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HW Solutions Ex2 2007

# HW Solutions Ex2 2007 - CHAPTER 4 1 Given a wire of initial...

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CHAPTER 4 1. Given a wire of initial length L o and an initial cross-section A o , a load F is applied to stretch this wire. Note that as the wire is stretched, the cross-section area will decrease because of the Poisson effect. Given that the wire material has a Young’s modulus E , Poisson ratio ν and electrical resistivity ρ and assuming completely elastic behavior, show that the resistance change Δ R is related to strain ε ( << 1) according to the following. This is the basis of strain gauges. ) 2 1 ( + = Δ o o A L R Hint: You need to determine the physical dimensions of the wire with an elastic strain of . Solution Resistance R is equal to L / A, where L is the wire length and A wire cross-section area after load application. We can write: ln R = ln + ln L ln A dR R = dL L dA A Given that L = L o (1 + ) and A = A o (1 – εν ) 2 A o (1 – 2 ), we have: dL L and dA A ≈ − 2 Therefore, dR R = + 2 , which can be written as: dR = L o A o (1 + 2 ) 2. (a) For an iron bar of length 10 cm and cross-section area of 1 cm 2 , calculate the force required in newtons (N) to produce an elongation of 0.1 mm, assuming elastic deformation and a Young’s modulus of 200 GPa. Be careful about units.

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(b) For the same iron bar, we first apply stress of 400 MPa. This produces a total strain of 0.3%. The stress is then reduced to zero. The material sustains a permanent strain because the elastic limit is exceeded. Calculate the value of this permanent strain. (c) The same iron bar discussed above has surface cracks 2 mm long. The relevant fracture toughness is 20 MPa-m 1/2 (2 × 10 7 Pa-m 1/2 ). Assume the geometric factor Y to be 1.0. What is the maximum stress this iron bar can withstand before fracture? Solution (a) First, write down the expression relating force to the extension as follows: F = σ A = E ε A = E Δ L L A where symbols have their usual meanings. Substituting E = 200 × 10 9 Pa, A = 10 -4 m 2 , L = 0.1 m, and Δ L = 10 -4 m, we have: F = 200 × 10 9 10 4 10 1 10 4 = 2 × 10 4 N (b) Given that total = elastic + plastic = E + plastic , the plastic or permanent strain is equal to: plastic = total E = 0.003 400 200 = 0.003 0.002 = 0.001 = 0.1% (c) The relevant formula is K c = Y c π a , so that the critical stress is given by: c = K c Y a = 20 × 10 6 1 2 × 10 3 m = 252 × 10 6 = 252 MPa 11. For a given alloy, experiments show that the fatigue life in a certain testing configuration is 1 × 10 6 cycles at a plastic strain amplitude of 1 × 10 -3 . Noting that the Coffin-Manson exponent β is typically between –0.5 and –1.0 and assuming validity of the Coffin-Manson relation, make a
conservative estimate of the plastic strain amplitude one can apply to have fatigue life of (a) 10 7 cycles, and (b) 10 5 cycles.

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HW Solutions Ex2 2007 - CHAPTER 4 1 Given a wire of initial...

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