to get linear function
lnln(1/1-P)=mlnσ-mlnσ
Plot left vs lnσ on graph and gradient is m and y intercept is -mlnσ*.
If L variable then y intercept changes to add lnL-lnL0

Question 23

Obtaining corresponding P values to set of σ values

Answer 23

Obtain P values on basis of rank. Place N strength data in ascending order. Assign rank r from r=1 to N for each fibre. Calculate failure probability for each rank (fibre) using P=(r-0.5)/N. Then can plot graph and find m and σ*

Question 24

Does the arithmetic mean from Weibull distribution match the mean from the normal distribution?

Answer 24

No

Question 25

Cumulative probability vs failure stress for Weibull analysis compared to actual fibre strengths

Answer 25

Get this graph using the determined m and σ* in the Weibull equation
Pf=1-exp(-(σ/σ*)^m)
Both standard cumulative frequency shape. Excellent agreement at low strengths. Weibull underestimates cumulative probability at higher strengths but this doesn’t matter