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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 12 "Question 2.9" }}{PARA 0 "
" 0 "" {TEXT -1 128 ". A thin rectangular plate a=20 mm and b=12 mm is
 acted upon by a stress distribution   \n      resulting in the unifor
m strains," }{OLE 1 4104 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyy
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;B:1:" }{TEXT -1 65 " .\n      Determine the changes in length of diag
onals QB and AC.\n" }}{PARA 0 "" 0 "" {TEXT 257 9 "Solution:" }
{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 86 "Now, let us write the stress transformation equations for
 the strain in any direction " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }}
{PARA 0 "" 0 "" {TEXT -1 9 "which is:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 98 "epsilontheta:=(epsilonx+epsilony)/2+(cos(2*theta)*(ep
silonx-epsilony)/2)+(gammaxy/2)*sin(2*theta);" }{TEXT -1 0 "" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%-epsilonthetaG,*%)epsilonxG#\"\"\"\"\"#%)e
psilonyGF'*&-%$cosG6#,$%&thetaGF)F(,&F&F(F*!\"\"F(F'*&%(gammaxyGF(-%$s
inGF.F(F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "Now" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 60 "Let us first initailise the variables to their \+
given values." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "epsilonx:=
300;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)epsilonxG\"$+$" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "epsilony:=500;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%)epsilonyG\"$+&" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "gammaxy:=200;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(gammaxyG\"$+#" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 "\027" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 149 "We also need to find the lengths of the \+
diagonals QB and AC, which happen to be equal for a rectangle.( since,
diagonals of a rectangle are congruent)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "QB:=sqrt(b^2+a^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%#QBG*$-%%sqrtG6#,&*$)%\"bG\"\"#\"\"\"F.*$)%\"aGF-F.F.F." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "AC:=QB;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#ACG*$-%%sqrtG6#,&*$)%\"bG\"\"#\"\"\"F.*$)%\"aGF-F.F.
F." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "epsilontheta;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#,(\"$+%\"\"\"-%$cosG6#,$%&thetaG\"\"#!$+\"-%$sin
GF(\"$+\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "When we consider dia
gonal QB, the angle which it makes with the positive X axis is," }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "theta:=arctan(b/a);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG-%'arctanG6#*&%\"bG\"\"\"%\"a
G!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "b:=12;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"bG\"#7" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 6 "a:=20;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG\"#?" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "evalf(theta);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"+.]>/a!#5" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "evalf(epsilontheta);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#$\"+1Zw6W!\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 249 "The change i
n length is given by the strain multiplied by the original length, sin
ce the given values of strains are in microns, and the values of the l
engths are given in millimetres, the following expression will yield t
he change in length in \"mm\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 31 "evalf(epsilontheta*QB/1000000);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#$\"+6:**G5!#6" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "Now, we are c
onsidering diagonal AC, which makes the following angle with the posit
ive X axis." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "theta:=arcta
n(a/b)+Pi/2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG,&-%'arctanG6
##\"\"&\"\"$\"\"\"%#PiG#F,\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 13 "evalf(theta);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+aJ<,E!\"*
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "evalf(epsilontheta);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+B)eqk#!\"(" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 263 "Again, we have the change in length is given by the
 strain multiplied by the original length.since the given values of st
rains are in microns, and the values of the lengths are given in milli
metres, the following expression will yield the change in length in \"
mm\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "evalf(epsilontheta*
AC/1000000);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+k!\\R<'!#7" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT 258 12 "Final Answer" }{TEXT -1 45 ": del
ta-QB = 0.0102mm and delta-AC = 0.0061mm" }{MPLTEXT 1 0 0 "" }}}}
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