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User:Eml5526.s11.team2/hwk4

=Problem 4.1: Solve the PDE using a half Fourier Basis with the Weighted Residual Method =

Given: PDE, Boundary Conditions, and Basis Function
PDE:
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$$  \displaystyle 2\frac{d^2u}{dx^2}+3=0 $$     (4.1.1)
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Boundary Conditions:
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$$  \displaystyle u(x=1)=0 $$     (4.1.2)
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$$  \displaystyle -\frac{du}{dx}(x=0)=4 $$     (4.1.3)
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Half Fourier Basis:
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$$  \displaystyle b_j(x)={1, sin(x), sin(2x),...,sin(jx)} $$     (4.1.4)
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Find: Unknown Coefficients in $$ u^h(x) $$ expansion to approximate $$ u(x) $$ of the PDE.
We want to approximate the solution of Eq.4.1.1 by a series expansion.
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$$  \displaystyle u^h(x)=\sum_{i=0}^n{d_ib_i(x)} $$     (4.1.5) To solve for the unknown coefficients $$ d_i, $$ we must develop the same number of linearly independent equations as we have unknowns. The number of unknowns we generate with the expansion of $$ u^h(x) $$ is entirely up to us, i.e. is it dependent on the number of degrees of freedom we decide to use for the problem formulation. The following solutions are examples of using 3, 5 and 7 degrees of freedom.
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Develop 2 equations which satisfy the given Boundary Conditions.
First let's expand Eq.4.1.1 out to n=2 terms.
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$$  \displaystyle u^h(x)=d_0b_0+d_1b_1+d_2b_2 $$     (4.1.6) Now consider Eq.4.1.4, and substitute the appropriate terms into Eq.4.1.6.
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$$  \displaystyle u^h(x)=d_0(1)+d_1\Big(sin(x)\Big)+d_2\Big(sin(2x)\Big) $$     (4.1.7) Let's rewrite this equation in terms of a product between 2 vectors.
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$$  \displaystyle u^h(x)=\left[1\quad sin(x) \quad sin(2x)\right ]^T\begin{bmatrix} d_0\\d_1\\d_2\end{bmatrix} $$     (4.1.8)
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In order for us to solve for the unknown $$d_i's$$ in Eq.4.1.8, we must have 3 linearly independent equations describing linear combinations of the $$d_i's.$$Two of the equations are easily obtained directly from the given boundary conditions in Eq.4.1.3 and Eq.4.1.4. We want our approximation to closely resemble our exact solution, therefore we must require identical behavior between Eq.4.1.6 and Eq.4.1.1 at the boundaries.
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$$  \displaystyle \begin{align} u^h(x=1)=d_0(1)+d_1sin(1)+d_2sin(2)&=0\\ \end{align} $$     (4.1.9)
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$$  \displaystyle \frac{du}{dx}(x=0)=d_1cos(0)+2d_2cos(0)=-4 $$     (4.1.11)
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Use the Weighted Residual Method to develop a $$ 3^{rd} $$ linearly independent equation.
Now we just need one more linearly independent equation and we can then readily solve for the $$d_i's.$$ To develop another set of equations which are linearly independent of Eq.4.1.10 and Eq.4.1.12, we take the inner product between, Eq.4.1.15 and our basis set of functions Eq.4.1.4. Before getting to the inner product first, let's define our linear differential operator as the following
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$$  \displaystyle P(\cdot)=\frac{d^2}{dx^2}+\frac{3}{2}=0 $$     (4.1.13) Now, using our operator in Eq.4.1.13, we can operate on our approximation in Eq.4.1.8.
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$$  \displaystyle \begin{align} P(u^h)=\frac{d^2}{dx^2}\left[1\quad sin(x)\quad sin(2x)\right ]^T\begin{bmatrix} d_0\\d_1\\d_2\end{bmatrix}+\frac{3}{2}&=0\\ \left[0\quad -sin(x)\quad -4sin(2x)\right ]^T\begin{bmatrix} d_0\\d_1\\d_2\end{bmatrix}+\frac{3}{2}&=0 \end{align} $$     (4.1.14) Which we can then rewrite Eq.4.1.14 as
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$$  \displaystyle \left[0\quad -sin(x)\quad -4sin(2x)\right ]^T\begin{bmatrix} d_0\\d_1\\d_2\end{bmatrix}=-\frac{3}{2} $$     (4.1.15) Now, we are in a position to take the inner product of Eq.4.1.15(our approximation) with the basis set in Eq.4.1.4 with $$ n=2.$$
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$$  \displaystyle \left \langle b_j, P(u^h) \right \rangle =\int_0^1b_jP(u^h)=0 $$     (4.1.16)
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$$  \displaystyle \int_0^1\begin{bmatrix} 1\\sin(x)\\sin(2x)\end{bmatrix} \left[0\quad -sin(x)\quad -4sin(2x)\right ]^T\,dx\begin{bmatrix} d_0\\d_1\\d_2\end{bmatrix}=-\frac{3}{2}\int_0^1\begin{bmatrix} 1\\sin(x)\\sin(2x)\end{bmatrix}\,dx $$     (4.1.17)
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$$  \displaystyle \int_0^1\begin{bmatrix} 0&-sin(x)&-4sin(2x)\\0&-sin^2(x)&-4sin(x)sin(2x)\\0&-sin(x)sin(2x)&-4sin^2(2x)\end{bmatrix}\,dx\begin{bmatrix} d_0\\d_1\\d_2\end{bmatrix}=-\frac{3}{2}\int_0^1\begin{bmatrix} 1\\sin(x)\\sin(2x)\end{bmatrix}\,dx $$     (4.1.18)
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$$  \displaystyle \begin{bmatrix} 0&-1&2cos(2)-2\\0&sin(2)/4-1/2&2sin(3)/3-2sin(1)\\0&sin(3)/6-sin(1)/2&sin(4)/2-2\end{bmatrix}\begin{bmatrix} d_0\\d_1\\d_2\end{bmatrix}=\begin{bmatrix} -3/2\\-3/2\\-3sin^2(1)/2\end{bmatrix} $$     (4.1.19) We can use any one of the above equations as our $$ 3^{rd} $$ equation. For simplicity, let's choose the $$ 1^{st} $$ equation in the above matrix expression. This would be
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(4.1.20)
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Now we are set to expression our formulated problem in Matrix Form.

Display in Matrix Form

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$$  \displaystyle \begin{bmatrix} 1&sin(1)&sin(2)\\0&1&2\\0&-1&2\Big(cos(2)-1 \Big)\end{bmatrix}\begin{bmatrix} d_0\\d_1\\d_2\end{bmatrix}=\begin{bmatrix} 0\\-4\\-3/2\end{bmatrix} $$     (4.1.21)
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Solve for unknown d.
Our stiffness matrix is
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$$  \displaystyle \underline{\mathbf{K}}=\begin{bmatrix} 1&sin(1)&sin(2)\\0&1&2\\0&-1&2\Big(cos(2)-1 \Big)\end{bmatrix} $$     (4.1.22) Using MATLAB to aid in our solution we find the inverse of our stiffness matrix to be
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$$  \displaystyle \underline{\mathbf{K}}^{-1}=\begin{bmatrix} 1&0&0\\1327/1577&1&-1\\401/441&2&-2644/1443\end{bmatrix} $$     (4.1.23) Therefore, we find $$ {\mathbf{d}} $$ to be
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$$  \displaystyle {\mathbf{d}} = {\mathbf{K}}^{-1}{\mathbf{f}} $$     (4.1.24)
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$$  \displaystyle \begin{bmatrix} d_0\\d_1\\d_2\end{bmatrix}=\begin{bmatrix} 1&0&0\\1327/1577&1&-1\\401/441&2&-2644/1443\end{bmatrix} \begin{bmatrix}0\\-4\\-3/2\end{bmatrix} $$     (4.1.25)
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Construct,Plot and find the error between the approximate and exact solutions for u(x).
A plot of both the approximate and exact solutions can be found below.

We can see that our approximate solution is though very close to exact, does not converge as quickly as we saw the polynomial basis function converge. This to be due to the slower converging half Fourier Basis when approximating a polynomial, versus using a polynomial basis. When using a polynomial basis of degree n, it will provide us with the exact solution of any polynomial of degree n or less. This can be easily proven through the use of the Integral Mean Value Theorem. Let's move on to using 5 degrees of freedom (n=4) and see our or approximate solution can improve.
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$$  \displaystyle \Delta \mathbf{E_{abs}}=u^h(x=0.5)-u(x=0.5) $$     (4.1.27)
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Or in other words approximately 3%. Not to bad but lets see if we can get better.

Construct,Plot and find the error between the approximate and exact solutions for u(x)..
Being that we have extensively explained the mathematics for case 2 above, we will just show the MATLAB code that I have written to solve this problem while focus more on the results of increasing the degrees of freedom, i.e. the number of coefficients that we expand to in our approximate solution. This not only makes it easier on the writer, but on the reader as well. The following is the Plot for the case n=4.

We see that we have a little further away with the 2 extra degrees of freedom. Let's calculate our error below.
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$$  \displaystyle \begin{align} \Delta \mathbf{E_{abs}}=u^h(x=0.5)-u(x=0.5) \end{align} $$     (4.1.29) Or in other words we have increased our error with an increase in expansion terms of our approximate solution.
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Computational Difficulties
This case proved to be too computational demanding for my computer. I let the code, which worked for both cases 2 and 4 above run for 45 min and still MATLAB said it was busy. Therefore, I force quit my program and can conclude our best result is obtained with the n=2 case.

=Problem 4.2: Heat Conduction Fish & Belytschko Page 73 Problem 3.5 =

Given: The Following Boundary Conditions
Where (4.2.1) is the essential boundary condition defined at $$ \displaystyle x = 0 $$ and (4.2.2) is the natural boundary condition defined at $$ \displaystyle x = 10 $$ on the domain $$ \Omega = ]0,10[ \quad $$

Find:The Weak Form for 1-D Heat Conduction From the Strong Form
Obtain the weak form for the equation of the heat conduction with the boundary conditions stated above. The condition on the natural boundary condition is a convection condition.

Solution: Obtaining the Weak Form
The strong form has the following equation for a 1-D heat conduction body.

Assuming a steady-state from for the heat conduction and a constant area ( 4.2.3) reduces to

Multiplying by an arbitrary weighting function $$ \displaystyle w(x) $$ and integrating with respect to x across the domain leads to the following form

Making use of Integration by Parts on the first term of (4.2.5) leads to the following equation.

Using the requirement of the essential boundary condition on the weighting function satisfying the essential boundary location implies that $$ \displaystyle w(0) = 0 $$ and also using the natural boundary condition (4.2.2) $$ k \frac{\partial T}{\partial x}\bigg|_{x=0} = hT $$ leads to the following

Therefore the weak form can be represented finding a $$ \displaystyle T(x) $$ such that for all smooth $$ \displaystyle T(x) $$ with $$ \displaystyle T(0) = 100 $$ (4.2.8) is satisfied.

= Problem 4.3: Construct weak form of heat problem and find solution =

Given: Strong Form and Boundary Conditions
Statement:- Given the strong form for the heat conduction problem in a circular plate:
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$$  \displaystyle k\frac{d}\left( {r\frac} \right) + rs = 0,0 < r \leqslant R $$ (4.3.1) natural boundary condition :$$ \frac\left( {r = 0} \right) = 0$$
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essential boundary condition :$$T\left( {r = R} \right) = 0$$

where R is the total radius of the plate, s is the heat source per unit length along the plate radius, T is the temperature and k is the conductivity. Assume that k, s and R are given:

a. Weak Form
Construct the weak form for the above strong form.

b. Solution Using Quadratic Polynomial Trial Function
Use quadratic trial (candidate) solutions of the form $$ \displaystyle T = {\alpha _0} + {\alpha _1}r + {\alpha _2}{r^2}$$ and weight functions of the same form to obtain a solution of the weak form.

c. Solve the Differential Equation
Solve the differential equation with the boundary conditions and showthat the temperature distribution along the radius is given by:

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$$  \displaystyle T = \frac{s}\left( {{R^2} - {r^2}} \right) $$     (4.3.2)
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a. Weak Form Derivation
Strong Form Can be written as:

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$$  \displaystyle p\left( T \right) = 0 $$
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$$  \displaystyle p\left( T \right) = k\frac{d}\left( {r\frac} \right) + rs $$
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$$w\left( r \right)$$ is weight function Weighted Residual Form can be written as

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$$  \displaystyle \int\limits_0^R {w\left( r \right)p\left( T \right)dr = 0} $$     (4.3.3)
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$$  \displaystyle \int\limits_0^R {w\left( r \right)\left[ {k\frac{d}\left( {r\frac} \right) + rs} \right]dr} = 0 $$
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$$  \displaystyle \int\limits_0^R {w\left( r \right)k\frac{d}\left( {r\frac} \right)dr + } \int\limits_0^R {w\left( r \right)rsdr} = 0 $$
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Write $$w\left( r \right) = w$$ Integrating the first term by parts.

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$$  \displaystyle \left[ {wkr\frac} \right]_0^R - \int\limits_0^R {\left( {\frackr\frac} \right)} dr + \int\limits_0^R {wrsdr} = 0 $$
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Assuming $$w\left( R \right) = 0$$ and using natural boundary condition $$\frac\left( {r = 0} \right) = 0$$

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$$  \displaystyle 0 - 0 - \int\limits_0^R {\left( {\frackr\frac} \right)} dr + \int\limits_0^R {wrsdr} = 0 $$
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$$  \displaystyle \int\limits_0^R {\left( {\frackr\frac} \right)} dr = \int\limits_0^R {wrsdr} $$     (4.3.4)
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Writing in standard form:

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b. Solution using quadratic polynomial trial function
let $$T$$ and $$w$$ are of the following
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$$  \displaystyle T = {\alpha _0} + {\alpha _1}r + {\alpha _2}{r^2} $$ and $$ \displaystyle w = {\beta _0} + {\beta _1}r + {\beta _2}{r^2} $$     (4.3.6)
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Now using essential Boundary Condition i.e $$T\left( R \right) = 0$$

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$$  \displaystyle {\alpha _0} + {\alpha _1}R + {\alpha _2}{R^2} = 0 $$     (4.3.7)
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$$  \displaystyle {\alpha _0} = - \left( {{\alpha _1}R + {\alpha _2}{R^2}} \right) $$     (4.3.8)
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Now using Natural boundary condition i.e. $$\frac\left( {r = 0} \right) = 0$$

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$$  \displaystyle {\alpha _1} + 2{\alpha _2}\left( 0 \right) = 0 $$     (4.3.9)
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$$  \displaystyle {\alpha _1} = 0 $$     (4.3.10)
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$$  \displaystyle {\alpha _0} = - {\alpha _2}{R^2} $$     (4.3.11)
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Now using homogeneous boundary condition $$w\left( R \right) = 0$$
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$$  \displaystyle {\beta _0} + {\beta _1}R + {\beta _2}{R^2} = 0 $$     (4.3.12)
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$$  \displaystyle {\beta _0} = - \left( {{\beta _1}R + {\beta _2}{R^2}} \right) $$     (4.3.13)
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Also

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$$  \displaystyle \frac = {\alpha _1} + 2{\alpha _2}r $$     (4.3.14)
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As $${\alpha _1} = 0$$
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$$  \displaystyle \frac = 2{\alpha _2}r $$     (4.3.15)
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Now differentiating $$w$$ with respect to $$r$$
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$$  \displaystyle \frac = {\beta _1} + 2{\beta _2}r $$     (4.3.16)
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Putting values of $$w,\frac,\frac$$ in the standard weak form.

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$$  \displaystyle \int\limits_0^R {kr\left( {{\beta _1} + 2{\beta _2}r} \right)} \left( {2{\alpha _2}r} \right)dr = \int\limits_0^R {sr\left( {{\beta _0} + {\beta _1}r + {\beta _2}{r^2}} \right)} dr $$
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$$  \displaystyle k\int\limits_0^R {\left( {2{\beta _1}{\alpha _2}{r^2} + 4{\beta _2}{\alpha _2}{r^3}} \right)dr} = s\int\limits_0^R {\left( {{\beta _0}r + {\beta _1}{r^2} + {\beta _2}{r^3}} \right)dr} $$
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$$  \displaystyle \left[ {\frac{3} + k{\beta _2}{\alpha _2}{r^4}} \right]_0^R = \left[ {\frac{2} + \frac{3} + \frac{4}} \right]_0^R $$
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Putting value of $$$$ from equation

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$$  \displaystyle \frac{3} + k{\beta _2}{\alpha _2}{R^4} = \frac{2} + \frac{3} + \frac{4} $$
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Rearranging terms

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$$  \displaystyle {\beta _1}\left[ {\frac{3} + \frac{2} - \frac{3}} \right] + {\beta _2}\left[ {k{\alpha _2}{R^4} + \frac{2} - \frac{4}} \right] = 0 $$
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As $${\beta _1},{\beta _2}$$ are arbitrary they can't be zero, putting bracketed terms equal to zero both equation give same value of $${\alpha _2}$$.

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$$  \displaystyle {\alpha _2} = \frac $$     (4.3.17)
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Using equation (4.3.10) $${\alpha _0}$$
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$$  \displaystyle {\alpha _0} = - {\alpha _2}{R^2} = \frac $$     (4.3.18)
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Putting these values in $$T = {\alpha _0} + {\alpha _1}r + {\alpha _2}{r^2}$$

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$$  \displaystyle T = \frac - \frac $$
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Now checking this solution by putting in given strong form given in equation.....

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$$  \displaystyle k\frac{d}\left( {r\frac{d}\left( {\frac{s}\left( {{R^2} - {r^2}} \right)} \right)} \right) + rs = 0 $$
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$$  \displaystyle k\frac{d}\left( { - \frac} \right) + rs = 0 $$
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$$  \displaystyle k\left( { - \frac{k}} \right) + rs = 0 $$
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$$  \displaystyle - \cancel{sr} + \cancel{sr} = 0 $$     (4.3.20)
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Comments
As we are not getting any particular interval of 'r' by putting the weak form solution in the given differential equation, so this solution is valid on given domain of 'r'.

c. Exact solution for given differential equation
Starting from 4.3.1 and rearranging we get

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$$  \displaystyle d\left( {r\frac} \right) = - \frac{s}{k}rdr $$     (4.3.21)
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Integrating this we get
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$$  \displaystyle r\frac = - \frac{s}{r^2} + {c_1} $$     (4.3.22)
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Using the boundary condition and solving
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$$  \displaystyle \frac(r = 0) = 0 $$     (4.3.23)
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$$  \displaystyle \Rightarrow {c_1} = 0 $$     (4.3.24)
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Next we rearrange again and integrate
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$$  \displaystyle dT = - \frac{s}rdr $$     (4.3.25)
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$$  \displaystyle T = - \frac{s}{r^2} + {c_2} $$     (4.3.26)
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Again using the boundary condition to solve for the constant of integration
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$$  \displaystyle T(R) = 0 $$     (4.3.27)
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$$  \displaystyle \Rightarrow {c_2} = \frac{s}{R^2} $$     (4.3.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We arrive at the exact solution to the strong form which was found from the weak form
 * {| style="width:100%" border="0"

$$  \displaystyle T(r) = \frac{s}\left( {{R^2} - {r^2}} \right) $$     (4.3.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution from weak form and exact solution from the differential equation are same for given heat conduction problem.

=Problem 4.4: Deriving the Weak Form from the Strong Form for the Torsion of a Bar Fish & Belytschko Page 73 Problem 3.7 =

Given: The Strong Form
Given the strong form for the circular bar in torsion (Fig. 1):

Where the natural boundary condition is

Where the essential boundary condition is

Where the essential boundary condition $$ \displaystyle m(x) $$ is a distributed moment per unit length, M is the torsion moment, $$ \displaystyle \phi $$ is the angle of rotation, $$ \displaystyle G $$ is the shear modulus and $$ \displaystyle J $$ is the polar moment of intertia given by $$ \displaystyle J = \frac{\pi C^4}{2} $$, where $$ \displaystyle C $$ is the radius of the circular shaft.

b. Determine Constants of Integration
Assume that $$ \displaystyle m(x) = 0 $$ and integrate the differential equation given above. Find the integration constants using the boundary conditions.

(a) Obtaining the Weak Form
Multiplying (4.4.1) by an arbitrary weighting function $$ \displaystyle w(x) $$ and integrating with respect to x across the domain leads to the following form

Making use of Integration by Parts on the first term of (4.4.4) leads to the following equation.

Using the requirement of the essential boundary condition on the weighting function satisfying the essential boundary location implies that $$ \displaystyle w(0) = 0 $$ and also using the natural boundary condition (4.4.2) $$ JG \frac{\partial \phi}{\partial x}_{x=l} = M $$ leads to the following

Therefore the weak form can be represented finding a $$ \displaystyle \phi(x) $$ such that for all smooth $$ \displaystyle \phi(x) $$ with $$ \displaystyle \phi(0) = 0 $$ (4.4.7) is satisfied.

(b)Obtaining the Constants of Integration
Applying equation (4.4.1) and with $$ \displaystyle m(x) = 0 $$ leads to the following form.

Integrating (4.4.8) with respect to x generates a constant of integration and provides the following form.

Applying the natural boundary condition from (4.4.2) leads to the following

Solving (4.4.9) for $$ \displaystyle \frac{\partial \phi}{\partial x} $$ and substituting in $$ \displaystyle C_1 = -M $$ leads to the following

Integrating (4.4.11) with respect to x generates a constant of integration and provides the following form, which is the well known equation for angle of twist of a bar.

Applying the essential boundary condition from (4.4.3) to (4.4.12) leads to the following

Therefore the constants and the angle of twist are as follows.

=Problem 4.5: Constraint Breaking Solution (CBS) for Multiple Families of Basis Functions=

Given: Multiple Families of Basis Functions on a Domain
For

The possible basis function families:

e) Exponential Family:
Where the above families satisfy the the CBS with a domain of

1) Show the Following and Plot $$ \displaystyle b_j $$ and $$ \displaystyle \bar{b_j} $$ for $$ \displaystyle j=1,2,3$$
i) $$ \displaystyle \overline  (x) = const $$ ii) And plot $$\displaystyle {b_j} $$ and $$\displaystyle \overline   $$ for $$\displaystyle j=1,2,3 $$

a) Polynomial Family
Satisfying the CBS for the zero term:

Satisfying the CBS for the remaining terms:

The CBS family solution:

Plot of basis functions and CBS basis functions vs. the domain of -2 to 4:

b) Cosine Family
Satisfying the CBS for the zero term:

Satisfying the CBS for the remaining terms:

The CBS family solution:

Plot of basis functions and CBS basis functions vs. the domain of -2 to 4:

c) Sine Family
Satisfying the CBS for the zero term:

Satisfying the CBS for the remaining terms:

The CBS family solution:

Plot of basis functions and CBS basis functions vs. the domain of -2 to 4:

d) Fourier Family
Satisfying the CBS for the zero term:

Satisfying the CBS for the remaining terms:

The CBS family solution:

Plot of basis functions and CBS basis functions vs. the domain of -2 to 4:

e) Exponential Family
Satisfying the CBS for the zero term:

Satisfying the CBS for the remaining terms:

The CBS family solution:

Plot of basis functions and CBS basis functions vs. the domain of -2 to 4:

Conclusion
It seems that the strength or effective representation of the basis family would be expected to follow the trend: (exponential family) << (polynomial family)<< (sine family) ~ (cosine family) < (fourier family). Although this trend may not follow for every case(small j's and linear or simple curvature), it demonstrates that as j becomes significantly large, the fourier family of basis functions tends to become more accurate.

Part 2: Show that the exponential family is linearly independent on the domain
To be linearly independent, the gram matrix of the basis functions must not equal zero:

The gram matrix consists of the dot products of basis functions as follows:

However, with a given domain, the Gram Matrix can also be written as:

where,

A few examples:

etc...

The gram matrix of the exponential family becomes:

The determinate of the exponential gram matrix does not zero:

Therefore, the exponential family of basis functions is linearly independent on the domain

=Problem 4.6: Elastic bar with a variable distributed spring force p(x)=

Given: bar of length l, cross-sectional area A(x), Young’s modulus E(x) with body force b(x)
Consider an elastic bar with a variable distributed spring p(x) along its length as shown in Figure 1. The distributed spring imposes an axial force on the bar in proportion to the displacement. Consider a bar of length l, cross-sectional area A(x), Young’s modulus E(x) with body force b(x) and boundary conditions as shown in Figure 1

a. Derivation of Strong Form
Let $$\displaystyle \vec{n}(x) $$ be defined as the unit vector having positive orientation in the direction of $$ \sigma $$ on the element we are considering. This means that the sign convention of our unit vectors will be

Which corresponds to the convention depicted in the free-body diagram shown above. Let us also define $$ \displaystyle \sigma(x) $$ as the force for a given area as,

Upon the rearrangement of terms we arrive at

Classical mechanics allows us to write the sum of the forces in static system is zero

Using the free-body diagram, let us now consider all of the forces acting on the elastic bar $$\displaystyle b(x),p(x), N(x), $$ and $$\displaystyle  N(x+dx) $$. Balancing the forces acting on our finite element results in the following relation,

Next we will substitute (1.5) and (1.6) into (1.10) to get

Recall that a Taylor's Series expansion (See:Taylor Series) about $$ a $$ near $$ x $$ has the following form:

Next we will need to determine a relationship for the displaced area and stress, namely $$\displaystyle \sigma(x+dx)$$ and $$ \displaystyle   A(x+dx)$$. To accomplish this we will performing a Taylor Series expansion choosing $$\displaystyle a=x+dx $$ and neglect all higher order terms.

Taking the product of (1.13) and (1.14) gives us

Substitute (1.15) into (1.11) results in

This can be simplified by canceling the like terms and dividing through by $$ \displaystyle dx $$ as shown below,

to get the following simplified relation,

Next we must apply the product rule in the reverse directions. Meaning we will consider (1.19) the expansion obtained by applying the product rule and recombine terms to get the unexpanded form shown here,

tensile stress according to Hooke's Law

Substituting (1.22) into our simplified and unexpanded form of our force balance shown in (1.20) we arrive at the solution,
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{d}[E(x)\cdot A(x)\frac] + b(x)-p(x) = 0 $$     (4.6.19) Standard form of the strong form can be written as
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

b. Derivation of weak form
Now strong form can be written as
 * {| style="width:100%" border="0"

$$  \displaystyle P\left( u \right) = 0 $$     (4.6.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Where
 * {| style="width:100%" border="0"

$$  \displaystyle p\left( u \right) = \frac{d}\left[ {E\left( x \right)A\left( x \right)\frac} \right] + \left( {b\left( x \right) - p\left( x \right)} \right) $$     (4.6.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Let $${w\left( x \right)}$$ is a weight function, then writing Weighted Residual Form
 * {| style="width:100%" border="0"

$$  \displaystyle \int\limits_0^l {w\left( x \right)\left[ {P\left( u \right)} \right]dx = 0} $$     (4.6.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \int\limits_0^l {w\left( x \right)\left[ {\frac{d}\left[ {E\left( x \right)A\left( x \right)\frac} \right] + \left( {b\left( x \right) - p\left( x \right)} \right)} \right]dx = 0} $$     (4.6.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Writing $$w\left( x \right) = w,E\left( x \right) = E,A\left( x \right) = A,b\left( x \right) = b,p\left( x \right) = p$$
 * {| style="width:100%" border="0"

$$  \displaystyle \int\limits_0^l {w\left[ {\frac{d}\left[ {EA\frac} \right] + \left( {b - p} \right)} \right]dx = 0} $$     (4.6.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Expanding integral
 * {| style="width:100%" border="0"

$$  \displaystyle \int\limits_0^l {w\frac{d}\left[ {EA\frac} \right]dx} + \int\limits_0^l {w\left( {b - p} \right)dx}  = 0 $$     (4.6.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \left[ {wEA\frac} \right]_0^l - \int\limits_0^l {EA\frac\fracdx} + \int\limits_0^l {w\left( {b - p} \right)dx}  = 0 $$     (4.6.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now assuming $$w\left( 0 \right) = 0$$ and also from natural boundary condition $$E(l)\frac = \bar t$$ above integral will take form as
 * {| style="width:100%" border="0"

$$  \displaystyle w\left( l \right)A\bar t - 0 - \int\limits_0^l {EA\frac\fracdx} + \int\limits_0^l {w\left( {b - p} \right)dx}  = 0 $$     (4.6.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Rearranging terms

 * {| style="width:100%" border="0"

$$  \displaystyle \int\limits_0^l {EA\frac\fracdx} = \int\limits_0^l {w\left( {b - p} \right)dx}  + w\left( l \right)A\bar t $$ (4.6.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now writing in standard format

Reference
=Problem 4.7: Calculix User's Guide Containing Illustrative Examples For Building FEA FBDs=

First Step: Download
Calculix is a open source finite element suite. This software contains its own pre- and post processing algorithms which enables the user to prepare and solve a model after defining the appropriate boundary conditions. The solver, CCX, was written and is maintained by Guido Dhondt. The pre- and post processor, CGX, was written and is maintained by Klaus Wittig. Because CalculiX is open source users can download the latest version from Convergent Mechanical Solution's website. Convergent has done an excellent job integrating an installation wizard with the distribution which makes the installation process a breeze for experts and novice alike.

 * Users intending on downloading CalculiX with a Windows operating system should visit: http://www.bconverged.com/products.php
 * The built-in GUI makes it simple to follow the on screen prompts for a quick pain-free installation.
 * Users will be prompted to select a working directory which will be used as the starting file path whenever opening CalculiX editor.
 * The source code, Linux builds, documentation, tests, and other resources for Calculix : http://www.calculix.de/

Second Step: Test
Before starting Calculix we need to accomplish one more step.

 * {| style="width:100%" border="0"

$$  \displaystyle [Start]--[All Programs]--[bConverged]--[Calculix]--[Test Calculix *.*] $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Using Calculix

 * To open Calculix in the user friendly windows version go to

 * {| style="width:100%" border="0"

$$  \displaystyle [Start]--[All Programs]--[bConverged]--[Calculix]--[Calculix Command] $$
 * style="width:95%" |
 * style="width:95%" |
 * }

 * Doing this will open the Calculix command window.
 * The command window will automatically open in the working direction that we have defined during the installation process.
 * Next we introduce the program command line parameters which serve as options when creating or edit a file.

 * {|style="width:100%" border="0"

$$  \displaystyle Command: cgx [-a|-b|-c|-duns2d|-duns3d|-isaac2d|-isaac3d|-foam|-step|-stl]  filename $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Now we proceed to show an example of a pre-built FEA problem. Examples can be download from Calculix website or you can access them once bConverged has been installed and the pre-build samples have been tested. Because of the amount of sample files CalculiX must test, something initiated by the user after installation, the time will very for the completion of the initial tests. To find the examples simply go to the program directory containing bConverged. The location will very slightly depending on if you are running a 32 or 64-bit version of windows.

Since we are in the working directory when opening CalculiX we will need to copy the appropriate examples to our working directory to open them. Or you can simply change your current directory to a directory of your choosing based on where your want the examples to be placed. In the command line type ' cgx -b sphere.fbd '. Important !!!: Always make sure that the mouse is positioned on the graph window before writing commands. This point is extremely important because failing to do so will result in commands not being entered in the command window. Initially this is somewhat cumbersome because we have trained ourselves to point the mouse to the location where we will be inserting text. After a few minutes you will get the feel for doing this. Position your command window such that you can simultaneously view your entered commands and geometric figure.

Command line $$ \displaystyle \Rightarrow $$ C:\..........\examples\basic>cgx -b sphere.fbd

By using mouse you can rotate, zoom in out and out, or reposition the figure. Rotating the object is controlled by the left mouse button, zooming in and out by the middle mouse button and translating the object is controlled by the right mouse button. Before moving on try these operations with the mouse in order to become well acquainted with how these movements orient the figure.

Learning By Example
Lets learn more by doing an example on Calculix. We can broaden our knowledge on commands by doing examples. We can take modelling disc as an example.

Building Your First Model

 * Run Calculix Command
 * Type cgx -b discmodel.fbd (You probably get an error message. Because there is no exists file name: discmodel)
 * To start with modelling we have to remember the construction procedure.
 * First we will construct nodes, lines, surfaces, and the bodies.
 * Lets look at point command.

 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} Command:\quad 'pnt' <name(char<9)>& |'!' [<x> <y> <z>]| \\ & [   ]| \\                                  &  [<P1> <P2>  ]| \\ & [<setname(containing nodes)>]
 * style="width:95%" |
 * style="width:95%" |

\end{align} $$
 * }

Creating Points

 * In order to type commands in to the command window please remember to press and release to the graph window with left button mouse button.
 * Type pnt y  -0.00000        1.00000        0.00000  (recognizing that py is a dummy with three coordinates on xyz plane)
 * Type pnt P0  -0.00000       -0.00000        0.00000
 * Type pnt P001 0.70711       -0.00000       -0.70711
 * Type pnt P003 -0.00000      -0.00000       -1.00000
 * Type pnt P005 -0.70711      -0.00000       -0.70711
 * Type pnt P006 -1.00000      -0.00000        0.00000
 * Type pnt P009 -0.70711       -0.00000       0.70711
 * Type pnt P00A 0.00000       -0.00000        1.00000
 * Type pnt P00G 0.70711       -0.00000        0.70711
 * Type pnt P00I 1.00000       -0.00000       -0.00000

 * Even though you have entered the points into the console you will not be able to see the point attributes until you enter a command to display them.
 * To see the points with attributes(labels) enter the following command.
 * Type plot pa all (get the same view with graph)
 * The general command used here is plot and has all of the available options shown here.
 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} Command: 'plot' & ['n'|'e'|'f'|'p'|'l'|'s'|'b'|'S'|'L']&['a'|'d'|'p'|'q'|'v'] -> \\ & 'w'|'k'|'r'|'g'|'b'|'y'|'m'
 * style="width:95%" |
 * style="width:95%" |

\end{align} $$
 * }

Creating Lines

 * Now we will look at how to generate lines.
 * Type line L001 P00I P001 p0 4
 * This will create the line connecting p001 to p00I.
 * To show this line
 * Type plus la all
 * The plus command is used when one wants to retain the previous plot and add addition
 * {| style="width:100%" border="0"

$$  \displaystyle Command: 'line' <name(char<9)>|'!' <p1> <p2> <cp|seq>
 * style="width:95%" |
 * style="width:95%" |

$$
 * }

 * As you see above we just created arc for a portion of disc.
 * Even if the command's name is line we also create arcs with the same command.
 * L001 is a dummy name.
 * P00I and P001 are limits.
 * p0 is the center.
 * And 4 is the number of divisions.
 * To see the line attributes use the following:

 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} Command: 'plus'  & ['n'|'e'|'f'|'p'|'l'|'s'|'b'|'S'|'L']&['a'|'d'|'p'|'q'|'v'] -> \\ & 'w'|'k'|'r'|'g'|'b'|'y'|'m'|'i'
 * style="width:95%" |
 * style="width:95%" |

\end{align} $$
 * }

 * Plus performs the same job as plot.
 * Plus retains all previous visuals along with the initial plot filters.
 * Type line L002 P001 P003 p0 4
 * Type line L003 P003 p0 8
 * Type line L004 p0 P00I 8
 * Type line L005 P003 P005 p0 4
 * Type line L006 P005 P006 p0 4
 * Type line L007 P006 p0 8
 * Type line L009 P006 P009 p0 4
 * Type line L00A P009 P00A p0 4
 * Type line L00C P00A p0 8
 * Type line L00G P00A P00G p0 4
 * Type line L00I P00G P00I p0 4

Creating Surfaces
To finish the model we need to add surfaces. To do so we will use the following generalized syntax.

 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} Command: 'gsur' &<name(char<9)>|'!' '+|-' 'BLEND| ' '+|-' <line|lcmb> '+|-' -> \\ &<line|lcmb>
 * style="width:95%" |
 * style="width:95%" |

\end{align} $$
 * }

 * This keyword is used to define or redefine a surface.
 * Surfaces are defined with boundary lines.
 * Type gsur A001 + BLEND - L003 - L002 - L001 - L004
 * Type gsur A002 + BLEND - L007 - L006 - L005 + L003
 * Type gsur A003 + BLEND - L00C - L00A - L009 + L007
 * Type gsur A004 + BLEND + L004 - L00I - L00G + L00C

Creating Meshes

 * In the final part of this example we will create the meshes.
 * The options tr3, qu4, and qu8 are for meshing surfaces.
 * The options he8 and he20 are used to mesh bodies.

 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} Command: 'elty' &[ ] ['be2'|'be3'|'tr3'|'tr3u'|'tr6'|'qu4'|-> \\ &'qu8'|'he8'|'he8f'||'he8i'|'he8r'|'he20'|'he20r']
 * style="width:95%" |
 * style="width:95%" |

\end{align} $$
 * }

 * Type elty all QU4
 * Type mesh all
 * We just have meshed the surface with an elements of qu4.

 * Type plot m all

Basic Models
Shown in the figures below are the example models that come installed with the CalculiX suite. We were able to regenerate these models by applying similar commands to the one present in the above tutorial. After playing around with this software we were finally able to obtain the right combination of commands the display the colorful models with attributes shown below.

Turbine: Illustrating the Use of Colors and Surface Element Attribute Specifications

=Problem 4.8: The Mass Matrix in Weak Form Using F&B problem 3.4=

Given: PDE and Corresponding Parameters
where the weak form is A trial solution for u(x) is suggested as and w(x) as

Find: The Mass Matrix in Weak Form
with n=3 and for dynamics:

Find $$\underline {F}(t)$$

State the Weak Form of the dynamic PDE for elasticity
Our given basis is The discretized forms of our trial solution and our weighting functions respectively are

Substitute back the operator definitions
=Contributing Members & Referenced Lecture=

title: User:Eml5526.s11.team2/hwk4


license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: File:Fe1.s11.team1.hw3.fig2.1.jpg


text: File:Hw 3.11 1.png

title: File:Hw 3.11 1.png


text: File:Hw 3.11 2.png

title: File:Hw 3.11 2.png


title: File:Fe1.s11.team1.hw3.fig3.1.jpg


text:
Talk:Introduction to Latin/Simple Grammatical Concepts

Thank you so much for going over basic grammar and sentence structure! I appreciate keeping things simple, but some elaboration on Direct/Indirect Objects and Transitive/Intransitive Verbs would help, especially those of us who've been out of school for more than a few years (a little refresher never goes amiss).

Also, by providing a link to the answer key instead of having others contact you, I believe would further simplify matters for you and others who are helping to construct this Latin course. I feel the answer key should not only give the correct answers, but also go over common mistakes made; however, with elaboration, one could discuss the common mistakes made and why.

Just some thoughts.

--Serenewhisper02 00:37, 16 February 2011 (UTC)


text:
User talk:Serenewhisper02

 Hello Serenewhisper02, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 00:44, 16 February 2011 (UTC)

title: User talk:Serenewhisper02


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User talk:Lokeshdahiya

 Hello Lokeshdahiya, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 00:44, 16 February 2011 (UTC)

title: User talk:Lokeshdahiya


text:
User:Eml5526.s11.team7.Srilalithkumar

HW 1 HW 2.5 HW 3.5 HW 4.1 HW 4.7 HW 5.5 HW 6.4 HW 6.6 Calculix Verification HW 7.1(a) HW 7.3

title: User:Eml5526.s11.team7.Srilalithkumar


text:
User:Eml5526.s11.team3.sahin/Homework 310

=Problem3.10 Quadratic Trial Solution of Weak Form =

Problem Statement
Refer to lecture notes 17.2

This problem is extracted Problem 3.4 from the textbook of Fish J., and Belytschko T.,

Given
Problem 3.1 given in Fish and Belytschko

The strong form is

The weak form is

Find
Consider a trial (candidate) solution of the form $$u\left(x\right)=\alpha_0+\alpha_1(x-3)+\alpha_2(x-3)^2$$ and a weight function of the same form.

1) Obtain a solution to the weak form in Problem 3.1.

2) Check the equilibrium equation in the strong form in Problem 3.1; is it satisfied?

3) Check the natural boundary condition; is it satisfied?

1)
The trial solutions $$u\left(x\right)$$ must satisfy the essential boundary condition $$u\left(x=3\right)= 0.001$$ so Therefore, it is weight function $$w\left(x=3\right)= 0$$ as the weight function must vanish on the essential boundaries

To find a trial solution, we need to put Eq 10.3 and Eq 10.4 into the weak form Eq 10.2

Evaluating the integrals with the assumption of cross area $$ \begin{align} & \qquad A \end{align} $$ and Young’s modulus $$ \begin{align} & \qquad E \end{align} $$ are constants

LHS was integrated by WolframMathematica and $$\begin{align} \qquad \quad \quad \quad \quad \end{align} $$ RHS was integrated by WolframMathematica

and factoring out $$ \begin{align} & \qquad \beta_1 \end{align} $$ and $$ \begin{align} & \qquad \beta_2 \end{align} $$  gives As it must hold for all $$ \begin{align} & \qquad \beta_1\end{align} $$ and $$ \begin{align} & \qquad \beta_2\end{align} $$ because of weight functions, the term in the parentheses must vanish, so So we obtain linear algebraic equation in $$ \begin{align} & \qquad \alpha_1 \end{align} $$ and $$ \begin{align} & \qquad \alpha_2 \end{align} $$:

Matlab Code: The solution of linear algebraic equation has been calculated above codes in Matlab,

With rearranging the both equations, then we get $$ \begin{align} & \qquad \alpha_1 \end{align} $$ and $$ \begin{align} & \qquad \alpha_2 \end{align} $$

Substituting this result into Eq 10.3 gives quadratic trial solution of the weak solution,

2)
We know that To check the equilibrium equation in the strong form in Problem 3.1, we need to substitute Eq 10.14 into Eq 10.1a

3)
We know that the natural boundary condition is

To check the natural boundary condition; we need to substitute Eq 10.14 into Eq 10.1b

Problem solved by sahin and Ushnish


text:
User:Eml5526.s11.team6/hwk3

Given
General one dimensional Model 1.0/Data set 2 on page 1 of Mtg 12:

where $${a_2} = 2$$,  f = 3

Consider $$ {b_j}\left( x \right) = (x+k)^j $$ and k=1 to avoid $${b_j}'\left( 0 \right)=0 $$

To Find
A) If n=2, What is the D.O.F. of this problem?

B) Find two equations that enforce boundary conditions for $$ {u^h}\left( x \right) = \sum\limits_{j = 0}^n $$

C) Find one more equation to solve for $$ {\mathbf{d}} = {\left\{ \right\}_{3 \times 1}} $$ by projecting the residue $$ {\mathbf{P}}\left(  \right) $$ on a basis function $$ b_k\left( x \right)     $$ with  k = 0,1,2 such that the additional equation is linear independent from the equations found in the solution of the (B)

D) Display 3 equations in matrix form $$ {\mathbf{Kd}} = {\mathbf{F}} $$ and observe symmetric property of $$ {\mathbf{K}} $$

E) Solve for $$ {\mathbf{d}} $$

F) Construct $$ u_n^h\left( x \right) $$ and plot $$ u_n^h\left( x \right) \ vs \ u\left( x \right) $$ ....(where $$ u\left( x \right) $$ is the exact solution)

G) Repeat (A) to (F) for n=4 and n=6

H) Compare $$ u_n^h\left( {x = 0.5} \right)   for n = 2,4,6 $$

If error $$ {e_n}\left( {0.5} \right) = u\left( {0.5} \right) - {u^h}\left( {0.5} \right) $$ ,Plot $$ {e_n}\left( {0.5} \right) \ vs \ n $$

Find the degrees of freedom
Since for this problem,'n' starts from 0;

D.O.F.=n+1

D.O.F.=3

Find the equations that enforce the boundary conditions
From (Eq 3.1.2) we can get $$ \sum\limits_{j = 0}^2 {{d_j}{b_j}\left( 1 \right)} = 0 $$

Then rewrite the equation in matrix form.

where

From (3.1.3) we can get $$ \sum\limits_{j = 0}^2 {{d_j}b_j^{\left( 1 \right)}\left( 0 \right)} =  - 4 $$

Then rewrite the above equation in matrix form:

where

Equations 3.1.4 and 3.1.6 are the two equations that enforce the boundary conditions for $$ {u}^h\left( x \right) $$

Project the residue
Because the residue $$ P = \frac{d}\left( {{a_2}\frac{d}{b_j}\left( x \right)} \right) - f\left( x \right)$$, after projecting the residue we get:

$$ \sum\limits_{j = 0}^2 {\left\{ {\int_0^1 {{b_k}\left( x \right)\left[ {\frac{d}\left( {{a_2}\frac{d}{b_j}\left( x \right)} \right)} \right]dx} } \right\}} {d_j} = - \int_0^1 {{b_k}\left( x \right)f\left( x \right)dx} $$

Then rewrite the above equation in matrix form:

where

$$ {K_{kj}} = 2\int_0^1 {{b_k}\left( x \right)b_j^{\left( 2 \right)}\left( x \right)dx} $$

$$         = 2\int_0^1 (x+1)^k. j.(j-1).(x+1)^{j-2} dx $$

It is clear that matrix K is not symmetric.

Matrix d
In order to simplify the computation, the following fortran code was used based on equations (3.1.8),(3.1.9) and (3.1.10) for n=1-10

Using the above codes we can get solutions for all the possible inputs as follows:

Note: There is a slight difference in final answers when using fortran code and matlab code.

The exact solution of eq.3.1.1 by enforcing the boundary condition is

$$ u(x)= -\frac{3x^2}{2}-8x+\frac{19}{2} $$

Plot the exact and approximate solutions
The results are compared with the exact solution as following:

From the above figure it can be seen that the approximate solution almost equals to exact solution for n=4 and above.

Error function

The above figure shows the value of error function with respect to value of n.

From the figure it is clear that error function follows random path for lower values on n and for higher values error function decreases with increase in n.

Solution can be considered to be converging after n=8.

Compare to result using Matlab codes
Integration can be obtained using following matlab code:

Important notes:

From the Matlab results, the solution is exactly the same as the exact solution because we use second second order polynomial to approximate the original function which is also a second order polynomial. The error will be zero since n=2 and the vector d won't change when increasing n. This phenomenon will occur again in HW 3.7 later since both problems have very similar basis with a second order polynomial. The reason that we have some differences between our Fortran results and Matlab results is we do have some numerical errors when solving linear eqautions using Gauss's method by Fortran codes, since the error is so small and results are also very close to what we got from Matlab codes.

=Problem 3.2= For the spring system shown in the figure below in given data.

Find
A. Number the elements and nodes.

B. Assemble the global stiffness and force matrix.

C. Partition the system and solve for the nodal displacements.

D. Compute the reaction forces.

Given
Spring system with closed ends as shown in the figure

Solution
A: The elements and nodes are numbered as follows: Element numbers are indicated in parentheses and node numbers are indicated at the bottom of the figure.

B: The elements of the stiffness matrices are:

Where global numbers corresponding to the element nodes are indicated above each column and to the right of each row.

By direct assembly, we get the global stiffness matrix as follows

The displacement and force matrices are as follows

Combined force matrix can be written as

C: Now the global system of equations is

As first two displacements are prescribed, partition after two rows and two columns. We get,

Where,

From eq.(3.2.10), we can write

From equations (3.2.12), (3.2.13), (3.2.14), (3.2.15), we get,

and

Solving above two equations, we get

D:Calculation of the reaction forces

From eq.(3.2.10), we can also write

From equations (3.2.11), (3.2.13), (3.2.14), (3.2.19), we get,

and

Solving above two equations, we get

=Problem 3.3=

Given
Referred from - Pg 37, Reference (1)

Consider the truss shown in the following FIGURE 3.3.1

-Nodes A and B are fixed.

-A force F = 10 N acts in the positive x- direction at node C. Coordinates of the joints are given in meters.

-Young's modulus E = 1011 Pa

-Cross sectional area for all bars, A = '''2. 10-2 m2'''

Find
a. Number the elements and nodes

b. Assemble the global stiffness and force matrix

c. Partition the system and solve for the nodal displacements

d. Compute the stresses & reactions

Number the Element and Nodes
Node and element numbering is done in the following fashion. Refer to the FIGURE 3.3.2 as given below-

Assemble the Global Stiffness and Force Matrix
We use the Following formula Referred from - Pg 30, Reference (1) for Elemental Stiffness Matrix of each element,

$$\displaystyle (Eq. 3.3.1) $$

Where, $${k^{e}} = \frac{l^{e}}$$ and the superscript e denotes the element number.

Now in this case as seen from the given data, A and E for all the 4 elements have same but distinct values. So we can eliminate the superscript e for A and E of each element.

Element 1

 * Element 1 goes from Node 1 to Node 3

 * Orientation : 90 degrees to the Positive X-direction, $$\phi^{(1)} =90^{o}$$ . Hence, $$cos(90^{o})=0,sin(90^{o})=1$$ and $${{l}^{(1)}}=1$$

Using Eq. 3.3.1 for Element 1,

Element 2

 * Element 2 goes from Node 2 to Node 4

 * Orientation : 90 degrees to the Positive X-direction, $$\phi^{(2)} =90^{o}$$ . Hence, $$cos(90^{o})=0,sin(90^{o})=1$$ and $${{l}^{(2)}}=1$$

Using Eq. 3.3.1 for Element 2,

Element 3

 * Element 3 goes from Node 2 to Node 3

 * Orientation : 135 degrees to the Positive X-direction, $$\phi^{(3)} =135^{o}$$ . Hence, $$cos(135^{o})=\frac{-1}{\sqrt{2}},sin(135^{o})=\frac{1}{\sqrt{2}}$$ and $$l^{(3)}=\sqrt{2}$$

Using Eq. 3.3.1 for Element 3,

Element 4

 * Element 4 goes from Node 3 to Node 4

 * Orientation : 0 degrees to the Positive X-direction,$$\phi^{(4)} =0^{o}$$, $$cos(0^{o})=1,sin(0^{o})=0$$, $$l^{(4)}=1$$,

Using Eq. 3.3.1 for Element 4,

GLOBAL STIFFNESS MATRIX
Using above Elemental Matrices, we construct the Global Stiffness Matrix for the given Truss System as follows,

FORCE MATRIX
The force matrix contains reactions and externally applied force components at various nodes.

Let Rx = Reaction at node in x -direction, Ry = Reaction at node in y -direction, subscripts denote the node number

Also node 3 is free to move in both x- and y- direction and is free of any external forces.

And node 4 is free to move in both x- and y- direction and has an external force of 10 N applied in the x- direction as per the given data in the problem statement.

The Force Matrix is assembled as follows -

NODAL DISPLACEMENT MATRIX
The noddal displacement matrix contains displacement components at various nodes.

Let ux = Displacement at node in x -direction, uy = Displacement at node in y -direction, subscripts denote the node number

Nodes 1 and 2 are fixed so they have all the components of displacement as zeros (0).

Also node 3 and 4 are free to move in both x- and y- direction.

The Nodal Displacement Matrix is assembled as follows -

Partition the system and solve for the Nodal Displacements.
Now, using the Global stiffness matrix (K), Force matrix (F) and the Nodal displacement matrix (d), we can write the Global System as follows,

$$\displaystyle (Eq.3.3.2) $$

As the displacements for first two nodes are prescribed, we partition the systems after four rows and columns.

System Partition
Also we use the following equation from - Pg 23, 'A first course in finite elements' - Fish & Belytschko -2007 and compare our global system with it to get the nodal reactions and elemental stresses using $$\displaystyle (Eq.3.3.2) $$ & $$\displaystyle (Eq.3.3.3)  $$ -

$$\displaystyle (Eq.3.3.3) $$

On comparison then we get the following partitioned matrices -

Nodal Displacements
We solve the following system of equations to get the nodal displacement matrix,

But $$\bar{d}_{E}$$ is a Zero Matrix, which will yield, $$d_{F}=K_{F}^{-1}.f_{F}$$ Solving using the above equation, the unknown nodal displacements are

$$\displaystyle (Eq.3.3.5) $$ -

Nodal reactions
Using the above equation the unknown reaction values are

$$\displaystyle (Eq.3.3.7) $$

Elemental Stresses
The stresses in the elements are given by the following expression from page 34 of the reference (1)

For element 1:

 * $$ cos90^{\circ} = 0,\quad sin90^{\circ}=1, $$

\mathbf{d}^{(3)}=l^{(3)}/AE\begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{2x}\\ u_{2y} \end{bmatrix}, \qquad \sigma ^{(3)}=1/A\begin{bmatrix} 0 & -1 &0 &1 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 38.2843\\ 10 \end{bmatrix} = 500N/m^{2}

$$

For element 2:

 * $$ cos135^{\circ} = -0.707,\quad sin135^{\circ}=0.707, $$

\mathbf{d}^{(2)}=l^{(2)}/AE\begin{bmatrix} u_{2x}\\ u_{2y}\\ u_{3x}\\ u_{3y} \end{bmatrix}, \qquad \sigma ^{(2)}=1/A\begin{bmatrix} 0 & -1 &-0 &1 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 48.2843\\ 0 \end{bmatrix} = 0 N/m^{2}

$$

For element 3:

 * $$ cos90^{\circ} = 0,\quad sin90^{\circ}=1, $$

\mathbf{d}^{(1)}=l^{(1)}/AE\begin{bmatrix} u_{2x}\\ u_{2y}\\ u_{4x}\\ u_{4y} \end{bmatrix}, \qquad \sigma ^{(1)}=1/A\begin{bmatrix} 0.5 & -0.5 &-0.5 &0.5 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 38.2843\\ 10 \end{bmatrix} = -1000N/m^{2}

$$

For element 4:

 * $$ cos0^{\circ} = 1,\quad sin0^{\circ}=0, $$

\mathbf{d}^{(4)}=l^{(4)}/AE\begin{bmatrix} u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix}, \qquad \sigma ^{(4)}=1/A\begin{bmatrix} -1& 0 &1 &0 \end{bmatrix}\begin{bmatrix} 38.2843\\ 1\\ 48.2843\\ 0 \end{bmatrix} = 10/(2*10^{-2})= 500N/m^{2}

$$

Plot exact and approximate solutions

As you can see the approximate solutions is same as exact solution

Note:Since problems 3.1 and 3.7 have similar basis, the graph of solution follows similar path. Approximate solution equals exact solution for n=2 and then result remains same for higher values of n.

Plot the error function
As error is 0 throughout, graph will be same as x-axis.

=Problem 3.8=

Given
Show that the weak form of

With Essential and Natural Boundary Conditions given by is given by

Solution
Choose $$ w\left ( x \right ) $$ such that $$ w \left ( 3 \right )=0 $$ Multiply the governing equation and the natural boundary condition by an arbitrary weight function:

Use integration by parts to simplify (Eq.3.8.5)

where therefore, (Eq.3.8.5) simplifies to Given that $$ w\left ( 3 \right )=0 $$, (Eq.3.8.9) reduces to Finally, apply the natural boundary condition (Eq.3.8.2) to reduce (Eq. 3.8.10) to Rearrange the terms in (Eq.3.8.11) to yield If (Eq.3.8.12) is equivalent to (Eq.3.8.4), then (Eq.3.8.1-3) from the Given section are equivalent to (Eq.3.8.4).

=Problem 3.9=

Given
Consider a trial (candidate) solution of the form $$u\left ( x \right )=\alpha _{0}+\alpha _{1}\left ( x-3 \right )$$ and a weight function of the same form. Obtain a solution to the weak form in Problem 3.8 (Eq.3.8.4). Check the equilibrium equation in the strong form in Problem 3.8; is it satisfied? Check the natural boundary condition; is it satisfied?

Solution
Given the candidate solution,

Assume,

The weight function must disappear at x=3, therefore

For the trial solution to be admissable, the function must satisfy the essential Boundary Conditions (Eq.3.8.3),

Substitute (Eq.3.9.4) and (Eq.3.9.5) into (Eq.3.9.1) and (Eq.3.9.2) to get

Also, we will need the derivatives of (Eq.3.9.6) and (Eq.3.9.7) in order to obtain the solution to the weak equation given.

Substitute (Eq.3.9.3,6-9) into the weak form (Eq.3.8.4)

Reduce and isolate $$ \beta _{1} $$

Solution must hold for all $$ \beta _{1} $$, therefore, $$ 2\ast 10^{5}\alpha _{1} + 6.47=0 $$

Substitute (Eq.3.9.12) into (Eq.3.9.6) to obtain the linearly approximated solution, i.e.,

Given

Next, substitute (Eq.3.9.13-14) into the Equilibrium Equation to determine whether or not the Equilibrium Equation is satisfied.

$$2x\neq 0 \; \therefore $$ The Equilibrium Equation is not satisfied.

Finally, we check to see if the Natural Boundary Conditions are satisfied.

$$ \sigma \left ( 1 \right )=\left ( E\frac{du}{dx} \right )_{x=1}=0.1\neq -3.23 \; \therefore $$ The Natural Boundary Conditions are not satisfied.

To summarize,

=Problem 3.10= Refer to lecture slide [[media:Fe1.s11.mtg17.djvu|17-2]] for problem assignment.

Given
Strong form:

Weak form:

Find
Consider a trial (candidate) solution and weight functions of the form

Find a solution to the weak form, then check if the equilibrium equation and natural boundary condition in the strong form are satisfied.

Solution
From $$\displaystyle (Eq. 3.10.2)$$ $$w(3) = 0$$ to satisfy the homogeneous essential boundary condition.

Likewise, to satisfy the essential boundary condition of the strong form, $$u(3) = 0.001$$

Next we will evaluate the derivatives of the trial solution and the weight function.

Plugging $$\displaystyle (Eq. 3.10.7)$$ and $$\displaystyle (Eq. 3.10.8)$$ into $$\displaystyle (Eq. 3.10.2)$$ and assuming A and E are uniform in the x-direction (i.e. not functions of x):

For legibility, each term in $$\displaystyle (Eq. 3.10.9)$$ will be evaluated separately.

Now, substituting equations $$\displaystyle (Eq. 3.10.10)$$,$$\displaystyle (Eq. 3.10.11)$$, and $$\displaystyle (Eq. 3.10.12)$$ back into $$\displaystyle (Eq. 3.10.9)$$:

Rearranging and isolating $$\beta_1$$ and $$\beta_2$$:

Equation $$\displaystyle (Eq. 3.10.14)$$ must hold for any arbitrary weight function $$w(x)$$, so long as it satisfies the homogeneous essential boundary condition. It then also follows that it must also hold for any arbitrary $$\beta_1$$ and $$\beta_2$$, leading to the following algebraic equation:

Pre-multiplying both sides $$\displaystyle (Eq. 3.10.15)$$ by the inverse of the coefficient matrix on the left side yields:

Substituting $$\displaystyle (Eq. 3.10.17)$$ back into $$\displaystyle (Eq. 3.10.3)$$ yields the quadratically approximated solution to the weak form $$\displaystyle (Eq. 3.10.2)$$.

Substituting $$\displaystyle (Eq. 3.10.19)$$ back into the equilibrium equation from the strong form $$\displaystyle (Eq. 3.10.1)$$

Therefore the equilibrium equation is not satisfied over the whole interval ]1,3[.

Now using $$\displaystyle (Eq. 3.10.19)$$ to check the natural boundary condition

Therefore, the solution $$\displaystyle u^{\text{quad}}(x)$$ satisfies the natural boundary condition only in the case where $$ A = \frac{40}{3}$$.

=Problem 3.11=

Given
General one dimensional Model 1.0/Data set 2 on page 1 of Mtg 12:

where $${a_2} = 2$$,  f = 3

Consider $$ {b_j}\left( x \right) = 1+{a_j}cos(jx)+{b_j}sin(jx) $$

To find
A) If n=2, What is the D.O.F. of this problem?

B) Find two equations that enforce boundary conditions for $$ {u^h}\left( x \right) = \sum\limits_{j = 0}^n $$

C) Find one more equation to solve for $$ {\mathbf{d}} = {\left\{ \right\}_{3 \times 1}} $$ by projecting the residue $$ {\mathbf{P}}\left(  \right) $$ on a basis function $$ b_k\left( x \right)     $$ with  k = 0,1,2 such that the additional equation is linear independent from the equations found in the solution of the (B)

D) Display 3 equations in matrix form $$ {\mathbf{Kd}} = {\mathbf{F}} $$ and observe symmetric property of $$ {\mathbf{K}} $$

E) Solve for $$ {\mathbf{d}} $$

F) Construct $$ u_n^h\left( x \right) $$ and plot $$ u_n^h\left( x \right) \ vs \ u\left( x \right) $$ ....(where $$ u\left( x \right) $$ is the exact solution)

G) Repeat (A) to (F) for n=4 and n=6

H) Compare $$ u_n^h\left( {x = 0.5} \right)   for n = 2,4,6 $$

If error $$ {e_n}\left( {0.5} \right) = u\left( {0.5} \right) - {u^h}\left( {0.5} \right) $$ ,Plot $$ {e_n}\left( {0.5} \right) \ vs \ n $$

Find the degrees of freedom
Since for this problem,'n' starts from 0;

D.O.F.=n+1

D.O.F.=3

Find the equations that enforce the boundary condition
From (Eq 2.9.2) we can get $$ \sum\limits_{j = 0}^2 {{d_j}{b_j}\left( 1 \right)} = 0 $$

Then rewrite the above equation in matrix form:

where

From (3.1.3) we can get $$ \sum\limits_{j = 0}^2 {{d_j}b_j^{\left( 1 \right)}\left( 0 \right)} =  - 4 $$

Then rewrite the above equation in matrix form:

where

Equations 3.1.4 and 3.1.6 are the two equations that enforce the boundary conditions for $$ {u}^h\left( x \right) $$

Matrix d
Following matlab code is used to compute integration

Plot exact and approximate solutions

The above figure compares the exact solution and the approximated solution

It can be seen that at n=5, approximate solution almost equals exact solution

Plot error function

The above figure compares error between the exact and approximate solution at x=0.5

=Contributing Team Members= Eml5526.s11.team6.vork 03:54, 16 February 2011 (UTC)

Eml5526.s11.team6.deshpande 05:36, 16 February 2011 (UTC)

Eml5526.s11.team6.joglekar 06:17, 16 February 2011 (UTC)

Eml5526.s11.team6.tupsakhare 08:49, 16 February 2011 (UTC)

Eml5526.s11.team6.gravois 16:48, 16 February 2011 (UTC)

Eml5526.s11.team6.kurth 18:21, 16 February 2011 (UTC)

eml5526.s11.team6.lee 20:31, 16 February 2011 (UTC)

title: User:Eml5526.s11.team6/hwk3


title: File:Fe1.s11.team3.HW3.11.fig1.jpg


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title: File:Fe1.s11.team3.HW3.11.fig7.jpg


text:
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title: File:Fem.tm7.3 11 fig 1.jpg


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User:JoshuaZ

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text: File:Eml5526.s.11.tm7.3 11 fig 2.jpg

title: File:Eml5526.s.11.tm7.3 11 fig 2.jpg


text:
Template:INDEX

__INDEX__

title: Template:INDEX


title: File:Fe1.s11.team3.HW3.2.fig1.jpg


text:
User:Eml5526.s11.team3.ushnish12/Homework 3

The strong form of the governing differential equation is:-
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\frac{d}{dx}(AE\frac{du}{dx})+2x=0 ; 1
 * }

The boundary conditions are given by the equations:-
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\sigma (1)={{(E\frac{du}{dx})}_{x=1}}=0.1

$$ (Eq )<2>
 * }

 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

u(3) = 0.001

$$ (Eq )<3>
 * }

To form the weak form, we first consider the strong form and multiply the governing equation Eq(2) by an arbitrary function $$\displaystyle w(x)$$ and integrating over the domain on which they hold.

The function $$\displaystyle w(x)$$ is called the weight function.

For the governing equation, the pertinent domain is the interval [1,3].

The resulting equation after multiplying the governing equation with $$\displaystyle w(x)$$ takes the form -

 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\int\limits_{1}^{3}{w[\frac{d}{dx}(AE\frac{du}{dx})+2x]}\text{dx} = 0 ; \forall w

$$ (Eq )<4>
 * }

In the above, $$\forall w$$ denotes that $$\displaystyle w(x)$$ is an arbitrary function which holds for all functions $$\displaystyle w(x)$$. Furthermore, it is convenient to consider that the weight function to satisfy the condition

 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

w(3)=0

$$ (Eq )<5>
 * }

For convenience, we write Eq(4) into the equivalent form :

 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\int\limits_{1}^{3}{w[\frac{d}{dx}(AE\frac{du}{dx})]\text{dx}+\int\limits_{1}^{3}{2wx\text{dx}=0}}

$$ (Eq )<6>
 * }

To the solve the first term of the above integral we use Integration by parts method:

The stress $$ \sigma\left (x\right)$$ can be expressed in terms of elastic modulus $$ E\left (x\right)$$ and strain$$(\frac{du}{dx})$$ as

 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\sigma (x)=E(x)\frac{du}{dx}

$$ (Eq )<9>
 * }

Using Eq(9), Eq(8) is written as

 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

{{(wA\sigma )}_{x=3}}-{{(wa\sigma )}_{x=1}}- \int\limits_{1}^{3}{\frac{dw}{dx}(AE\frac{du}{dx})\text{dx}} +\int\limits_{1}^{3}{2wx\text{dx}=0}

$$ (Eq )<10>
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Now   $${{(w)}_{x=3}}=0\quad and\quad{{(\sigma )}_{x=1}}=0.1$$ as defined earlier.

Applying these conditions to Eq(10), we obtain

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$$\displaystyle
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-0.1{{(wA)}_{x=1}} - \int\limits_{1}^{3}{\frac{dw}{dx}(AE\frac{du}{dx})\text{dx}} +\int\limits_{1}^{3}{2wx\text{dx}=0} $$ (Eq )<11>
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Rearranging the terms in Eq(11) we obtain the strong form which is given by:-

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$$\displaystyle
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\int\limits_{1}^{3}{\frac{dw}{dx}(AE\frac{du}{dx})\text{dx}} = -0.1{{(wA)}_{x=1}}+\int\limits_{1}^{3}{2wx\text{dx}=0} \quad \forall w\quad\text{with}\quad w(3)=0

$$
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title: File:Fe1.s11.team3.HW3.7.fig3.jpg


text: File:HW 3.2.jpg

title: File:HW 3.2.jpg


text:
Psycholinguistics/Prosody

Introduction
Emotion is a very important part of our conveyance and understanding of language. Research has shown that our understanding and memory of sentences and speech can be altered by emotional aspects. Emotions in speech can be conveyed from the words that we say or by the way that they are said. This idea works on a number of levels. For instance neutral sentences such as “I wanted cake” could have different meanings depending on the tone of voice of the speaker. Further words with seemingly positive or negative attributes can also be used with different tones to convey different meaning. The word ‘love’ typically has a positive emotional bias behind it, however if used in a sarcastic tone it can have negative meanings.

This idea is further explained in the 2008 study by Nygaard and Queen in which they state that listeners of speech not only obtain information from a speaker’s words but also the “non-linguistic properties of the utterances.” In similar terms Nygaard and Queen state that to understand completely the message conveyed by a speaker, the listener must understand what the speaker is saying (meaning of words and phrases being expressed) and how the speaker is saying it (most frequently classified through tone of voice). In this chapter we explore the aspects of emotional speech and comprehension to better understand the use and importance of emotion in language.

Expression of Emotion
To begin our discussion of emotion in language we will start with a discussion of the emotional aspects of speech. When discussing the expression of emotion an important starting point is prosody. Prosody refers to a wide range of ideas in language. In one sense it can be considered an aspect of the speech with regards to the speech cues such as pitch, loudness and so on or it could refer to the perception and understanding of the emotional cues that a speaker is trying to portray to a listener. Here we will refer to these two concepts as affective prosody and vocal emotion, which further explores the expression and recognition of emotion in speech.

Before we dive into our discussion of emotion in speech through the use of verbal cues, we should recognize that it is not the only way to allow for the expression of emotion in speech. As Bostanov and Kotchoubey explain their 2004 study on the recognition of affective prosody:

"In summary, there are three important features of nonverbal emotional exclamations. First, they convey the speaker’s affective state much better than words. Second, they most probably abstract some key prosodic features, primarily the voice quality, of verbal emotional speech. Third, the underlying physiological mechanisms are similar in all primates" This is a great explanation of the different ways that emotional aspects are present in speech. However in our discussion we will stick to the verbal cues of emotion and their importance rather than the aspects of nonverbal emotional expression.

Affective Prosody
Our discussion of emotional speech will start with affective prosody. Affective prosody is the “suprasegmental aspects of speech that contain emotional as well as linguistic information.” To further explain this concept we can say that affective prosody relates to the prosodic cues and features of speech including pitch, loudness and articulatory rates. In a sense this relates heavily to how sentences are expressed verbally compared to graphically. In speaking, a listener cannot rely on such cues as periods, commas, question marks or other pieces of punctuation to help them know when people have finished one thought (sentence or phrase) and move on to another. This therefore has to be conveyed through the use of pauses in speech, stress on certain words and other verbal cues. As stated by Cultler, Dahan and van Donselaar, prosodic structure has been noted as an important aspect in the recognition of speech. Therefore it is no wonder that it becomes important when discussing the idea of emotional conveyance in speech. In the sense of emotional speech, prosody allows for cues that allow us to stress the importance of particular words and phrase that represent our emotions certain topics. Reading a sentence such as I can’t believe he did that, will have very different meanings depending on where pauses and emphasis is placed. For instance if you were to pause after the word "he" compared to the word "that" the sentence would change in meaning from that idea that you are surprised at him compared to you are surprised at his action.

There is more to affective prosody than simply pauses and stress of certain words. Affective prosody also includes aspects of tone, pitch and loudness of speech. These are all important aspects that convey further meaning of speech to listeners. It is importantly noted here by Dupuis and Pichora-Fuller that even when using seemingly positive or negative words (positive words include compassion, love while negative words include hatred, anger) these words can have mixed meaning when used in different ways, such as sarcastic remarks.

Vocal Emotions
Our second topic in the expression of emotional language is that of vocal expression of emotion. This idea refers to not only the speaker’s expressions but also the reception and understanding of these ideas by a listener. In a sense, vocal expression of emotion does not relate to the prosodic cues and ideas of tone, loudness and articulatory rates that we discussed with affective prosody, but instead it focuses on the aspects of speech that relate to purely the emotional side of speech without any effect on the meaning of the words being conveyed.

When discussing the vocal expression of emotion, it is the interaction between speakers and listeners that needs to be carefully observed. It is this relationship that makes the expression of emotion important. This is because expression of emotion is based more off of understanding the speaker’s emotional state than actually understanding what they are saying. This idea becomes important when considering the theory that we base our understanding of what we are hearing from the emotional state of the person who is speaking to us. In the study done by Nygaar and Queen, not only was it found that emotional state of the speaker relates to the understanding of what is being said, but also mixed emotions between state and meaning of words can cause increases in the time it takes for a participant to understand the ideas being conveyed to them. In other words, when a speakers emotional expression was not the same as what they were saying (they were sad but were telling a happy story) the participants would take longer to understand what they speaker was saying.

Due to the great importance of understanding that comes from telling the emotion state of a speaker it is sometimes a mystery that humans can typically understand what is being said to them. However this is explained in a 2004 study by Bostanov and Kotchoubev in which they found that participants could identify all the emotions of speakers correctly with the primary use of voice quality. Further, although not overly important to this discussion, was the concept that of the nonverbal cues mentioned early. These are another important aid that can increase the understanding of a speaker’s emotional state.

To summarize emotional expression it is important to note the importance of affective prosody and vocal expression of emotion when trying to understand the information being provided to us by a speaker. It is also an important idea to note that these concepts occur simultaneously over the course of a conversation and are not aspects of speech that we have to think about turning on one while turning off the other.

Comprehension
The conveying of ideas is a completely useless tool without being able to understand what is being told to us. We will find in this section that there are many forms of comprehension that help us in the understanding of language. Even though there is more that can be discussed, for this section we will consider sentence comprehension to have two main types; emotional comprehension and non-emotional comprehension. By using these categories we will easily be able to distinguish between these two topics in this section.

Non Emotional Comprehension
To consider the effects of emotion on the comprehension of language we first must explore the ways in which language is typically comprehended outside of emotion. The two main ways non-emotional memory are verbatim memory and the gist effect.

Verbatim memory is the ability to perfectly recall a sentence that has been told or read by a person. This form of memory has been found to be less important than other types. This is because the use of knowing exactly what has been said is not necessary. In general the exact wording of what has been said is not that important. What is important is for humans to know the meaning conveyed in the sentence that they have just seen or heard. Imagine if whenever you were asked a question you could only give the answer within the context that you learned the answer. This would cause two problems, one you would take a longer to answer all questions and information would only be learned in certain contexts. Instead humans generally remember the main ideas and important aspects of things. This is called gist memory and it is the type of memory that focuses on the understanding of the meaning of the sentence that has been said, but not the exact wording.

The importance of these memory types has been explored by Sachs in the 1967 study which tested the ability of participants to remember the words of a sentence over time. What was found was participants could remember the meaning of the statement or gist memory, but could not remember the exact words that made up the statement over time. For instance if you read "The dog caught the ball in the park" you will be able to remember in a few minutes that the dog caught the ball but you will not remember for certain if you read that "The ball was caught by the dog in the park" or the first version of the sentence. The important thing to note is that verbatim memory is not overly important and as long as you know what was said being able to recite the exact sentence back is not normally necessary.

Emotional Attachment
Emotional memory refers to memory that is bolstered by emotional reaction to a certain topic. However, how far does this idea go? Can we understand and remember emotional ideas better than non-emotional ones? Is it possible to convey important ideas with emotional aspects to allow ourselves to remember them better? And are certain emotions better for memory than others?

While not all of these ideas will be answered in this section we can begin by saying that in general, emotional involvement in language makes the information being conveyed more memorable to a recipient. That is to say that when comparing speech that has no emotional connection to speech with emotional connection, the research has shown that emotional memory has a more lasting impression on the person obtaining the speech.

One study in which this is seen is the Kintsch and Bates study of 1977. In this study participants were able to recognize “extraneous sentences” such as jokes or other off the topic comments better than they could recognize the “descriptive statements” of a lecture. In this case both types of statements had a semantic meaning and were based off of gist memory rather than verbatim causing these issues to not matter. Therefore, this study is a great example of how emotionally charged statements (in this case with a positive, happy emotion) are better remembered than other semantically based statements. Furthermore, other research has agreed that language with a greater degree of prosodic aspects have an influence on the amount of linguistic processing that will occur.

Aprosodia
While it may seem automatic to be able to detect emotion in the context of a person’s speech, this is not an ability held by all people. Aprosodia is the inability to comprehend emotional cues in language. Further people with aprosodia also have do not have the ability to produce emotional aspects of language, such as the prosodic cues that we have discussed earlier. Aprosodia tends to be caused by damage to a variety of brain areas. While areas of the right hemisphere, especially to the anterior frontal lobe, are more widely recognized and studied, there is also evidence of damage to the homologous left cerebral structure, Broca’s area, causing aprosodia.

Research has been occurring in the last few years in the area of treatment for people with aprosodia. Even though there does not exist a set treatment at this time, there are a few treatments that are being tested. Two interesting techniques are the motoric-imitative and the cognitive-linguistic treatments. These treatments are based off two different causes for aprosodia. Motoric-imitative treatment is based from the concept that aprosodia is caused due to a motor impairment of the person effected. Therefore the treatment gets patients to say emotional statements (i.e “We won the game” [Happy] and “He was hit by a car” [Sad]) with the proper prosodic cues in the sentences. During the treatment help from the clinician would slowly lower as the patient became more comfortable using the emotional cues. In cognitive-linguistic treatment the patient learns the characteristics of different emotions to learn how they are expressed. This is done because of the belief that aprosodia is based from a decreased access to emotional words and cues by the patient.

Summary
Overall the importance of this chapter is to show the effect of emotion in language understanding and conveyance. In the end it should be recognized that emotion has an effect on the way in which language is portrayed to those around us both in the meaning of what is being conveyed and the memory processing that it will obtain.

Learning Exercise
1)	Our first section of this chapter dealt with the vocal expression of emotion. One way it was said to be done was with prosodic cues that relate to a variety of speech features including pitch, articulatory rate, etc.  Here we are going to explore these features and their ability to alter the meaning of sentences.  For this exercise you will be asked to say the sentence “I have a test tomorrow that I have been studying all day for.”  Each time you see the sentence different words will have either a 't' or 'p' written after it in brackets.  When a word has a 't' after it then there is a change in the tone of the speaker and when a 'p' is written it means that there is a pause after that word.  Say each sentence out loud and think about the change in meaning each sentence can have.

a.	I have a test tomorrow that I have been studying all day for.

b.	I have a test tomorrow (t,p) that I have been studying all day for.

c.	I have a test (t,p) tomorrow that I have been studying all day for.

d.	I have a test tomorrow that I have been studying (p) all day for.

e.	I have a test tomorrow that I have been studying all day (t) for.

As you read through the sentence you should have noticed a slight change in meaning for each one. For instance during sentence (b) the change of tone in the word ‘tomorrow’ would lead to the thought that the speaker is worried about the test being so soon. In sentence (c) however, the change of tone at the word ‘test’ would seem that the person is always worried by tests.

2)The inability to detect emotion, aprosodia, has been discussed in this chapter but at times it can be hard to imagine. One of the best ways to understand this idea is to hear it.  There exist many computer programs now that can read text, one of these will be needed for the following question.  The useful part of these programs is that they are not good at conveying prosody.  Therefore we can get an insight into what patients with aprosodia hear.  Below are some sample sentences that can be used that tend to be conveyed with much emotion try putting them into a text to speech program or use whatever you want.  Take not of the lack of prosody and how the meaning of the sentence isn’t as easily conveyed.

a.	Your sister just had a baby.

b.	John just landed in France.

c.	I just finished writing my test.

d.	Yesterday we went to the park, when we were met by a group of dogs.

You should notice that some of these sentences do not have a clear meaning without the proper emotion cues. Sentence (c) may be stating that you are happy to have finished a test, but it may also convey the fact that you did badly on it if you have negative cues in your speech. Sentence (d) may be happy in the fact that you met a group of fun dogs at the park, or it could be angered or sad if the dogs were bad behaving. Again try this experiment with other sentences that may have different meanings.

Answers to the following question can be found in this chapter:

1)	Name the three types of memory from this chapter and state the differences between them.

2)	What is aprosodia? Which parts of the brain are typically damaged?

3)	What are the techniques used for patients with aprosodia?

4)	What are the differences between verbatim and gist memory? Which one is used more in our day-to-day life?

5)	State the different features of affective prosody.

6)	Differentiate between affective prosody and the vocal expression of emotion.

title: Psycholinguistics/Prosody


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title: File:HW 3.2 sol.jpg