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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 13 "Question.8.5:" }}{PARA 0 
"" 0 "" {TEXT -1 410 "\011Two thick walled, closed -ended cylinders of
 the same dimensions are subjected to internal and external pressure, \+
respectively. The outer diameter of each is twice the inner diameter. \+
What is the ratio of the pressures for the following case?(a) The maxi
mum tangential stress has the same absolute value in each cylinder.(b)
 The maximum tangential strain has the same absolute value in each cyl
inder. Take  \n\n" }{TEXT 257 8 "Solution" }{TEXT -1 2 ":\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 78 "We first write down the formulae for tangential st
ress for both the cylinders." }}{PARA 0 "" 0 "" {TEXT -1 123 "First, f
or the cylinder with internal pressure;where \"ip\" is the internal pr
essure, and \"a\"and \"b\" are the inner and outer " }}{PARA 0 "" 0 "
" {TEXT -1 35 "radii  of the cylinder respectively" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "sigmatheta1
:=(a^2*ip)/(b^2-a^2)*(1+(b/r)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%,sigmatheta1G*&*()%\"aG\"\"#\"\"\"%#ipG\"\"\",&F,F,*&*$)%\"bGF)F*F**$
)%\"rG\"\"#F*!\"\"F,F,F*,&*$F0F*F,*$F'F*!\"\"F6" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 35 "This value will be maximum at r = a" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 5 "r:=a;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"r
G%\"aG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "sigmatheta1max:= \+
sigmatheta1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%/sigmatheta1maxG*&*(
)%\"aG\"\"#\"\"\"%#ipG\"\"\",&F,F,*&*$)%\"bGF)F*F**$)F(\"\"#F*!\"\"F,F
,F*,&*$F0F*F,*$F'F*!\"\"F5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 124 "  \+
 Now, for the cylinder with external pressure.where \"op\" is the exte
rnal pressure, and \"a\"and \"b\" are the inner and outer " }}{PARA 0 
"" 0 "" {TEXT -1 120 "radii  of the cylinder respectively.Here, we hav
e to re-assign \"r\",so that we can use it in the formula for sigmathe
ta2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "r:='r';" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%\"rGF$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 45 "sigmatheta2:=-(op*b^2)/(b^2-a^2)*(1+(a/r)^2);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%,sigmatheta2G,$*&*(%#opG\"\"\")%\"bG\"\"#\"\"\",
&F)F)*&*$)%\"aGF,F-F-*$)%\"rG\"\"#F-!\"\"F)F)F-,&*$F*F-F)*$F1F-!\"\"F7
F;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "We again note that this too
 is maximum at r = a." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "r:=
a;" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG%\"aG" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "   sigmatheta2max:=sigmatheta2;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%/sigmatheta2maxG,$*&*&%#opG\"\"\")%\"bG\"\"#\"\"\"F-,
&*$F*F-F)*$)%\"aGF,F-!\"\"!\"\"!\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 187 "Now as per the problem statement, we are to compare the absolu
te values of these stresses, so we define the max to be the absolute v
alue so as to drop the negative sign in sigmatheta2max." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "sigmatheta2max:=abs(sigmatheta2max)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%/sigmatheta2maxG,$-%$absG6#*&*&
%#opG\"\"\")%\"bG\"\"#\"\"\"F0,&*$F-F0F,*$)%\"aGF/F0!\"\"!\"\"F/" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "Now, as per the problem statement
, we solve for ratio of ip/op for these maximum stresses to be equal a
nd the data that b = 2a" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "b:=2*a;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"bG,$%\"aG\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 36 "eqn1:=sigmatheta2max=sigmatheta1max;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%%eqn1G/,$-%$absG6#%#opG#\"\")\"\"$,$%#ipG#\"\"&F
-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "solve(eqn1,ip);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%$absG6#%#opG#\"\")\"\"&" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "This is the first part of the solu
tion." }}{PARA 0 "" 0 "" {TEXT -1 35 "Now, we find the tangential stra
in." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 7 "r:='r';" }{TEXT -1 44 "We again re-assign 'r' to its
 variable form." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rGF$" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "epsilontheta1:=(1/E)*(sigmat
heta1-(nu*sigmar1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%.epsilonthet
a1G*&,&*&%#ipG\"\"\",&F)F)*&*$)%\"aG\"\"#\"\"\"F0*$)%\"rG\"\"#F0!\"\"
\"\"%F)#F)\"\"$*&%#nuGF)%(sigmar1GF)!\"\"F0%\"EGF5" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "sig
matheta1:=(a^2*ip)/(b^2-a^2)*(1+(b/r)^2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%,sigmatheta1G,$*&%#ipG\"\"\",&F(F(*&*$)%\"aG\"\"#\"\"
\"F/*$)%\"rG\"\"#F/!\"\"\"\"%F(#F(\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "nu:=1/3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#nuG#\"
\"\"\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "sigmar1:=(a^2*
ip)/(b^2-a^2)*(1-(b/r)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(sigma
r1G,$*&%#ipG\"\"\",&F(F(*&*$)%\"aG\"\"#\"\"\"F/*$)%\"rG\"\"#F/!\"\"!\"
%F(#F(\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "epsilontheta
1;" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*&,&*&%#ipG\"\"\",&F'F'*&*$)%\"aG\"\"#\"\"\"F.*$)%\"rG
\"\"#F.!\"\"\"\"%F'#F'\"\"$*&F&F.,&F'F'F)!\"%F'#!\"\"\"\"*F.%\"EGF3" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "We again note that this too is max
imum at r = a." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 5 "r:=a;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG
%\"aG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "epsilontheta1max:=
epsilontheta1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%1epsilontheta1maxG
,$*&%#ipG\"\"\"%\"EG!\"\"\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 
"Now, we consider the second cylinder," }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "r:='r';" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rGF$" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "sigmatheta2:=-(op*b^2)/(b^
2-a^2)*(1+(a/r)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,sigmatheta2G
,$*&%#opG\"\"\",&F(F(*&*$)%\"aG\"\"#\"\"\"F/*$)%\"rG\"\"#F/!\"\"F(F(#!
\"%\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "sigmar2:=-(op*b
^2)/(b^2-a^2)*(1-(a/r)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(sigma
r2G,$*&%#opG\"\"\",&F(F(*&*$)%\"aG\"\"#\"\"\"F/*$)%\"rG\"\"#F/!\"\"!\"
\"F(#!\"%\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 7 "r:='r';" }{TEXT -1 44 "We again re-assign \+
'r' to its variable form." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rGF$
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "epsilontheta2:=(1/E)*(s
igmatheta2-nu*sigmar2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%.epsilont
heta2G*&,&*&%#opG\"\"\",&F)F)*&*$)%\"aG\"\"#\"\"\"F0*$)%\"rG\"\"#F0!\"
\"F)F)#!\"%\"\"$*&F(F0,&F)F)F+!\"\"F)#\"\"%\"\"*F0%\"EGF5" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "simplify(epsilontheta2);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#,$*&*&%#opG\"\"\",&*$)%\"rG\"\"#\"\"\"F'*$)%
\"aGF,F-F,F'F-*&%\"EG\"\"\")F+\"\"#F-!\"\"#!\")\"\"*" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 21 "evalf(epsilontheta2);" }{TEXT -1 0 "" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&*&%#opG\"\"\",&$F'\"\"!F'*&*$)%\"a
G\"\"#\"\"\"F0*$)%\"rG\"\"#F0!\"\"F'F'$!+LLLL8!\"**&F&F0,&F)F'F+$!\"\"
F*F'$\"+WWWWW!#5F0%\"EGF5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 
"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "We again note that this to
o is maximum at r = a." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "r;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%\"rG" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 5 "r:=a;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG%\"aG
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "epsilontheta2max:=abs(e
psilontheta2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%1epsilontheta2maxG
,$-%$absG6#*&%#opG\"\"\"%\"EG!\"\"#\"\")\"\"$" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 40 "eqn2:=epsilontheta2max=epsilontheta1max;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eqn2G/,$-%$absG6#*&%#opG\"\"\"%\"EG
!\"\"#\"\")\"\"$,$*&%#ipGF,F-F.\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "solve(eqn2,ip);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$
*&-%$absG6#*&%#opG\"\"\"%\"EG!\"\"\"\"\"F+F-#\"\"%\"\"$" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
258 12 "Final Answer" }{TEXT -1 6 ": (a) " }{OLE 1 3592 1 "[xm]Br=WfoR
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::::::::::::::::::::::::::::::::::::::::::::::::::::::j:b:;B:<::::::::
::=Jyyy;d:yayQZHQ:>:;`:Z@w<fccWflwgmwg`_hZfbkgh[_hcwgh?^mn_jB=BKaDBETV
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;Z::::::::JMTjAN=yyyxY:>\\:B:;xyyQ]yyyyYjT:GY:<j?<:G;Sj`@Pt\\Pd`QrP`:>
FeZ:Vy<>jx]:JBAj:<J:<J<DZJ^dcgg_WhZnc_whZNdigg[oGkcGC:US:>;N`DFg;f\\?F
:<Jr>;N@NG=:[mDjw?^yM:<:[>^::::ML<jPF:C:[Y:>;N@YX:;Y;=Z:vGK:_;_S:=J:FG
e:qQ:[:JB?:<J<:::>AjP>:CB:>X=J>JS>f;fx=FZ:B:gK>JSjDIj:JQ<J>JSjdEj:>:]C
:[B:<jw;<:[>DZ<>ZJVdscRYEU^:f?=J<Jr?:K:_c<Vt;FZ:>Z:fAO:G;Ojysy=:;JBB:q
QBv:<J:^q<Z:FZ<NZ\\VdsWhngfgcUC:UK:^:>X=j;B:;:::::::::1:" }{TEXT -1 
57 "when maximum tangential stress in both cylinders is equal" }}
{PARA 0 "" 0 "" {TEXT -1 26 "                       (b)" }{OLE 1 3588 
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G;Sj`@Pt\\Pd`QrP`:>FeZ:Vy<>jx]:JBAj:<J:<J<DZJ^dcgg_WhZnc_whZNdigg[oGSL
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:AZ:>::::::::3:" }{TEXT -1 59 " when maximum tangential strain  in bot
h cylinders is equal" }}}}{MARK "52 0 2" 2 }{VIEWOPTS 1 1 0 1 1 1803 }