Q1: Answer with true or false for the following?

1. Copper alloys, aluminum alloys, thermoplastics, glasses are non-ferrous metals. F

2. Ductility, elasticity, Oxidation and density are mechanical properties. F

3. Within the elastic deformation range, if the load is released, the material will return back, like a spring, to its original size. T

4. Most materials are much stronger in compression than in tension. T

5. Torsion is a twisting force, It introduces a shear stress in the material. T

6. Fatigue is a process in which damage accumulates due to the repetitive application of loads that may be well above the yield point. F

7. Creep is distinguished from low temperature deformation by its time dependence. T

8. Fortunately the reactivity of a metal and the rate at which it corrode is not related. T

9. Corrosion takes place in these regions of high energy. T10. Engineering stress: is a load divided by actual area in the necked-

down region. F

Q2: Aluminum alloy bar tested by tensile tester, the initial gauge length of sample was 50 mm and initial diameter was 10 mm, the applied force was 32 KN. The diameter reduced after test to 9.5 mm and the length of bar increased to 60 mm. Calculate the engineering stress and true stress?

Ans:

Engineering stress=σe = F/Ao =32000/ 78.5mm2 =407.6 N/mm2

σT = 32000/70.84 = 451.7 N/mm2

Q3: A crankshaft in a diesel engine fails. Examination of the crankshaft reveals no plastic deformation. The fracture surface is smooth. In addition, several other cracks appear at other locations in the crankshaft. What type of failure mechanism would you expect?

SOLUTION

Since the crankshaft is a rotating part, the surface experiences cyclical loading. We should immediately suspect fatigue. The absence of plastic deformation supports our suspicion. Furthermore, the presence of other cracks is consistent with fatigue; the other cracks didn’t have time to grow to the size that produced catastrophic failure. Examination of the fracture surface will probably reveal beach marks or fatigue striations.

Q4. Ni-base superalloys that are used for jet turbine applications exhibit Qcreep= 320 kJ/mol and n=5.

1- What is the creep rate at 925 oC and 350 MPa if C=1.7x10-7

and R=8.314 J/mol-oC ?2- What would the creep rate be if the stress were increased by

25 MPa ?Solution:

ºƹss

= C σn

exp (- Q / RT)

ºƹss

= 1.7x10-7

3505

exp( -320,000

/8.314 x 1198 K

)

= 1.7x10-7

x 5.25x1012

x 11.1x10-15

=99.0675*10-10

= 10-8

sec-1

Increasing by 25 MPa :

εss-1

= C σn

exp (- Q / RT1

)/{εss-2

= C σn

exp (- Q /

RT2

)}

εss-1

/ εss-2

= (σ1/ σ2 )n = (350/375)5 = 0.708

10-8 sec-1/ εss-2

=0.708

εss-2

= 1.4 x 10-8 mm/mm.sec