Mechanical Metallurgy Questions and Answers – Creep & Stress Rupture – Super Plasticity


This set of Mechanical Metallurgy Multiple Choice Questions & Answers (MCQs) focuses on “Creep & Stress Rupture – Super Plasticity”.

1. As the temperature increases in polycrystalline metal, the nature of fracture change from _____________ to __________
a) intergranular, transgranular
b) transgranular, intergranular
c) tuctile, brittle
d) creep, fatigue
View Answer

Answer: b
Explanation: At lower temperature, most of metal have brittle fracture tendency of at least some percentage of brittle nature in fracture, so the fracture travels across the grain boundary. But as the temperature increases, the fracture traversals along the grain boundary becomes intergranular.

2. The equicohesive temperature is defined as the temperature at which both the grain and grain boundary have the same strength.
a) True
b) False
View Answer

Answer: a
Explanation: At lower temperature, grains are weaker than grain boundary, and at higher temperature, the grain boundary is more fragile than grain. So the temperature after which this transition takes place is known as the equicohesive temperature.

3. The equicohesive temperature is a material property and always fixed for a given material.
a) True
b) False
View Answer

Answer: b
Explanation: The equicohesive temperature is similar to recrystallization, which depends on the material history. Decreasing the strain rate decreases the equicohesive temperature, hence the tendency of intergranular fracture increases.

4. The larger grain material will have _________ equicohesive temperature(ECT), while the fine grain material will have ____________ ECT.
a) lower, higher
b) higher, lower
c) lower, lower
d) higher, higher
View Answer

Answer: b
Explanation: The larger grain material will have less grain boundary area, so the relative amount of sliding is less, hence grain boundary remains intact at the higher temperature. The inverse of this is also true.

5. The metal with ________ stacking fault energy has higher creep resistance.
a) low
b) high
c) average
d) alternate
View Answer

Answer: a
Explanation: The metal with low stacking fault energy have higher creep resistance because the climb is the chief mechanism for cross slip in metal under creep. But the cross slip is difficult in metal with low stacking fault energy, so creep resistance is high.

6. Which of the following strengthening mechanism is most effective in increasing the creep resistance of the material?
a) Precipitation hardening
b) Strain hardening
c) Dispersion strengthening
d) Solid solution strengthening
View Answer

Answer: c
Explanation: In case of dispersion strengthening, the other particles of the high melting point are added to the matrix mechanically. So, at high temperature also, the second phase remains rigid and offers the resistance to dislocation movement and creep of the material.

7. Which of the following is preferred as solute for high-temperature application for metals?
a) Interstitial solute with low valance
b) Interstitial solute with high valance
c) Substitutional solute with high valance
d) Substitutional solute with low valance
View Answer

Answer: c
Explanation: The solid solution strengthening having high valance is most effective in improving the creep strength because the solid solution decreases the stacking fault energy of the material.

8. The Suzuki Interaction in the crystal lattice is defined as the interaction between _______
a) dislocations
b) solute atom and dislocation
c) stacking faults
d) grain boundary and solute atom
View Answer

Answer: c
Explanation: The interaction between the two stacking faults which strengthens the creep resistance of the material at high temperature is known as the Suzuki interaction.

9. Which of the following is not a high-temperature application precipitate?
a) VC
b) TiC
c) Cr23C6
d) Cu3Al
View Answer

Answer: d
Explanation: For high-temperature carbide, precipitates are most stable and provides strength up to significantly high temperature. Cu3Al is used in a structural application for building light aluminum alloy and precipitates dissolves back in matrix at high temperature.

10. The Inconel is an alloy which consists of _____________
a) Cr, Co, Mo
b) Cr, W, Ti
c) Cr, Ni, Fe
d) Ti, Al, Co
View Answer

Answer: c
Explanation: The Inconel is a commercially very important alloy having a very low coefficient of thermal expansion. It contains mainly Cr-15%, Ni- 75%, Fe-7%, and other minor alloying elements.

11. Which of the following is not a nickel-based high-temperature application alloy?
a) Astroloy
b) Inconel
c) Udimet
d) Vitallium
View Answer

Answer: d
Explanation: The Astroloy (Ni, Mo, Co), Inconel (Cr, Ni, Fe), Udimet ( Cr, Co, Ni) are nickel-based alloys with major alloying element being nickel, whereas the Vitallium is a cobalt-based alloy with major alloying element being Cr and molybdenum.

12. The high-temperature alloys are fabricated by which of the following method?
a) Rolling
b) Forging
c) Precision casting
d) Extrusion
View Answer

Answer: c
Explanation: The high-temperature application is highly alloyed, constituting so many elements that to achieve uniform composition of all the element, high precision casting is a major method. Other methods include the powder metallurgy route.

13. According to Monkman and Grant, the relationship between the minimum creep rate and the rupture time is given as ________ (where the C and K are constant, and the tr is time to rupture and ε minimum creep rate).
a) tr + Cε = K
b) tr + C log ε = K
c) log tr + C log ε = K
d) tr + Cε = log K
View Answer

Answer: c
Explanation: The log-log relationship between the tr and ε shows a straight line behaviour. So, according to this relationship, it is easier to extrapolate the data to large time or stress values.

14. For a creep strain of 1% in 1000hr, determine the strain rate in sec-1 for the given material?
a) 10-2 s-1
b) 10-5 s-1
c) 2.8*10-9 s-1
d) 10-9 s-1
View Answer

Answer: c
Explanation: 1% strain means = 1/100 in the fraction.
-> In 1/100 in 1000 hr
-> So 1/100000 in 1 hr
-> 10-5/3600 in 1 sec
10-5/3600 = 2.8*10-9 s-1.

15. The Larser-Miller parameter is used to predict the creep strength and the life of the component in long time by comparing the results obtained at elevated temperature in shorter time.
a) True
b) False
View Answer

Answer: a
Explanation: The creep test runs for years for sometimes for some materials. Hence predicting the material property becomes quite impossible in short time. So the same analysis is carried out at elevated temperature to get result in shorter time.

16. According to Larser-Miller parameter, the relation between the log of rupture time and the inverse of temperature is shown in which curve in the following figure?
a) A
b) B
c) C
d) D
View Answer

Answer: c
Explanation: According to the Laser-Miller parameter, the curve follows the path as:
-> T(ln t + C1) = P1
-> ln t + C1 = P1/T
So the relationship between the ln t and 1/T follows an inverse linear relationship. So, it will have straight line with decreasing slope.

17. The Laser miller parameter is defined as ____________
a) T(ln t+C1) = P1
b) ln T(ln t+C1) = P1
c) T(t+C1) = P1
d) T/t+C1 = P1
View Answer

Answer: a
Explanation: The Larsen Miller parameter defines the relationship between rupture time and the temperature. This relationship is given as the T(ln t+C1) = P1 where P is the Larsesn-Miller parameter and C is constant.

18. If the Larsen-Miller parameter is 50*103 at the temperature of 600 degrees. The C1 is equal to 50. Find the fatigue life of the component in hr?
a) 1400 hr
b) 500 hr
c) 1442 hr
d) 54841 hr
View Answer

Answer: c
Explanation: The Larser-Miller relationship states that:
-> T(ln t+C1) = P1
-> where T=600+273=873 k,
-> 873 (ln t + 50)=50*103
-> ln t=7.27
-> t = 1441.97 hr.

Sanfoundry Global Education & Learning Series – Mechanical Metallurgy.

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