Calculate strength coefficient:
A copper specimen of 64 mm gauge length and 12.80 mm dia. was tested in tension. Following two diameters were recorded in the plastic range of deformation.
Load = 25.75 kN, d_{1} = 12.176 mm
Load = 24.25 kN, d_{2} = 8.581 mm
Calculate strength coefficient and strain hardening exponent.
Solution
Original area of cross-section,
A_{0} = 128.6144 mm^{2}
Area of cross-section at 25.75 kN
A_{1} = 116.3802 mm^{2}
Area of cross-section at 24.25 kn
A_{2} = 57.8023 mm^{2}
∴ True stress at 25.75 kN,
True stress at 24.25 kN,
Note from Eq. (1.15) true strain ε′ = ln A_{0}
And
Thus two pairs of values are obtained. Use these values of σ′ and ε′ in Eq. (1.15), i.e. ln σ′ = ln k + n ln ε′
∴ ln 221.25 = ln k + n ln 0.0999 and
ln 419.53 = ln k + n ln 0.7998
Subtract first equation from second
6.039 - 5.399 = n (- 0.2234) - n (- 2.3036)
i.e. 0.64 = 2.0802 n
or, n = 0.64 /2.0802= 0.3077 ------------ (i)
Using this value in one of above equations
5.399 = ln k + 0.3077 (- 2.3036)
∴ ln k = 5.399 + 0.7088 = 6.1078
k = 449.35 N/mm^{2} ---------------- (ii)
∴ True stress-true strain relationship for copper is
σ′ = 449.35 ε′0.3077 ------------(iii)