FEA Solution Final and Complete

8/14/2019 FEA Solution Final and Complete

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Question No 1

As a general procedure, we take the node numbering as shown in the right side figure.Moreover, all the equations are derived using same node number arrangement. So, asa first step, we will transform the node numbering so as to make them the same for

which the derivation of element stiffness matrices and shape functions has been carriedout.

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inches

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2

Now we will write the results in the original node numbering which was given in questionstatement

4.35645.554960.931465.946371.1192

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Question No 2

Part (a)

As a first step, we will determine Aatrix [D].

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Part (b)

In the given question statement, the displacements U 1, U 2U 13 are the valuesof global displacements. We first draw each element separately and establishrelationship between local displacements of that particular element with the global

displacements. Subsequently, we will determine the respective elements of ElementStiffness Matrix for each element. We will then use the principle of superposition to therequired elements of Structural Stiffness Matrix [K]. Now taking each element one byone, we follow the above mentioned procedure.

Figure 2 (a)

1 2

34

1 2

34

1 2

34

1 2

34

Elea 1

U1

U2

U3

U4

U5

2

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U7

U8

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U4

U5

1

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2

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U11

U12

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U7

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U9

U10

U11

U12

U13


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Figure 2(b)

Element K U2U2

The displacement U 2 is shared by Element 1 and 2. So the effect of KU2U2 will come

from both the elements. So, considering both the elements one by one.Element 1:Global U 2 ------- Local u 2 So KU2U2 ------- K u2u2 for element 1From stiffness matrix [K] (given in question No 1) we find

3 3 Element 2:Global U 2 ------- Local u 1

So KU2U2 ------- K u1u1 for element 2From stiffness matrix [K] (given in question No 1) we find

3 3 3 3 3 3 2 32.967 33 411.538 43 311.538 33 4 10 24.967 10 Element K U6U7 The displacement U 6 and U 7 are shared by Elements 2 and 3. So the effect of KU6U7 will

come from both the elements. So, considering both the elements one by one.

Element 2:Global U 6 ------- Local u 2 Global U 7 ------- Local v 2 So KU6U7 ------- K u2v2 for element 2From stiffness matrix [K] (given in question No 1) we find

4

Element 3:Global U 6 ------- Local u 1 Global U 7 ------- Local v 1 So KU6U7 ------- K u1v1 for element 3From stiffness matrix [K] (given in question No 1) we find


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4

4 4

0Element K U7U6 The displacement U 6 and U 7 are shared by Elements 2 and 3. So the effect of KU7U6 willcome from both the elements. So considering both the elements one by one.

Element 2:Global U 6 ------- Local u 2

Global U 7 ------- Local v 2 So KU7U6 ------- K v2u2 for element 2From stiffness matrix [K] (given in question No 1) we find

4 Element 3:Global U 6 ------- Local u 1 Global U 7 ------- Local v 1 So KU7U6 ------- K v1u1 for element 3From stiffness matrix [K] (given in question No 1) we find

4

4 4 0Element K U5U12

The displacements U 5 and U 12 does not have any common element. U 5 belongs toelements 1 and 2, while U 12 is associated with element 3. Therefore, in the StructuralStiffness Matrix, the place of element K U5U12 will be empty. Hence we can conclude that

0


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Part (c)

In order to find the nodal loads, we will have to convert the distributed loading toits equivalent nodal loading. For this purpose, as the distributed load is applied toelement 3 only, for simplicity purpose, we will consider only this element in our

calculations.

,

0234 0,0,4, 44 00 0 0 00 00 00 0 0 00 00

0 0 0 0 04234044234 0 0 0 0 0 416 0 238 416

00000806


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Question No 3

As per the principle of Virtual Work, the work done by the body forces aust beequal to the work done by the concentrated applied loads in the absence of any other type of loading.

,

1 1

2 10,0 2,0 3, 40, : 1111 00 00 000

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1

1 0 0 0110 010 011 11 1

1


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0, ,


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Q No 4

Part (a)

In this question, the center of the coordinate system has been shifted to the

centroide of the rectangle. Therefore, the corresponding shape matrix will be differentfor this case. Hence, before going to the matlab programming we will first find matrix [A]which will subsequently be used in matlab code for calculation of shape matrix.

1 1 2

x

y

12

34

a

b

12

34

a

b

qx2

qy2qy1


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12,2 2 2 4 22,2 2 2 4 32,2 2 2 4

42,2 2 2 4 : 122412 2 4122 4122 4

122412 2 4122 4122 4


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Part (b)

Now, we will use this new [A] for calculation of our shape matrix using matlab

The Matlab code is as follows:

%---------- SOLUTION CODE FOR Q NO 4 ---------------%----------(ASSIGNMENT FEA)-------------------------clcclear allsyms x yP=1000;% DEFINING a AND ba=2.0;

b=2.0;% DEFININT MATERIAL PROPERTIESv=0.3;E=70000000000;%A=[1 -a/2 -b/2 a*b/4;1 a/2 -b/2 -a*b/4;1 a/2 b/2 a*b/4;1 -a/2 b/2 -a*b/4];phi=[1 x y x*y];N=phi*(inv(A));N1=N(1,1);N2=N(1,2);N3=N(1,3);N4=N(1,4);NN=[N1 0 N2 0 N3 0 N4 0;0 N1 0 N2 0 N3 0 N4];B=[diff(N1,x) 0 diff(N2,x) 0 diff(N3,x) 0 diff(N4,x) 0;0 diff(N1,y) 0 diff(N2,y) 0 diff(N3,y) 0diff(N4,y);diff(N1,y) diff(N1,x) diff(N2,y) diff(N2,x) diff(N3,y) diff(N3,x) diff(N4,y)diff(N4,x)];% DEFINING MATRICES [D] AND [B]D=(E/((1+v)*(1-2*v))).*[1-v v 0;v 1-v 0;0 0 (1-2*v)/2];Bt=transpose(B); %TRANSPOSE OF [B]k1=Bt*D*B;K=int(int(k1,x,-a/2,a/2),y,-b/2,b/2);% DEFINING Kff -----------------%% FREE NODES -- U1, U3, V3, U4, V4Kff=[K(1,3) K(1,5) K(1,6) K(1,7) K(1,8); K(5,3) K(5,5) K(5,6) K(5,7) K(5,8);K(6,3) K(6,5)K(6,6) K(6,7) K(6,8);K(7,3) K(7,5) K(7,6) K(7,7) K(7,8);K(8,3) K(8,5) K(8,6) K(8,7)K(8,8)];F=[0;0;-P;0;0] % DEFINING LOAD VECTOR {F}Uf=(inv(Kff))*F;%


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%%---- PART (b)-------%U=[0;0;Uf(1,1);Uf(2,1);Uf(3,1);Uf(4,1);0;0];U=[Uf(1,1);0;0;0;Uf(2,1);Uf(3,1);Uf(4,1);Uf(5,1)]

Epsilon=B*USigma=D*Epsilon

The Resultant displacement and stress values are as follows:

Displacement ValueU1 5.57143E 09V1 0U2 0V2 0U3 3.03333E 08

V3 3.77619E 08U4 2.47619E 08V4 1.17619E 08

Part (c)

The resulting relations for stresses are as follows:

Stress Value x 262.5y 262.5 500x y 112.5y 612.5 1166.667xXY 75x 333.333y


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Part (d)

The same problem was modeled and run in Ansys. The model with boundary conditionsand displacement plot are shown below:

(Figure 1- Force and Boundary Conditions)

(Figure 2- Displacement Plot)


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In the following table, Ansys results are shown in comparison with Matlab results:

PARAMETER MATLAB RESULTS ANSYS RESULTSU1 5.57143E 09 -5.234E-9V1 0 0

U2 0 0V2 0 0U3 3.03333E 08 2.9123E-8V3 3.77619E 08 -3.567E-8U4 2.47619E 08 -2.442E-8V4 1.17619E 08 -1.23E-8


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Question No 5

1 2

3 , , 200 00 ,3 4

12664 10000 10 121223 1000010230.12 0.60.6 4

4 320001003 55 1200

P=20

B

X

Y

v2, 2

v1, 1


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1320 2 12662 20003 0.12 0.60.6 2 20003 0.12 0.60.6 21320 20200 200 1

23300

1

300 Part (b)

The matlab code for the beam problem is as follows:

%---------- SOLUTION CODE FOR Q NO 5 ---------------%----------(ASSIGNMENT FEA)-------------------------clcclear allP=20; %FORCE AT THE FREE ENDM=0; %MOMENT AT THE FREE END% DEFINING GeometryNumberofElement=10;L=10;b=1;h=2;% DEFININT MATERIAL PROPERTIESE=10000;I=(b*h^3)/12;m=2*NumberofElement;%% CONSTRUCTING Kff %n=m;temp=1;for i=1:m

for j=1:n


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Kff(i,j)=0;end

endfor i=1:m

for j=1:n

if i==jif temp==1if (m-i)>1

Kff(i,j)=2*E*I/L*(12/L^2);else Kff(i,j)=E*I/L*(12/L^2);end

endif temp==0

if (m-i)>1Kff(i,j)=2*E*I/L*4;

else Kff(i,j)=E*I/L*4;

endendendif abs(i-j)==1

Kff(i,j)=E*I/L*(6/L);endendif temp==0

temp=1;else temp=0;end

end%% CONSTRUCTING qf %for i=1:m

qf(i)=0;endqf(m-1)=P;qf(m)=M;

qf=transpose(qf);

uf=-inv(Kff)*qf vc=uf(m-1)thetac=uf(m)%----------------------------------------------------

FEA Solution Final and Complete

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