note set d:

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Module 3: Fracture

note set d:

• Strength of Brittle Solids. • Probabilistic Fracture. • Examples.

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Strength of Brittle Materials:

• Simple Logic: Concrete has a Young modulus of 50 GPa. If we assume that concrete deforms elastically up to a strain of 0.001 and fractures, then the tensile strength of concrete should be 50GPa * 0.001 = 50 MPa.

C

K

a

Y







• Reality Check: The tensile strength of concrete is only 2 MPa!!!

• Puzzle: So, what’s the problem? Why is concrete so weak in tension?

• Answer: Probably, it contains flaws/cracks that cause it to fail at low stresses. To check this idea, let’s do a simple calculation:

using K_c =0.2MN/m3/2 _an_ _{=2 MPa, we find that a}

c = 5.6 cm!

conclusion: concrete is very weak in tension because it contains very large defects that form during mixing and curing.

S

tr

ess

Strain X

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Strength of Brittle Materials:

• Another Puzzle: So, why is concrete strong in compression? It is still full of large cracks after all!!

• Answer: Recall that crack propagation depends strongly on the mode of

loading!

- In compression, the crack grows in a stable manner parallel to the direction of loading.

- The compressive (crushing) strength, corresponds to the stress at which many cracks have intersected causing the

material to start falling apart. The compressive strength is given by:

C

a

K

C







1

average crack size

compressive (crushing) strength

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Strength of Brittle Materials:

• A third Puzzle: So, we are saying that the strength of a brittle material depends on the size of flaws/cracks inside it.

Consider a company which is producing 1,000,000 units/month of a component made out of ceramic. According to our arguments, the strength of each unit will depend on the size of the flaws inside it. Obviously, we can’t check each unit to find out the exact size of the defects inside it and then determine the strength. So, what can we do? How can we specify the strength of a brittle material?

W. Weibull: 1887-1979 Weibull Statistics:

• Let P_S0 be the probability that a specimen of volume V₀ will survive if a stress, , is applied to it.

• Note how P_S0 is a function of the applied stress.

- If the stress is 0, then P_S0 =1.

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Strength of Brittle Materials:

• How do we determine P_S0?

P_S0 is a function of the applied stress, P_S0() If the stress is 0, then P_S0 =1.

If the stress is very large (), then P_S0=0

To determine how P_S0() varies between these extremes we do the following

experiment:

-We choose stress levels in the range that we are interested in (e.g. 0 to 500 MPa). -For each stress level, we test N samples which have exactly the same volume, V₀. -For each stress, we determine the

probability of survival.

-Next we plot P_S0 vs.  and derive a function that relates them.

Stress used in the test

Number of samples that

survived

P_S0

500 MPa 0 0

400 MPa 4 4/N

300 MPa 10 10/N

……….

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Strength of Brittle Materials:

0 0.2 0.4 0.6 0.8 1 1.2

0 100 200 300 400 500

Stress (MPa)

PS

0

0 0.2 0.4 0.6 0.8 1 1.2

0 100 200 300 400 500

Stress (MPa)

PS

0

    

  

       

m

S P

0

0 exp _



m is called the Weibull modulus. When m is large the curve will be steep (like a step function). When m is small, the curve will be shallow.

₀ is the stress at which 1/e or 37% of the samples survive. 0

0.2 0.4 0.6 0.8 1 1.2

0 100 200 300 400 500

Stress (MPa)

PS

0

    

  

       

m

S P

0

0 exp _



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Strength of Brittle Materials:

• The weakest link argument:

What if the volume of the actual part is not the same as the volume of the samples we tested?

The probability of this chain surviving is: (P_S0)(P_S0)(P_S0)(P_S0)(P_S0)=(P_S0)n

where n is the number of links; n = V/V₀.

• Let’s define P_SV as the probability of survival of the actual sample (i.e. the chain in the above example). Then P_SV is related to P_S0 by the simple relation:

P_S0 P_S0

P_S0

0

/ 0

V V S SV

P

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• Are we there yet?

- Almost! We just have to find P_SV





























































m

SV

m

S

S SV

V V S SV

V

P

V

P

V

P

0 0

/ 0

exp

)

ln(

but

)

ln(

exp

)

ln(

)

ln(

0



Strength of Brittle Materials:

Size Effect:

•The strength of a brittle specimen depends on its size. The smaller the specimen, the stronger it is!!

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• Weibull Plots

 



 

)

ln(

)

ln(

1 ln

ln

1 ln

ln

1 ln

exp

0 0 0 0 0 0 0



m

x

m

y

P

V

P

V

P

SV m SV m SV m SV m SV



































































































































Strength of Brittle Materials:

intercept slope

y = 5.0163x - 28.586 R2 = 0.9987

-16 -14 -12 -10 -8 -6 -4 -2 0 2 4

0 1 2 3 4 5 6 7

ln()

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Strength of Brittle Materials:

Example 1:

•The 1,000,000 components that we discussed earlier are made out of a

ceramic with a Weibull modulus of 8 and a ₀ of 200 MPa.

•The company is proud of its “6-sigmas” standard which means that 99.9997% of the parts must survive.

•What is the highest stress rating that the company can put on its products?

(assume V/V₀ =1).

MPa

8 .

40 MPa

200

1 exp

999997

.

0 exp

8 0 0





























































m SV

V

P

Example 2:

•Instead of testing the actual

components, the manufacturer decided to test little coupons that are made from the same material and using the same processing method. The coupons’ volume is ¼ the volume of the actual component. If the reference stress, ₀, and Weibull modulus determined by

testing the coupons are 200 MPa and 8. Determine the stress rating of the

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• Three point bending Test:

• The most common way of measuring the tensile strength of brittle materials is using the 3-point bend test.

• We can’t use our simple equation to analyze the 3-point bend test because the stress during bending is not constant.

• Instead, we use the generalized equation:                         





dV

V P V V P m m SV m

SV _ _ 

 0 0 0 0 1 exp exp

Strength of Brittle Materials:

2 max max 3 , 2 / Now, bd FL d y

x  

     



d b                                 



m d _m m m m m d _m m SV V V m dy y d V bL dy bL V P 0 max 0 2 / 0 0 0 max 2 / 0 0

0 2( 1)

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• We could carry out a number of 4 point bending tests. The idea is very similar to the 3 point-bend test. The main

advantage is that the 4-point test gives a larger region of more or less uniform maximum stress.

• The main difference between demo 4 and what we have done in the notes is the method used to determine P_S0. The method that we described utilizes many tests at each stress level to determine P. In demo 4 we use a more efficient

statistical method:

-Let’s say we have N samples. The 4 point test is used to determine the stress at which each sample fails.

-The samples are then ranked from strongest to weakest.

-Probabilities are assigned as shown

A practical approach:

P

_S0

……….

F



(min)

1



2



i



(max)

N



2 1

1

1 



N

2 1

2

1 



N

2 1

1 



N

i

2 1

1 



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Final Word: Origin & Detection of Cracks

• The strength of brittle materials

depends on the existence of defects or flaws. What is the origin of these? and how can we detect them?

• Sources of flaws/cracks:

- Porosity (especially in castings and materials prepared by sintering).

- Sintering Cracks and agglomerates. - Quenching and welding cracks (both result from large thermal stresses).

- Radiation exposure. - Corrosion.

- Surface impact. - Fatigue.

- Deformation.

p

ro

c

es

s

in

g

S

er

vi

ce

Technique Defect

Location

Size sensitivity

SEM Surface _{>1 um}

Dye

penetrant

Surface _{25-250 um}

Ultrasonics Subsurface _{>50 um.}

Optical microscopy

Surface _{0.1 -0.5}

mm

Acoustic emission

Surface,

subsurface >0.1 mm

Radio-graphy

Subsurface _{>2% of}