# paramath

## * B8

The Relaxation function problem of an orthotropic cylinder

Co. H. Tran.
Faculty of Mathematics, University of Natural Sciences - VNU-HCM

June   06   2007

NOTE:

This worksheet demonstrates Maple's  capabilities in researching  the numerical and graphical solution of the relaxation function problem of  an orthotropic cylinder .

All rights reserved.  Copying or transmitting of this material without the permission of the authors is not allowed .

Abstract

The worksheet presents some thoughts about the plane strain problem of the viscous orthotropic composite materials cylinder under internal and external pressure with

respect to using the direct method . To compute the interior stress , from the elastic solution we use the correspondence principle  and the  inverse Laplace transform .

1. Analysis of  the composite orthotropic cylinder :

We examine an orthotropic viscoelastic composite material cylinder which has the horizontal section within limit of 2 circles : r = a , r = b ( a < b ) .

Choosing the cylindrical coordinates r , , z ( the axial z is along with the cylinder ) . The components of stress and deformation

are functions of   r , t   respectively . The two components of deformation-tensor :
and the differential equation of equilibrium :
The boundary conditions :

2. Direct method  :

The direct method is an approximate inversion technic based on  the direct relation between the time  dependence and the transformed solution . If the plot of the viscoelastic solution has small curvature when plotted  with  variables logt  then  :

(1)
where       C is Euler's constant ..
(1) is exact if        ,        is proportional to  logt  .
(1) can be rewritten  :

(2)
Note that  (2) is used when   ,        has small  curvature with respect to logt .
From the correspondence principle   we obtain  the viscoelastic solution .

(3)
(4)
(5)
(6)
The  operator moduli :
(7)

We consider the relaxation test  , in which    ,

is  a constant   at     t  =  0  (8)  .   We  have    ,    ,        (9)
By the similar way , we find  out :  (10)
Assume that the relaxation moduli  have power form :     (11)
where     are  constants .
By applying the Laplace transfom for  (11)  ,  we  obtain  the operator moduli :         (12)
with the values of Gamma function :    ;           (13)

3. Parameters - The Numerical and Graphical Solution :

 > restart; cycrstrecom:=proc(T,Gamma1,c1,P1,Q1,M1,d1) global P,Q,sigmaat1,sigmaat2,sigmabt2,sigmabt1,sigmaatisotropic,sigmabtisotropic ; local To,E,E1,M,d,j,Gamma,Gamma_form,gamma; with(inttrans):with(plottools):with(plots): print(" PARAMETERS DEFINITION : "); print( T=To,gamma=Gamma1,c=c1); print(" REPRESENTATION OF STRESS : "); sigma[theta](at):=(gamma*P*(1+c^(2*gamma))-gamma*Q*2*c^(gamma-1))/(1-c^(2*gamma)); print(sigma[theta](a)=sigma[theta](at)); sigma[theta](bt):=(gamma*P*2*c^(gamma+1)-Q*(1+c^(2*gamma)))/(1-c^(2*gamma)); print(sigma[theta](b)=sigma[theta](bt)); P:=P1;Q:=Q1; To:=T;E[rt]:=(100*(t/To)^(-0.5)+1)*E[e];E[thetat]:=(100*(t/To)^(-0.1)+1)*E[e]; print(E[r]=E[rt]); print(E[theta]=E[thetat]); print(" LAPLACE TRANSFORM OF MODULI : "); E1[rp]:=p*evalf(laplace(E[rt],t,p),3); print(E1[r]=E1[rp]); E1[thetap]:=p*evalf(laplace(E[thetat],t,p),3); print(E1[theta]=E1[thetap]); Gamma_form:=sqrt(E1[theta]/E1[r]); print(" EXPRESSION OF  : ",gamma=Gamma_form); Gamma:=evalf(sqrt(E1[thetap]/E1[rp]),5): Gamma:=simplify(Gamma); print(gamma=Gamma); sigma[theta](a):=(Gamma*P1*(1+c1^(2*Gamma))-Gamma*Q1*2*c1^(Gamma-1))/(1-c1^(2*Gamma)); sigma[theta](b):=(Gamma*P1*2*c1^(Gamma+1)-Q1*(1+c1^(2*Gamma)))/(1-c1^(2*Gamma)); print(sigma[Theta](a)=sigma[theta](a));;print(sigma[Theta](b)=sigma[theta](b)); sigma[theta](at):=(gamma*P*(1+c^(2*gamma))-gamma*Q*2*c^(gamma-1))/(1-c^(2*gamma)); print(" SUBSTITUTE ",c=c1 ,p =1/(2*t),gamma=Gamma); sigmaat1:=evalf(subs(c=(1/2),p=(1/(2*t)),gamma=Gamma,P=P1,Q=Q1,sigma[theta](a)),3); sigmaat1:=evalf(simplify(sigmaat1),2); print(sigma[Theta](a)=sigmaat1); sigmaat2:=subs(t=10^(s)*To,sigmaat1): sigmaat2:=evalf(simplify(sigmaat2),2)/P1; sigmabt1:=evalf(subs(c=(1/2),p=(1/(2*t)),gamma=Gamma,P=P1,Q=Q1,sigma[theta](b)),3); sigmabt1:=evalf(simplify(sigmabt1),2); print(sigma[Theta](b)=sigmabt1); sigmabt2:=subs(t=10^(s)*To,sigmabt1): sigmabt2:=evalf(simplify(sigmabt2),2); print(" CHANGE THE PRESENTATION OF TIME INTO  LOG(t/To) "); print(sigma[Theta](a)=sigmaat2); print(sigma[Theta](b)=sigmabt2); sigmaatisotropic:=subs(s=0,sigmaat2); sigmaatisotropic:=evalf(simplify(sigmaatisotropic),2); print(sigmaa_isotropic=sigmaatisotropic); sigmabtisotropic:=subs(s=0,sigmabt2): sigmabtisotropic:=evalf(simplify(sigmabtisotropic),2); print("   OUTPUT   DATA  "); M:=M1; d:=d1; printf("    s=log(t/To)         sigma[Theta](a)(s)/P       nn"); for j from 0 to M do printf("%10.1f                %10.4f    n", -d*(10-j), subs(s=-d*(10-j),sigmaat2)); end do; for j from 1 to M do printf("%10.1f                %10.4f    n", d*j, subs(s=d*j,sigmaat2)); end do; print(" NUMERICAL AND GRAPHICAL SOLUTION "); printf("n%s","   KET THUC BAI TOAN ONG TRU COMPOSITE DAN NHOT TRUC HUONG BANG PHUONG PHAP TRUC TIEP "); plot([sigmaat2,sigmaat2,sigmaatisotropic],s=-10..30,y=0.85..5.2,color=[grey,black,black],style=[line,point,point],thickness=1,symbol=[cross,diamond,cross],linestyle=1,axes=boxed,labels=["logt/To","sigma(a,t)/P"],legend=[`sigma(a,t)/P`,`sigma(a,t)/P`,`Isotropic solution`],title="Numerical solution"); end:

 > cycrstrecom(1, .83, 1/2, 1, 0, 30, 1);

s=log(t/To)         sigma[Theta](a)(s)/P

-10.0                    1.4286
-9.0                    1.4286
-8.0                    1.4287
-7.0                    1.4289
-6.0                    1.4294
-5.0                    1.4305
-4.0                    1.4335
-3.0                    1.4409
-2.0                    1.4595
-1.0                    1.5056
0.0                    1.6182
1.0                    1.8804
2.0                    2.4264
3.0                    3.3300
4.0                    4.3240
5.0                    4.8725
6.0                    4.8419
7.0                    4.5127
8.0                    4.1110
9.0                    3.7260
10.0                    3.3828
11.0                    3.0857
12.0                    2.8323
13.0                    2.6184
14.0                    2.4396
15.0                    2.2913
16.0                    2.1691
17.0                    2.0691
18.0                    1.9877
19.0                    1.9217
20.0                    1.8684
1.0                    1.8804
2.0                    2.4264
3.0                    3.3300
4.0                    4.3240
5.0                    4.8725
6.0                    4.8419
7.0                    4.5127
8.0                    4.1110
9.0                    3.7260
10.0                    3.3828
11.0                    3.0857
12.0                    2.8323
13.0                    2.6184
14.0                    2.4396
15.0                    2.2913
16.0                    2.1691
17.0                    2.0691
18.0                    1.9877
19.0                    1.9217
20.0                    1.8684
21.0                    1.8255
22.0                    1.7910
23.0                    1.7634
24.0                    1.7413
25.0                    1.7237
26.0                    1.7096
27.0                    1.6984
28.0                    1.6894
29.0                    1.6823
30.0                    1.6766

KET THUC BAI TOAN ONG TRU COMPOSITE DAN NHOT TRUC HUONG BANG PHUONG PHAP TRUC TIEP

REFERENCES

[1] Ngo Thanh Phong , Nguyen Thoi Trung , Nguyen Dình Hien , Ap dung
phap gan dung bien doi Laplace nguoc de giai bai toan bien dang phang trong
lieu  composite dan nhot truc huong , Tap chí phat trien KHCN , tap 7 , so 4 &
in Vietnamese ) , 2002 .

[2] R.A. Schapery , Stress Analysis of Viscoelastic Composite Materials ,
Edited by G.P.Sendeckyj ,Academic Press , Newyork –London , 1971 .

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B8.COHONGTRAN-REFUCYORT