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HW3_solution_2008

# HW3_solution_2008 - EGM 5533 HW#3 – Solution Page 1 of 6...

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Unformatted text preview: EGM 5533 HW#3 – Solution Page 1 of 6 EGM 5533 HW#3 – Solution Case B) If we take the x‐y axis oriented as shown below (assuming angle between x‐axis and strain gage ‘a’ is 45°, and strain gage ‘b’ coincides with y‐axis). The direction cosines of arms a, b, and c are, (la, ma, na) = (cos 45, sin 45, 0) = (0.7071, 0.7071, 0) (lb, mb, nb) = (cos 90, sin 90, 0) = (0, 1, 0) (lc, mc, nc) = (cos 135, sin 135, 0) = (‐0.7071, 0.7071, 0) From Eq 2.61 in text, εa = la2εxx + ma2εyy + 2lamaεxy, εb = lb2εxx + mb2εyy + 2lbmbεxy, εc = lc2εxx + mc2εyy + 2lcmcεxy, Solving these simultaneously, we get, εxx = – εb + (εa + εc)/ 2la2 = 0.0011 εyy = εb = 0.0001 εxy = (εa ‐ εc)/ 4la2 = 0.0002 Since the surface is free of applied forces, σ zz , σ yz , σ xz = 0 and σ xx = E (ε xx + υε yy ) = 120.17 MPa (1 − υ 2 ) σ yy = E (ε yy + υε xx ) = 34.33 MPa (1 − υ 2 ) σ xy = E ε xy = 17.17 MPa (1 + υ ) Page 2 of 6 EGM 5533 HW#3 – Solution Page 3 of 6 EGM 5533 HW#3 – Solution Page 4 of 6 EGM 5533 HW#3 – Solution Additional assignment: P 3.21] Comparison of Poisson’s ratio and Young modulus : If we compare cubic material stress‐strain relation given in the problem with isotropic material stress‐strain relation (Eq 3.32 in the textbook), we can derive, C1 = E (1 −ν ) (1 + ν )(1 − 2ν ) C2 = E.ν (1 + ν )(1 − 2ν ) C3 = E 2(1 + ν ) Poisson’s ratio and Young modulus can be derived from these equations based on which 2 constants we use. Table below details the comparison of using different constants, Constant 1 Constant 2 E (GPa) ν C1 C2 (C1 + 2C2 )(C1 − C2 ) = 64.71 (C1 + C2 ) C2 = 0.348 (C1 + C2 ) C1 C3 C3 (4C3 − 3C1 ) = 72.69 (C3 − C1 ) 2C3 − C1 = 0.317 2(C3 − C1 ) C3 6C2C3 + 4C32 = 73.58 2(C2 + C3 ) C2 = 0.333 2(C2 + C3 ) C2 Page 5 of 6 EGM 5533 HW#3 – Solution P 2] Strain rosette problem from HW#2 ‐ find the principal stresses using E=70GPa, and check that the principal stresses and the principal strains satisfy Hooke’s law. From the solution for HW#2, The state of strain : {εxx, εyy, εzz, εyz, εxz, εxy} = {0.001, ‐0.0006, ‐0.00017, 0, 0, 0.0006} Principal strains are : ε1= 0.0012, ε2= ‐0.00017, ε3= ‐0.0008; (2.1) (2.2) For the case of plane stress from Eq 3.32a in the text and (2.1) above we have, σ xx = E (ε xx + υε yy ) = 63.07 MPa (1 − υ 2 ) σ yy = E (ε yy + υε xx ) = ‐23.07 MPa (1 − υ 2 ) σ xy = E ε xy = 32.31 MPa (1 + υ ) σ zz , σ yz , σ xz = 0 From the state of stress, principal stresses can be calculated as : σ1 = 73.84 MPa, σ2 = 0 , σ3 = ‐33.84 MPa From principal stresses, principal strains can be calculated as : ε1 = 1 (σ 1 − υσ 2 − υσ 3 ) = 0.0012 E ε2 = 1 (σ 2 − υσ 1 − υσ 3 ) = ‐0.00017 E ε3 = 1 (σ 3 − υσ 1 − υσ 2 ) = ‐0.0008 E As these matches exactly with (2.2) above, Principal stresses and strains satisfy the Hooke’s law. Page 6 of 6 ...
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