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Unformatted text preview: EGM 5533 HW#2 – Solution Page 1 of 8 EGM 5533 HW#2 – Solution Page 2 of 8 EGM 5533 HW#2 – Solution Page 3 of 8 EGM 5533 HW#2 – Solution Page 4 of 8 EGM 5533 HW#2 – Solution Page 5 of 8 EGM 5533 HW#2 – Solution Page 6 of 8 EGM 5533 HW#2 – Solution Non‐text book problems – P 1] a) For a rectangular strain rosette the strains are given as : ε xx = ε a = 0.001 ε yy = ε b = ‐0.0006 1
2 ε xy = ε c − (ε a + ε b ) = 0.0008 – ½ (0.001 ‐ 0.0006) = 0.0006 The strain measurements are done on a thin walled plate. Hence it is state of plane stress, with σzz = σxz = σyz = 0. The remaining strain components can be calculated as follows: 1
σ xz = 0 2G
1
ε yz =
σ yz = 0 2G
−υ
1
−υ
(ε xx + ε yy ) … substitute eq 3.32a in ε zz = (σ zz − υσ xx − υσ yy ) = (σ xx + σ yy ) = (1 − υ )
E
E ε xz = eq. 3.30 ∴ ε zz = −0.3
(0.001 − 0.0006) = −1.71e − 4 (1 − 0.3) The state of strain : {εxx, εyy, εzz, εyz, εxz, εxy} = {0.001, 0.0006, 0.00017, 0, 0, 0.0006} b) Principal strains are calculated using Matlab as follows : ε1= 0.0012, ε2= ‐0.00017, ε3= ‐0.0008; Principal strain directions are : n1 = 0 l1 = ‐0.9487, m1 = ‐0.3162, m2 = 0, n2= 1 l2 = 0, n3 = 0 l3 = 0.3162, m3 = ‐0.9487, Page 7 of 8 EGM 5533 HW#2 – Solution P2] The nonzero stress components are σ =3Mpa, σ =‐4Mpa, τ =τ =τ =5Mpa. xx zz xz yz xy The unit normal to x‐y plane (normal in the negative z direction) has direction cosines, l = 0, m= 0, n= ‐1 N = ‐1k a) The traction vector can be calculated as, σP= σPxi + σPyj + σPzk where, σPx = l σxx + m τyx + n τzx = ‐1 (5) = ‐5 σPy = l τxy + m σyy + n τzy = ‐1 (5) = ‐5 σPz = l τxz + m τyz + n σzz = ‐1(‐4) = 4 σP = ‐5i ‐ 5j + 4k b) The shear stress on x‐y plane is given as, σPS = sqrt(σ2P ‐ σ2PN) σPN = l2σxx +m2σyy + n2σzz + 2mn τyz + 2ln τxz + 2lm τxy = (‐1)2(‐4) = ‐4 Mpa hence, σPS = sqrt(σ2P ‐ σ2PN) = sqrt((‐5)2+(‐5)2+(4)2‐(‐4)2) = sqrt(50) σPS = 7.07 Mpa c) To check whether given triads can be principal stresses, we can compare the stress invariants, principal stress values and maximum shear stress values. (7, ‐3.5, ‐3.5): not possible because the first stress invariant I1 does not match (3.5, 0, ‐4.5): not possible because it gives a maximum shear stress that is smaller than the largest shear stress in the stress matrix Page 8 of 8 ...
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 Spring '08
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 Strain

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