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The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 48400 0.11 60400 0.19 What true stress is necessary to produce a true plastic strain of 0.26

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1 vote

Answer:

70,900

Step-by-step explanation:

Given :

True stress (psi) _____ True strain (psi)

48400 ______________ 0.11

60400 ______________ 0.19

Using ratio simplification :

Let :

s = True stress ; t = true strain

s1 = 48400

s2 = 60400

t1 = 0.11

t2 = 0.19

True stress, s0 ; needed to produce a True plastic strain, tp = 0.26

(s0 - s1) / (s2 - s1) = (tp - t1) / (t2 - t1)

(s0 - 48400)/(60400 - 48400) = (0.26 - 0.11)/(0.19 - 0.11)

(s0 - 48400)/12000 = 0.15/0.08

Cross multiply :

0.08(s0 - 48400) = 0.15 * 12000

0.08s0 - 3872 = 1800

0.08s0 = 1800 + 3872

0.08s0 = 5672

s0 = 5762 / 0.08

s0 = 70,900

The true stress required to produce a true plastic strain of 0.26 is 70,900

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