MATHEMATICS_THE RELAXATION FUNCTION PROBLEM OF AN ORTHOTROPIC CYLINDER

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The Relaxation function problem of an orthotropic cylinder Co. H. Tran.

Faculty of Mathematics, University of Natural Sciences - VNU-HCM coth123@math.com & cohtran@math.com Copyright 2007

By CO HONG TRAN at 5:20 pm, Jun 19, 2009

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 (1) is exact if (1) can be rewritten :

C is Euler's constant .. ,

is proportional to logt .

(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

By the similar way , we find out :

,

,

(9)

(10)

(11)

Assume that the relaxation moduli have power form : where

are constants .

By applying the Laplace transfom for (11) , we obtain the operator moduli : 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));

(12)

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 \n\n"); 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=[ 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 -9.0 -8.0 -7.0 -6.0 -5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0

1.4286 1.4286 1.4287 1.4289 1.4294 1.4305 1.4335 1.4409 1.4595 1.5056 1.6182 1.8804 2.4264 3.3300 4.3240 4.8725 4.8419 4.5127

8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0 21.0 22.0 23.0 24.0 25.0 26.0 27.0 28.0

4.1110 3.7260 3.3828 3.0857 2.8323 2.6184 2.4396 2.2913 2.1691 2.0691 1.9877 1.9217 1.8684 1.8804 2.4264 3.3300 4.3240 4.8725 4.8419 4.5127 4.1110 3.7260 3.3828 3.0857 2.8323 2.6184 2.4396 2.2913 2.1691 2.0691 1.9877 1.9217 1.8684 1.8255 1.7910 1.7634 1.7413 1.7237 1.7096 1.6984 1.6894

29.0 30.0

1.6823 1.6766

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

By CO HONG TRAN at 5:20 pm, Jun 19, 2009

REFERENCES

[1] Ngo Thanh Phong , Nguyen Thoi Trung , Nguyen Dinh 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 chi 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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Application Details Author: Application Type: Publish Date: Created In: Language: Category:

Dr. Co Tran Maple Document July 23, 2007 Maple 11 Maple 10 English Engineering: Mechanical

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Digitally signed by COHONGTRAN DN: cn=COHONGTRAN, c=US, o=MMI, ou=NCU HUI, email=cohtran@math.com Reason: I am the author of this document Location: HCMC Date: 2009.06.19 17:16:47 +07'00'