In a compression test, a steel test specimen (modulus of elasticity 30 106 lb/in2 ) has a starting height 2.0 in and diameter 1.5 in. The metal yields (0.2% offset) at a load 140,000 lb. At a load of 260,000 lb, the height has been reduced to 1.6 in. Determine (a) yield strength and (b) fl ow curve parameters (strength coeffi cient and strain-hardening exponent). Assume that the cross-sectional area increases uniformly during the test.

A) σ_y = 79096 lb/in² = 79.1 ksi

B) strain-hardening exponent = 0.102

(strength coefficient = 137838.78 lb/in²

Explanation:

A) Formula for volume is;

V = πd²h/4

We are given;

height 2.0 in and diameter 1.5 in

Thus;

V = (π × 1.5² × 2)/4

V = 3.53 in³

Area is;

A = πd²/4

A = (π × 1.5²)/4

A = 1.77 in²

Yield strength is gotten from the formula;

σ_y = Force/Area

We are given load = 140,000 lb

Thus;

σ_y = 140000/1.77

σ_y = 79096 lb/in²

B) We are given

modulus of elasticity: E = 30 × 10^(6) lb/in²

Formula for strain is;

ε = σ_y/E

ε = 79096/(30 × 10^(6))

ε = 0.00264

The metal yields (0.2% offset), thus;

strain offsets = 0.00264 + 0.002

strain offsets: ε1 = 0.00464

Thus;

(h_i - h_o)/h_o = 0.00464

(h_i/h_o) - 1 = 0.00464

(h_i/h_o) = 1.00464

h_i = h_o(1.00464)

h_o = 2 in

Thus; h_i = 2(1.00464) = 2.00928 in

Area = Volume/height = 3.53/2.00928 = 1.757 in²

True stress is;

σ = force/area = 140000/1.757

σ1 = 79681.27 lb/in²

At a load of 260,000 lb, the height has been reduced to 1.6 in. Thus;

Area = 3.53/1.6 = 2.206 in²

True stress is;

σ2 = 260000/2.206

σ2 = 117860.38 lb/in²

True strain;

ε2 = In(2/1.6)

ε2 = 0.223

From flow curve;

σ = kεⁿ

Thus;

σ1 = k(ε1)ⁿ

79681.27 = k(0.00464ⁿ) - - - (eq 1)

Also for σ2 = k(ε2)ⁿ;

117860.38 = k(0.223ⁿ) - - - - - (eq 2)

From eq 1,

k = 79681.27/0.00464ⁿ

Putting this for k in eq2 to get;

117860.38 = (0.223ⁿ) × 79681.27/0.00464ⁿ

117860.38/79681.27 = 0.223ⁿ/0.00464ⁿ

Solving for n, we have ≈ 0.102

Thus,K is;

k = 79681.27/0.00464^(0.102)

k = 137838.78 lb/in²