A sample of gage length 2 inches has received an elongation to 2.2 inches in tension and compressed to 1.818 inches in compression. The engineering stress in tension was 6 Mpsi and 6.2 Mpsi in compression. The area of cross section initially is 2 square inches. Requried:
a. The true stress in tension is:
b. The true stress in compression is:
c. The true strains in both tension and compression are :

a. 6.6 Mpsi

b. 5.64 Mpsi

c. i. 0.095  ii. -0.095

Explanation:

a. The true stress in tension is:

The true stress in tension, σ = σ'(1 + ε) where σ' = engineering stress in tension = 6 Mpsi and ε = engineering strain under tension = (L₁ - L₀)/L₀ where L₀ = gage length = 2 inches and L₁ = length at elongation under tension = 2.2 inches. So,  ε = (L₁ - L₀)/L₀ = (2.2 - 2)/2 = 0.2/2 = 0.1

So, σ = σ'(1 + ε)

σ = 6 Mpsi(1 + 0.1)

σ = 6 Mpsi(1.1)

σ = 6.6 Mpsi

b. The true stress in compression is:

The true stress in tension, σ₁ = σ"(1 + ε) where σ' = engineering stress in tension = 6 Mpsi and ε' = engineering strain under tension = (L₂ - L₀)/L₀ where L₀ = gage length = 2 inches and L₂ = length at elongation under compression = 1.818 inches. So,  ε' = (L₂ - L₀)/L₀ = (1.818 - 2)/2 = -0.182/2 = -0.091

So, σ₁ = σ'(1 + ε')

σ₁ = 6.2 Mpsi[1 + (-0.091)]

σ₁ = 6.2 Mpsi[1 - 0.091)

σ₁ = 6.2 Mpsi(0.909)

σ₁ = 5.64 Mpsi

c. The true strains in both tension and compression are :

i. The true strain in tension

The true strain in tension ε" = ㏑(1 + ε) where ε = engineering strain in tension = 0.1

So, ε" = ㏑(1 + ε)

ε" = ㏑(1 + 0.1)

ε" = ㏑(1.1)

ε" = 0.095

ii. The true strain in compression

The true strain in tension ε₁ = ㏑(1 + ε') where 'ε = engineering strain in compression = -0.091

So, ε₁ = ㏑(1 + ε')

ε₁ = ㏑[1 + (-0.091)]

ε₁ = ㏑(1 - 0.091)

ε₁ = ㏑(0.909)

ε₁ = -0.095

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explanation:

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