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Calculate the stiffness matrix for the Q8 element shown below. Make use of the shape functions in natural coordinates and the Jacobian calculated in previous problems. Use 2x2 Gaussian numerical integration. Use plane strain conditions (for unit thickness) with E=200E9 (MPa) and nu=0.25. The final answers is F1= 0.18452e12. I have attached my matlab code below. Please look thru it and correct it. Thanks. u=sym('u')  %%zta v=sym('v')  %%eta x1=1 x2=2 x3=3.5 x4=3.6 x5=4.0 x6=2.3 x7=1.2 x8=0.8 y1=1 y2=1 y3=1.3 y4=2.3 y5=3.3 y6=3.2 y7=3.0 y8=2.0 X=[x1;x2;x3;x4;x5;x6;x7;x8] Y=[y1;y2;y3;y4;y5;y6;y7;y8] %% diff of zta DN1Du=1/4*(1-v)*(2*u+v); DN2Du=-u*(1-v); DN3Du=1/4*(1-v)*(2*u-v); DN4Du=1/2*(1-v^2); DN5Du=1/4*(1+v)*(2*u+v); DN6Du=-u*(1+v); DN7Du=1/4*(1+v)*(2*u-v); DN8Du=-1/2*(1-v^2); %% diff of eta DN1Dv=1/4*(1-u)*(2*v+u); DN2Dv=-1/2*(1-u^2); DN3Dv=1/4*(1+u)*(2*v-u); DN4Dv=-v*(1+u); DN5Dv=1/4*(1+u)*(2*v+u); DN6Dv=1/2*(1-u^2); DN7Dv=1/4*(1-u)*(2*v+u); DN8Dv=-v*(1-u); %% find dx/du dx/dv, dy/du and dy/dv a=[DN1Du, DN2Du, DN3Du, DN4Du, DN5Du, DN6Du,DN7Du, DN8Du] dxdu=a*X b=[DN1Dv, DN2Dv, DN3Dv, DN4Dv, DN5Dv, DN6Dv, DN7Dv,DN8Dv] dxdv=b*X dydu=a*Y dydv=b*Y %% Jacobian  J=dxdu*dydv-dxdv*dydu %% define DNiDx DN1Dx=1/J*(dydv*DN1Du-dydu*DN1Dv) DN2Dx=1/J*(dydv*DN2Du-dydu*DN2Dv) DN3Dx=1/J*(dydv*DN3Du-dydu*DN3Dv) DN4Dx=1/J*(dydv*DN4Du-dydu*DN4Dv) DN5Dx=1/J*(dydv*DN5Du-dydu*DN5Dv) DN6Dx=1/J*(dydv*DN6Du-dydu*DN6Dv) DN7Dx=1/J*(dydv*DN7Du-dydu*DN7Dv) DN8Dx=1/J*(dydv*DN8Du-dydu*DN8Dv) %% defrine DniDx DN1Dy=1/J*(-dxdv*DN1Du+dxdu*DN1Dv) DN2Dy=1/J*(-dxdv*DN2Du+dxdu*DN2Dv) DN3Dy=1/J*(-dxdv*DN3Du+dxdu*DN3Dv) DN4Dy=1/J*(-dxdv*DN4Du+dxdu*DN4Dv) DN5Dy=1/J*(-dxdv*DN5Du+dxdu*DN5Dv) DN6Dy=1/J*(-dxdv*DN6Du+dxdu*DN6Dv) DN7Dy=1/J*(-dxdv*DN7Du+dxdu*DN7Dv) DN8Dy=1/J*(-dxdv*DN8Du+dxdu*DN8Dv) %% B matrix B=[DN1Dx, 0, DN2Dx, 0, DN3Dx, 0, DN4Dx, 0, DN5Dx, 0, DN6Dx, 0, DN7Dx, 0, DN8Dx, 0;    0, DN1Dy, 0, DN2Dy, 0, DN3Dy, 0, DN4Dy, 0, DN5Dy, 0, DN6Dy, 0, DN7Dy, 0, DN8Dy;    DN1Dy, DN1Dx, DN2Dy, DN2Dx, DN3Dy, DN3Dx,DN4Dy,DN4Dx,DN5Dy,DN5Dx,DN6Dy,DN6Dx, DN7Dy,DN7Dx,DN8Dy,DN8Dx;]  BT=transpose(B) %% define D strain E=200*10^9 nu=0.25 Q=200*10^9/((1+0.25)*(1-2*0.25)) R=[1-0.25, 0.25, 0;     0.25, 1-0.25, 0;     0, 0, (1-2*0.25)/2;] D=Q*R %% u1=-1/sqrt(3); v1=-1/sqrt(3); u2=1/sqrt(3); v2=1/sqrt(3); w1=1; w2=1; % Substitute u and v with u1 and v1 J_sub = subs(J, [u, v], [u1, v1]); B_sub = subs(B, [u, v], [u1, v1]); BT_sub = subs(BT, [u, v], [u1, v1]); F1=BT_sub*D*B_sub*J_sub % Substitute u and v with u1 and v2 J_sub2 = subs(J, [u, v], [u1, v2]); B_sub2 = subs(B, [u, v], [u1, v2]); BT_sub2 = subs(BT, [u, v], [u1, v2]); F2=BT_sub2*D*B_sub2*J_sub2 % Substitute u and v with u2 and v1 J_sub3 = subs(J, [u, v], [u2, v1]); B_sub3 = subs(B, [u, v], [u2, v1]); BT_sub3 = subs(BT, [u, v], [u2, v1]); F3=BT_sub3*D*B_sub3*J_sub3 % Substitute u and v with u2 and v2 J_sub4 = subs(J, [u, v], [u2, v2]); B_sub4 = subs(B, [u, v], [u2, v2]); BT_sub4 = subs(BT, [u, v], [u2, v2]); F4=BT_sub4*D*B_sub4*J_sub4 F=(F1*w1*w2)+(F2*w1*w2)+(F3*w1*w2)+(F4*w1*w2);

Calculate the stiffness matrix for the Q8 element shown below. Make use of the shape functions in natural coordinates and the Jacobian calculated in previous problems. Use 2x2 Gaussian numerical integration. Use plane strain conditions (for unit thickness) with E=200E9 (MPa) and nu=0.25. The final answers is F1= 0.18452e12. I have attached my matlab code below. Please look thru it and correct it. Thanks. u=sym('u')  %%zta v=sym('v')  %%eta x1=1 x2=2 x3=3.5 x4=3.6 x5=4.0 x6=2.3 x7=1.2 x8=0.8 y1=1 y2=1 y3=1.3 y4=2.3 y5=3.3 y6=3.2 y7=3.0 y8=2.0 X=[x1;x2;x3;x4;x5;x6;x7;x8] Y=[y1;y2;y3;y4;y5;y6;y7;y8] %% diff of zta DN1Du=1/4*(1-v)*(2*u+v); DN2Du=-u*(1-v); DN3Du=1/4*(1-v)*(2*u-v); DN4Du=1/2*(1-v^2); DN5Du=1/4*(1+v)*(2*u+v); DN6Du=-u*(1+v); DN7Du=1/4*(1+v)*(2*u-v); DN8Du=-1/2*(1-v^2); %% diff of eta DN1Dv=1/4*(1-u)*(2*v+u); DN2Dv=-1/2*(1-u^2); DN3Dv=1/4*(1+u)*(2*v-u); DN4Dv=-v*(1+u); DN5Dv=1/4*(1+u)*(2*v+u); DN6Dv=1/2*(1-u^2); DN7Dv=1/4*(1-u)*(2*v+u); DN8Dv=-v*(1-u); %% find dx/du dx/dv, dy/du and dy/dv a=[DN1Du, DN2Du, DN3Du, DN4Du, DN5Du, DN6Du,DN7Du, DN8Du] dxdu=a*X b=[DN1Dv, DN2Dv, DN3Dv, DN4Dv, DN5Dv, DN6Dv, DN7Dv,DN8Dv] dxdv=b*X dydu=a*Y dydv=b*Y %% Jacobian  J=dxdu*dydv-dxdv*dydu %% define DNiDx DN1Dx=1/J*(dydv*DN1Du-dydu*DN1Dv) DN2Dx=1/J*(dydv*DN2Du-dydu*DN2Dv) DN3Dx=1/J*(dydv*DN3Du-dydu*DN3Dv) DN4Dx=1/J*(dydv*DN4Du-dydu*DN4Dv) DN5Dx=1/J*(dydv*DN5Du-dydu*DN5Dv) DN6Dx=1/J*(dydv*DN6Du-dydu*DN6Dv) DN7Dx=1/J*(dydv*DN7Du-dydu*DN7Dv) DN8Dx=1/J*(dydv*DN8Du-dydu*DN8Dv) %% defrine DniDx DN1Dy=1/J*(-dxdv*DN1Du+dxdu*DN1Dv) DN2Dy=1/J*(-dxdv*DN2Du+dxdu*DN2Dv) DN3Dy=1/J*(-dxdv*DN3Du+dxdu*DN3Dv) DN4Dy=1/J*(-dxdv*DN4Du+dxdu*DN4Dv) DN5Dy=1/J*(-dxdv*DN5Du+dxdu*DN5Dv) DN6Dy=1/J*(-dxdv*DN6Du+dxdu*DN6Dv) DN7Dy=1/J*(-dxdv*DN7Du+dxdu*DN7Dv) DN8Dy=1/J*(-dxdv*DN8Du+dxdu*DN8Dv) %% B matrix B=[DN1Dx, 0, DN2Dx, 0, DN3Dx, 0, DN4Dx, 0, DN5Dx, 0, DN6Dx, 0, DN7Dx, 0, DN8Dx, 0;    0, DN1Dy, 0, DN2Dy, 0, DN3Dy, 0, DN4Dy, 0, DN5Dy, 0, DN6Dy, 0, DN7Dy, 0, DN8Dy;    DN1Dy, DN1Dx, DN2Dy, DN2Dx, DN3Dy, DN3Dx,DN4Dy,DN4Dx,DN5Dy,DN5Dx,DN6Dy,DN6Dx, DN7Dy,DN7Dx,DN8Dy,DN8Dx;]  BT=transpose(B) %% define D strain E=200*10^9 nu=0.25 Q=200*10^9/((1+0.25)*(1-2*0.25)) R=[1-0.25, 0.25, 0;     0.25, 1-0.25, 0;     0, 0, (1-2*0.25)/2;] D=Q*R %% u1=-1/sqrt(3); v1=-1/sqrt(3); u2=1/sqrt(3); v2=1/sqrt(3); w1=1; w2=1; % Substitute u and v with u1 and v1 J_sub = subs(J, [u, v], [u1, v1]); B_sub = subs(B, [u, v], [u1, v1]); BT_sub = subs(BT, [u, v], [u1, v1]); F1=BT_sub*D*B_sub*J_sub % Substitute u and v with u1 and v2 J_sub2 = subs(J, [u, v], [u1, v2]); B_sub2 = subs(B, [u, v], [u1, v2]); BT_sub2 = subs(BT, [u, v], [u1, v2]); F2=BT_sub2*D*B_sub2*J_sub2 % Substitute u and v with u2 and v1 J_sub3 = subs(J, [u, v], [u2, v1]); B_sub3 = subs(B, [u, v], [u2, v1]); BT_sub3 = subs(BT, [u, v], [u2, v1]); F3=BT_sub3*D*B_sub3*J_sub3 % Substitute u and v with u2 and v2 J_sub4 = subs(J, [u, v], [u2, v2]); B_sub4 = subs(B, [u, v], [u2, v2]); BT_sub4 = subs(BT, [u, v], [u2, v2]); F4=BT_sub4*D*B_sub4*J_sub4 F=(F1*w1*w2)+(F2*w1*w2)+(F3*w1*w2)+(F4*w1*w2);

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Calculate the stiffness matrix for the Q8 element shown below. Make use of the shape functions in natural coordinates and the Jacobian calculated in previous problems. Use 2x2 Gaussian numerical integration. Use plane strain conditions (for unit thickness) with E=200E9 (MPa) and nu=0.25.

The final answers is F1= 0.18452e12. I have attached my matlab code below. Please look thru it and correct it. Thanks.

u=sym('u')  %%zta
v=sym('v')  %%eta

x1=1
x2=2
x3=3.5
x4=3.6
x5=4.0
x6=2.3
x7=1.2
x8=0.8

y1=1
y2=1
y3=1.3
y4=2.3
y5=3.3
y6=3.2
y7=3.0
y8=2.0
X=[x1;x2;x3;x4;x5;x6;x7;x8]
Y=[y1;y2;y3;y4;y5;y6;y7;y8]
%% diff of zta
DN1Du=1/4*(1-v)*(2*u+v);
DN2Du=-u*(1-v);
DN3Du=1/4*(1-v)*(2*u-v);
DN4Du=1/2*(1-v^2);
DN5Du=1/4*(1+v)*(2*u+v);
DN6Du=-u*(1+v);
DN7Du=1/4*(1+v)*(2*u-v);
DN8Du=-1/2*(1-v^2);

%% diff of eta
DN1Dv=1/4*(1-u)*(2*v+u);
DN2Dv=-1/2*(1-u^2);
DN3Dv=1/4*(1+u)*(2*v-u);
DN4Dv=-v*(1+u);
DN5Dv=1/4*(1+u)*(2*v+u);
DN6Dv=1/2*(1-u^2);
DN7Dv=1/4*(1-u)*(2*v+u);
DN8Dv=-v*(1-u);
%% find dx/du dx/dv, dy/du and dy/dv
a=[DN1Du, DN2Du, DN3Du, DN4Du, DN5Du, DN6Du,DN7Du, DN8Du]
dxdu=a*X
b=[DN1Dv, DN2Dv, DN3Dv, DN4Dv, DN5Dv, DN6Dv, DN7Dv,DN8Dv]
dxdv=b*X
dydu=a*Y
dydv=b*Y

%% Jacobian 
J=dxdu*dydv-dxdv*dydu

%% define DNiDx

DN1Dx=1/J*(dydv*DN1Du-dydu*DN1Dv)
DN2Dx=1/J*(dydv*DN2Du-dydu*DN2Dv)
DN3Dx=1/J*(dydv*DN3Du-dydu*DN3Dv)
DN4Dx=1/J*(dydv*DN4Du-dydu*DN4Dv)
DN5Dx=1/J*(dydv*DN5Du-dydu*DN5Dv)
DN6Dx=1/J*(dydv*DN6Du-dydu*DN6Dv)
DN7Dx=1/J*(dydv*DN7Du-dydu*DN7Dv)
DN8Dx=1/J*(dydv*DN8Du-dydu*DN8Dv)

%% defrine DniDx

DN1Dy=1/J*(-dxdv*DN1Du+dxdu*DN1Dv)
DN2Dy=1/J*(-dxdv*DN2Du+dxdu*DN2Dv)
DN3Dy=1/J*(-dxdv*DN3Du+dxdu*DN3Dv)
DN4Dy=1/J*(-dxdv*DN4Du+dxdu*DN4Dv)
DN5Dy=1/J*(-dxdv*DN5Du+dxdu*DN5Dv)
DN6Dy=1/J*(-dxdv*DN6Du+dxdu*DN6Dv)
DN7Dy=1/J*(-dxdv*DN7Du+dxdu*DN7Dv)
DN8Dy=1/J*(-dxdv*DN8Du+dxdu*DN8Dv)

%% B matrix

B=[DN1Dx, 0, DN2Dx, 0, DN3Dx, 0, DN4Dx, 0, DN5Dx, 0, DN6Dx, 0, DN7Dx, 0, DN8Dx, 0;
   0, DN1Dy, 0, DN2Dy, 0, DN3Dy, 0, DN4Dy, 0, DN5Dy, 0, DN6Dy, 0, DN7Dy, 0, DN8Dy;
   DN1Dy, DN1Dx, DN2Dy, DN2Dx, DN3Dy, DN3Dx,DN4Dy,DN4Dx,DN5Dy,DN5Dx,DN6Dy,DN6Dx, DN7Dy,DN7Dx,DN8Dy,DN8Dx;] 

BT=transpose(B)

%% define D strain E=200*10^9 nu=0.25
Q=200*10^9/((1+0.25)*(1-2*0.25))
R=[1-0.25, 0.25, 0;
    0.25, 1-0.25, 0;
    0, 0, (1-2*0.25)/2;]
D=Q*R
%%
u1=-1/sqrt(3);
v1=-1/sqrt(3);
u2=1/sqrt(3);
v2=1/sqrt(3);

w1=1;
w2=1;

% Substitute u and v with u1 and v1
J_sub = subs(J, [u, v], [u1, v1]);
B_sub = subs(B, [u, v], [u1, v1]);
BT_sub = subs(BT, [u, v], [u1, v1]);

F1=BT_sub*D*B_sub*J_sub

% Substitute u and v with u1 and v2
J_sub2 = subs(J, [u, v], [u1, v2]);
B_sub2 = subs(B, [u, v], [u1, v2]);
BT_sub2 = subs(BT, [u, v], [u1, v2]);

F2=BT_sub2*D*B_sub2*J_sub2

% Substitute u and v with u2 and v1
J_sub3 = subs(J, [u, v], [u2, v1]);
B_sub3 = subs(B, [u, v], [u2, v1]);
BT_sub3 = subs(BT, [u, v], [u2, v1]);

F3=BT_sub3*D*B_sub3*J_sub3

% Substitute u and v with u2 and v2
J_sub4 = subs(J, [u, v], [u2, v2]);
B_sub4 = subs(B, [u, v], [u2, v2]);
BT_sub4 = subs(BT, [u, v], [u2, v2]);

F4=BT_sub4*D*B_sub4*J_sub4

F=(F1*w1*w2)+(F2*w1*w2)+(F3*w1*w2)+(F4*w1*w2);

The image is a labeled graph plotted on a Cartesian coordinate system with the x-axis and y-axis. The graph appears to depict a polygonal shape formed by connecting a series of vertices. Each vertex is represented as a black dot and is labeled with both a number and its corresponding coordinates in the format (x, y). The vertices are labeled and listed in order, as follows: 1. Vertex 1: (1.0, 1.0) 2. Vertex 2: (2.0, 1.0) 3. Vertex 3: (3.5, 1.3) 4. Vertex 4: (3.6, 2.3) 5. Vertex 5: (4.0, 3.3) 6. Vertex 6: (2.3, 3.2) 7. Vertex 7: (1.2, 3.0) 8. Vertex 8: (0.8, 2.0) The vertices are sequentially connected with lines to form an enclosed shape. The diagram indicates a somewhat irregular polygon spanning from x = 0.8 to x = 4.0 along the x-axis and from y = 1.0 to y = 3.3 along the y-axis. The direction of increasing values for each axis is indicated with arrows.
Transcribed Image Text:The image is a labeled graph plotted on a Cartesian coordinate system with the x-axis and y-axis. The graph appears to depict a polygonal shape formed by connecting a series of vertices. Each vertex is represented as a black dot and is labeled with both a number and its corresponding coordinates in the format (x, y). The vertices are labeled and listed in order, as follows: 1. Vertex 1: (1.0, 1.0) 2. Vertex 2: (2.0, 1.0) 3. Vertex 3: (3.5, 1.3) 4. Vertex 4: (3.6, 2.3) 5. Vertex 5: (4.0, 3.3) 6. Vertex 6: (2.3, 3.2) 7. Vertex 7: (1.2, 3.0) 8. Vertex 8: (0.8, 2.0) The vertices are sequentially connected with lines to form an enclosed shape. The diagram indicates a somewhat irregular polygon spanning from x = 0.8 to x = 4.0 along the x-axis and from y = 1.0 to y = 3.3 along the y-axis. The direction of increasing values for each axis is indicated with arrows.
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