Question

5. The following data were obtained during the true-stress-true-strain test of a nickel specimen: Diameter, in Load, Diameter, in 3,440 3,580 3,670 3,710 3,720 0.252 0.250 0.245 0.240 0.235 0.230 3.570 3,500 3.350 3,150 2,950 2.800 0.201 0.200 0.190 0.180 0.170 0.149 (a) Plot the true-stess-true-strain curve. (b) For the data beyond the maximum load correct for necking by means of Bridgman's correction (c) Determine the following: (1) true stress at maximum load: (2) true fracture strain: (3) true uniform strain

Answer 1

True stress is given by, $\sigma$ = F / A

where, A = cross-section area = $\pi$ (d/2)²

True strain is given by, $\epsilon$ = ln (d₀ / d_f)²

(a) To plot the true stress vs. true strain curve which can be shown below as :

Load (lb) Load (kN) Diameter (inch) Diameter (mm) Area (mm^2) True Stress (N/mm^2) True Strain 6.4 3440 3580 3670 3710 3720 3570 3500 3350 3150 2950 2800 15.3 15.92 16.32 16.5 16.54 15.88 15.56 14.9 14.01 13.12 12.45 0.252 0.25 0.245 0.24 0.235 0.23 0.201 0.2 0.19 0.18 0.17 0.149 6.35 6.223 6.096 5.969 5.842 5.105 5.08 4.826 4.572 4.318 3.784 32.1 31.6 30.3 29.1 27.9 26.7 20.4 20.2 18.2 0 484.1 525.4 560.8 591.3 619.4 778.4 770.2 818.6 854.2 898.6 1111.6 0.0156 0.056 0.097 0.139 0.182 0.452 0.461 0.564 0.672 0.787 1.051 16.4 14.6 11.2

(c) Determine the following :

i. True stress at maximum load which will be given by -

$\sigma$_max = F_max / A

where, F_max = maximum load = 3720 lb = 16547.3 N

A = cross-section area = $\pi$ (d/2)² = $\pi$ [(0.230 in) / 2]²

then, we get

$\sigma$_max = (16547.3 N) / {(3.14) [(0.005842 m) / (2)]²}

$\sigma$_max = [(16547.3 N) / (2.67 x 10^-5 m²)]

$\sigma$_max = 6.197 x 10⁸ Pa

$\sigma$_max = 619.7 MPa

ii. True fracture strain which will be given by -

$\epsilon$_fracture = ln (d₀ / d_f)²

where, d₀ = initial diameter = 0.252 in = 0.0064008 m

d_f = final diameter = 0.149 = 0.0037846 m

then, we get

$\epsilon$_fracture = ln [(0.0064008 m) / (0.0037846 m)]²

$\epsilon$_fracture = ln (2.8604)

$\epsilon$_fracture = 1.051

iii. True unifrom strain which will be given by -

$\epsilon$_uniform = ln (d₀ / d_f)²

where, d₀ = initial diameter = 0.252 in = 0.0064008 m

d_f = final diameter = 0.230 = 0.005842 m

then, we get

$\epsilon$_uniform = ln [(0.0064008 m) / (0.005842 m)]²

$\epsilon$_uniform = ln (1.2004)

$\epsilon$_uniform= 0.182