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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.
S7P3A31 (Force - Pull).
Pj Problem of Interest is of type force (pull).
Formulas of interest:
Conventional (also called nominal, or engineering) stress, σ = Load/Ao
Where Ao is original cross-section area of member
Conventional(also called nominal, or engineering) strain, ε = δ/lo
Where l0 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' = loge(Ao/A') = loge(1 + ε)
(a) σ = Load/Ao ; σ' = Load/A'
So, σ' = σ(Ao/A')
ε' = loge(Ao/A') = loge(1 + ε)
So, (Ao/A') = (1 + ε) = 1.350
So, σ' = 14,815(1.350) = 20,000 psi
So, true stress = 20,000 psi
ε' = loge(1 + ε) = loge(1.350) = 0.300
So, true strain = 0.300.
(b) Ao = π(0.505/2)2.
A' = π(0.388/2)2.
So, Ao/A' = 0.5052/0.3882
ε' = loge(Ao/A') = loge(1 + ε)
So, true strain, ε' = loge((0.5052/0.3882) = 0.527.
So,conventional strain, ε = e0.527 - 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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