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Two opposite sign dislocations are separated by 100 nm in iron with grain size 4 mm. Determine the attractive force and total force on the two dislocations? Shear modulus of iron is 80 GPa, and the lattice parameter is 2.86 Ao

Two opposite sign dislocations are separated by 100 nm in iron with grain size 4 mm. Determine the attractive force and total force on the two dislocations? Shear modulus of iron is 80 GPa, and the lattice parameter is 2.86 Ao Correct Answer 312.59*10-3 N/m, 1.25*10-9 N

The attractive force between the dislocation will be: F=Gb2/2πr 80*109(b)2/2(3.14)(100*10-10) The burger vector for iron=BCC=a0/2(111), So half the distance of ao direction. Length of (111) = √3 a0 = √3*2.86*10-10 = 4.95*10-10 80*109(4.95*10-10)2/2(3.14)(100*10-10) 312.59*10-3 N/m The total force on the dislocation will be: 312.59*10-3*4*10-3 = 1.25*10-9 N.

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