HW2_solution_2008

# HW2_solution_2008 - EGM 5533 HW#2 – Solution Page 1 of 8...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online