麻省理工学院：《polymer engineering》lec 13

Relaxation modulus of a pol urthowre log(t, min) Erel(t),ps 6 56,280 -5 22.880 4,450 957 578 481 0 480

The W'iecket从 艺 G-∈ 2w如:飞→只=E、巴 Enel 七/ E叭C=k+ave

Shaper Col location Glassy and rubbery moduli E_g:=91100;Ex:=480 E_g:=91100 E厂:=480 Arrays of time and relaxation time >t:= array(1..6,[10^(-6),10^(-5),10^(-4),10^(-3),10^(-2),10^(-1)]); 100000010000010000100010010 >tau:= array(1..6,[10^(-6),10^(-5),10^(-4),10^(-3),10^(-2),10^(-1)1) 100000100010000100010010 Coefficient matrix A in Ak=B >ke:=Er;A:= array(1..6,1..6) ke:=480 A:=aay(1.6.1.6.|1) for i from 1 to 6 do for 3 from 1 to 6 do A[i, 31: = exp(-t[il/tau[3])if (evalf (A[i, 31) A inv: =evalf(map(eval, inverse(A)))i 27186.6829.127-11.831533-19.97 02.7186.6829.127-11.8315.33 A∥v 0 02.71866829.127-11.83 00027186.6829.127 0 0 027186.682 0 0 0 02.718 rhs vector b: >Er:= vector(6,[56280,22880,445C067,578,4811);

Br:=156280.22880.4450.957.578.4811 =eval(Er-ke)i B:=155800.22400.3970.477.98.1 multiply Inverse by B to get k values >k: array(1..6); k:=array(1.6. I D) k: =eval(A inv &* B)i k:=|34070.37560.8485.650.3.259.7.2.718 Correct for model undershoot undershoot: =E g-(ke+sum(k[i]','i=1.6))i k[1]: =k[1]+undershoot k1=436 > 'k final'sevalm (k)i k final=143660.37560.8485650.3.259.7.2.718 Examine and plot final model formulation >E rel: =ke+sum('k[3]'*exp(-10log t/'tau[3]'),'3'=1.6)i Ene=480+4360-1000+37560-10m0+8485-m +6503e (-10010 +259.7e -02 +2.718e plot (log10(E rel),log t=-8.0)i 4日 46 42 3.B 34 28 2 log t

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