Tensile test engineering and true values

Given data

$d_0 = 12\ \text{mm},\ L_0 = 50\ \text{mm},\ L_f = 64\ \text{mm},\ d_f = 8.4\ \text{mm}$

Initial area:

$$A_0 = \frac{\pi}{4}(12)^2 = 113.10\ \text{mm}^2$$

Area at fracture:

$$A_f = \frac{\pi}{4}(8.4)^2 = 55.42\ \text{mm}^2$$

(a) Yield strength

$$\sigma_y = \frac{P_y}{A_0} = \frac{28\,500}{113.10} = 252.0\ \text{MPa}$$

(b) Ultimate tensile strength

$$\sigma_{UTS} = \frac{P_{max}}{A_0} = \frac{42\,400}{113.10} = 374.9\ \text{MPa}$$

(c) Percentage elongation

$$\%EL = \frac{L_f - L_0}{L_0}\times 100 = \frac{64-50}{50}\times 100 = 28\%$$

(d) Percentage reduction in area

$$\%RA = \frac{A_0 - A_f}{A_0}\times 100 = \frac{113.10 - 55.42}{113.10}\times 100 = 51\%$$

(e) True fracture stress and strain

True stress uses the instantaneous (smallest) area:

$$\sigma_{f} = \frac{P_f}{A_f} = \frac{33\,000}{55.42} = 595.5\ \text{MPa}$$

True strain at fracture (volume constant, so $L/L_0 = A_0/A$):

$$\varepsilon_f = \ln\frac{A_0}{A_f} = \ln\frac{113.10}{55.42} = 0.713$$

The true fracture stress (595 MPa) is much higher than the engineering stress at fracture ($33\,000/113.1 = 291.8$ MPa) because of the large reduction of area at the neck.

Answer: $\sigma_y = 252$ MPa, UTS $= 374.9$ MPa, elongation $= 28\%$, reduction in area $= 51\%$, true fracture stress $= 595.5$ MPa, true fracture strain $= 0.713$.