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Question Bank 2 Answers Problem 1 The components of the displacement vector are u 1 = 0 , u 2 = AX 3 , u 3 = AX 2 The strain tensor becomes ε = 0 0 0 0 0 A 0 A 0 Problem 2: Assignment 2 Solutions Problem 3 H = 0 0 e 2 - 1 0 0 e 2 - e - 2 0 0 e 2 - 1 Problem 4 H = X 2 3 0 2 X 1 X 3 2 X 1 X 2 X 2 1 0 0 2 X 2 X 3 X 2 2 ε = X 2 3 X 1 X 2 X 1 X 3 X 1 X 2 X 2 1 X 2 X 3 X 1 X 3 X 2 X 3 X 2 2 Problem 5 Let v be the vector joining the two points A’ and B’ in the deformed configuration. Then, v = 32 e 1 + 13 e 2 + 29 e 3 Problem 6 The normal strain in direction v at point P is - 6 / 81. Problem 7 The normal strain in AB was calculated in solutions to Assignment 2. The increase in length of the AB is, Δ AB = k 2 2 Problem 8 Correction: Part (a) asks for elongation “per unit length” of the infinitesimal vector which is the normal strain along that direction.
Note an infinitesimal vector is a vector with a very very small or infinitesimal magnitude (such as dX 1 or dX 2 in this problem).

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