Question Finite Element Analysis IPlease solve the temperatures of the right bottom corner and the right top corner using MATRICES. Will give a thumbs up. its very important!! Problem 4 of 4: The left side of a long metal slab of a rectangular cross section (2 x 1 m2) is maintained at temperature 350°C. The front, back and bottom sides are thermally insulated. The top and right sides are exposed to an atmosphere at To = 18°C. Heat conduction and heat convection coefficients are Kxx = Kyy = 28.0 W/°C.m and h = 10.0 W/°C m2. Determine temperatures at the right bottom corner (point 2) and right top corner (point 3) of the slab's cross section, using two 2D triangular finite elements as shown. To -18°C 350°C 300°C 2 m 777777777777777

D4XXVI The Asker · Mechanical Engineering
Finite Element Analysis I
Please solve the temperatures of the right bottom corner and the right top corner using MATRICES. Will give a thumbs up. its very important!!

Transcribed Image Text: Problem 4 of 4: The left side of a long metal slab of a rectangular cross section (2 x 1 m2) is maintained at temperature 350°C. The front, back and bottom sides are thermally insulated. The top and right sides are exposed to an atmosphere at To = 18°C. Heat conduction and heat convection coefficients are Kxx = Kyy = 28.0 W/°C.m and h = 10.0 W/°C m2. Determine temperatures at the right bottom corner (point 2) and right top corner (point 3) of the slab's cross section, using two 2D triangular finite elements as shown. To -18°C 350°C 300°C 2 m 777777777777777
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Transcribed Image Text: Problem 4 of 4: The left side of a long metal slab of a rectangular cross section (2 x 1 m2) is maintained at temperature 350°C. The front, back and bottom sides are thermally insulated. The top and right sides are exposed to an atmosphere at To = 18°C. Heat conduction and heat convection coefficients are Kxx = Kyy = 28.0 W/°C.m and h = 10.0 W/°C m2. Determine temperatures at the right bottom corner (point 2) and right top corner (point 3) of the slab's cross section, using two 2D triangular finite elements as shown. To -18°C 350°C 300°C 2 m 777777777777777
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For element 1 (Node 1,2,4)x_(1)=0,y_(1)=0,x_(2)=2,y_(2)=0,x_(3)=0,y_(3)=1.The area A of cross-section{:[{:[2A=(x_(1)y_(2)-x_(2)y_(1))+(x_(2)y_(3)-x_(3)y_(2))+(x_(3)y_(1)-x_(1)y_(3))],[=2cm^(2)]:}],[a_(1)=x_(2)y_(3)-x_(3)y_(2)=2xx1-0xx0=2],[a_(2)=x_(3)y_(1)-x_(1)y_(3)=0xx0-0xx1=0],[a_(3)=x_(1)y_(2)-x_(2)y_(1)=0xx0-2xx0=0.],[b_(1)=y_(2)-y_(3)=0-1=-1quadc_(1)=x_(3)x_(2)=0-2=-2],[b_(2)=y_(3)-y_(1)=1-0=1quadc_(2)=x_(1)-x_(3)=0-0=0],[b_(3)=y_(1)-y_(2)=0-0=0quadc_(3)=x_(2)-x_(1)=2-0=2.],[" For elemont "2(" Node "2","3","4)quada_(3)=2xx1-2xx0],[x_(1)=2","y_(1)=0","x_(2)=2","y_(2)=1","x_(3)=0","y_(3)=1.],[2A=2cm^(2)],[a_(1)=2xx1-0xx1=2],[a_(2)=0xx0-2xx1=-2]:}quad{:" a ":}View image!{:[b_(1)=1-1=0,c_(1)=0-2=-2],[b_(2)=1-0=1,c_(2)=2-0=2],[b_(3)=0-1=-1,c_(3)=2-2=0.]:}Shope functions:{:[" Eumet 1: "N_(1)=(1)/(2)(2-x-2y)],[N_(2)=(1)/(2)(x)],[N_(4)=(1)/(2)(2y)],[{T_(e)^(1)}={[1-0.5 x-y,0.5 x,y]}{[T_(1)],[T_(2)],[T_(4)]}]:}Eloment 2: N_(2)=(1)/(2)(2-2y){:[N_(3)=(1)/(2)(-2+x+2y).],[N_(4)=(1)/(2)(2-x)]:}View image!Elem ... See the full answer