User:Eml5526.s11.t1.ab/HW3

=Problem 2=

1.1 Number the Element and Nodes.
The Spring Systems is shown in the figure.

Superscripts to each of the stiffness values denotes the element number. Furthermore, Node Numbers are also shown in the figure. As shown in the figure, there are 4 nodes.

1.2 Assemble the Global Stiffness and Force Matrix.
For Element Number 1,$$I=1 $$ and $$J=4$$,

$$K^{(1)}=k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} k& 0 & 0 & -k\\ 0 & 0 & 0 & 0\\ 0& 0 & 0 & 0\\ -k& 0 & 0 & k \end{bmatrix}$$

For Element Number 2,$$I=4 $$ and $$J=3$$,

$$K^{(2)}=2k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} 0& 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0& 0 & 2k& -2k\\ 0& 0 & -2k & 2k \end{bmatrix}$$

For Element Number 3,$$I=3 $$ and $$J=2$$,

$$K^{(3)}=k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} 0& 0 & 0 & 0\\ 0 & k & -k & 0\\ 0& -k & k& 0\\ 0& 0 & 0& 0 \end{bmatrix}$$

For Element Number 4,$$I=1 $$ and $$J=3$$,

$$K^{(4)}=3k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} 3k& 0 & -3k & 0\\ 0 & 0& 0& 0\\ -3k& 0 & 3k& 0\\ 0& 0 & 0& 0 \end{bmatrix}$$

Assembled Global Systems Matrix will be given by,
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$$K=\sum_{e=1}^{4}\tilde{K}^{e}=\begin{bmatrix} k+3k & 0 & -3k & -k\\ 0 & k & -k & 0\\ -3k & -k &6k &-2k \\ -k & 0 & -2k & 3k \end{bmatrix}$$
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Where Superscript $$'e'$$ denotes the Element Number.

The Displacement and Global Force Matrices for this system are given by,

1.3 Partition the system and solve for the Nodal Displacements.
Eventually, The Global System of Equations is Given by,

$$\begin{bmatrix} k+3k & 0 & -3k & -k\\ 0 & k & -k & 0\\ -3k & -k &6k &-2k \\ -k & 0 & -2k & 3k \end{bmatrix}\begin{bmatrix} 0\\ 0\\ u_{3}\\ u_{4} \end{bmatrix}=\begin{bmatrix} r_{1}\\ r_{2}\\ 50\\ 0 \end{bmatrix}$$ $$\displaystyle (Eq.2.1) $$

As the first two displacements are prescribed, We can Partition the system after 2 Rows and 2 Columns,

And Can form the Matrix System of Equivalent to,

$$\begin{bmatrix} K_{E} & K_{EF}\\ K_{EF}^{T} & K_{F} \end{bmatrix}\begin{bmatrix} \bar{d}_{E}\\ d_{F} \end{bmatrix}=\begin{bmatrix} r_{E}\\ f_{F} \end{bmatrix}$$

Where,

$$K_{E}=\begin{bmatrix} 4k & 0\\ 0 & K \end{bmatrix}, K_{EF}=\begin{bmatrix} -3k & -k\\ -k& 0 \end{bmatrix},K_{EF} ^T=\begin{bmatrix} -3k & -k\\ -k & 0 \end{bmatrix}, K_{F}=\begin{bmatrix} 6k & -2k\\ -2k & 3K \end{bmatrix}, \bar{d}_{E}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}, d_{F}=\begin{bmatrix} u_{3}\\ u_{4} \end{bmatrix},r_{E}=\begin{bmatrix} r_{1}\\ r_{2} \end{bmatrix}, f_{F}=\begin{bmatrix} 50\\ 0 \end{bmatrix}$$

Solving the Following Reduced System of Equations,

$$K_{E}\bar{d}_{E}+K_{EF}d_{F}=r_{E}, K_{EF}^T\bar{d}_{E}+K_{F}d_{F}=f_{F}$$

Using $$\displaystyle (Eq.2.1) $$ we can form the Following Simultaneous Equations,

$$ 6k.u_{3}-2k.u_{4}=50$$ $$\displaystyle (Eq.2.2) $$

$$ -2k.u_{3}-3k.u_{4}=0$$ $$\displaystyle (Eq.2.3) $$

Solving These 2 Simultaneous Equations, We get,

1.4 Compute the Reaction Forces.
Now, Utilizing the First row of the math>\displaystyle (Eq.2.1) we can get

And Further, Utilizing the Second row of the math>\displaystyle (Eq.2.1) we can get,

=Problem 3=

Given
===1.1 The truss Structure Shown in the Figure. Nodes A & B are fixed. A Force of 10N Acts in the Positive x-Direction at node C. Young's Modulus is $$E=10^{11}Pa$$ and the Cross Sectional Area for all bars are $$ A= 2.10^{-2}$$===

1.1 Number the Element and Nodes.
The Truss System Along with the Element Numbers and the Node Numbers Written on it, is Shown in the figure.

1.2Assemble the Global Stiffness and Force Matrix.
For Element Number 1,$$I=1 $$ and $$J=3$$, Also Element 1 is inclined at 90 deg to the Positive X-direction,$$\phi^{(1)} =90^{0}$$, $$cos(90)=0,sin(90)=1$$, $$l=1$$,

Using the Following formula for obtaining the Elemental Stiffness Matrix for all the elements,

$$K^{e}=\frac{AE}{l}\begin{bmatrix} (cos\phi ^{e})^{2} & cos\phi ^{e}.sin\phi ^{e} & -(cos\phi ^{e})^{2} & -cos\phi ^{e}.sin\phi ^{e} \\ cos\phi ^{e}.sin\phi ^{e} & (sin\phi ^{e})^{2} & -cos\phi ^{e}.sin\phi ^{e} &-(sin\phi ^{e})^{2} \\ -(cos\phi ^{e})^{2} & -cos\phi ^{e}.sin\phi ^{e} & (cos\phi ^{e})^{2} & cos\phi ^{e}.sin\phi ^{e}\\ -cos\phi ^{e}.sin\phi ^{e} & -(sin\phi ^{e})^{2} & cos\phi ^{e}.sin\phi ^{e} & (sin\phi ^{e})^{2} \end{bmatrix}$$ $$\displaystyle (Eq.3.1) $$

Utilizing above equation for Element 1,

$$K^{(1)}=\frac{AE}{l}\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 &1 &  0&-1 \\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 \end{bmatrix}$$

For Element Number 2,$$I=2$$ and $$J=4$$, Also Element 2 is inclined at 90 deg to the Positive X-direction,$$\phi^{(2)} =90^{0}$$, $$cos(90)=0,sin(90)=1$$, $$l=1$$, Therefore,

$$K^{(2)}=\frac{AE}{l}\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 &1 &  0&-1 \\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 \end{bmatrix}$$

For Element Number 3,$$I=2$$ and $$J=3$$, Also Element 3 is inclined at 135 deg to the Positive X-direction,$$\phi^{(3)} =135^{0}$$, $$cos(135)=\frac{-1}{\sqrt{2}},sin(135)=\frac{1}{\sqrt{2}}$$, $$l=\sqrt{2}$$, Therefore,

$$K^{(3)}=\frac{AE}{l}\begin{bmatrix} \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}}\\ -\frac{1}{2\sqrt{2}} &\frac{1}{2\sqrt{2}} &  \frac{1}{2\sqrt{2}}&-\frac{1}{2\sqrt{2}} \\ -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}\\ \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} \end{bmatrix}$$

For Element Number 4,$$I=3$$ and $$J=4$$, Also Element 4 is inclined at 90 deg to the Positive X-direction,$$\phi^{(2)} =0^{0}$$, $$cos(0)=1,sin(0)=0$$, $$l=1$$, Therefore,

$$K^{(4)}=\frac{AE}{l}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 &0 &  0&0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Making the Assembly of all these 4 Elemental Matrices, we get the following Global Stiffness Mtria for the given Truss System,

$$\displaystyle (Eq.3.2) $$ Now, Only force is acting in positive X-Direction at C, and there will be Reactions at A & B.

Therefore, The Force Matrix Can be Assembled as,

$$\displaystyle (Eq.3.3) $$

1.3Partition the system and solve for the Nodal Displacements.
From $$\displaystyle (Eq.3.2) $$ & $$\displaystyle (Eq.3.3)  $$ The Global System os Equation will be,

$$AE \begin{bmatrix} 0 & 0 &0 & 0 & 0 & 0 &  0& 0\\ 0 & 1 & 0 & 0 & 0 & -1 & 0 &0 \\ 0 & 0 & \frac{1}{2\sqrt{2}} & \frac{-1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} &0  &0 \\ 0 & 0 & -\frac{1}{2\sqrt{2}} &1+\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & -1\\ 0 & 0 & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & 1+\frac{1}{2\sqrt{2}} &-\frac{1}{2\sqrt{2}} & -1 &0 \\ 0 &-1 &\frac{1}{2\sqrt{2}}  & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} &1+\frac{1}{2\sqrt{2}}  & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 \end{bmatrix}.\begin{bmatrix} 0\\ 0\\ 0\\ 0\\ u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix}=\begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ 0\\ 10\\ 0 \end{bmatrix}$$

As the Displacements are prescribed untill second node, Partitioning the systems after second row and column and further comparing with the following equation,

$$\begin{bmatrix} K_{E} & K_{EF}\\ K_{EF}^{T} & K_{F} \end{bmatrix}\begin{bmatrix} \bar{d}_{E}\\ d_{F} \end{bmatrix}=\begin{bmatrix} r_{E}\\ f_{F} \end{bmatrix}$$

we get, $$\bar{d_{E}}=\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix},\ d_{F}=\begin{bmatrix} u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix},\ r_{E}=\begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y} \end{bmatrix},\ f_{F}=\begin{bmatrix} 0\\ 0\\ 10\\ 0 \end{bmatrix},\ K_{F}=\begin{bmatrix} 1+\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -1 &0 \\ -\frac{1}{2\sqrt{2}} &1+\frac{1}{2\sqrt{2}} &0  & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 &1 \end{bmatrix},\ K_{EF}=\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 &-1 &0  & 0\\ -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} &  0& 0\\ \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} &0 &-1 \end{bmatrix}$$

Eventually, we can obtain,

$$d_{F}=K_{F}^{-1}(f_{F}-K_{EF}^T.\bar{d}_{E})$$

But $$\bar{d}_{E}$$ is a Zero Matrix, which will yield, $$d_{F}=K_{F}^{-1}.f_{F}$$ Solving this system in MATLAB, Using the following code, we get,

 Matlab Code: 

$$\displaystyle (Eq.3.4) $$

1.4Compute the Stresses & Reactions.
Now the Reactions can be calculated as,

$$r_{E}=K_{E}.\bar{d}_{E}+K_{EF}.d_{F}$$

But again $$\bar{d}_{E}$$ is a Zero Matrix, Which will give, $$r_{E}=K_{EF}.d_{F}$$.

Using the Following MATLAB Code to obtain the Reaction Matrix,

 Matlab Code: 

$$\displaystyle (Eq.3.5) $$

Now We can calculate the Stress in each element using the following formula,

$$\sigma ^e=\frac{E^e}{l^e}\begin{bmatrix} -cos\phi ^{e} &-sin\phi ^{e} & cos\phi ^{e} & sin\phi ^{e} \end{bmatrix}.d^e$$ $$\displaystyle (Eq.3.6) $$

Now we can utilize the above equation to calculate the Stresses for all the elements,

For element 1, $$\phi =90^0, l=1$$ &

$$d^1=\begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{3x}\\ u_{3y} \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 38.2843\\ 10 \end{bmatrix}$$ Using the Following MATLAB code,  Matlab Code: 

Therefore,

In the similar fashion, For Element 2, Utilizing $$\displaystyle (Eq.3.6) $$ we get,

For Element 3, Utilizing $$\displaystyle (Eq.3.6) $$ we get,

Finally, For Element 4, Utilizing $$\displaystyle (Eq.3.6) $$ we get,