Materials 1 MCQ Tutorial 4- Solutions PDF

Preview

Summary

Download Materials 1 MCQ Tutorial 4- Solutions PDF

Description

MEE1021/Tutorial 4 Materials 1, Tutorial 4, Solutions 1. Define what is meant by the  of a material A. B. C. D.

The point at which cracking occurs The appearance of a necked region in a tensile specimen The point at which the material undergoes permanent deformation The maximum strain that the material can withstand

2. A polymer that is well below its glass transition temperature will tend to fail in which of one the following manners: A. B. C. D.

Brittle facture Limited plasticity Cold drawing Viscous flow

3. Which of the following statements is TRUE about ceramics A. B. C. D.

They tend to fail in a ductile manner The atomic bonding tends to be Van der Waal They have low melting points They are stronger in compression than tension

4. Figure 1 depicts which of the following crystal imperfections

Figure 1 A. B. C. D.

Vacancies Solute atoms on interstitial and substitutional sites Edge dislocation Grain boundary

5. The lattice resistance of copper, like that of most FCC metals, is small. When 10% of nickel is dissolved in copper to make a solid solution, the strength of the alloy is 150 MPa. What would you expect the strength of an alloy with 20% nickel to be?

MEE1021/Tutorial 4

A. B. C. D.

300 MPa 160 MPa 212 MPa 322.25 MPa

The contribution of solid solution to the yield strength is directly proportional to dislocation shear strength:

τss ∝ C1/2 

that is, it scales as  where is the concentration of solute. Thus if 10% gives a strength of 150 MPa, 20% will give a strength of 212 MPa. 1/2

τss is directly proportional to c when the ratio or τss /c 150

=

1/2

1/2

is constant – therefore

x 1/2

20

10

x = 212 MPa

6. A metal matrix composite consists of aluminum containing hard particles of silicon carbide SiC with a mean spacing of 3 microns. The composite has a strength of 180 MPa. If a new grade of the composite with a particle spacing of 2 microns were developed, what would you expect its strength to be? A. B. C. D.

270 MPa 385 MPa 180 MPa 76 MPa

The contribution of particles to the yield strength is directly proportional to dislocation shear strength:

τptt ∝1/L that is, it scales as where is the particle spacing. If particles of spacing 3 microns gives a strength of 180 MPa, particles of spacing 2 microns will give a strength of 270 MPa. 180

=

(1/3) x = 270 MPa

x (1/2)

MEE1021/Tutorial 4

7. If the contribution of grain boundary strengthening in an alloy with grains of 0.1 microns is 20 MPa, what would you expect it to be if the grain size were reduced to 0.01 microns? A. B. C. D.

12.5 MPa 18 MPa 63 MPa 1.95 MPa

The contribution of grain boundaries to the yield strength is directly proportional to dislocation shear strength:

τgb ∝1/D1/2 

that is, it scales as  where is the grain size. If 0.1 micron grains give a contribution to strength of 20 MPa, 0.01 micron grains will give a contribution of 63 MPa. 20

= 1/2

1/(0.1 )

x 1/2

1/(0.01 )

x = 63 MPa

8. A polycarbonate polymer (yield strength 70 MPa) is blended with polyester polymer (yield strength 55 MPa) in the ratio 30%/70% respectively. If the strength of blend follows a rule of mixtures, what would you expect the yield strength of this blend to be? A. B. C. D.

130 MPa 59.5 MPa 62.5 MPa 155 MPa

Using a rule of mixtures the strength of the blend is

σ ≈ 0.3 x strength of PC + 0.7 x strength of PET = 59.5 MPa

9. Which of the following statements is FALSE regarding the work hardening of metals due to dislocation motion A. B. C. D.

The strength decreases in proportion to the increase in square root of dislocation density The strength increases due to buildDup of dislocations The strength increases in proportion to the increase in square root of dislocation density Annealing can be used to remove work hardening

MEE1021/Tutorial 4 10. As shown in Figure 2, a uniform tensile stress creates a shear stress on crystal slip planes that lie at an angle θ to the tensile axis. If the angle θ is 52° calculate the shear stress on this plane caused by an applied tensile stress of 65 MPa

Figure 2 A. B. C. D.

32.5 MPa 31.5 MPa 25 MPa 0 MPa...