3. The crystal structure of TiC (illustrated below) is the same as that exhibited by NaCl, with Ti atoms in the corners and faces. 0.4436 nm In this structure, the atoms are in contact along the edges of the unit cell. (a) Determine the radius of the carbon atoms in this structure assuming the size (radius) of the Ti ions is 0.745 Á. For an applied load of 1,000 N along the [211] direction, what would be the engineering strain and true strain in this crystal. The cross- sectional area of the crystal is circular and of 10-cm diameter. (b) S11 = 0.21 x 10-11 Pa-1, S12 = - 0.036 x 10-11 Pa1, S44 = 0.561 x 10-11 Pa1 %3D %D

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The crystal structure of TiC (illustrated below) is the same as that exhibited by
NaCl, with Ti atoms in the corners and faces.
3.
0.4436 nm
In this structure, the atoms are in contact along the edges of the unit cellI.
(a)
Determine the radius of the carbon atoms in this structure assuming the
size (radius) of the Ti ions is 0.745 Å.
(b)
For an applied load of 1,000 N along the [211] direction, what would be
the engineering strain and true strain in this crystal. The cross-
sectional area of the crystal is circular and of 10-cm diameter.
S11 = 0.21 x 10-11 Pa-1, S12 = - 0.036 x 10-11 Pa1, S44 = 0.561 x 10-11 Pa1
%3D
Transcribed Image Text:The crystal structure of TiC (illustrated below) is the same as that exhibited by NaCl, with Ti atoms in the corners and faces. 3. 0.4436 nm In this structure, the atoms are in contact along the edges of the unit cellI. (a) Determine the radius of the carbon atoms in this structure assuming the size (radius) of the Ti ions is 0.745 Å. (b) For an applied load of 1,000 N along the [211] direction, what would be the engineering strain and true strain in this crystal. The cross- sectional area of the crystal is circular and of 10-cm diameter. S11 = 0.21 x 10-11 Pa-1, S12 = - 0.036 x 10-11 Pa1, S44 = 0.561 x 10-11 Pa1 %3D
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