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Home : User Community : Application Center : Engineering : Mechanical : 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

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 \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=["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 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 .

Legal Notice: The copyright for this application is owned by the author(s). Neither Maplesoft nor the author are responsible for any errors contained within and are not liable for any damages resulting from the use of this material. This application is intended for non-commercial, non-profit use only. Contact the author for permission if you wish to use this application in for-profit activities.

Application Details

Author: Dr. Co Tran Application Type:

Maple Document

Publish Date: July 23, 2007 Created In: Maple 11

Maple 10 Language: English Category: Engineering: Mechanical

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