paramath

* B9

This worksheet demonstrates Maple's  capabilities in researching  the numerical and graphical solution of the variable boundary problem of  a thick-walled cylinder of material enclosed in a thin metallic shell .                                     

The relations between stress and strain in linear viscoelastic theory are discussed from the viewpoint of application to problems of stress analysis. This consideration includes some important differences from the estimation of linear viscoelastic laws for the representation of material properties , and the integral operators expressing the creep function or relaxation function can be applied . By using of the differential operator for the  relation  between stress and strain  it is usually most convenient to solve some problems which have the variable boundary . 

Problem Definition 

We consider a thick-walled cylinder of material enclosed in a thin metallic shell .  ( Fig. 1)  

 

( Fig 1 ) .  

 

The outer radius of cylinder : b ;  the thickness of the metallic shell :  h  .  

The inner radius  a  is assumed  by  :  a(t) ,  satisfied the condition   

 . The varying pressure influences  upon the inner surface is a given function  q(t)  .   

Notice that the tube and  shell are under the plain strain conditions and the material is incompressible.  

By  setting   : the radical stress  ,      :  the circular stress   ,   

    : the radical displacement  , 

      are the deformation respectively  . 

 

The diferential equation of equilibrium: 

 

Bywhat we introduce in (I)  it follows that the diferential equation of equilibrium can be written consequently  : 

                                                                                                     (1)                          

In this case we have the law of viscoelasticity relation :    

 

                                                                                            (2)    

At the inner boundary  :   r = a   ,       =   -q(t)      

The circular deformation of  the cylinder of material obtained by integrating  : 

                                                                                        (3)     

Here   E  and    are the elastic constants of the shell material  . 

The second boundary condition  :   e =   (b)   

Because of incompressibility of material and zero axial strain  :   

    it follows that  :    ,    

                            ,                               (4.29) 

                               

From  (2)    we obtain   :                         (4)       

By substituting   (4)  to   (1)  ,  the deduction can be rewritten  :                      (5)      

Integrating (5)  gives    :                      (6)      

By using the condition :   r = a  ,      =   - q (t)     

 

then                    and                 (7)      

 The relation (6) becomes    : 

                   (8)     

Consider the second boundary condition  :                   , then we arrive at the integral equation for  k(t)   :        

                                                                      (9)    

inwhich we  set up :  

               (10)             

.  The equation  (9)  can be considered as especial case of the general formation   :    

 

                                                                            (11)        

 

The authors of [2] have considered the simplest case of a viscoelastic body  expressed by formular  : 

                                                                                        (12)          

Here       are the given constants defined by experiments on material . The equation (9)  now becomes  differential  for     k(t)  . 

 

          (13)     

From (13) , we find out the solution  k(t)  correspondently with  choosing the functions    a(t)   and   q(t)   . 

For examples , some authors have taken   :  

                     (14)        

Here   q  =   const  .  [     =  E   in case of elasticity  ,      =  3   in case of  Newton's  fluid ]  .  By integrating (13)  we obtain the solution of the viscoelasticity problem  . In generally the differential equation for k(t)  would be established in symbolic form by using the operator  :  

                            (15) 

 

Numerical and graphical solution: 

 

Here we have used the parameters  :   

 

 

  (13)  canbe rewitten as the following  : 

 

 

The analytic solution :  

 

 

 

 

 

 

 

 

 

 

 

>	restart; with(DEtools): a(t)=a[0]/sqrt(1-delta*t); eq:=alpha*(1/a(t)^2-1/b^2)*diff(k(t),t)+(beta*(1/a(t)^2-1/b^2)-1/(b^2*c))*k(t)=q(t); eq:=subs(a(t)=a[0]/sqrt(1-delta*t),eq); eq:=subs(q(t)=q[0],eq); nok:=dsolve(eq,k(t));

 

 

 

 

 

(3.1)

 

>

restart; with(DEtools): cof := 1000: eq := alpha*(1/a(t)^2-1/b^2)*(diff(k(t), t))+(beta*(1/a(t)^2-1/b^2)-1/(b^2*c))*k(t) = q(t);eq := subs(a(t) = a[0]/sqrt(1-delta*t), eq); eq := subs(q(t) = q[0], eq); nok := dsolve(eq, k(t)); nok := evalf(subs(c = 1.2, delta = 0.5e-2, alpha = .5, b = 1.5, a[0] = 1, beta = 2, q[0] = 1, nok), 3); u := k(t)/(r*cof);

 

 

 

 

 

 

 

(3.2)

 

>	k(t):=(int(-4.50*(-1.25+.112e-1*s)^147.*exp(4.00*s),s=0..t)+.22013e14)*exp(-4.00*t)/(-1.25+.112e-1*t)^148.: plot(k(t),t=0..6,labels=[t,"k(t)"],color=blue,style=line,linestyle=4,thickness=2);

 

 

>	plot3d(u, r = 0 .. 1.5, t = 0 .. 6, axes = boxed, labels = ["r", "t(s)", "u(r,t)"], grid = [40, 40]);

 

 

>	epsilon[r] := -k(t)/r^2; plot3d(epsilon[r]/cof, r = 0 .. 1.5, t = 0 .. 6, axes = boxed, labels = ["r", "t(s)", "e[r](r,t)"], grid = [40, 40]);

 

 

>	epsilon[theta] := k(t)/r^2; plot3d(epsilon[theta]/cof, r = 0 .. 1.5, t = 0 .. 6, axes = boxed, labels = ["r", "t(s)", "e[th](r,t)"], grid = [40, 40]);

 

>

 

>	restart; k(t) = (Int(-4.50*(-1.25+0.112e-1*t)^147.*exp(4.00*t), t)+_C1)*exp(-4.00*t)/(-1.25+0.112e-1*t)^148.;

 

(3.3)

 

>	k(t):=(int(-4.50*(-1.25+.112e-1*tau)^147.*exp(4.00*tau),tau=0..t)+.22013e14)*exp(-4.00*t)/(-1.25+.112e-1*t)^148.:;k0:=subs(t=0,k(t));

 

(3.4)

 

>	edk := k0 = .1; C1 = evalf(solve(edk, _C1), 5);

 

 

(3.5)

 

>	u:=k(t)/r/10^3: epsilon[theta]:=k(t)/r^2/10^3: epsilon[rt]:=-k(t)/r^2/10^3: print(" NUMERICAL AND GRAPHICAL SOLUTION "); print("   OUTPUT   DATA  "); N:=10; M:=6; d:=1;

 

>

for m from 1 to N  do r:=evalf(1.25*m/N,3); u:=k(t)/r/10^3: epsilon[theta]:=k(t)/r^2/10^3: epsilon[rt]:=-k(t)/r^2/10^3: printf("        t           k(t)            u(r,t)          e[theta]           nn"); for j from 0 to M do printf("%10.1f    %10.5f       %10.5f       %10.5f       n", d*j, subs(t=d*j,k(t)), subs(t=d*j,u), subs(t=d*j,epsilon[theta])); end do; end do;

 

 

 

 

 

(3.6)

 

>

plot3d(u,r=0..1.5,t=0..6,axes=boxed,labels=["r","t(s)","u(r,t)"],grid=[40,40]); cof:=10^3: epsilon[r]:=-k(t)/r^2: plot3d(epsilon[r]/cof,r=0..1.5,t=0..6,axes=boxed,labels=["r","t(s)","e[r](r,t)"],grid=[40,40]); epsilon[theta]:=k(t)/r^2: plot3d(epsilon[theta]/cof,r=0..1.5,t=0..6,axes=boxed,labels=["r","t(s)","e[th](r,t)"],grid=[40,40]); with(plots): N:=3; for j from 0 to N do a[0]:=1/(j+1): plot([a[0]/sqrt(1-delta*t),a[0]/sqrt(1-delta*t),1.25,1.5],t=0..(1.95+j)*3600,color=[blue,blue,black,red],thickness=[1,2,2,2],style=[line,point,point,point],labels=["t(s)","a(t)"],symbol=[diamond,cross],linestyle=4,legend=[`a(t)`,`a(t)`,`b-h = 1.25`,`b = 1.5`]); od;

 

 

 

 

 

 

 

 

 

 

 

 

 

>

 

>	func_bankinhtrong(50, 5.2, 2, 7, Hh);

 

 

REFERENCES 

 

[1] ю.н. РАБОТНОВ ,  Пoлзyчесть  элементов  конструкций  , издательство “наука”  , MOCKBA 1966 

[2] Lee E .H ., Radok J.R.M., Woodward W.B., Stress Analysis for Linear Viscoelastic Materials , Trans. Soc. Rheol.,  3, 41-59 (1959)  

 

 

 

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