HW_4_solution_2008

HW_4_solution_2008 - EGM 5533 HW#4 – Solution Page 1 of 9...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EGM 5533 HW#4 – Solution Page 1 of 9 EGM 5533 HW#4 – Solution Additional problem : For Elastic‐perfectly plastic case, stress‐strain relation is given by, σ = Eε ε ≤ εY (0.001195) (elastic stress‐strain) σ = Y ε ≥ εY (0.001195) (inelastic stress‐strain) For u = 2 mm, εBD = 0.000833, εAD = εCD = 0.000533 σBD = 211400 (0.000833) = 176.17 MPa σAD = σCD = 211400 (0.000533) = 112.75 MPa hence P = (176.17 + (1.6) 112.75) 645 = 229.98 KN For u = 4 mm, εBD = 0.001667, εAD = εCD = 0.001067 σBD = Y = 252.6 MPa σAD = σCD = 211400 (0.001067) = 225.49 MPa hence P = (252.6 + (1.6) 225.49) 645 = 395.63 KN Page 2 of 9 EGM 5533 HW#4 – Solution For u = 4.481 mm, εBD = 0.001867, εAD = εCD = 0.001195 σBD = Y = 252.6 MPa σAD = σCD = Y = 252.6 MPa hence P = (252.6 + (1.6) 252.6) 645 = 423.61 KN For u = 8 mm, εBD = 0.003333, εAD = εCD = 0.002133 σBD = Y = 252.6 MPa σAD = σCD = Y = 252.6 MPa hence P = (252.6 + (1.6) 252.6) 645 = 423.61 KN Load ‐ Deflection curve P (KN) Elastic‐perfectly plastic Elastic‐strain hardening 500 400 300 200 100 0 0 1 2 3 4 5 6 7 8 9 u (mm) Page 3 of 9 EGM 5533 HW#4 – Solution Page 4 of 9 EGM 5533 HW#4 – Solution Using strain energy criterion, the effective stress is given as σ e = σ 12 + σ 2 2 + σ 32 − 2ν (σ 1σ 2 + σ 1σ 3 + σ 2σ 3 ) = 248.58 MPa SF = Y/σe = 490/248.58 = 1.97 Page 5 of 9 EGM 5533 HW#4 – Solution 1.18 Page 6 of 9 EGM 5533 HW#4 – Solution Page 7 of 9 EGM 5533 HW#4 – Solution Page 8 of 9 EGM 5533 HW#4 – Solution Additional problem : Mohr‐Coulomb yield criterion is given as, f = σ1 ‐ σ3 + (σ1 + σ3)sinø – 2c.cosø If we impose uni‐axial tension until yield occurs (σ1 = YT, σ2 = σ3 = 0) we can derive, YT = (2c.cosø) / (1+sinø) Similarly, if we impose uni‐axial compression (σ3 = ‐YC, σ1 = σ2 = 0) we get, YC = (2c.cosø) / (1‐sinø) Rearranging Eq (1), and using the limit state (f = 0) f = 0 = σ1 + σ1sinø ‐ σ3 + σ3sinø – 2c.cosø = σ1 (1+ sinø) ‐ σ3 (1‐sinø) – 2c.cosø 2f.c.cosø = σ1 /[2c.cosø/(1+ sinø)] ‐ σ3 /[2c.cosø/(1‐sinø)] – 1 0 = σ1 /YT ‐ σ3 /YC – 1 This show, Eq (1) and Eq (4) are equivalent. (1) (2) (3) (4) Page 9 of 9 ...
View Full Document

Ask a homework question - tutors are online