12.17 Consider a material with a uniaxia! stress-strain curve of the Ramberg-Osgood form, Eq. 12.12, subjected to the st

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12 17 Consider A Material With A Uniaxia Stress Strain Curve Of The Ramberg Osgood Form Eq 12 12 Subjected To The St 1 (94.25 KiB) Viewed 45 times

12 17 Consider A Material With A Uniaxia Stress Strain Curve Of The Ramberg Osgood Form Eq 12 12 Subjected To The St 2 (7.21 KiB) Viewed 45 times

12 17 Consider A Material With A Uniaxia Stress Strain Curve Of The Ramberg Osgood Form Eq 12 12 Subjected To The St 3 (135.57 KiB) Viewed 45 times

12 17 Consider A Material With A Uniaxia Stress Strain Curve Of The Ramberg Osgood Form Eq 12 12 Subjected To The St 4 (239.94 KiB) Viewed 45 times

12 17 Consider A Material With A Uniaxia Stress Strain Curve Of The Ramberg Osgood Form Eq 12 12 Subjected To The St 5 (72.52 KiB) Viewed 45 times

12.17 Consider a material with a uniaxia! stress-strain curve of the Ramberg-Osgood form, Eq. 12.12, subjected to the state of stress Oj = 02, σ3 = ασι (-1<a<1) where 01, 02, and o3 are the principal normal stresses and a is a constant. (a) Derive an equation for the principal normal strain & as a function of 91, X, and materials constants. (b) Assume that the material is the 7075-T651 aluminum of Ex. 12.1, with Poisson's ratio v = 0.33. Plot the family of curves resulting from a = -1, -0.5,0,0.5, and 1. Continue your curves well beyond yielding where this occurs. Then comment on the trends observed.
1/n E = Ep + Ep; E = + E (6)". (12.12) H
Example 12.1 Some test data points on the monotonic stress-strain curve of 7075-T651 aluminum for uniaxial stress are given in Table E12.1. Obtain values of the constants for a stress-strain curve of the Ramberg-Osgood form, Eq. 12.12, that fits these data. Use E = 71 GPa. Table E12.1 Test Data Calculation o Ep = 8 0, MPa E E 433 451 469 487 505 7.40 x 10-3 8.95 x 10-3 1.28 x 10-2 2.29 x 10-2 4.57 x 10-2 1.301 x 10-3 2.598 x 10-3 6.194 x 10-3 1.604 x 10-2 3.859 x 10-2
Solution In addition to E as given, constants H and n for Eq. 12.12 are needed. These can be found by fitting the data to Eq. 12.11, 0 = He'. To proceed, plastic strains Ep for each data point are required. These values are calculated by subtracting elastic strains, &e =0/E, from 1000 (a) 700 H = 585.5 MPa o = Hep" 500 B v 1.0 h o, Stress, MPa 300 DV V DA hD = 0.04453 200 o Ep from data 100 DH LLLLLLLL 10-1 Ep, Plastic Strain 10-3 10-2 1 600 (b) 500 400 Data Equation o, Stress, MPa 300 200 7075-T651 AI 100 0 0.01 0.04 0.05 0.02 0.03 e, Strain Figure E12.1
total strains e, as indicated in Table E12.1. Taking logarithms of both sides of Eq. 12.11 gives logo = n log ep + log H This is a straight line on a log-log plot; that is, y = mx + b where y = logo x = log Ep b = log H Performing a linear least-squares fit on this basis gives mon=0.04453 Ans. b=2.7675, H = 10% = 585.5 MPa Ans. mn,