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Unformatted text preview: MEEN 222502 Fall 2008 Homework #4 Due: 31 October 2008 at the beginning of class 1. Callister 6.5 This problem asks us to compute the elastic modulus of aluminum. For a square cross section, A = b 2 , where b is the edge length. Combining Equations 6.1, 6.2, and 6.5 and solving for E , leads to psi) 10 (10.4 GPa 71.2 N/m 10 71.2 m) 10 (0.43 m) 10 (16.5 m) 10 N)(125 (66,700 / / 6 2 9 3 2 3 3 2 × = × = × × × = ∆ = ∆ = = − − − l b Fl l l A F E ε σ 2. Callister 6.20 (a) This part of the problem asks that we ascertain which of the metals in Table 6.1 experience an elongation of less than 0.072 mm when subjected to a tensile stress of 50 MPa. The maximum strain that may be sustained, (using Equation 6.2) is just 4 10 4.8 mm 150 mm 0.072 − × = = ∆ = l l ε Since the stress level is given (50 MPa), using Equation 6.5 it is possible to compute the minimum modulus of elasticity which is required to yield this minimum strain. Hence GPa 104.2 10 4.8 MPa 50 4 = × = = − ε σ E Which means that those metals with moduli of elasticity greater than this value are acceptable candidatesnamely, Cu, Ni, steel, Ti and W. (b) This portion of the problem further stipulates that the maximum permissible diameter decrease is 2.3 x 103 mm when the tensile stress of 50 MPa is applied. This translates into a maximum lateral strain ε x (max) as 4 3 10 53 . 1 . 15 10 3 . 2 (max) − − × − = × − = ∆ = mm mm d d x ε But, since the specimen contracts in this lateral direction, and we are concerned that this strain be less than 1.53 x 104 , then the criterion for this part of the problem may be stipulated as −∆ d/d <1.53 x 104 . MEEN 222502 Fall 2008 Now, Poisson’s ratio is defined by Equation 6.8 as z x v ε ε − = For each of the metal alloys let us consider a possible lateral strain, ε x = ∆ d/d . Furthermore, since the deformation is elastic, then, from Equation 6.5, the longitudinal strain, ε z is equal to ε z = σ / E. Substituting these expressions for ε x and ε z into the definition of Poisson’s ratio we have E d d v z x / / σ ε ε ∆ = − = which leads to the following: E v d d σ = ∆ − Using values for ν and E found in Table 6.1 for the six metal alloys that satisfy the criterion for part (a), and for σ = 50 MPa, we are able to compute a −∆ d/d for each alloy as follows: 4 2 9 2 6 4 2 9...
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This note was uploaded on 06/24/2009 for the course MEEN 222 taught by Professor Radovic during the Fall '08 term at Texas A&M.
 Fall '08
 RADOVIC

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