# Mechanical Metallurgy Questions and Answers – Strengthening Mechanisms – Grain Boundaries & Deformation – 2

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This set of Mechanical Metallurgy Questions and Answers for Experienced people focuses on “Strengthening Mechanisms – Grain Boundaries & Deformation – 2”.

1. Considering Hall-Patch relationship, the Red-colored marked region is called __________ a) yield strength
b) critical diameter
c) friction stress
d) hall-patch constant

Explanation: According to Hall-Patch relation:
-> σoi+kD-1/2
The intercept on the y-axis will be equal to = σi
This σi is physically interpreted as the yield strength when single grain exist. It is called lattice friction stress.

2. The Li model to predict the yield strength is based on dislocation density, which states that:
-> σoi+αGρ-1/2
Where α is constant, and ρ is dislocation density.
a) True
b) False

Explanation: The Hall-Patch relationship predicts the yield strength of material is almost equal to theoretical strength. When the grain diameter is minimal (4 nm) it is not correct. So to address this indiscreetly in strength prediction, the Li model of dislocation density was proposed.

3. The yield strength of a material is 200MPa when the grain size is 80 micron. If the frictional stress is 80MPa, find the strength when the grain diameter is 160 micron?
a) 100 MPa
b) 150MPa
c) 250MPa
d) 180 MPa

Explanation: According to the hall-patch relationship:
σoi+kD-1/2
-> 200=80+k(80)-1/2
-> k=1252
So at D=160
σo=80+1252(160)-1/2
σo=178.9 MPa.

4. The mean intercept method of grain size measurement ______________
a) overestimates the grain size
b) underestimates the grain size
c) predicts the accurate grain size
d) no defined relation

Explanation: Mean intercept method measure the grain size by counting the number of grain boundaries/grain intersecting a line along a certain direction. This method does not give an idea about the shape of grain and underestimate the grain size.

5. ASTM Grain size measurement method counts the __________________
a) number of grain in 1 mm2 at magnification of 1X
b) number of grain in 1 mm2 at magnification of 100X
c) number of grain in 100 mm2 at magnification of 1X
d) number of grain in 100 mm2 at a magnification of 100X

Explanation: ASTM Grain size measurement G method counts the number of grain in 1 mm2 at a magnification of 1X=na :
Grain size is given by:
-> G=-2.9542+1.4427 ln na.

6. As the ASTM grain size number increases, the grain diameter will ______________
a) increase
b) decrease
c) remain constant
d) increase first, then decrease

Explanation: The ASTM grain size number is given by:
-> G=-2.9542+1.4427 ln na
So as the value of G increases, na will also increase (number of grain in per mm square increase). So, the grain size will decrease.

7. As the ASTM grain size number increases, the number of grain per unit volume will ______________
a) increase
b) decrease
c) remain constant
d) increase first, then decrease

Explanation: The ASTM grain size number is given by:
-> G=-2.9542+1.4427 ln na
So as the value of G increases, na will also increase (number of grain in per mm square increase). So the number of grain per unit volume will also increase.

8. Determine the grain diameter if ASTM grain size number is 5?
a) 9.32*10-5 m
b) 6.32*10-3 m
c) 44.7*10-6 m
d) 44.7*10-8 m

Explanation: The ASTM grain size number is given by:
-> G=-2.9542+1.4427 ln na
G=5
So, 5=-2.9542+1.4427 ln na
na=exp{(5+2.95)/1.44}
na=249.84
So at 1mm square in 1X, there is 249.84 grain. So in the 1meter square, it will be 249.84*106
Grain diameter is roughly related to the number of grain as:
-> D=$$\sqrt{(1/n_a)}$$
D=$$\sqrt{(1/249.84*10^6)}$$
D= 6.32*10-5 m.

9. If there are 40 grains in 1 mm2 at 100X, how many grains were there in the actual grain of 1 mm2 of the surface?
a) 200
b) 200000
c) 0.4
d) 400000

Explanation: 40 grains at 100X in 1mm2
So, at 1X (the actual size of the sample) the number of grain will be;
40*100*100=400000.

10. If there are 10000 grains in 1 mm2 at 50X, how many grains will be there in the same area at 100X?
a) 20000
b) 5000
c) 2500
d) 100000

Explanation: 10000 grains -> at 50X
So at 1X, it will be 10000*50*50
And at 100X it will be -> 10000*50*50/100*100=2500.

11. What is the equilibrium spacing between the edge dislocation along a low angle grain boundary of 1.5-degree misorientation? (Given that the burger’s vector of dislocation is equal to 3 A°).
a) 10 A°
b) 1550 A°
c) 30 A°
d) 115 A°

Explanation: The misorientation, along with the grain boundary, is given in terms of θ:
-> θ=b/D
-> where D is equilibrium spacing
So, (1.5*3.14)/180= 3*10-10/D
-> D= 115.38*10-10
-> 115 A°.

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