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21a – Fracture and Fatigue Revision Examples

EG2101 / EG2401 March 2015 Dr Rob Thornton Lecturer in Mechanics of Materials

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Semester 1 example problems

• The following examples are similar to those covered in lectures last semester

• Parameters (material properties, geometries etc.) have been changed and only incomplete solutions and numerical answers provided:

– Shows good ‘structure’ for answers

– Gives you target values to aim for

– Always show your working

• ‘Model’ answers of questions of the same type can be found in the previous lectures identified

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Example L11a – AA7074-T8

𝐾 = 𝑌𝜎 𝜋𝑎

• A 50mm wide sample plate of 7074-T8 aluminium alloy contains a central through-crack of length 2a

• For 7074-T8: – Kc = 22.2 MN m-3/2; σy = 520MPa

1. Under an applied stress of 200 MPa, determine if the plate will fail by fracture with a crack half-length a of: – 1 mm; 5 mm; 10 mm

2. Determine the critical crack size ac below which the plate will not fracture under the applied stress, using an appropriate value of Y

3. Determine the limiting crack size ay below which the plate will fail by yielding (assume Y = 1)

σ

σ

2a

W

𝑌 = cos𝜋𝑎

𝑊

−12

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Solution L11a – AA7074-T8

1. Is Y significant? a = 10 mm; Y = ? Kc = 22.2 MN m-3/2 Is 𝐾 = 𝑌𝜎 𝜋𝑎 < 𝐾𝑐? a = 1 mm K = 11.2 MN m-3/2 Safe as K ? Kc a = 5 mm K = ? MN m-3/2 Safe? a = 10 mm K = ? MN m-3/2 Safe?

2. Setting K = ?; σ = ? MPa; Y = ? ac = ? [ans = 3.73 mm]

3. Setting K = ?; σ = ? MPa; Y = 1 ay = ? [ans = 0.58 mm]

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Example L11b – High-strength steel

• A high-strength steel has the following material properties: – E = 206 GPa; σy = 1400 MPa; Gc = 15.5 kJ m-2

• On inspection of an industrial storage tank, a thin plate made of this steel, with a width of 85mm, is found to contain an edge-crack, with length a

1. Plot a graph of stress intensities against crack lengths, assuming an applied stress of 150MPa and crack lengths of 0.5, 10, 20, 25 mm

2. Estimate the critical crack size for fracture under the applied stress [ans = 20.8 mm]

3. For a = 20mm, calculate the radius of plasticity around the crack tip under the applied stress [ans = 0.24 mm]

4. Estimate the critical crack size below which yielding will always occur before fracture [ans = 0.41 mm]

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Example L11b – Tips

1. Values for Y can be found in Ashby and Jones, Engineering Materials 1, Ch13 p198-203 (accessible online through the Library) Try plotting Y against a/W to understand how the correction factor changes

2. Estimate critical crack size for fracture using your graph

3. Try plotting these for each stress intensity as well

4. Estimate the critical crack size for yielding by plotting stress intensities against small crack sizes (<1 mm) when the applied stress is equal to the yield stress

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Example L11b - Graphs

• Note non-linear trend between Y and a/W

• Often we approximate Y by linear interpolation: – What effect does this

have on ac or Kc calculations?

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Paris Law integration (1)

• Paris’ Law describes the steady-state crack growth typically seen under cyclic loading

• Definition: d𝑎

d𝑁= 𝐴 ∆𝐾 𝑚

• We can therefore calculate fatigue life by rearranging and integrating this relationship:

– 𝑁𝑓 = d𝑁𝑁𝑓0

= d𝑎

𝐴 ∆𝐾 𝑚𝑎𝑓𝑎0

• Ignoring the fact that 𝑌 = 𝑓 𝑎 (as it is often an empirical relationship):

– 𝑁𝑓 = d𝑎

𝐴 𝑌∆𝜎 𝜋𝑎𝑚

𝑎𝑓𝑎0

=1

𝐴 𝑌∆𝜎 𝜋𝑚 𝑎

−𝑚

2𝑎𝑓𝑎0

d𝑎

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Paris Law integration (2)

• Carry out the Paris Law integration:

– 𝑁𝑓 =1

𝐴 𝑌∆𝜎 𝜋𝑚 𝑎

−𝑚

2𝑎𝑓𝑎0

d𝑎 (1)

• Show that you can obtain:

– 𝑁𝑓 =1

𝐴 𝑌∆𝜎 𝜋𝑚𝑎−𝑚 2 +1

−𝑚 2 +1

𝑎𝑓

𝑎0

– 𝑁𝑓 =1

𝐴 𝑌∆𝜎 𝜋𝑚

1

−𝑚 2 +1𝑎𝑓−𝑚

2+1 − 𝑎0

−𝑚

2+1 (2)

• This can be rearranged into a more convenient form:

– 𝑁𝑓 =1

𝐴 𝑌∆𝜎 𝜋𝑚

1

𝑚 2 −1

1

𝑎0

𝑚

2−1−

1

𝑎𝑓

𝑚

2−1

(3)

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Paris Law integration (3)

• Did you get it?

• If you didn’t, try again.

• If you did, turn over the page and do it again.

• Carry out the Paris Law integration:

– 𝑁𝑓 =1

𝐴 𝑌∆𝜎 𝜋𝑚 𝑎

−𝑚

2𝑎𝑓𝑎0

d𝑎 (1)

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Paris Law integration (4)

• How about this time?

• They say that practice makes perfect…

• …and seeing as good Engineers listen to wise words…

• …turn over the page / cover your notes and…

• …you guessed it, try again.

• Carry out the Paris Law integration:

– 𝑁𝑓 =1

𝐴 𝑌∆𝜎 𝜋𝑚 𝑎

−𝑚

2𝑎𝑓𝑎0

d𝑎 (1)

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Paris Law integration (5)

• You might be getting the idea by now that being able to do this integration is important.

• It is.

• In fact, you should aim to be able to do this in your sleep.

• Once you begin to have nightmares pleasant dreams about integrating to calculate fatigue lives, you’ll know you’re ready.

• Actually, while you’ve been reading this your memory might’ve started to fade a bit so…

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Paris Law integration (6)

• …try again.

• Carry out the Paris Law integration:

– 𝑁𝑓 =1

𝐴 𝑌∆𝜎 𝜋𝑚 𝑎

−𝑚

2𝑎𝑓𝑎0

d𝑎 (1)

• Got it now?

• Great.

• Trying not to refer to your previous 10 pages of integrations, have a go at the following example…

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Example L13 – Engine crankshaft (1)

• A large steel crankshaft (Kc = 45 MN m-3/2) undergoes cyclic tensile (225 MPa) and compressive (60 MPa) stresses during use

• Prior to use, it was inspected using ultrasonic techniques, from which the largest surface crack found was 2.5 mm in length

• For the steel in question, the Paris Law constants are:

– A = 1.5 x 10-12 m/(MN m-3/2)m per cycle

– m = 2.5

• Calculate the number of cycles to failure

Units!!!

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Example L13 – Engine crankshaft (2)

• Kc = 45MN m-3/2

σmax = 225MPa (tensile), σmin = 60MPa (compressive) a0 = 2.5mm A = 1.5 x 10-12 m/(MN m-3/2)m per cycle, m = 2.5

1) Calculate critical crack length (ac = af): 𝐾𝑐 = 𝑌𝜎 𝜋𝑎𝑐 (assume Y = 1) [ans = 12.7 mm]

2) Determine number of cycles to failure (Nf): d𝑎

d𝑁= 𝐴 ∆𝐾 𝑚

𝑁𝑓 = d𝑁𝑁𝑓0

[ans = 1.26 x 106 cycles)

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Example L14a – Variable cyclic loading (1)

• An aircraft panel made of a 0.4%C steel is subjected to a fully reversed variable cyclic stress during operation:

σ

Δσ

1

N1

Δσ

2

N2

Δσ

3

N3 • Stress amplitudes:

– Δσ1/2 = 360MPa

– Δσ2/2 = 500MPa

– Δσ3/2 = 440MPa

• Cycles:

– N1 = 5.44 x 105

– N2 = 5.35 x 103

– N3 = 3.24 x 104 𝑁𝑖𝑁𝑓𝑖𝑖

= 1

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Example L14a – Variable cyclic loading (2)

0.4%C steel

2000 series Al-Cu

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Example L14a – Variable cyclic loading (3)

Δσi /MPa Ni Nfi

𝑁𝑖𝑁𝑓𝑖%

𝑁𝑖𝑁𝑓𝑖

𝑖

%

360 5.44 x 105 ? ? ?

500 5.35 x 103 ? 26.8 ?

440 3.24 x 104 6 x 104 ? ?

Complete the table:

1. Using the S-N diagram determine the fatigue life of the 0.4%C steel under the three different stress regimes

2. Using Miner’s Rule, calculate the fractional damage caused by each stress regime to the fatigue life of the component and thus determine under which stress regime it is predicted to fail

3. How many cycles will the component sustain under the failure stress regime and therefore how many cycles in total will the component have sustained when it fails? [ans = 27,600; 577,000 ]

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Example L14b – Fatigue + Fracture

∆𝐾 = 𝑌∆𝜎 𝜋𝑎;d𝑎

d𝑁= 𝐴 ∆𝐾 𝑚; 𝑁𝑓 = d𝑁

𝑁𝑓0

• A 65.4 mm wide aircraft inspection panel is made of 7074-T651 aluminium alloy. The material properties are: – Fracture toughness, Kc = 25.8 MN m-3/2 – Yield stress, σy = 505 MPa

• Following inspection, an edge through-crack of length 6.4 mm is found. During flight, this plate is subjected to a cyclic stress of 90 ± 30 MPa.

1. Ignoring the effect of the mean stress, calculate the number of cycles to failure (Nf) using Paris’ Law and the material constants (in their standard units): A = 1.2 x 10-12 / m = 2.8 [ans = 4.33 x 106]

2. Briefly describe what effect would the mean stress have on the fatigue life of the component?

σ

σ

W

a

Y a/W

1.12 0.0

1.37 0.2

2.11 0.4

2.83 0.5

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Example L14b – Tips

• Q1: Calculate fatigue life

– The process to get to the fatigue life is much the same as Example L13…

– …except that you must use linear interpolation for the geometry factor (Y) using the values in the table provided

– In this example you can assume a constant value of Y as the crack grows, but what effect would it have if you accounted for its variation with crack length (a)?

σ

σ

W

a

Y a/W

1.12 0.0

1.37 0.2

2.11 0.4

2.83 0.5

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0

+

-

Mean stress

Example L14b – Tips

• Q2: Effect of positive mean stress? – Refer back to your

lecture notes – Sketch some typical

curves for the different mean stress values indicated to the right

0

+

ΔK

σm3 > σm2 > σm1

σa

Nf

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Exam reminder

• Two and a half hours:

– 6 questions in two sections; answer 4 in total

– Part A – Answer one question out of two

– Part B – Answer three questions out of four

• Questions in Part A do not follow style of previous years’ examples:

– New lecturer, new content – new example slides!

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Good luck!