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, d₁ = 12.176 mm

Load = 24.25 kN, d₂ = 8.581 mm

Calculate strength coefficient and strain hardening exponent.

Solution

Original area of cross-section,

A₀  = 128.6144 mm²

Area of cross-section at 25.75 kN

A₁  = 116.3802  mm²

Area of cross-section at 24.25 kn

A₂  = 57.8023 mm²

∴          True stress at 25.75 kN,

True stress at 24.25 kN,

Note from Eq. (1.15) true strain ε′ = ln A₀

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/mm²       ---------------- (ii)

∴          True stress-true strain relationship for copper is

σ′ = 449.35 ε′0.3077 ------------(iii)