The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 49300 0.11 61300 0.21 What true stress is necessary to produce a true plastic strain of 0.25

64640.92 psi

Explanation:

True stress ( psi )       True strain

49300                          0.11

61300                           0.21

Determine the true stress necessary to produce a true plastic strain of 0.25

бT1 = 49300

бT2 = 61300

бT3 = ?

∈T1 = 0.11

∈T2 = 0.21

∈T3  = 0.25

note : бTi = k ∈Ti^h

∴ 49300 = k ( 0.11 )^h ( 1 )

   61300 = k ( 0.21)^h ( 2 )

solving equations 1 and 2 simultaneously

49300/61300 = ( 0.11 / 0.21 )^h

0.804 = (0.52 )^h

next step : apply logarithm

log  ( 0.804 ) = log(0.52)^h

h = log 0.804 /  log (0.52)

  =  0.33

back to equation 1

49300 = k ( 0.11 )^0.33

k = 49300 / (0.11)^0.33

 = 102138

therefore бT3 = K (0.25)^h

                        = 102138 ( 0.25 )^ 0.33  

                         = 64640.92

twist boundary

explanation:

more information

explanation: