{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 }{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 "Text Output" -1 2 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 0 0 0 0 0 1 3 0 3 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "War ning" 2 7 1 {CSTYLE "" -1 -1 "" 0 1 0 0 255 1 0 0 0 0 0 0 1 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Error" 7 8 1 {CSTYLE "" -1 -1 " " 0 1 255 0 255 1 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 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 14 "Question 8.30:" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 345 "Consider a steel rotating disk o f hyperbolic cross-section with a = 0.125 m , b= 0.625 m, ti=0.125m, a nd to = 0.0625m.Determine the maximum tangentail force that can occur \+ at the outer surface in newtons per meter of circumference if the maxi mum stress at the bore is not to exceed 140MPa.Assume that outer and i nner edges are free of presuure." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 256 9 "Solution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "First , we write the profile eqaution for the hyperbolic cross-section" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "t:=t1*(r^(-s));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"tG*&%#t1G\"\"\")%\"rG,$%\"sG!\"\"F'" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "Now, we substitute the values of \+ thickness at the inner and outer radii , so as to obtain the values of \"t1\" and\"s\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "eqn1:= 0.125=t1*(0.125^(-s));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eqn1G/$\" $D\"!\"$*&%#t1G\"\"\")F&,$%\"sG!\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "eqn2:=0.0625=t1*(0.625^(-s));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eqn2G/$\"$D'!\"%*&%#t1G\"\"\")$F'!\"$,$%\"sG!\"\"F+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eqnset:=\{eqn1,eqn2\}; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'eqnsetG<$/$\"$D\"!\"$*&%#t1G\" \"\")F',$%\"sG!\"\"F,/$\"$D'!\"%*&F+\"\"\")$F3F)F.F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "varset:=\{t1,s\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'varsetG<$%#t1G%\"sG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "solnset:=solve(eqnset,varset);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(solnsetG<$/%#t1G$\"+(>,Z5&!#6/%\"sG$\"+\"elnI%!#5" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "assign(solnset);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "s;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+\"elnI%!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 136 "Now, we write the auxiliary equation, and solve it to get the value o f the parameter \"m\", which will be used in evaluating the stresses. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eqn3:=m^2+s*m-(1+nu*s)= 0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eqn3G/,**$)%\"mG\"\"#\"\"\"\" \"\"F)$\"+\"elnI%!#5!\"\"F,%#nuG$!+\"elnI%F/\"\"!" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 10 "for steel:" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 8 "nu:=0.3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %#nuG$\"\"$!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solnset :=solve(eqn3,m);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(solnsetG6$$!+rp d*H\"!\"*$\"+IT+*o)!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "m 1:=solnset[1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#m1G$!+rpd*H\"!\"* " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "m2:=solnset[2];" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#m2G$\"+IT+*o)!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "for steel:" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "rho:= 7.8*10^3; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$rhoG$\"&+!y!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "The expressions for the radial and the tangential stresses are: " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "sigm ar:=(c1/t1)*(r^(m1+s-1))+(c2/t1)*(r^(m2+s-1))-(rho*(w*r)^2)*(3+nu)/(8- (3+nu)*s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'sigmarG,(*&%#c1G\"\" \"*$)%\"rG$\"+8/!*o=!\"*F(!\"\"$\"+@&y*e>!\")*&%#c2G\"\"\")F+$\"*rpd*H F.F(F0*&)%\"wG\"\"#F()F+F " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "sigmath eta:=m1*(c1/t1)*(r^(m1+s-1))+m2*(c2/t1)*(r^(m2+s-1))-(rho*(w*r)^2)*(1+ 3*nu)/(8-(3+nu)*s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+sigmathetaG, (*&%#c1G\"\"\"*$)%\"rG$\"+8/!*o=!\"*F(!\"\"$!+sL%ea#!\")*&%#c2G\"\"\") F+$\"*rpd*HF.F($\"+Ys:-F($!+P;q_A!\"'" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "we find the radial stress at the i nner radius and label it \"sigmari\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "r:=0.125;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG$\" $D\"!\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "sigmari:=sigmar ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(sigmariG,(%#c1G$\"+%*p&ea*!\"( %#c2G$\"+mWr]5!\")*$)%\"wG\"\"#\"\"\"$!+tsT8hF-" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "we fin d the radial stress at the outer radius and label it \"sigmaro\" " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "r:=0.625;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"rG$\"$D'!\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "sigmaro:=sigmar;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %(sigmaroG,(%#c1G$\"+zGI:Z!\")%#c2G$\"+g$)o, " 0 "" {MPLTEXT 1 0 31 "eqnset2:=\{sigmari=0,sigmaro =0\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(eqnset2G<$/,(%#c1G$\"+%*p& ea*!\"(%#c2G$\"+mWr]5!\")*$)%\"wG\"\"#\"\"\"$!+tsT8hF/\"\"!/,(F($\"+zG I:ZF/F,$\"+g$)o, " 0 "" {MPLTEXT 1 0 17 "varset2:=\{c1,c2\};" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%(varset2G<$%#c1G%#c2G" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "solnset:=solve(eqnset2,varset2);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%(solnsetG<$/%#c1G,$*$)%\"wG\"\"#\"\"\"$!+v]FO&*!#5/%#c2G,$F)$\"+ib kX#*!\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "assign(solnset) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "c1;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#,$*$)%\"wG\"\"#\"\"\"$!+v]FO&*!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalf(c2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$)%\"wG\"\"#\"\"\"$\"+ibkX#*!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 162 "Now, we find both stresses at the inner radius,then equa te the maximum of the two values to the allowable value of 140 MPa to \+ solve for maximum allowable speed w." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "r:=0.125;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG$\" $D\"!\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "sigmatheta;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$)%\"wG\"\"#\"\"\"$\"+,w#>*>!\"'" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "sigmar;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#,$*$)%\"wG\"\"#\"\"\"$!\"$!\"(" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 27 "eqn3:=sigmatheta=140000000;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%%eqn3G/,$*$)%\"wG\"\"#\"\"\"$\"+,w#>*>!\"'\"*+++S\" " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "varset3:=\{w\};" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%(varset3G<#%\"wG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solnset:=solve(eqn3,w);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%(solnsetG6$$!+>p5^E!\"($\"+>p5^EF(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "assign(solnset);" }}{PARA 8 "" 1 " " {TEXT -1 36 "Error, (in assign) invalid arguments" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "w:=solnset[1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"wG$!+>p5^E!\"(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "c2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+X)z\")\\'! \"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 149 "Now, to find the tangenti al force per meter of circumfernce, we multiply the tangential stress at the outer radius by the the thickness of the disk." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "r:=0.625;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG$\"$D'!\"$" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "sigmathetao:=sigmatheta;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,sigmathetaoG$\"+a#>U$Q!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "0.0625*sigmathetao/1000;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+MqQ'R#!\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 " We divide by 1000, so as to get the force in KN/m" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 13 "Final Answer:" }{TEXT -1 119 " The maximum tangentail force th at can occur at the outer surface in newtons per meter of circumferenc e is 2396.38KN/m." }{MPLTEXT 1 0 0 "" }}}}{MARK "54 0 0" 9 }{VIEWOPTS 1 1 0 1 1 1803 }