Question

Question 2. (Total 18 marks) 350 350 300 300 Same plot, but with expanded x axis 250 250 200 Stress (MPa) Stress (MPa) 150 $150 100 100 0 0 0 0.02 0.04 0.06 0.12 0.14 0.16 0.18 0 0.08 0.1 Strain 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 Strain (a) Determine the % elongation, yield stress and ultimate tensile strength of the material tested above (3 marks) (b) Calculate the elastic modulus of the material tested above (2 marks) + t was (c) If a 200mm cylindrical rod of the material tested above, with radius 20mm, subjected to a tensile load of 200kN, what would the length be? (3 marks) (d) An underground wastewater steel pipe with 2mm walls carries an ammonia solution of 40 g/m3. The pipe is in contact with groundwater (assume 0 g/m3 ammonia). Determine the diffusion rate of ammonia through the pipe at 27°C. Do = 8 x 10-6 m2/s, Qe= 0.7eV Assume the system is at steady state (5 marks). (e) An aluminium alloy is to be used as a structural member and the Architect has specified a maximum applied stress of 200 MPa. The fracture toughness of the alloy is 47 MPa m12. Determine the maximum external crack length which can be permitted during the manufacturing process. Assume Y = 1 (3 marks) (f) For an applied stress of 200MPa, calculate the stress at the tip of a surface crack of length 0.7mm with tip radius 1.9 x 10-2 mm. (2 marks)-

Answer 1

Guiven. - Stress - Strain Plot- 320 250 Stress (MPa) 0.14 0.16 0 0.02 0.04 -> E strain Solin - (a) % Elongation - lo From given Stours- strain diagram - % elongation = 0.16 Ě 16% yield stores! - by drawing 0.002 strain offert on strass- strain Curve (+) 250 1 270 150 0 0.005 0.002 0.008 E 0.002 strain offert geius approximately 250 mpa story yuld stress (ayx) = 250 MPa 80 CS Scanned with CamScanner

Ultimate Tensile strength :- Mam strues 330 MPa Sut = 330 MPa (le) Elastic Moduls: - [under elastic limit E-E, E= 150-o - 30,000 MPa 0.005 - E= 30 G Pa (c) Lengthin li= 200 mm r=20mm; d = 40 mm P=200kn solun, - Р Applied stress = E Р axd? A 200x10 こ 159.154 MPa x (40)? at this o=159.154 MPa from strain-strain Curve E= 0.0051 E=0.0051 = lt - 200 lo 200 If=201102 mm CS Scanned with CamScanner

a) Given! X₂-21= 2mm = 0.002 m C2 = 40g/m3 'C,=09/m3 - J=? T = 27°C 3ook Do = 8 x 180 m2/8 Qd = 0.7 eu R= 8.62x105 D = Do e { - Old D= 8x156 1 2 1 2 7 5x300 $= -17 1.4 x 10 m2 Now J =D (-a 40 1.4x1017 Х N₂- 0.002 J= -13 2.8 x 10 %% m=8 -16 = 2.8x10 (e) Grivent T= 200 MPa Kc = 47 MPa um Y=1 solyn:- Ke=YT raa 47 = IX 200 ſta CS Scanned with So CamScanner la= 0.01757m 17.57mm

(f ) T= 200 MPa 200 MPa , a=0.7mm a=0.7 mm " ht=1.94102 mm soly. - We know - 1/2 an cm = 2a SA VO am=2x200 0.7 1.9x102 m = 2427.9 MPa Se CS ed with canner