Mechanical Behaviour Questions and Answers – Toughness


This set of Mechanical Behaviour Multiple Choice Questions & Answers (MCQs) focuses on “Toughness”.

1. The property of a material which enables it to absorb energy and deform plastically without fracture is ______
a) Stiffness
b) Toughness
c) Hardness
d) Resilience
View Answer

Answer: b
Explanation: Toughness is material’s resistance against fracture. While stiffness is its resistance to elastic deflection and hardness is resistance to indentation. Resilience is the ability of a material to absorb energy when it is deformed elastically.

2. What is the SI unit of tensile toughness?
a) N/m
c) Jm-3
d) Jm3
View Answer

Answer: c
Explanation: Toughness is calculated by area under the stress-strain curve. Its unit is stress (N/m2) × strain (%) = N/m2. Which can also be written as Jm-3.

3. What is fracture toughness range of ceramics?
a) 2-8 MN m-3/2
b) 1-10 MN m-3/2
c) 10-30 MN m-3/2
d) 30-40 MN m-3/2
View Answer

Answer: b
Explanation: fracture toughness range for ceramics is 1 to 10 MN m-3/2. Polymers have low toughness 2-8 MN m-3/2. Metals have high fracture toughness of 10-30 MN m-3/2.
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4. What is fracture toughness of magnesia in terms of stress intensity factor?
a) 1 MPa m1/2
b) 2 MPa m1/2
c) 3 MPa m1/2
d) 5 MPa m1/2
View Answer

Answer: c
Explanation: KIc of MgO is 3 MPa m1/2. High density Polyethylene and low density Polyethylene has KIc 2 and 1 MPa m1/2 respectively. Alumina has KIc 5 MPa m1/2.

5. Match the following.

Material KIc (MPa m1/2)
i) Titanium alloy p) 3-4
ii) Aluminium q) 0.2-1.4
iii) Concrete r) 44-66
iv) Silicon carbide s) 14-28

a) i-r, ii-s, iii-p, iv-q
b) i-s, ii-r, iii-q, iv-p
c) i-s, ii-r, iii-p, iv-q
d) i-r, ii-s, iii-q, iv-p
View Answer

Answer: d
Explanation: Titanium alloys have very high KIc (44-66 MPa m1/2). Aluminium alloys have KIc 14-28 MPa m1/2. Concrete has KIc 0.2-1.4 and SiC has 3-4 MPa m1/2.

6. The toughness of material depends on _____ and _____ of material.
a) Ductility, yield strength
b) Ductility, tensile strength
c) Stiffness, tensile strength
d) Hardness, yield strength
View Answer

Answer: b
Explanation: For a material to be tough, it should have a combination of good ductility and tensile strength. Materials which don’t undergo plastic deformation lacks toughness. For example, ceramics have poor toughness.

7. Which factor decreases toughness of material?
a) Alloying
b) Temperature
c) Grain refinement
d) Strain rate
View Answer

Answer: d
Explanation: As the rate of loading (strain rate) increases, toughness of material decreases. With increasing temperature ductility and toughness increase. Alloying and grain refinement also improve the toughness of a material.

8. What is the reason of low toughness of grey cast iron?
a) Strain rate
b) Notch effect
c) Grain size effect
d) Crystal structure
View Answer

Answer: b
Explanation: Grey cast iron has very poor toughness. Its major reason is notch like an effect of graphite flakes. These flakes increase stress concentration in it lowering toughness.

9. Which material, would you choose for high toughness property on lower temperatures?
a) Steel
b) Aluminium
c) Zinc
d) SiC
View Answer

Answer: b
Explanation: Aluminium has FCC structure so it doesn’t show Ductile-Brittle transition. While steel and zinc show this phenomenon. SiC is ceramic which show low toughness.

10. Stress intensity factor is not a function of _______
a) Loading
b) Crack size
c) Volume
d) Structural geometry
View Answer

Answer: c
Explanation: Stress intensity factor is a function of loading, crack size and structural geometry. The stress required to propagate crack, changes with crack size. It also changes with rate of loading.

11. Material has tensile strength 40 MPa and fracture strain 0.39. What will be approximate value of toughness?
a) 10.2 MJ/m3
b) 15.6 MJ/m3
c) 16.7 MJ/m3
d) 18.9 MJ/m3
View Answer

nswer: b
Explanation: Toughness of ductile materials is approximated as-
UT ≈ σu εf
≈ 40 × 0.39
≈ 15.6 MJ/m3
Here, σu = tensile stress, εf = fracture strain.

12. A material has tensile strength 210 MPa, yield strength 117 MPa and fracture strain 0.45. What will be approximate value of toughness?
a) 54.8 MJ/m3
b) 67.4 MJ/m3
c) 73.6 MJ/m3
d) 89.2 MJ/m3
View Answer

Answer: c
Explanation: Toughness of material may be given as-
UT ≈ ((σu + σy) εf)/2
≈((210 + 117) 0.45)/2
≈ 73.6 MJ/m3
Here, σu = tensile stress, σy = yield stress, εf = fracture strain.

13. Cast materials are tougher than wrought materials.
a) True
b) False
View Answer

Answer: b
Explanation: Cast materials have lower ductility than wrought materials. It is because cast materials are more non-uniform. That’s why cast materials have lower toughness.

14. Which statement is false?
a) A matrix of thermoplastic provides higher toughness than thermoset in the composite.
b) Microcracking can improve toughness in a composite
c) Tempering of martensite decreases toughness of steel
d) Copolymerization can improve the toughness of polymer
View Answer

Answer: c
Explanation: Tempering of martensite is done to improve the toughness of steel. Tempering reduces excess hardness by heating. It improves ductility and hence toughness.

15. The toughness of the composite depends on the toughness of the matrix.
a) True
b) False
View Answer

Answer: a
Explanation: The toughness of the composite depends on the composition of a matrix. The tougher its matrix, the tougher is the composite. That’s why the metal matrix is chosen for high toughness.

Sanfoundry Global Education & Learning Series – Mechanical Behaviour & Testing of Materials.

To practice all areas of Mechanical Behaviour & Testing of Materials, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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