HW 3, Due Oct 16, 2014

The objective function is $\frac{1}{2}\left \Vert d\right \Vert ^{2}$ where $d$ is the distance from origin the plane. Hence $Q\left ( y\right ) =\frac{1}{2}\left ( y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right ) $. The constraint $R=y_{1}+2y_{2}+2y_{3}-18$. Therefore, the Lagrangian is\begin{align*} L\left ( y,x\right ) & =Q\left ( y\right ) +xR\\ & =\frac{1}{2}\left ( y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right ) +x\left ( y_{1}+2y_{2}+2y_{3}-18\right ) \end{align*}

Now we set up the optimization problem\begin{align*} \frac{\partial L}{\partial y_{1}} & =y_{1}+x=0\\ \frac{\partial L}{\partial y_{2}} & =y_{2}+2x=0\\ \frac{\partial L}{\partial y_{3}} & =y_{3}+2x=0\\ \frac{\partial L}{\partial x} & =y_{1}+2y_{2}+2y_{3}-18 \end{align*}

In Matrix form\begin{equation} \fbox{$\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 2\\ 1 & 2 & 2 & 0 \end{pmatrix}\begin{pmatrix} y_{1}\\ y_{2}\\ y_{3}\\ x \end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\\ 18 \end{pmatrix} $}\tag{1} \end{equation} Comparing the above to the standard form given\[\begin{pmatrix} I & A\\ A^{T} & 0 \end{pmatrix}\begin{pmatrix} y\\ x \end{pmatrix} =\begin{pmatrix} 0\\ 18 \end{pmatrix} \] We see that $A=\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} $. Now we solve (1) using Gaussian elimination\[\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 2\\ 1 & 2 & 2 & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 2\\ 0 & 2 & 2 & -1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 2\\ 0 & 0 & 2 & -5 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & -9 \end{pmatrix} \] Hence $U=\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & -9 \end{pmatrix} $ and $L=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 1 & 2 & 2 & 1 \end{pmatrix} $. Therefore $Lc=x$ or\[\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 1 & 2 & 2 & 1 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\\ c_{4}\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\\ 18 \end{pmatrix} \] Hence $c_{1}=0,c_{2}=0,c_{3}=0,\,c_{4}=18$. Now solving $Ux=c$\[\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & -9 \end{pmatrix}\begin{pmatrix} y_{1}\\ y_{2}\\ y_{3}\\ x \end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\\ 18 \end{pmatrix} \] Hence solution is Solution is: \[\begin{pmatrix} y_{1}\\ y_{2}\\ y_{3}\\ x \end{pmatrix} =\begin{pmatrix} 2\\ 4\\ 4\\ -2 \end{pmatrix} \] So the Lagrangian multiplier is $x=-2$. Now we can calculate the distance\begin{align*} d & =\sqrt{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}\\ & =\sqrt{2^{2}+4^{2}+4^{2}}\\ & =6 \end{align*}

3.3.2 Problem 2.2.8

3.3.2.1 Part(i)

Let us assume that \[ y=k\times \left [ 1,2,2\right ] \] where $k$ is this multiple. This means $y_{1}=k,y_{2}=2k,y_{2}=2k.$ In other words, the vector is \[ y=\left [ k,2k,2k\right ] \] But since the constraint is $y_{1}+2y_{2}+2y_{3}=18$ this substituting the values of each $y_{i}$ in the constraint gives\begin{align*} k+2\left ( 2k\right ) +2\left ( 2k\right ) & =18\\ 9k & =18 \end{align*}

Hence\[ \fbox{$k=2$}\] Using this $k$, the vector is \begin{align*} y & =\left [ k,2k,2k\right ] \\ & =\left [ 2,4,4\right ] \end{align*}

Hence the norm of the vector is\begin{align*} \left \Vert y\right \Vert & =\sqrt{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}\\ & =\sqrt{2^{2}+4^{2}+4^{2}}\\ & =6 \end{align*}

3.3.2.2 Part(ii)

Using \begin{align} 18 & \leq 3\left \Vert y\right \Vert \tag{1}\\ \left \Vert y\right \Vert & \geq 6\nonumber \end{align}

In (1), $18=f$ from the equation $A^{T}y=f$ and $3=\left \Vert A\right \Vert $. This means the \[ y_min= \frac{f}{\left \Vert A\right \Vert } \]

3.3.3 Problem 2.2.9

The primal problem is minimization of $Q\left ( y\right ) $ over $y$ (unconstrained optimization), and the dual problem is maximization of $-P\left ( x\right ) $ over $x$. The minimum of $Q\left ( y\right ) $ is the maximum of $-P\left ( x\right ) $. This is the weak duality. In this problem, the point on the line must also be on a point on the plane since the line is constrained to be on the plane.

So the distance to the plane can not be larger than the distance to the line. The distance to the plane is represented by $-P\left ( x\right ) $ and the distance to the the line is represented by $Q\left ( y\right ) $. So this leads to \[ -P\left ( x\right ) \leq Q\left ( y\right ) \]

3.3.4 Problem 2.2.10

To ﬁnd distance to $S$, we need to solve\begin{align*} \left ( \text{distance to S}\right ) ^{2} & =\min _{x}\left ( Ax-b\right ) ^{T}\left ( Ax-b\right ) \\ & =\min _{x}\left ( \begin{pmatrix} 2\\ 1 \end{pmatrix} x-\begin{pmatrix} 15\\ 10 \end{pmatrix} \right ) ^{T}\left ( \begin{pmatrix} 2\\ 1 \end{pmatrix} x-\begin{pmatrix} 15\\ 10 \end{pmatrix} \right ) \\ & =\min _{x}\left ( x-10\right ) ^{2}+\left ( 2x-15\right ) ^{2}\\ & =\min _{x}5x^{2}-80x+325 \end{align*}

Hence $\frac{d}{dx}\left ( 5x^{2}-80x+325\right ) =10x-80$ hence $x=\frac{80}{10}=8$. Therefore \[ Ax=\begin{pmatrix} 16\\ 8 \end{pmatrix} \] To ﬁnd $y$ we need to solve\begin{align*} \left ( \text{distance to T}\right ) ^{2} & =\min _{A^{T}y=0}\left \Vert b-y\right \Vert ^{2}=\min _{A^{T}y=0}y^{T}y-2b^{T}y+b^{T}b\\ & =\min _{A^{T}y=0}\begin{pmatrix} y_{1}\\ y_{2}\end{pmatrix} ^{T}\begin{pmatrix} y_{1}\\ y_{2}\end{pmatrix} -2\begin{pmatrix} 15\\ 10 \end{pmatrix} ^{T}\begin{pmatrix} y_{1}\\ y_{2}\end{pmatrix} +\begin{pmatrix} 15\\ 10 \end{pmatrix} ^{T}\begin{pmatrix} 15\\ 10 \end{pmatrix} \\ & =y_{1}^{2}-30y_{1}+y_{2}^{2}-20y_{2}+325 \end{align*}

Need to minimize the above subject to $A^{T}y=0$ or $\begin{pmatrix} 2 & 1 \end{pmatrix}\begin{pmatrix} y_{1}\\ y_{2}\end{pmatrix} =0$, or $2y_{1}+y_{2}=0$. Therefore, we setup an optimization problem\begin{align*} L & =Q+xR\\ & =y_{1}^{2}-30y_{1}+y_{2}^{2}-20y_{2}+325+x\left ( 2y_{1}+y_{2}\right ) \end{align*}

And\begin{align*} \frac{\partial L}{\partial y_{1}} & =2y_{1}-30+2x=0\\ \frac{\partial L}{\partial y_{2}} & =2y_{2}-20+x=0\\ \frac{\partial L}{\partial x} & =2y_{1}+y_{2}=0 \end{align*}

Hence\[\begin{pmatrix} 2 & 0 & 2\\ 0 & 2 & 1\\ 2 & 1 & 0 \end{pmatrix}\begin{pmatrix} y_{1}\\ y_{2}\\ x \end{pmatrix} =\begin{pmatrix} 30\\ 20\\ 0 \end{pmatrix} \] Solving gives \[\begin{pmatrix} y_{1}\\ y_{2}\\ x \end{pmatrix} =\begin{pmatrix} -1\\ 2\\ 16 \end{pmatrix} \] Hence \[ y=\begin{pmatrix} y_{1}\\ y_{2}\end{pmatrix} =\begin{pmatrix} -1\\ 2 \end{pmatrix} \] Since now we know the optimal $Ax$ and $y$, we can ﬁnd the lengths.\[ \fbox{$\left \Vert Ax\right \Vert =\left \Vert \begin{pmatrix} 16\\ 8 \end{pmatrix} \right \Vert =8\sqrt{5}$}\] and \[ \fbox{$\left \Vert y\right \Vert =\left \Vert \begin{pmatrix} -1\\ 2 \end{pmatrix} \right \Vert =\sqrt{5}$}\] and \[ \fbox{$\left \Vert b\right \Vert =\left \Vert \begin{pmatrix} 15\\ 10 \end{pmatrix} \right \Vert =5\sqrt{13}$}\] Therefore \begin{align*} \left ( 8\sqrt{5}\right ) ^{2}+\left ( \sqrt{5}\right ) ^{2} & =\left ( 5\sqrt{13}\right ) ^{2}\\ 325 & =325 \end{align*}

3.3.5 Problem 2.2.16

The constraint is $x_{1}+x_{2}+\cdots +x_{m}=1$ and the objective function is $\frac{1}{2}\left \Vert d\right \Vert ^{2}=\frac{1}{2}\left ( x_{1}^{2}+x_{2}^{2}+\cdots +x_{m}^{2}\right ) $. Hence\[ L=\frac{1}{2}\left ( x_{1}^{2}+x_{2}^{2}+\cdots +x_{m}^{2}\right ) +x\left ( x_{1}+x_{2}+\cdots +x_{m}-1\right ) \] Setting up\begin{align*} \frac{\partial L}{\partial x_{1}} & =x_{1}+x=0\\ \frac{\partial L}{\partial x_{2}} & =x_{2}+x=0\\ & \vdots \\ \frac{\partial L}{\partial x_{n}} & =x_{n}+x=0\\ \frac{\partial L}{\partial x} & =x_{1}+x_{2}+\cdots +x_{m}-1=0 \end{align*}

Or in matrix form\[\begin{pmatrix} 1 & 0 & 0 & \cdots & 1\\ 0 & 1 & 0 & \cdots & 1\\ 0 & 0 & 1 & \cdots & 1\\ 0 & 0 & \cdots & 1 & 1\\ 1 & 1 & 1 & 1 & 0 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ \vdots \\ x_{m}\\ x \end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\\ \vdots \\ 1 \end{pmatrix} \] Solving, for speciﬁc $m$ to be able to see the pattern gives for $m=3$\[\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\\ 1 & 1 & 1 & 0 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x \end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\\ 1 \end{pmatrix} \] Solution is: \[\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x \end{pmatrix} =\begin{pmatrix} \frac{1}{3}\\ \frac{1}{3}\\ \frac{1}{3}\\ -\frac{1}{3}\end{pmatrix} \] So $x_{i}=\frac{1}{m}$ so the distance is \begin{align*} \left \Vert x_{1}^{2}+x_{2}^{2}+\cdots +x_{m}^{2}\right \Vert & =\sqrt{m\left ( \frac{1}{m}\right ) ^{2}}\\ & =\sqrt{m} \end{align*}

3.3.6 Problem 2.4.1

$m$ is number of bars, and $N$ is number of nodes. Truss is stable if $m\geq 2N-r$ where $r$ is the number of constraints. For determining rigid motion and mechanism, we need to solve $Ax=0$ and look at the solutions.


	$N\left ( \text{nodes}\right ) $	$m($bar$)$	$r$	$n=2N-r$	determinate? $m=n$	indeterminate? $m>n$	stable?

1	4	5	4	4	No	yes	stable

2	4	4	4	4	Yes	No	stable

3	4	3	4	4	No	No	mechanism

For case (3), since it is neither determinate nor indeterminate, we need to look at $Ax=0$. But it is clear that the truss in (3) will not move as a rigid body, but will deform. It is not stable. The table below summarizes the results.

3.3.7 Problem 2.4.4

The $A$ matrix is found from $A^{T}y=f$. where $f$ is a column vector of length $4$ since there are 2 nodal forces, and each has 2 components. This represents a force at each node. So we ﬁrst ﬁnd $A^{T}$. To do this, we resolve internal forces $y$ to balance the external nodal forces $f$. We assume there are nodal forces only on nodes $1,2$ in the above diagram and that $f_{3}=f_{4}=0$.

Clearly $f_{1V}=y_{1}$ to make forces balance in the vertical direction at node $1$ and that $f_{2V}=y_{3}$ for similar reason on node 2. On node 1, assuming $y_{2}$ is in positive, so in tension, then $-f_{1H}=y_{2}$ and $+f_{2H}=y_{2}$. If we had assumed $y_{2}$ is in the negative direction then we will get same result but signs reversed.

Therefore\begin{align*} f_{1V} & =y_{1}\\ f_{2V} & =y_{3}\\ f_{1H} & =-y_{2}\\ f_{2H} & =y_{2} \end{align*}

Hence $A^{T}y=f$ becomes\[ \fbox{$\overset{A^T}{\overbrace{\begin{pmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} }}\begin{pmatrix} y_{1}\\ y_{2}\\ y_{3}\end{pmatrix} =\begin{pmatrix} f_{1H}\\ f_{1V}\\ f_{2H}\\ f_{2V}\end{pmatrix} $}\] Hence \[ A=\begin{pmatrix} 0 & 1 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \] The matrix $A$ has rank $3$ and the same for $A^{T}$. For $A^{T}y=f$ to have solution, then $f$ must be in the column space of $A^{T}$. For solution, (equilibrium) we need $\sum f_{iH}=0$ and $\sum f_{iV}=0$ and moments about a point zero.

3.3.8 Problem 2.4.10

With the new truss as above, the number of bars $m=7$, and the number of nodes is $N=5$. The number of constraints $r=4$ (two from each support). Hence\begin{align*} n & =2N-r\\ & =10-4\\ & =6 \end{align*}

3.3.9 Problem 2.4.11

Work over the ﬁrst bar, of say length $L_{1}$ is \begin{align*} W_{i} & =\int \sigma _{1}\epsilon _{1}dV\\ & =\int \frac{y_{1}}{A_{1}}\frac{e_{1}}{L_{1}}A_{1}dL\\ & =y_{1}\frac{e_{1}}{L_{1}}\int dL\\ & =y_{1}\frac{e_{1}}{L_{1}}L_{1}\\ & =y_{1}e_{1} \end{align*}

Therefore, the sum all the truss is $y_{1}e_{1}+y_{2}e_{2}+\cdots +y_{m}e_{m}$ or\begin{align} W_{total} & =\begin{bmatrix} y_{1} & y_{2} & \cdots & y_{m}\end{bmatrix}\begin{bmatrix} e_{1}\\ e_{2}\\ \vdots \\ e_{m}\end{bmatrix} \nonumber \\ & =y^{T}e\tag{1} \end{align}

But \begin{align} A^{T}y & =f\nonumber \\ \left ( A^{T}y\right ) ^{T} & =f^{T}\nonumber \\ y^{T}A & =f^{T}\nonumber \\ y^{T} & =f^{T}A^{-1}\tag{2} \end{align}

Substituting (2) into (1) gives\begin{equation} W_{total}=f^{T}A^{-1}e\tag{3} \end{equation} But \[ e=Ax \] Hence (3) becomes\begin{align*} W_{total} & =f^{T}A^{-1}Ax\\ & =f^{T}x \end{align*}

This is an expression of the work done by external forces at nodes. So this says the internal work equals the external work.

3.3.10 Problem 2.4.12

The potential energy is $P\left ( x\right ) =\frac{1}{2}x^{T}A^{T}CAx-f^{T}x$. This is minimum at $A^{T}CAx=f$. Hence\begin{align*} P_{\min }\left ( x\right ) & =\frac{1}{2}x^{T}A^{T}CAx-\left ( A^{T}CAx\right ) ^{T}x\\ & =\frac{1}{2}x^{T}A^{T}CAx-x^{T}A^{T}C^{T}Ax \end{align*}

But $C=C^{T}$ since diagonal matrix, then\begin{align*} P_{\min }\left ( x\right ) & =-\frac{1}{2}x^{T}A^{T}CAx\\ -P_{\min }\left ( x\right ) & =\frac{1}{2}x^{T}A^{T}CAx \end{align*}

But strain energy is the quadratic term in $P\left ( x\right ) $, which is $\frac{1}{2}x^{T}A^{T}CAx$. Hence they are the same, which is what we are asked to show.

3.3.11 Problem 2.4.17

If we have a bar $1$, then the elongation is due to total motion of bar two nodes due to motion of all bar attached as was shown on page 124 of the text, which is\[ e_{1}=x_{1}\cos \theta _{1}-x_{3}\cos \theta _{1}+x_{2}\sin \theta _{1}-x_{4}\sin \theta _{1}\] The second bar $2$ which could have one joint common with the bar $1$, say $\left ( x_{3},x_{4}\right ) $ displacement, will then add to these when bar $2$ itself deforms. Hence for bar $2$ we have\[ e_{2}=x_{5}\cos \theta _{2}-x_{3}\cos \theta _{2}+x_{6}\sin \theta _{2}-x_{4}\sin \theta _{2}\] Where in the above $x_{3},x_{4}$ are kept the same as bar $1$ since the joint is common. Now if bar $3$ had joint $\left ( x_{1},x_{2}\right ) $ common with bar $1$, it will have\[ e_{3}=x_{1}\cos \theta _{3}-x_{7}\cos \theta _{3}+x_{2}\sin \theta _{3}-x_{8}\sin \theta _{3}\] When assembling the $Ax$ matrix the pattern given should result using trigonometric relations.

3.3 HW 3, Due Oct 16, 2014

3.3.1 Problem 2.2.7