Failure Analysis- Vaibhav Shah (120100015)

Ductile & Brittle Fracture-Mohit Agrawal (120040089) Question: A material has Poisson’s ratio= 0.3, Shear modulus G= 76.92 GPa.

Fracture Surface Energy ᵧs=1J/m2. Crack length= 5*10-10m.

Find Griffith’s Stress. Ans:

Therefore, E=200GPa.

a0= crack length/2

Using Griffith’s Equation,Griffith’s Stress, σGR= {2ᵧE/[πao (1-υ2)]} ½

Plugging in values,

We get σGR= 23.66 GPa

Principles of Fracture Mechanics- Parvezsh Ahamed

(120100090) A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane

strain fracture toughness of 98.9 mpa(m)^1/2and a yield strength of 860 MPa. The flaw size resolution

limit of the flaw detection apparatus is 3.0 mm. If the design stress is one half of the yield strength and

the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

Ans.:

From the geometry of the specimen, w/h = (40 mm)/(20mm) = 2.0; furthermore, the r/h ratio is(4

mm)/(20 mm) = 0.20. Using the w/h= 2.0 curve in Figure 8.2cW, the Kt value at r/h= 0.20 is 1.8. And

since Kt = σm / σo, then σm= Kt*σo = (1.8)(140 MPa) = 252 Mpa

Now it is necessary to determine how much r must be increased to reduce σm by25%;this reduction

corresponds to a stress of (0.75)(252MPa) = 189 MPa . The value of Kt is therefore 189/140=1.35. Using

the w/h= 2.0curve inFigure8.2cW, the value of r/h for Kt= 1.35 is about 0.60.Therefore, r = (0.60)h =

(0.60)(20 mm) = 12.0 mm.

Principles of Fracture Mechanics- Sanchit Dhole

(120100025) Some aircraft component is fabricated from an aluminium alloy (E=393 Gpa)that has a plane strain

fracture toughness of It has been determined that fracture results at a stress of 300 MPa (43,500 psi)

when the maximum (or critical) internal crack length is 4.0 mm (0.16 in.). For this same component and

alloy, will fracture occur at a stress level of 260 MPa (38,000 psi) when the maximum internal crack

length is 6.0 mm (0.24 in.)? Why or why not?

Ans.:

The critical stress required for the propagation of an internal crack in aluminium oxide using the below

formula taking the value of 393 GPa as the modulus of elasticity we get

σc =root(2Eγs/πa)

Factors that affect fatigue life- Rohan Maheshwari

(120110042)

Q) You are given rods of 3 different materials of same cross-secional area and length such that:

αa=αb/2=αc/2

2Ea=Eb=Ec/2 E=young’s modulus

On a temperature change from T to 2T, which one will have least fatigue life?

Ans)

S=αE(2T-T)

Now, Sa= α0E0T/2

Sb=2α0E0T

Sc=α0E0T

Sb>Sc>Sa

Thus, Sb will have least fatigue life because stress in inversely proportional to fatigue life.

Fundamentals of fracture- Ajinkya Deshpande (120110019) Q) A relatively large plate of a glass is subjected to a tensile stress of 40 MPa. If the specific surface energy and modulus of elasticity for this glass are 0.3 J/m2

and 69 GPa, respectively, determine the maximum length of a surface flaw that is possible without fracture. Ans) S= 40MPa

γs= 0.3J/m2

E=69GPa

a= 2E γs/πS2

Substituting values we get a=8.2 µm

Crack Initiation and Propogation- Rahul poonia (120100038)

Q) S-N curve for beass alloy is given

a) Determine the fatigue strength at 4 x106 cycles.

b) Determine the fatigue life for 120 MPa.

Ans)

a) As indicated by one set of dashed lines on the plot, the fatigue strength at

4 x 106 cycles [log (4 x 106) = 6.6] is about 100 MPa.

b) As noted by the other set of dashed lines, the fatigue life for 120 MPa is

about 6 x 105 cycles (i.e., the log of the lifetime is about 5.8).

Crack propogation rate – V Sarath Kumar (120100071)

Q) A mild steel plate is subjected to constant amplitude uniaxial fatigue loads to

produce stresses varying from max=180 MPa to min=-40 MPa. The static

properties of the steel are o = 500, MPa, E=207 GPa, Kc=100 MPa.m1/2.

If the plate contains an initial through thickness edge crack of 0.5mm, how

many fatigue cycles will be required to break the steel?

Assume an infinite wide plate with Y=1.12. For ferritic-stainless steels a

general correlation gives

= 6.9x10-12 ( ∆K)3 (MPa m)3

Ans. Thus, A= 6.9 x 10-12 MPam0.5, m=3.0 and r = 180-0 (since compressive

stresses are ignored and we shall neglect the small influence of mean

stress on the crack growth.

ai = 0.0005 m,

af = ()2 = 0.078m = 78mm

Nf = = [(0.078)-0.5 – (0.0005)-0.5 ]/[-157.4 x 10-6 ]

Nf = 261,000 cycles.

Crack Propogation Rate – Suraj Poonia (120100033)

Q) Consider a flat plate of some metal alloy that is to be exposed to repeated tensile-compressive

cycling in which the mean stress is 25 MPa. If: ao = 0.25 mm, ac = 5.0 mm, m = 4.0, A = 5 * 10-15, Y = 2.0,

and Nf =3.2 * 105 cycles. Estimate the maximum tensile stress to yield the fatigue life prescribed.

Ans)

Creep-Divyanshu sharma (120100047) Question 1: (a)Creep tests on a stainless steel at 550oC produced a strain of 0.12 after 300 hours when subjected to a stress of 350 MN/m2 and a strain of 0.08 after 1200 hours when stressed to 245 MN/m2. Assuming steady state creep, calculate the time to produce 0.1 % strain in a link bar of the same material when stressed to 75 MN/m2 at 550oC. Solution: At 550oC and stress of 350 MPa, the creep rate is 0.12/300 = 4x10-4 /hr

At 550oC and stress of 245 MPa, the creep rate is 0.08/1200 = 6.67x10-5 /hr The relationship between strain rate, dσ/dt, and stress, σ at a constant temperature is given by: dσ/dt = C*σ n, where C and n are constant 4 x10-4 = C x 350n ……………………….(1) 6.67x10-5 = C x 245n…………………….(2) Apply log for both eq (1) and (2), then subtract (1) and (2), then n = 5.04 log C = - 16.12 For the stress of 75 MPa, we get the strain rate as 1.67x10-7 /hr Therefore, the time needed to produce 0.1 % strain, or 0.001 strain is 0.001/1.67x10-7 = 5988 hrs. = ANS

(b) A specimen 1015 mm long of a low C-Ni alloy is to be exposed to a tensile stress of 70 MPa at 427oC. Determine its elongation after 10000 hrs. Assume that the total of both instantaneous and primary creep elongations is 1.3mm.(Take the steady state creep rate dєs/dt 4.7*10-7 h-1 at 70 MPa) Solution: the steady state creep rate dєs/dt is 4.7x 10-7 h-1 at 70 MPa. The steady state creep strain, s, therefore, is just the product of dєs/dt and time єs= dєs/dt * time = (4.7 x 10-7 h-1)(10,000 h)= 4.7 x10-3 Strain and elongation relation are Δls=lo * єs Δls =(1015 mm)(4.7 x 10-3)=4.8 mm Finally, the total elongation is just the sum of this Δls and the total of both instantaneous and primary creep elongations [i.e., 1.3 mm]. Therefore, the total elongation is 4.8 mm + 1.3 mm =6.1 mm = ANS

Topic: cyclic stress and the S-N curve

Prahlad Kumar

120100085

Q. During fatigue testing, the cyclic loading on a specimen is varied with the number of cycles as follows:

Stress amplitude S1 = 100 MPa for N< 10^5 cycles

Stress amplitude S2 = 400 MPa for 10^5 <= N < 2x10^5 cycles

Stress amplitude S3 = 300 MPa for 2x10^5 <= N < 5x10^5 cycles, and then the specimen fractures.

Assume the stress distribution as a function of number of cycles (N) to be sinusoidal with the given

Stress amplitudes and a constant mean stress to be 25 MPa. Also it is given that fatigue life of the

specimen only corresponding to S2 = 4x10^5 cycles, S3 = 5x10^5 cycles

(a) What is the stress ratio corresponding to each of the stress amplitudes? 2 marks

(b) Find the fatigue life of the specimen only corresponding to S1. 2 marks

(c) Had the fatigue limit of the specimen been 200 MPa, how would the result of part (a) been

different? 2 marks

(d) Due to which transition of the Stress amplitude (S1 to S2 or S2 to S3) during the test do you

think the result obtained in part (a) is more likely to be deviate from the real experiment

results? Explain. 1 + 3 = 4 marks

Solution:

a. stress ratio = min stress/max stress

for repeated stress cycle as this case mean stress = 25 MPa

thus stress ratios for S1, S2, S3 are -75/125, -375/425, -275/325

b. using Miner’s rule,

n1/N1 +n2/N2 +n3/N3 =1

implies N1 = around 6.67x10^5 cycles

n corresponds to no. of cycles the specimen undergoes under that stress amplitude

N corresponds to fatigue life of the specimen only corresponding to that stress amplitude

c. fatigue limit being 200 MPa implies that N1 should tend to infinity as it would be impossible

for the specimen to fracture at a stress below 200 MPa. In that case the question itself would

not be justified.

d. Transition from S1 to S2.

This can be explained based on the limitation of the Miner’s rule –

That is, it does not consider the effect of overload or high stress which may result in a compressive

residual stress. In some circumstances, cycles of low stress followed by high stress cause more

damage than would be predicted by the rule. High stress followed by low stress may have less

damage due to the presence of compressive residual stress.