Micromechanics- Intro and Brittle Materials Flashcards
What is a composite?
A material made from two or more constituent phases
The phases have different properties
Phases remain separate within the composite
What is well defined and what varies for a brittle material?
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.
Two types of flaws
Unintentional flaws- created during manufacturing, processing or handling.
Intrinsic flaws- unavoidable and the result of the material’s properties
Weakest link theory
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.
Probability vs flaw size graph
Like the normal distribution curve
Failure stress relative to crack size
σf proportional to 1/rt(c)
Which flaws influence strength?
Smaller flaws don’t initiate failure so don’t influence strength.
Only large flaws need to be considered as they control the strength
What’s wrong with normal (Gaussian) distribution?
Doesn’t accurately characterise the large flaws/low strengths.
Method used for fibre strength
Weibull. Estimate probability of failure at known strength then extrapolate to desired failure probability to give maximum allowed stress
How do Weibull statistics of fibre failure work?
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)
Consequence of very small lengths in Weibull statistics
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
Formula for risk function
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
What is the Weibull modulus?
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.
Is the Weibull distribution symmetrical?
No
Assumptions for fibres
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
What is σ*?
σ*=σ0/L^(1/m)
Is the characteristic failure stress of a set of fibres with length L
What happens to formulae if length is a variable?
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
What happens to probability of failure vs failure stress graph as m increases?
Peak gets higher and curves more steep and narrow
Cumulative probability of failure vs failure stress graph
Like normal cumulative frequency graph. As m increases the middle part gets steeper. All lines of different m intersect at σ=σ*
Relating characteristic strength to arithmetic mean strength
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
Comparing Weibull and normal probability vs stress graphs
Weibull has higher peak slightly right. Comes back down on the right and crosses normal graph before reaching zero
How to determine Weibull modulus and characteristic strength
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
Obtaining corresponding P values to set of σ values
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*
Does the arithmetic mean from Weibull distribution match the mean from the normal distribution?
No
Cumulative probability vs failure stress for Weibull analysis compared to actual fibre strengths
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
