True Stress - True Strain

**Strings (S _{i}P_{j}A_{jk}) = S_{7}P_{3}A_{31} Base Sequence = 12735 String Sequence = 12735 - 3 - 31 **

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True Stress - True Strain

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(a) The conventional strain in a member subjected to a tensile stress of 14,815 psi is 0.350. Calculate the true stress and the true strain. Assuming constant volume.

(b) The original diameter of a tension specimen is 0.505 inches (figure 115.4). At a certain load, the diameter is found to be 0.388 inches. Calculate the true and conventional strain at this point. Assuming constant volume.

**The strings**:
S_{7}P_{3}A_{31} (Force - Pull).
**The math**:

Pj Problem of Interest is of type *force* (pull).

Formulas of interest:

Conventional (also called nominal, or engineering) stress, σ = Load/A_{o}

Where A_{o} is original cross-section area of member

Conventional(also called nominal, or engineering) strain, ε = δ/l_{o}

Where l_{0} is original length of member

True stress, σ' = Load/A'

Where A' is actual area of the cross-section corresponding to the given load.

True (also called natural) strain, ε' = log_{e}(A_{o}/A') = log_{e}(1 + ε)

(a) σ = Load/A_{o} ; σ' = Load/A'

So, σ' = σ(A_{o}/A')

ε' = log_{e}(A_{o}/A') = log_{e}(1 + ε)

So, (A_{o}/A') = (1 + ε) = 1.350

So, σ' = 14,815(1.350) = 20,000 psi

So, true stress = 20,000 psi

ε' = log_{e}(1 + ε) = log_{e}(1.350) = 0.300

So, true strain = 0.300.

(b) A_{o} = π(0.505/2)^{2}.

A' = π(0.388/2)^{2}.

So, A_{o}/A' = 0.505^{2}/0.388^{2}

ε' = log_{e}(A_{o}/A') = log_{e}(1 + ε)

So, true strain, ε' = log_{e}((0.505^{2}/0.388^{2}) = 0.527.

So,conventional strain, ε = e^{0.527} - 1 = 0.694.

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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