SATHYABAMA UNIVERSITY

ME: CAD Subject Code: SPRX5002

Mechanical Behaviour of Engineering Materials

Time: 2 hours Model Test-IV Max.Marks: 100

Note: (i) For every numerical answer, clearly state the UNITS

(ii) Answer all as Assignment-IV and submit on 14-10-2014

(Test IV is on 28-09-2014)

PART AAnswer All Questions

(6x5 = 30 marks)1. a) Sketch the three modes of fracture. (3 marks)

b) State Griffith’s equation and discuss the correction made by Irwin and Orawan (2 marks)

2.a) Where in design fracture mechanics is important? (1 mark)

b) Derive the expression for G for Tearing. (4 marks)

3. a) Explain COD (2013 Full Time Qn) (3 marks) b) Explain J-contour integral (2013 Full Time Qn) (2 marks)

4. a) Where does creep property become relevant? (1 mark)b) Sketch a typical creep curve for two stresses and mark on them the relevant parameters. (2 marks)c) Sketch the creep recovery relation and mark the various components on it. (2 marks)5. a) Explain the difference between creep and relaxation. (3 marks)b) Sketch typical creep relaxation curves for two initial stresses and mark on them the relevant parameters. (2 marks)

6. a) Explain parameters affecting creep(2013 Full Time Qn) (3 marks) b) Give an expression for creep rate as a function of temperature and high temperature. (2013 Full Time Qn) ( 2 marks)

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PART B(70 marks)

7. The creep-time-stress relation for a material is given by

e(t,S) = S/E + k (S/So)6 . t

where, t = time in days. S = stress in MPa.So = 6.89 MPa, k = 5x10-12 mm/mm/day and E = 200 GPaA rod made of this material has a length of 1000 mm and is subjected to a constant axial load 106 N.. It is required that the elongation should not exceed 5.0 mm at the end of 5 years. a) Find the minimum diameter required. (10 marks)b) At two temperatures 700deg and 800 deg celcius at the same stress 100 MPa, the two creep strain rates are: 10-4 and 10-5 per sec respectively. Determine the activation energy. (2013 Full Time Qn) (10 marks)

8. The creep-time-stress relation of a material is given by

e(t,S) = S/E + k (S/So)6. t

where, t = time in days. S = stress in MPa.E = 200 GPa, So = 6.89 MPa and k = 8x10-10 mm/mm/day.

A bolt made of this material is used to keep two parts together with an initial tensile stress Si. It was maintained at a constant temperature. After 2 years, it was found that the stress has relaxed to 80% of Si.a) Find Si. (10 marks)b) For this initial stress Si, find the stress after 1 year. (5 marks)

9. In a fracture mechanics test of a strip of thickness 2 mm, the central crack of 2 mm started opening at a load of 100 N, when the displacement was 1.0 mm. When the load was increased to 150 N, the displacement was 2 mm and the crack length 4 mm. The modulus of elasticity is 10 GPa.a )Compute elastic stain energy release rate G. (10 marks)b) Compute the stress intensity factor K for 2a = 2 mm (2 marks)

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c) If the finite width of the strip is 100 mm, what will be the corrected value of K? (3 marks)

10 a) In a symmetric double cantilever beam used to measure the strain energy release rate G of a material, the thickness of the specimen B = 4 mm, depth of each limb of the cantilever D = 8 mm, the crack length a = 100 mm. The equilibrate tip force P = 100 N. For a modulus of elasticity E = 200 GPa

i) Compute G . (7 marks)ii) Compute the stress intensity factor K. (3 marks)

b)An infinite steel plate with a through thickness crack of length 2a = 20 mm is subjected to 400 MPa normal to the crack. The yield strength of the steel is 1500 MPa. Find (i) Stress intensity factor and ( 5marks) (ii) the plastic zone size. (2013 Full Time Qn) (5 marks)

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