HW Solutions Ex2 2007

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(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
Background image of page 2
conservative estimate of the plastic strain amplitude one can apply to have fatigue life of (a) 10 7 cycles, and (b) 10 5 cycles.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/21/2011 for the course MAT SCI 201 taught by Professor Matsci during the Spring '10 term at Northwestern.

Page1 / 11

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online