{VERSION 3 0 "IBM INTEL NT" "3.0" }
{USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 
1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 
0 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 }
{CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 }{CSTYLE "" -1 257 
"" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 
1 1 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }
{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{PSTYLE "Normal" 
-1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 
-1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "
" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }}
{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 257 14 "Question 1.3.\n" }{TEXT 
-1 458 "\011\011A square prismatic bar of 1300-sq.mm cross-sectional a
rea is composed of two pieces of wood glued together along the x' plan
e, which makes an angle q with the axial direction. The normal and the
 shearing stresses acting simultaneously on the joint are limited to 2
0 and 10 Mpa, respectively, and on the bar itself, to 56 and 28 Mpa re
spectively. Determine the maximum allowable axial load that the bar ca
n carry and the corresponding value of angle theta.\n" }}{PARA 0 "" 0 
"" {TEXT 256 8 "Solution" }{TEXT -1 203 "\n\011\011\nLet us denote the
 axial force applied by \"P\", the cross-sectional area of the bar by \+
\"A\", s,s',t,t' be the normal and shear stresses on the normal cross-
section of the bar and the joint respectively\n" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }
{TEXT -1 114 "This is done so that every time you run the program it d
oes not \"remember\" the \"previous\" values of the variables." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "a:=1300;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"aG\"%+8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
16 "j:=a/cos(theta);" }{TEXT -1 32 "Cross-section of the glued-joint" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"jG,$*&\"\"\"F'-%$cosG6#%&thetaG!
\"\"\"%+8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "Now, we write expres
sions for normal and shear stresses at the glued section" }{MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "sigmaj:=P*cos(thet
a)/j;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'sigmajG,$*&%\"PG\"\"\")-%$
cosG6#%&thetaG\"\"#\"\"\"#F(\"%+8" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "tauj:=P*sin(theta)/j;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%%taujG,$*(%\"PG\"\"\"-%$sinG6#%&thetaGF(-%$cosGF+F(#F(\"%+8" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "Equating the stresses to the al
lowable values, in order to get two equations which are solved to get \+
the values of P and theta" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 16 "eqn1:=sigmaj=20;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%%eqn1G/,$*&%\"PG\"\"\")-%$cosG6#%&thetaG\"\"#\"\"\"#F)\"%+8\"#
?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "eqn2:=tauj=10;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eqn2G/,$*(%\"PG\"\"\"-%$sinG6#%&the
taGF)-%$cosGF,F)#F)\"%+8\"#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 20 "eqnset:=\{eqn1,eqn2\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'eqn
setG<$/,$*&%\"PG\"\"\")-%$cosG6#%&thetaG\"\"#\"\"\"#F*\"%+8\"#?/,$*(F)
F1-%$sinGF.F*F,F*F2\"#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "
varset:=\{P,theta\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'varsetG<$%
\"PG%&thetaG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "solnset:=so
lve(eqnset,varset);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(solnsetG<$/%
\"PG\"&+D$/%&thetaG-%'arctanG6$-%'RootOfG6#,&*$)%#_ZG\"\"#\"\"\"\"\"&!
\"\"\"\"\",$F.F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "assign(
solnset);" }{TEXT -1 125 "this statement assigns the obtained values t
o the appropriate variables, so that they can be accessed later in the
 worksheet." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "P;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#\"&+D$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "
evalf(theta);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+X]%zn#!\"*" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "Now, convert the theta  from radi
ans into degrees, just for the output.Note that  all maple functions o
perate on radian values only." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 29 "evalf(180+(theta*180/3.142));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#$\"*L%\\eE!\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 111 "now that we have the values of P, theta, we shall check whethe
r the stresses in the bar are within safe limits." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Normal stress in \+
the bar," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 11 "sigma:=P/a;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sig
maG\"#D" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "This value of 25 is ve
ry well within the allowable stress value of 56 Mpa, hence, Safe." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "tau:= P/(2*a);" }{TEXT -1 0 
"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$tauG#\"#D\"\"#" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalf(tau);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"++++]7!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 89 "T
his value of 12.5 is very well within the allowable stress value of 28
 Mpa, hence, Safe." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT 258 12 "Final Answer" }{TEXT 260 3 "   " }{TEXT -1 1 "T
" }{TEXT 259 0 "" }{TEXT -1 0 "" }{TEXT -1 120 "he maximum allowable a
xial load that the bar can carry is 32.5 KN and the corresponding valu
e of angle is 26.56 degrees." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}
{MARK "24 0 5" 120 }{VIEWOPTS 1 1 0 1 1 1803 }