text:
Nonlinear finite elements/Homework 5/Solutions/Problem 2

Problem 2: Numerical simulation
 Given:

Consider the cantilever beam shown in Figure~22. The beam is made of 6061-T6 aluminum alloy. The Young's modulus of the beam is 69 MPa, the Poisson's ratio is 0.33, the tensile yield strength is 275 MPa.

The length of the beam is 5 m, the width and the thickness are both 10 cm. A point load of 10 kN is applied at the free end as shown in the figure.

Part 1
Perform a linear three-dimensional finite element analysis of the beam using 20 elements along the length and 3 elements across the thickness. Then,
 * 1) Plot the axial stress of the centroidal line as a function of distance from the fixed end.
 * 2) Plot the axial displacement of the centroidal as a function of distance from the fixed end.
 * 3) Plot the transverse displacement of the centroidal line as a function of distance from the fixed end.

A SOLID45 (3-D Structural) element is used for the model of this problem. This element has the following degree of freedoms: $$u_x,u_y,u_z$$. The built-in end conditions are $$u_x=u_y=u_z=0$$. Plots are illustrated in Figure 23, 24, and 25.

Part 2
Perform a  nonlinear three-dimensional finite element analysis of the beam using 20 elements along the length and 3 elements across the thickness. Then,
 * 1) Plot the axial stress of the centroidal line as a function of distance from the fixed end.
 * 2) Plot the axial displacement of the centroidal as a function of distance from the fixed end.
 * 3) Plot the transverse displacement of the centroidal line as a function of distance from the fixed end.

A SOLID45 (3-D Structural) element is used for the model of this problem. This element has the following degree of freedoms: $$u_x,u_y,u_z$$. The built-in end conditions are $$u_x=u_y=u_z=0$$. The effect of large deformations is included in the solution process. Plots are illustrated in Figure 23, 24, and 26.

Part 3
Perform a linear finite element analysis of the beam using 4 beam elements. Then,
 * 1) Plot the axial stress of the centroidal line as a function of distance from the fixed end.
 * 2) Plot the axial displacement of the centroidal as a function of distance from the fixed end.
 * 3) Plot the transverse displacement of the centroidal line as a function of distance from the fixed end.

A BEAM3 (2-D Beam) element is used for the model of this problem. This element has the following degree of freedoms: $$u_x,u_y,\theta_z$$. The built-in end conditions are $$u_x=u_y=\theta_z=0$$. Plots are illustrated in Figure 23, 24, and 25.

Part 4
Perform a  nonlinear finite element analysis of the beam using 4 beam elements. Then,
 * 1) Plot the axial stress of the centroidal line as a function of distance from the fixed end.
 * 2) Plot the axial displacement of the centroidal as a function of distance from the fixed end.
 * 3) Plot the transverse displacement of the centroidal line as a function of distance from the fixed end.

A BEAM3 (2-D Beam) element is used for the model of this problem. This element has the following degree of freedoms: $$u_x,u_y,\theta_z$$. The built-in end conditions are $$u_x=u_y=\theta_z=0$$. The effect of large deformations is included in the solution process. Plots are illustrated in Figure 23, 24, and 26.

Part 5
Comment on your results.

The axial and transverse displacement are not different by much between each analysis. Figure 23 shows that axial displacement is non zero in the nonlinear case while it is zero in the linear case.

The axial stress for each analysis yields very small values in comparison to maximum stress (Figure 25 and 26). In theory, the axial stress should be zero for each case. However, due to errors such as round off errors or the interpolations between elements, ANSYS computes non zero value for this stress. The stresses for SOLID45 are illustrated as Von Mises stress since ANSYS does not provide $$\hat{\sigma}_x$$ in local coordinate for use as the axial stresses.

Note: stresses given by ANSYS are in global cartesian coordinate. Hence, $$\sigma_x$$ does not represent axial stress.

There are significant difference between the maximum stresses obtained using BEAM3 and SOLID45 near the built-in end. SOLID45 model has constraints in all degree of freedoms at the fixed end. BEAM3 allows the fixed section to distort. Its results rely on the fact that the cantilever beam is comparatively long at considerable distance from the terminals. As one can see from Figure 25 and 26 that further from the built-in end, the stresses value from BEAM3 becomes closer to the value obtained using SOLID45 element.

A comparison of the displaceemnts is shown in the table below.

The ANSYS input file listing for the linear and nonlinear 3-D finite element analysis is given below.

Use the following commands according to the analysis type.

The postprocessing code is shown below

The input listing for the linear and nonlinear finite element analysis of the beam problem is given below.

Use the following commands according to the analysis type.


text:
Nonlinear finite elements/Homework 5/Solutions/Problem 1

Problem 1: Nonlinear Beam Bending
 Given:

The differential equations governing the bending of straight beams are

\begin{align} \cfrac{dN_{xx}}{dx} + f(x) & = 0 \\ \cfrac{dV}{dx} + q(x) & = 0 \\ \cfrac{dM_{xx}}{dx} -V + N_{xx}\cfrac{dw_0}{dx} & = 0 \end{align} $$

Part 1
Show that the weak forms of these equations can be written as
 * $$\begin{align}

\int_{x_a}^{x_b} \cfrac{dv_1}{dx}N_{xx}~dx & = \int_{x_a}^{x_b} v_1 f~dx + \left.v_1 N_{xx}\right|_{x_a}^{x_b}\\ \int_{x_a}^{x_b} \left[ \cfrac{dv_2}{dx}\left(\cfrac{dw_0}{dx}N_{xx}\right) - \cfrac{d^2v_2}{dx^2}M_{xx}\right]~dx & = \int_{x_a}^{x_b} v_2 q~dx + \left.v_2\left(\cfrac{dw_0}{dx}N_{xx}+\cfrac{dM_{xx}}{dx}\right) \right|_{x_a}^{x_b} - \left.\cfrac{dv_2}{dx}M_{xx} \right|_{x_a}^{x_b} \end{align}$$

First we get rid of the shear force term by combining the second and third equations to get
 * $$\begin{align}

\cfrac{dN_{xx}}{dx} + f(x) & = 0 \text{(1)} \qquad \\ \cfrac{d^2M_{xx}}{dx^2} + q(x) + \cfrac{d}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right) & = 0 \text{(2)} \qquad \end{align}$$ Let $$v_1$$ be the weighting function for equation (1) and let $$v_2$$ be the weighting function for equation (2).

Then the weak forms of the two equations are
 * $$\begin{align}

\int_{x_a}^{x_b} v_1\left(\cfrac{dN_{xx}}{dx} + f(x)\right)~dx & = 0 \text{(3)} \qquad \\ \int_{x_a}^{x_b} v_2\left[\cfrac{d^2M_{xx}}{dx^2} + q(x) + \cfrac{d}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right)\right]~dx & = 0 \text{(4)} \qquad \end{align}$$ To get the symmetric weak forms, we integrate by parts (even though the symmetry is not obvious at this stage) to get
 * $$\begin{align}

\left.(v_1~N_{xx})\right|_{x_a}^{x_b} & - \int_{x_a}^{x_b} \cfrac{dv_1}{dx} N_{xx}~dx + \int_{x_a}^{x_b} v_1~f(x)~dx = 0 \text{(5)} \qquad \\ \left.\left(v_2~\cfrac{dM_{xx}}{dx}\right)\right|_{x_a}^{x_b} & - \int_{x_a}^{x_b} \cfrac{dv_2}{dx}\cfrac{dM_{xx}}{dx}~dx + \int_{x_a}^{x_b} v_2~q(x)~dx + \\ & \left.\left(v_2~N_{xx}\cfrac{dw_0}{dx}\right)\right|_{x_a}^{x_b} - \int_{x_a}^{x_b} \cfrac{dv_2}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right)~dx = 0 \text{(6)} \qquad \end{align}$$ Integrate equation (6) again by parts, and get
 * $$\begin{align}

\left.\left(v_2~\cfrac{dM_{xx}}{dx}\right)\right|_{x_a}^{x_b} & - \left.\left(\cfrac{dv_2}{dx}~M_{xx}\right)\right|_{x_a}^{x_b} + \int_{x_a}^{x_b} \cfrac{d^2v_2}{dx^2}M_{xx}~dx + \int_{x_a}^{x_b} v_2~q(x)~dx + \\ & \left.\left(v_2~N_{xx}\cfrac{dw_0}{dx}\right)\right|_{x_a}^{x_b} - \int_{x_a}^{x_b} \cfrac{dv_2}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right)~dx = 0 \text{(7)} \qquad \end{align}$$ Collect terms and rearrange equations (5) and (7) to get
 * $$\begin{align}

\int_{x_a}^{x_b} \cfrac{dv_1}{dx} N_{xx}~dx & = \int_{x_a}^{x_b} v_1~f(x)~dx + \left.(v_1~N_{xx})\right|_{x_a}^{x_b} \text{(8)} \qquad \\ \int_{x_a}^{x_b} \left[ \cfrac{dv_2}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right) - \cfrac{d^2v_2}{dx^2}M_{xx}\right]~dx & = \int_{x_a}^{x_b} v_2~q(x)~dx +\\ &\left[\left(v_2~\cfrac{dM_{xx}}{dx}\right) - \left(\cfrac{dv_2}{dx}~M_{xx}\right) + \left(v_2~N_{xx}\cfrac{dw_0}{dx}\right)\right]_{x_a}^{x_b} \text{(9)} \qquad \end{align}$$ Rewriting the equations, we get

\begin{align} \int_{x_a}^{x_b} \cfrac{dv_1}{dx} N_{xx}~dx & = \int_{x_a}^{x_b} v_1 f~dx + \left.v_1 N_{xx}\right|_{x_a}^{x_b} \\ \int_{x_a}^{x_b} \left[ \cfrac{dv_2}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right) - \cfrac{d^2v_2}{dx^2}M_{xx}\right]~dx & = \int_{x_a}^{x_b} v_2 q~dx + \left[v_2\left(\cfrac{dM_{xx}}{dx} + N_{xx}\cfrac{dw_0}{dx}\right) \right]_{x_a}^{x_b} - \left[\cfrac{dv_2}{dx}~M_{xx}\right]_{x_a}^{x_b} \end{align} \text{(10)} \qquad $$ Hence shown.

Part 2
The  von Karman strains are related to the displacements by
 * $$\begin{align}

\varepsilon_{xx} & = \varepsilon^0_{xx} + z\varepsilon^1_{xx} \\ \varepsilon^0_{xx} & = \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \\ \varepsilon^1_{xx} & = -\cfrac{d^2w_0}{dx^2} \end{align}$$ The stress and moment resultants are defined as
 * $$\begin{align}

N_{xx} & = \int_A \sigma_{xx}~ dA \\ M_{xx} & = \int_A z\sigma_{xx}~ dA \end{align}$$ For a linear elastic material, the stiffnesses of the beam in extension and bending are defined as
 * $$\begin{align}

A_{xx} & = \int_A E~dA\qquad \leftarrow \qquad \text{extensional stiffness}\\ B_{xx} & = \int_A zE~dA\qquad \leftarrow \qquad \text{extensional-bending stiffness}\\ D_{xx} & = \int_A z^2E~dA\qquad \leftarrow \qquad \text{bending stiffness} \end{align}$$ where $$E$$ is the Young's modulus of the material.

Derive expressions for $$N_{xx}$$ and $$M_{xx}$$ in terms of the displacements $$u_0$$ and $$w_0$$ and the extensional and bending stiffnesses of the beam assuming a linear elastic material.

The stress-strain relations for an isotropic linear elastic material are

\begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{zz} \\ \sigma_{yz} \\ \sigma_{zx} \\ \sigma_{xy} \end{bmatrix} = \cfrac{E}{(1+\nu)(1-2\nu)} \begin{bmatrix} 1-\nu & \nu & \nu & 0 & 0 & 0 \\ \nu & 1-\nu & \nu & 0 & 0 & 0 \\ \nu & \nu & 1-\nu & 0 & 0 & 0 \\ 0& 0 & 0 & (1-2\nu) & 0 & 0 \\ 0& 0 & 0 & 0 & (1-2\nu) & 0 \\ 0& 0 & 0 & 0 & 0 & (1-2\nu) \end{bmatrix} \begin{bmatrix} \varepsilon_{xx} \\ \varepsilon_{yy} \\ \varepsilon_{zz} \\ \varepsilon_{yz} \\ \varepsilon_{zx} \\ \varepsilon_{xy} \end{bmatrix} $$ Since all strains other than $$\varepsilon_{xx}$$ are zero, the above equations reduce to

\sigma_{xx} = \cfrac{E(1-\nu)}{(1+\nu)(1-2\nu)}\varepsilon_{xx}; \qquad \sigma_{yy} = \cfrac{E\nu}{(1+\nu)(1-2\nu)}\varepsilon_{xx}; \qquad \sigma_{zz} = \cfrac{E\nu}{(1+\nu)(1-2\nu)}\varepsilon_{xx}~. $$ If we ignore the stresses $$\sigma_{yy}$$ and $$\sigma_{zz}$$, the only allowable value of $$\nu$$ is zero. Then the stress-strain relations become

\sigma_{xx} = E\varepsilon_{xx} ~. $$ Plugging this relation into the stress and moment of stress resultant equations, we get
 * $$\begin{align}

N_{xx} & = \int_A E\varepsilon_{xx}~ dA \\ M_{xx} & = \int_A zE\varepsilon_{xx}~ dA \end{align}$$ Plugging in the relations for the strain we get
 * $$\begin{align}

N_{xx} & = \int_A E(\varepsilon^0_{xx} + z\varepsilon^1_{xx})~ dA   = \int_A E\varepsilon^0_{xx}~dA + \int_A zE\varepsilon^1_{xx}~dA \\ M_{xx} & = \int_A zE(\varepsilon^0_{xx} + z\varepsilon^1_{xx})~ dA   = \int_A zE\varepsilon^0_{xx}~dA + \int_A z^2E\varepsilon^1_{xx}~dA ~. \end{align}$$ Since both $$\varepsilon^0_{xx}$$ and $$\varepsilon^1_{xx}$$ are independent of $$y$$ and $$z$$, we can take these quantities outside the integrals and get
 * $$\begin{align}

N_{xx} & = \varepsilon^0_{xx}\int_A E~dA + \varepsilon^1_{xx}\int_A zE~dA \\ M_{xx} & = \varepsilon^0_{xx}\int_A zE~dA + \varepsilon^1_{xx}\int_A z^2E~dA ~. \end{align}$$ Using the definitions of the extensional, extensional-bending, and bending stiffness, we can then write
 * $$\begin{align}

N_{xx} & = \varepsilon^0_{xx} A_{xx} + \varepsilon^1_{xx} B_{xx} \\ M_{xx} & = \varepsilon^0_{xx} B_{xx} + \varepsilon^1_{xx} D_{xx} ~. \end{align}$$ To write these relations in terms of $$u_0$$ and $$w_0$$ we substitute the expressions for the von Karman strains to get

\begin{align} N_{xx} & = \left[\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2\right] A_{xx} -\cfrac{d^2w_0}{dx^2} B_{xx} \\ M_{xx} & = \left[\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2\right] B_{xx} -\cfrac{d^2w_0}{dx^2} D_{xx} ~. \end{align} $$ These are the expressions of the resultants in terms of the displacements.

Part 3
Express the weak forms in terms of the displacements and the extensional and bending stiffnesses.

The weak form equations are
 * $$\begin{align}

\int_{x_a}^{x_b} \cfrac{dv_1}{dx} N_{xx}~dx & = \int_{x_a}^{x_b} v_1 f~dx + \left.v_1 N_{xx}\right|_{x_a}^{x_b} \text{(11)} \qquad \\ \int_{x_a}^{x_b} \left[ \cfrac{dv_2}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right) - \cfrac{d^2v_2}{dx^2}M_{xx}\right]~dx & = \int_{x_a}^{x_b} v_2 q~dx + \left[v_2\left(\cfrac{dM_{xx}}{dx} + N_{xx}\cfrac{dw_0}{dx}\right) \right]_{x_a}^{x_b} - \left[\cfrac{dv_2}{dx}~M_{xx}\right]_{x_a}^{x_b} \text{(12)} \qquad \end{align}$$ At this stage we make two more assumptions: From the first assumption, we have
 * 1) The elastic modulus is constant throughout the cross-section.
 * 2) The $$x$$-axis passes through the centroid of the cross-section.

B_{xx} = E \int_A z~dA ~. $$ From the second assumption, we get

\int_A z~dA = 0 \qquad \implies \qquad B_{xx} = 0 ~. $$ Then the relations for $$N_{xx}$$ and $$M_{xx}$$ reduce to
 * $$\begin{align}

N_{xx} & = \left[\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2\right] A_{xx} \text{(13)} \qquad \\ M_{xx} & =- \cfrac{d^2w_0}{dx^2} D_{xx} \text{(14)} \qquad~. \end{align}$$ Let us first consider equation (11). Plugging in the expression for $$N_{xx}$$ we get

\int_{x_a}^{x_b} \cfrac{dv_1}{dx} \left[\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2\right] A_{xx}~dx = \int_{x_a}^{x_b} v_1 f~dx + \left.v_1 N_{xx}\right|_{x_a}^{x_b} $$ We can also write the above in terms of virtual displacements by defining $$\delta u_0 := v_1$$, $$Q_1 := -N_{xx}(x_a)$$, and $$Q_4 := N_{xx}(x_b)$$. Then we get

\int_{x_a}^{x_b} \cfrac{d(\delta u_0)}{dx} \left[\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2\right] A_{xx}~dx = \int_{x_a}^{x_b} (\delta u_0) f~dx + \delta u_0(x_a) Q_1 + \delta u_0(x_b) Q_4 ~. $$ Next, we do the same for equation (12). Plugging in the expressions for $$N_{xx}$$ and $$M_{xx}$$, we get
 * $$\begin{align}

\int_{x_a}^{x_b} \left\{\cfrac{dv_2}{dx}\left(\left[\cfrac{du_0}{dx} + \cfrac{1}{2}~\left(\cfrac{dw_0}{dx}\right)^2\right] A_{xx} \cfrac{dw_0}{dx}\right) \right. & +    \left.\cfrac{d^2 v_2}{dx^2} \left(\cfrac{d^2 w_0}{dx^2}\right) D_{xx} \right\}~dx= \int_{x_a}^{x_b} v_2 q~dx +\\ & \left[v_2\left(\cfrac{dM_{xx}}{dx} + N_{xx}\cfrac{dw_0}{dx}\right)\right]_{x_a}^{x_b} - \left[\cfrac{dv_2}{dx}~M_{xx}\right]_{x_a}^{x_b} \end{align}$$ We can write the above in terms of the virtual displacements and the generalized forces by defining
 * $$\begin{align}

\delta w_0 & := v_2 \\ \delta \theta & := \cfrac{dv_2}{dx} \\ Q_2 & := -\left[\cfrac{dM_{xx}}{dx} + N_{xx}\cfrac{dw_0}{dx}\right]_{x_a} \\ Q_5 & := \left[\cfrac{dM_{xx}}{dx} + N_{xx}\cfrac{dw_0}{dx}\right]_{x_b} \\ Q_3 & := -M_{xx} (x_a) \\ Q_6 & := M_{xx} (x_b) \end{align}$$ to get

\begin{align} \int_{x_a}^{x_b} \left\{\cfrac{d(\delta w_0)}{dx} \right. &  \left[\cfrac{du_0}{dx} + \cfrac{1}{2}~\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{dw_0}{dx} A_{xx} + \left.\cfrac{d^2(\delta w_0)}{dx^2} \left(\cfrac{d^2w_0}{dx^2}\right) D_{xx} \right\}~dx= \\ & \int_{x_a}^{x_b} (\delta w_0) q~dx + \delta w_0(x_a) Q_2 + \delta w_0(x_b) Q_5 + \delta \theta(x_a) Q_3 + \delta \theta(x_b) Q_6 ~. \end{align} $$

Part 4
Assume that the approximate solutions for the axial displacement and the transverse deflection over a two noded element are given by
 * $$\begin{align}

u_0(x) & =u_1 \psi_1(x) + u_2 \psi_2(x) \text{(15)} \qquad \\ w_0(x) & =w_1 \phi_1(x) + \theta_1 \phi_2(x) + w_2 \phi_3(x) + \theta_2 \phi_4(x) \text{(16)} \qquad \end{align}$$ where $$\theta = -(dw_0/dx)$$.

Compute the element stiffness matrix for the element.

The weak forms of the governing equations are
 * $$\begin{align}

\int_{x_a}^{x_b} \cfrac{d(\delta u_0)}{dx} & \left[\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2\right] A_{xx}~dx = \int_{x_a}^{x_b} (\delta u_0) f~dx + \delta u_0(x_a) Q_1 + \delta u_0(x_b) Q_4 \text{(17)} \qquad \\ \int_{x_a}^{x_b} \left\{\cfrac{d(\delta w_0)}{dx} \right. &    \left[\cfrac{du_0}{dx} + \cfrac{1}{2}~\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{dw_0}{dx} A_{xx} + \left.\cfrac{d^2(\delta w_0)}{dx^2} \left(\cfrac{d^2w_0}{dx^2}\right) D_{xx} \right\}~dx = \\ & \int_{x_a}^{x_b} (\delta w_0) q~dx + \delta w_0(x_a) Q_2 + \delta w_0(x_b) Q_5 + \delta \theta(x_a) Q_3 + \delta \theta(x_b) Q_6 ~. \text{(18)} \qquad \end{align}$$

Let us first write the approximate solutions as
 * $$\begin{align}

u_0(x) & = \sum_{j=1}^2 u_j \psi_j \text{(19)} \qquad \\ w_0(x) & = \sum_{j=1}^4 d_j \phi_j \text{(20)} \qquad \end{align}$$ where $$d_j$$ are generalized displacements and

\{d_1, d_2, d_3, d_4\} \equiv \{w_1, \theta_1, w_2, \theta_2\} ~. $$

To formulate the finite element system of equations, we substitute the expressions from $$u_0$$ and $$w_0$$ from equations (19) and (20) into the weak form, and substitute the shape functions $$\psi_i$$ for $$\delta u_0$$, $$\phi_i$$ for $$\delta w_0$$.

For the first equation (17) we get

\int_{x_a}^{x_b} \cfrac{d\psi_i}{dx}\left[ \left(\sum_{j=1}^2 u_j\cfrac{d\psi_j}{dx}\right) + \frac{1}{2}\cfrac{dw_0}{dx} \left(\sum_{j=1}^4 d_j\cfrac{d\phi_j}{dx}\right)\right] A_{xx}~dx = \int_{x_a}^{x_b} \psi_i f~dx + \psi_i(x_a) Q_1 + \psi_i(x_b) Q_4 ~. $$ After reorganizing, we have

\begin{align} \sum_{j=1}^{2} \left[\int_{x_a}^{x_b} A_{xx}\cfrac{d\psi_i}{dx}\cfrac{d\psi_j}{dx}~dx\right]u_j & + \sum_{j=1}^{4} \left[\frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx}\cfrac{dw_0}{dx}\right) \cfrac{d\psi_i}{dx} \cfrac{d\phi_j}{dx}~dx\right]d_j = \\ & \int_{x_a}^{x_b} \psi_i f~dx + \psi_i(x_a) Q_1 + \psi_i(x_b) Q_4 ~. \end{align} $$ We can write the above as

\sum_{j=1}^{2} K_{ij}^{11} u_j + \sum_{j=1}^{4} K_{ij}^{12} d_j = F_i^1 $$ where $$i = 1, 2$$ and

\begin{align} K_{ij}^{11} & = \int_{x_a}^{x_b} A_{xx}\cfrac{d\psi_i}{dx}\cfrac{d\psi_j}{dx}~dx \\ K_{ij}^{12} & = \frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx}\cfrac{dw_0}{dx}\right) \cfrac{d\psi_i}{dx} \cfrac{d\phi_j}{dx}~dx\\ F_i^1 & = \int_{x_a}^{x_b} \psi_i f~dx + \psi_i(x_a) Q_1 + \psi_i(x_b) Q_4 ~. \end{align} $$

For the second equation (18) we get

\begin{align} \int_{x_a}^{x_b} \left\{\cfrac{d\phi_i}{dx} \right. &    \left[\left(\sum_{j=1}^{2} u_j\cfrac{d\psi_j}{dx}\right) + \frac{1}{2}\cfrac{dw_0}{dx} \left(\sum_{j=1}^{4} d_j\cfrac{d\phi_j}{dx}\right)\right] \cfrac{dw_0}{dx}A_{xx} + \left.\cfrac{d^2\phi_i}{dx^2} \left(\sum_{j=1}^{4} d_j \cfrac{d^2\phi_j}{dx^2}\right)D_{xx} \right\}~dx = \\ & \int_{x_a}^{x_b} \phi_i q~dx + \phi_i(x_a) Q_2 + \phi_i(x_b) Q_5 + \cfrac{d\phi_i}{dx}(x_a) Q_3 + \cfrac{d\phi_i}{dx}(x_b) Q_6 \end{align} $$ After rearranging we get

\begin{align} \sum_{j=1}^{2} \left[ \int_{x_a}^{x_b} \left(A_{xx} \cfrac{dw_0}{dx}\right) \cfrac{d\phi_i}{dx}\cfrac{d\psi_j}{dx}~dx \right] u_j& + \sum_{j=1}^{4} \left\{\frac{1}{2} \int_{x_a}^{x_b} \left[A_{xx} \left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_i}{dx}\cfrac{d\phi_j}{dx}~dx \right\} d_j\\ & + \sum_{j=1}^{4} \left( \int_{x_a}^{x_b} D_{xx}\cfrac{d^2\phi_i}{dx^2}\cfrac{d^2\phi_j}{dx^2}~dx    \right) d_j = \\ & \int_{x_a}^{x_b} \phi_i q~dx + \phi_i(x_a) Q_2 + \phi_i(x_b) Q_5 + \cfrac{d\phi_i}{dx}(x_a) Q_3 + \cfrac{d\phi_i}{dx}(x_b) Q_6 \end{align} $$ We can write the above as

\sum_{j=1}^{2} K_{ij}^{21} u_j + \sum_{j=1}^{4} K_{ij}^{22} d_j = F_i^2 $$ where $$i = 1, 2, 3, 4$$ and

\begin{align} K_{ij}^{21} & = \int_{x_a}^{x_b} \left(A_{xx} \cfrac{dw_0}{dx}\right) \cfrac{d\phi_i}{dx}\cfrac{d\psi_j}{dx}~dx\\ K_{ij}^{22} & = \int_{x_a}^{x_b}\left\{ \frac{1}{2}\left[A_{xx} \left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_i}{dx}\cfrac{d\phi_j}{dx} + D_{xx}\cfrac{d^2\phi_i}{dx^2}\cfrac{d^2\phi_j}{dx^2}\right\}~dx \\ F_i^2 & = \int_{x_a}^{x_b} \phi_i q~dx + \phi_i(x_a) Q_2 + \phi_i(x_b) Q_5 + \cfrac{d\phi_i}{dx}(x_a) Q_3 + \cfrac{d\phi_i}{dx}(x_b) Q_6 ~. \end{align} $$ In matrix form, we can write

\mathbf{K} = \begin{bmatrix} K_{11}^{11} & K_{12}^{11} & \vdots & K_{11}^{12} & K_{12}^{12} & K_{13}^{12} & K_{14}^{12} \\ K_{21}^{11} & K_{22}^{11} & \vdots & K_{21}^{12} & K_{22}^{12} & K_{23}^{12} & K_{24}^{12} \\ &&\dots &&&&\\ K_{11}^{21} & K_{12}^{21} & \vdots & K_{11}^{22} & K_{12}^{22} & K_{13}^{12} & K_{14}^{12} \\ K_{21}^{21} & K_{22}^{21} & \vdots & K_{21}^{22} & K_{22}^{22} & K_{23}^{12} & K_{24}^{12} \\ K_{31}^{21} & K_{32}^{21} & \vdots & K_{31}^{22} & K_{32}^{22} & K_{33}^{22} & K_{34}^{22} \\ K_{41}^{21} & K_{42}^{21} & \vdots & K_{41}^{22} & K_{42}^{22} & K_{43}^{22} & K_{44}^{22} \\ \end{bmatrix} $$ or

\mathbf{K} = \begin{bmatrix} \mathbf{K}^{11} & \vdots & \mathbf{K}^{12} \\ & \vdots &\\ \mathbf{K}^{21} & \vdots & \mathbf{K}^{22} \end{bmatrix} ~. $$ The finite element system of equations can then be written as
 * $$\text{(21)} \qquad

\begin{bmatrix} K_{11}^{11} & K_{12}^{11} & \vdots & K_{11}^{12} & K_{12}^{12} & K_{13}^{12} & K_{14}^{12} \\ K_{21}^{11} & K_{22}^{11} & \vdots & K_{21}^{12} & K_{22}^{12} & K_{23}^{12} & K_{24}^{12} \\ &&\dots &&&&\\ K_{11}^{21} & K_{12}^{21} & \vdots & K_{11}^{22} & K_{12}^{22} & K_{13}^{12} & K_{14}^{12} \\ K_{21}^{21} & K_{22}^{21} & \vdots & K_{21}^{22} & K_{22}^{22} & K_{23}^{12} & K_{24}^{12} \\ K_{31}^{21} & K_{32}^{21} & \vdots & K_{31}^{22} & K_{32}^{22} & K_{33}^{22} & K_{34}^{22} \\ K_{41}^{21} & K_{42}^{21} & \vdots & K_{41}^{22} & K_{42}^{22} & K_{43}^{22} & K_{44}^{22} \\ \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\\\ d_1 \\ d_2 \\ d_3 \\ d_4 \end{bmatrix} = \begin{bmatrix} F_1^1 \\ F_2^1 \\\\ F_1^2 \\ F_2^2 \\ F_3^2 \\ F_4^2 \end{bmatrix} $$ or

\begin{bmatrix} \mathbf{K}^{11} & \vdots & \mathbf{K}^{12} \\ & \vdots &\\ \mathbf{K}^{21} & \vdots & \mathbf{K}^{22} \end{bmatrix} \begin{bmatrix} \mathbf{u} \\\\ \mathbf{d} \end{bmatrix} = \begin{bmatrix} \mathbf{F}^1 \\ \\ \mathbf{F}^2 \end{bmatrix} ~. $$

Part 5
Show the alternate procedure by which the element stiffness matrix can be made symmetric.

The stiffness matrix is unsymmetric because $$\mathbf{K}^{12}$$ contains a factor of $$1/2$$ while $$\mathbf{K}^{21}$$ does not. The expressions of these terms are
 * $$\begin{align}

K_{ij}^{12} & = \frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx}\cfrac{dw_0}{dx}\right) \cfrac{d\psi_i}{dx} \cfrac{d\phi_j}{dx}~dx, \qquad i=1,2, ~\text{and}~ j=1,2,3,4 \\ K_{ij}^{21} & = \int_{x_a}^{x_b} \left(A_{xx} \cfrac{dw_0}{dx}\right) \cfrac{d\phi_i}{dx}\cfrac{d\psi_j}{dx}~dx, \qquad i=1,2,3,4, ~\text{and}~ j=1,2 ~. \end{align}$$

To get a symmetric stiffness matrix, we write equation (18) as

\begin{align} \int_{x_a}^{x_b} \left\{\cfrac{d(\delta w_0)}{dx} \right. & \left[\frac{1}{2}\cfrac{du_0}{dx}+\frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2+ {\frac{1}{2}\cfrac{du_0}{dx}}\right] \cfrac{dw_0}{dx} A_{xx} + \left.\cfrac{d^2(\delta w_0)}{dx^2} \left(\cfrac{d^2w_0}{dx^2}\right) D_{xx} \right\}~dx = \\ & \int_{x_a}^{x_b} (\delta w_0) q~dx + \delta w_0(x_a) Q_2 + \delta w_0(x_b) Q_5 + \delta \theta(x_a) Q_3 + \delta \theta(x_b) Q_6 ~. \end{align} $$ The quantity is green is assumed to be known from a previous iteration and adds to the $$\mathbf{K}^{22}$$ terms.

Repeating the procedure used in the previous question

\begin{align} \int_{x_a}^{x_b} \left\{\cfrac{d\phi_i}{dx} \right. & \left[\frac{1}{2}\left(\sum_{j=1}^{2} u_j\cfrac{d\psi_j}{dx}\right) + \frac{1}{2}\cfrac{dw_0}{dx} \left(\sum_{j=1}^{4} d_j\cfrac{d\phi_j}{dx}\right) \right] \cfrac{dw_0}{dx}A_{xx} + { \frac{1}{2}\cfrac{d\phi_i}{dx} \cfrac{du_0}{dx}\left(\sum_{j=1}^{4} d_j\cfrac{d\phi_j}{dx}\right)A_{xx} + }\\ &\left.\cfrac{d^2\phi_i}{dx^2} \left(\sum_{j=1}^{4} d_j \cfrac{d^2\phi_j}{dx^2}\right)D_{xx} \right\}~dx = \\ & \int_{x_a}^{x_b} \phi_i q~dx + \phi_i(x_a) Q_2 + \phi_i(x_b) Q_5 + \cfrac{d\phi_i}{dx}(x_a) Q_3 + \cfrac{d\phi_i}{dx}(x_b) Q_6 \end{align} $$ After rearranging we get

\begin{align} \sum_{j=1}^{2} & \left[\frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx} \cfrac{dw_0}{dx}\right) \cfrac{d\phi_i}{dx}\cfrac{d\psi_j}{dx}~dx \right] u_j + \sum_{j=1}^{4} \left\{\frac{1}{2} \int_{x_a}^{x_b} \left[A_{xx} \left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_i}{dx}\cfrac{d\phi_j}{dx}~dx \right\} d_j + \\ & { \sum_{j=1}^{4} \left[\frac{1}{2}\int_{x_a}^{x_b} A_{xx}\cfrac{du_0}{dx}\cfrac{d\phi_i}{dx} \cfrac{d\phi_j}{dx}~dx\right] d_j } + \sum_{j=1}^{4} \left( \int_{x_a}^{x_b} D_{xx}\cfrac{d^2\phi_i}{dx^2}\cfrac{d^2\phi_j}{dx^2}~dx \right) d_j = \\ & \int_{x_a}^{x_b} \phi_i q~dx + \phi_i(x_a) Q_2 + \phi_i(x_b) Q_5 + \cfrac{d\phi_i}{dx}(x_a) Q_3 + \cfrac{d\phi_i}{dx}(x_b) Q_6 \end{align} $$

We can write the above as

\sum_{j=1}^{2} K_{ij}^{21} u_j + \sum_{j=1}^{4} K_{ij}^{22} d_j = F_i^2 $$ where $$i = 1, 2, 3, 4$$ and

\begin{align} K_{ij}^{21} & = \frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx} \cfrac{dw_0}{dx}\right) \cfrac{d\phi_i}{dx}\cfrac{d\psi_j}{dx}~dx\\ K_{ij}^{22} & = \int_{x_a}^{x_b}\left\{ \frac{1}{2} A_{xx} \left[\cfrac{du_0}{dx}+\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_i}{dx}\cfrac{d\phi_j}{dx} + D_{xx}\cfrac{d^2\phi_i}{dx^2}\cfrac{d^2\phi_j}{dx^2}\right\}~dx \\ F_i^2 & = \int_{x_a}^{x_b} \phi_i q~dx + \phi_i(x_a) Q_2 + \phi_i(x_b) Q_5 + \cfrac{d\phi_i}{dx}(x_a) Q_3 + \cfrac{d\phi_i}{dx}(x_b) Q_6 ~. \end{align} $$ This gives us a symmetric stiffness matrix.

Part 6
Derive the element tangent stiffness matrix for the element.

Equation (21) can be written as

\mathbf{K}(\mathbf{U}) \mathbf{U} = \mathbf{F} $$ where
 * $$\begin{align}

U_1 & = u_1, ~ U_2= u_2, ~ U_3= d_1, ~ U_4= d_2, ~ U_5= d_3, ~ U_6= d_4 \\ F_1 & = F^1_1, ~ F_2= F^1_2, ~ F_3= F^2_1, ~ F_4= F^2_2, ~ F_5= F^2_3, ~ F_6= F^2_4 \end{align}$$ The residual is

\mathbf{R} = \mathbf{K} \mathbf{U} - \mathbf{F} ~. $$ For Newton iterations, we use the algorithm

\mathbf{U}^{r+1} = \mathbf{U}^r - (\mathbf{T}^r)^{-1} \mathbf{R}^r $$ where the tangent stiffness matrix is given by

\mathbf{T}^r = \frac{\partial \mathbf{R}^r}{\partial \mathbf{U}} ~. $$ The coefficients of the tangent stiffness matrix are given by

T_{ij} = \frac{\partial R_i}{\partial U_j}, \qquad i=1 \dots 6, j=1 \dots 6~. $$ Recall that the finite element system of equations can be written as

\begin{align} \sum_{p=1}^{2} K_{mp}^{11} u_p & + \sum_{q=1}^{4} K_{mq}^{12} d_q = F_m^1, \qquad m=1,2 \\ \sum_{p=1}^{2} K_{np}^{21} u_p & + \sum_{q=1}^{4} K_{nq}^{22} d_q = F_n^2, \qquad n=1,2,3,4 \end{align} $$ where the subscripts have been changed to avoid confusion.

Therefore, the residuals are

\begin{align} R_m^1 & = \sum_{p=1}^{2} K_{mp}^{11} u_p + \sum_{q=1}^{4} K_{mq}^{12} d_q - F_m^1, \qquad m=1,2 \\ R_n^2 & = \sum_{p=1}^{2} K_{np}^{21} u_p + \sum_{q=1}^{4} K_{nq}^{22} d_q - F_n^2, \qquad n=1,2,3,4 ~. \end{align} $$ The derivatives of the residuals with respect to the generalized displacements are

\begin{align} T^{11}_{mk} = \frac{\partial R_m^1}{\partial u_k} & = \sum_{p=1}^{2} \frac{\partial }{\partial u_k}(K_{mp}^{11} u_p) + \sum_{q=1}^{4} \frac{\partial }{\partial u_k}(K_{mq}^{12} d_q) , & & \qquad m=1,2; k=1,2 \\ T^{12}_{ml} = \frac{\partial R_m^1}{\partial d_l} & = \sum_{p=1}^{2} \frac{\partial }{\partial d_l}(K_{mp}^{11} u_p) + \sum_{q=1}^{4} \frac{\partial }{\partial d_l}(K_{mq}^{12} d_q) , & & \qquad m=1,2; l=1,2,3,4 \\ T^{21}_{nk} = \frac{\partial R_n^2}{\partial u_k} & = \sum_{p=1}^{2} \frac{\partial }{\partial u_k}(K_{np}^{21} u_p) + \sum_{q=1}^{4} \frac{\partial }{\partial u_k}(K_{nq}^{22} d_q) , & & \qquad n=1,2,3,4; k=1,2 \\ T^{22}_{nl} = \frac{\partial R_n^2}{\partial d_l} & = \sum_{p=1}^{2} \frac{\partial }{\partial d_l}(K_{np}^{21} u_p) + \sum_{q=1}^{4} \frac{\partial }{\partial d_l}(K_{nq}^{22} d_q) , & & \qquad n=1,2,3,4; l=1,2,3,4 ~. \end{align} $$ Differentiating, we get

\begin{align} T^{11}_{mk} & = \sum_{p=1}^{2} \left(u_p\frac{\partial K_{mp}^{11}}{\partial u_k} + K_{mp}^{11}\frac{\partial u_p}{\partial u_k}\right) + \sum_{q=1}^{4} \left(d_q\frac{\partial K_{mq}^{12}}{\partial u_k} + K_{mq}^{12}\frac{\partial d_q}{\partial u_k}\right), & &\qquad m=1,2; k=1,2 \\ T^{12}_{ml} & = \sum_{p=1}^{2} \left(u_p\frac{\partial K_{mp}^{11}}{\partial d_l} + K_{mp}^{11}\frac{\partial u_p}{\partial d_l}\right) + \sum_{q=1}^{4} \left(d_q\frac{\partial K_{mq}^{12}}{\partial d_l} + K_{mq}^{12}\frac{\partial d_q}{\partial d_l}\right), & &\qquad m=1,2; l=1,2,3,4 \\ T^{21}_{nk} & = \sum_{p=1}^{2} \left(u_p\frac{\partial K_{np}^{21}}{\partial u_k} + K_{np}^{21}\frac{\partial u_p}{\partial u_k}\right) + \sum_{q=1}^{4} \left(d_q\frac{\partial K_{nq}^{22}}{\partial u_k} + K_{nq}^{22}\frac{\partial d_q}{\partial u_k}\right), & &\qquad n=1,2,3,4; k=1,2 \\ T^{22}_{nl} & = \sum_{p=1}^{2} \left(u_p\frac{\partial K_{np}^{21}}{\partial d_l} + K_{np}^{21}\frac{\partial u_p}{\partial d_l}\right) + \sum_{q=1}^{4} \left(d_q\frac{\partial K_{nq}^{22}}{\partial d_l} + K_{nq}^{22}\frac{\partial d_q}{\partial d_l}\right), & &\qquad n=1,2,3,4; l=1,2,3,4 ~. \end{align} $$ These equations can therefore be written as

\begin{align} T^{11}_{mk} & = K^{11}_{mk} + \sum_{p=1}^{2} u_p\frac{\partial K_{mp}^{11}}{\partial u_k} + \sum_{q=1}^{4} d_q\frac{\partial K_{mq}^{12}}{\partial u_k}, & &\qquad m=1,2; k=1,2 \\ T^{12}_{ml} & = K^{12}_{ml} + \sum_{p=1}^{2} u_p\frac{\partial K_{mp}^{11}}{\partial d_l} + \sum_{q=1}^{4} d_q\frac{\partial K_{mq}^{12}}{\partial d_l}, & &\qquad m=1,2; l=1,2,3,4 \\ T^{21}_{nk} & = K^{21}_{nk} + \sum_{p=1}^{2} u_p\frac{\partial K_{np}^{21}}{\partial u_k} + \sum_{q=1}^{4} d_q\frac{\partial K_{nq}^{22}}{\partial u_k}, & &\qquad n=1,2,3,4; k=1,2 \\ T^{22}_{nl} & = K^{22}_{nl} + \sum_{p=1}^{2} u_p\frac{\partial K_{np}^{21}}{\partial d_l} + \sum_{q=1}^{4} d_q\frac{\partial K_{nq}^{22}}{\partial d_l}, & &\qquad n=1,2,3,4; l=1,2,3,4 ~. \end{align} $$ Now, the coefficients of $$\mathbf{K}^{11}$$, $$\mathbf{K}^{12}$$, and $$\mathbf{K}^{21}$$ of the symmetric stiffness matrix are independent of $$u_1$$ and $$u_2$$. Also, the terms of $$\mathbf{K}^{11}$$ are independent of the all the generalized displacements. Therefore, the above equations reduce to

\begin{align} T^{11}_{mk} & = K^{11}_{mk}, & &\qquad m=1,2; k=1,2 \\ T^{12}_{ml} & = K^{12}_{ml} + \sum_{q=1}^{4} d_q\frac{\partial K_{mq}^{12}}{\partial d_l}, & & \qquad m=1,2; l=1,2,3,4 \\ T^{21}_{nk} & = K^{21}_{nk} + \sum_{q=1}^{4} d_q\frac{\partial K_{nq}^{22}}{\partial u_k}, & & \qquad n=1,2,3,4; k=1,2 \\ T^{22}_{nl} & = K^{22}_{nl} + \sum_{p=1}^{2} u_p\frac{\partial K_{np}^{21}}{\partial d_l} + \sum_{q=1}^{4} d_q\frac{\partial K_{nq}^{22}}{\partial d_l}, & & \qquad n=1,2,3,4; l=1,2,3,4 ~. \end{align} $$

Consider the coefficients of $$\mathbf{T}^{12}$$:

T^{12}_{ml} = K^{12}_{ml} + \sum_{q=1}^{4} d_q\frac{\partial K_{mq}^{12}}{\partial d_l}, \qquad m=1,2; l=1,2,3,4 ~. $$ From our previous derivation, we have

K_{mq}^{12} =\frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx}\cfrac{dw_0}{dx}\right) \cfrac{d\psi_m}{dx} \cfrac{d\phi_q}{dx}~dx~. $$ Therefore,
 * $$\begin{align}

\frac{\partial K_{mq}^{12}}{\partial d_l} &= \frac{1}{2}\int_{x_a}^{x_b}\left[A_{xx}\frac{\partial }{\partial d_l}\left(\cfrac{dw_0}{dx}\right)\right] \cfrac{d\psi_m}{dx} \cfrac{d\phi_q}{dx}~dx \\ & = \frac{1}{2}\int_{x_a}^{x_b}\left[A_{xx}\left(\sum_{T=1}^{4}\frac{\partial d_T}{\partial d_l} \cfrac{d\phi_T}{dx}\right)\right] \cfrac{d\psi_m}{dx} \cfrac{d\phi_q}{dx}~dx \\ & = \frac{1}{2}\int_{x_a}^{x_b}\left[A_{xx}\cfrac{d\phi_l}{dx}\right] \cfrac{d\psi_m}{dx} \cfrac{d\phi_q}{dx}~dx \end{align}$$ The tangent stiffness matrix coefficients are therefore
 * $$\begin{align}

{ T^{12}_{ml}} & = K^{12}_{ml} + \sum_{q=1}^{4} d_q\left\{ \frac{1}{2}\int_{x_a}^{x_b}\left[A_{xx}\cfrac{d\phi_l}{dx}\right] \cfrac{d\psi_m}{dx} \cfrac{d\phi_q}{dx}~dx \right\} \qquad m=1,2; l=1,2,3,4 ~. \\ & = K^{12}_{ml} + \frac{1}{2}\int_{x_a}^{x_b} A_{xx}\cfrac{d\phi_l}{dx} \cfrac{d\psi_m}{dx}\left(\sum_{q=1}^{4}d_q\cfrac{d\phi_q}{dx}\right)~dx \\ & = K^{12}_{ml} + \frac{1}{2}\int_{x_a}^{x_b} A_{xx}\cfrac{d\phi_l}{dx} \cfrac{d\psi_m}{dx}\cfrac{dw_0}{dx}~dx\\ & = { 2K^{12}_{ml}} ~. \end{align}$$ Next, {consider the coefficients of $$\mathbf{T}^{21}$$:

T^{21}_{nk} = K^{21}_{nk} + \sum_{q=1}^{4} d_q\frac{\partial K_{nq}^{22}}{\partial u_k} ~. $$ The coefficients of $$\mathbf{K}^{22}$$ are

K_{nq}^{22}= \int_{x_a}^{x_b}\left\{ \frac{1}{2} A_{xx} \left[\cfrac{du_0}{dx}+\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_n}{dx}\cfrac{d\phi_q}{dx} + D_{xx}\cfrac{d^2\phi_n}{dx^2}\cfrac{d^2\phi_q}{dx^2}\right\}~dx ~. $$ Therefore, the derivatives are
 * $$\begin{align}

\frac{\partial K_{nq}^{22}}{\partial u_k} & = \int_{x_a}^{x_b} \frac{1}{2} A_{xx} \left[\sum_{T=1}^{2}\frac{\partial u_T}{\partial u_k}\cfrac{d\psi_T}{dx}+ 2\cfrac{dw_0}{dx}\left(\sum_{T=1}^{4}\frac{\partial d_T}{\partial u_k}\cfrac{d\phi_T}{dx} \right)\right] \cfrac{d\phi_n}{dx}\cfrac{d\phi_q}{dx}~dx \\ &= \frac{1}{2}\int_{x_a}^{x_b} A_{xx} \cfrac{d\psi_k}{dx} \cfrac{d\phi_n}{dx}\cfrac{d\phi_q}{dx}~dx \\ \end{align}$$ Therefore the coefficients of $$\mathbf{T}^{21}$$ are
 * $$\begin{align}

{ T^{21}_{nk}} & = K^{21}_{nk} + \sum_{q=1}^{4} d_q\left(\frac{1}{2}\int_{x_a}^{x_b} A_{xx} \cfrac{d\psi_k}{dx} \cfrac{d\phi_n}{dx}\cfrac{d\phi_q}{dx}~dx\right) \\ & = K^{21}_{nk} + \frac{1}{2}\int_{x_a}^{x_b} A_{xx}\cfrac{d\psi_k}{dx}\cfrac{d\phi_n}{dx} \left(\sum_{q=1}^{4} d_q\cfrac{d\phi_q}{dx}\right)~dx \\ & = K^{21}_{nk} + \frac{1}{2}\int_{x_a}^{x_b} A_{xx}\cfrac{d\psi_k}{dx}\cfrac{d\phi_n}{dx} \cfrac{dw_0}{dx}~dx \\ & = { 2K^{21}_{nk}} ~. \end{align}$$ Finally, for the $$\mathbf{T}^{22}$$ coefficients, we start with

T^{22}_{nl} = K^{22}_{nl} + \sum_{p=1}^{2} u_p\frac{\partial K_{np}^{21}}{\partial d_l} + \sum_{q=1}^{4} d_q\frac{\partial K_{nq}^{22}}{\partial d_l}, \qquad n=1,2,3,4; l=1,2,3,4 $$ and plug in the derivatives of the stiffness matrix coefficients
 * $$\begin{align}

K_{np}^{21} & = \frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx} \cfrac{dw_0}{dx}\right) \cfrac{d\phi_n}{dx}\cfrac{d\psi_p}{dx}~dx\\ K_{nq}^{22} & = \int_{x_a}^{x_b}\left\{ \frac{1}{2} A_{xx} \left[\cfrac{du_0}{dx}+\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_n}{dx}\cfrac{d\phi_q}{dx} + D_{xx}\cfrac{d^2\phi_n}{dx^2}\cfrac{d^2\phi_q}{dx^2}\right\}~dx ~. \end{align}$$ \\ The derivatives are
 * $$\begin{align}

\frac{\partial K_{np}^{21}}{\partial d_l} & = \frac{1}{2} \int_{x_a}^{x_b} \left[A_{xx} \frac{\partial }{\partial d_l}\left(\cfrac{dw_0}{dx}\right) \right] \cfrac{d\phi_n}{dx}\cfrac{d\psi_p}{dx}~dx\\ & = \frac{1}{2} \int_{x_a}^{x_b} A_{xx}\left[\sum_{T=1}^{4}\frac{\partial d_T}{\partial d_l}\cfrac{d\phi_T}{dx} \right] \cfrac{d\phi_n}{dx}\cfrac{d\psi_p}{dx}~dx \\ & = \frac{1}{2} \int_{x_a}^{x_b} A_{xx}\cfrac{d\phi_l}{dx}\cfrac{d\phi_n}{dx} \cfrac{d\psi_p}{dx}~dx \end{align}$$ and
 * $$\begin{align}

\frac{\partial K_{nq}^{22}}{\partial d_l} & = \int_{x_a}^{x_b} \frac{1}{2} A_{xx} \left[\sum_{T=1}^{2}\frac{\partial u_T}{\partial d_l}\cfrac{d\psi_T}{dx}+ 2\cfrac{dw_0}{dx}\left(\sum_{T=1}^{4} \frac{\partial d_T}{\partial d_l}\cfrac{d\phi_T}{dx}\right)\right] \cfrac{d\phi_n}{dx}\cfrac{d\phi_q}{dx}~dx \\ & = \int_{x_a}^{x_b} A_{xx} \cfrac{dw_0}{dx} \cfrac{d\phi_l}{dx} \cfrac{d\phi_n}{dx}\cfrac{d\phi_q}{dx}~dx \end{align}$$ Therefore, the coefficients of the tangent stiffness matrix can be written as
 * $$\begin{align}

{ T^{22}_{nl}} & = K^{22}_{nl} + \sum_{p=1}^{2} u_p \left[ \frac{1}{2} \int_{x_a}^{x_b} A_{xx}\cfrac{d\phi_l}{dx}\cfrac{d\phi_n}{dx} \cfrac{d\psi_p}{dx}~dx \right] + \sum_{q=1}^{4} d_q \left[ \int_{x_a}^{x_b} A_{xx} \cfrac{dw_0}{dx} \cfrac{d\phi_l}{dx} \cfrac{d\phi_n}{dx}\cfrac{d\phi_q}{dx}~dx \right]\\ & = K^{22}_{nl} + \frac{1}{2} \int_{x_a}^{x_b} A_{xx}\cfrac{d\phi_l}{dx}\cfrac{d\phi_n}{dx} \left(\sum_{p=1}^{2} u_p \cfrac{d\psi_p}{dx}\right)~dx+ \int_{x_a}^{x_b} A_{xx} \cfrac{dw_0}{dx} \cfrac{d\phi_l}{dx} \cfrac{d\phi_n}{dx} \left(\sum_{q=1}^{4} d_q \cfrac{d\phi_q}{dx}\right)~dx\\ & = K^{22}_{nl} + \frac{1}{2} \int_{x_a}^{x_b} A_{xx}\cfrac{d\phi_l}{dx}\cfrac{d\phi_n}{dx} \cfrac{du_0}{dx}~dx + \int_{x_a}^{x_b} A_{xx} \cfrac{dw_0}{dx} \cfrac{d\phi_l}{dx} \cfrac{d\phi_n}{dx} \cfrac{dw_0}{dx}~dx \\ & = { K^{22}_{nl} + \frac{1}{2} \int_{x_a}^{x_b} A_{xx}\left[\cfrac{du_0}{dx} + 2\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_l}{dx}\cfrac{d\phi_n}{dx}~dx }~. \end{align}$$

The final expressions for the tangent stiffness matrix terms are

\begin{align} T^{11}_{ij} & = K^{11}_{ij}, & &\qquad i=1,2; j=1,2 \\ T^{12}_{ij} & = 2 K^{12}_{ij}, & & \qquad i=1,2; j=1,2,3,4 \\ T^{21}_{ij} & = 2 K^{21}_{ij}, & & \qquad i=1,2,3,4; j=1,2 \\ T^{22}_{ij} & = K^{22}_{ij} + \frac{1}{2} \int_{x_a}^{x_b} A_{xx}\left[\cfrac{du_0}{dx} + 2\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_i}{dx}\cfrac{d\phi_j}{dx}~dx & & \qquad i=1,2,3,4; j=1,2,3,4 ~. \end{align} $$

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


text:
Nonlinear finite elements/Homework 5/Solutions

 * Nonlinear finite elements/Homework 5/Solutions/Problem 1
 * Nonlinear finite elements/Homework 5/Solutions/Problem 2


text: Category:Network administration

title: Category:Network administration


text:
User talk:JWSchmidt/3rd Qtr 2007 discussions

Note: this is an archive of old discussions. Please do not edit this page. Continue these discussions at User talk:JWSchmidt. There are other archives at User talk:JWSchmidt/Discussion archive.

Request
I am on a long term wikiversity break. Please block this account until October 30th 2009, in which is the day I will set to return. Student Galaxy 20:29, 1 July 2007 (UTC)

Hi am george and i done something on topic:astrochemistry, schol of chemistry and i want to know what i must to do there becouse i want to start my work!thanks!

but how i can do this?if u take a look u can see that i wrote something, what i knew about this!this is wrong?

ok i see the point!well when i found the topic it was free and empty and i thought......

"Bcp3" created pages...
Uh sorry... I guess that really isn't clear, is it. With the excepteion of BCP, all of these pages were created using one version or another of bcp3 :). --SB_Johnny | talk 23:50, 1 July 2007 (UTC)

Thanks for the welcome! I have a full schedule, but I'll try to help whenever I can. I'll be a "handyman" around here! Zark 03:29, 3 July 2007 (UTC)

good job i hope you become a coustodian

Really good job...
Can you upload this picture to Commons? I would like to use this gif at the German Wikiversity... It would be great, if you could add a CC-license, too. The German wikiversity is using CC-BY-SA 2.5 (en). The CC-BY-SA 2.5-license would allows us to use the gif more easier for marketing purposes (outside the Wikiversity). Thanks... --Michael Reschke 19:56, 3 July 2007 (UTC)

Cognition
Thank you for going to work on Thought 5501.01: Cognition. Somebody has to actually do something. (Your Servant has been too much of a Generalist.) :-) (s) Dionysios (talk), a Participant in the Wikiversity School of Advanced General Studies, Date: 2007-07-03 (July 3, 2007) Time: 210701 UTC

 * I'm not sure how you wanted to develop Thought 5501.01. --JWSchmidt 21:44, 3 July 2007 (UTC)

 * Your Servant has no particular plan or desire as to how Thought 5501.01 or any other Wikiendeavor should develop, but rather, he is pleased to see an organic growth as the Wikisphere does that which it does.

 * Your work pleases Your Servant very much.

 * (s) Dionysios (talk), Date: 2007-07-04 (July 4, 2007) Time: 1400 UTC

Pratheep
Hi. You seem to be an active member here, and I just registered several minutes ago. So, I ask you, what should be done about Pratheep? I'd move it to User:Pratheep, but it tells me my account is too new. Since there is no user named Pratheep, perhaps a deletion would be better, especially considering the contributions of that IP? — SheeEttin {T/C} 19:42, 7 July 2007 (UTC)

Hi there
Hi there. I'm a student teacher, and I recently got this IP address. I don't want to get an account, yet...

Anyway, is it OK to use Wikiversity as a platform for writing GCSE English related articles?? currently my IP is blocked on Wikibooks and Wikipedia and I don't know why - I've certainly never done any "vandalism"!

can you get it unblocked for me there?? Good site here, I'll try and use it more! --84.45.219.185 08:36, 9 July 2007 (UTC)

--84.45.219.185 08:39, 9 July 2007 (UTC)
 * should just add I see a message saying the IP is blocked, it's weird, I didn't vandalise, and don't know who could have. My colleague, who's an IT technician, knows what Wikimedia is though!

 * Feel free to contribute learning resources for GCSE. Be sure to coordinate your efforts with Wikibooks GCSE content....any resources that take the form of a textbook should be on pages at Wikibooks. About "84.45.219.185".......it seems likely that 84.45.219.185 will be blocked again in the future since there are vandals who edit from that IP address. You should create an account. --JWSchmidt 14:21, 9 July 2007 (UTC)

 * Just a heads up, the owner of this IP has a long history of requesting unblock, then creating more spambot and vandal accounts. CU on wikibooks (and wp, iirc) has shown nothing but vandal edits and accounts. It's hardblocked on wb, not sure what the current status is on wp. --SB_Johnny | PA! 12:40, 16 July 2007 (UTC)

Thanks
Hi..thanks for the message...that was my first time.. i've left my signature on the page

New Course Materials
I'm a TA, so in the last few semesters I've generated a large number of teaching and learning materials for coursework: Lab assignments and solutions, lecture notes, collections of references, timelines, etc. The course that I am working on now has strong robotics components, but it's for an introductory audience.

I'm going to start writing a book about programming the "Basic Stamp" processor, which is a common platform to teach about programming and robotics. I would like to start uploading some of my course materials here to work in conjunction with the book on Wikibooks. I just want to double check with you, because this is my first wikiversity project, and I want to make sure this is the right thing and the right place. --Whiteknight 14:34, 12 July 2007 (UTC)

 * Thanks for the welcome and the links! I guess what I'm trying to figure out here is what, exactly, i can do at wikiversity. I am going to write a textbook on the subject, but I have lots and lots of these learning materials, and I am making new materials every day of every week. Is this the kind of thing that I can upload to Wikiversity? I know that WV is focused on "collaborative learning", I just want to make sure that my interpretation of that statement is correct.
 * What I would like to do, Ideally, is to set up a timeline to give a temporal dimension to the learning. It's all well and good to have a book, it's a different thing entirely to have as schedule and say "Read these pages on day 1, do these exercises. etc" I would like to create is a syllabus organized on a per-day basis with reading assignments, lecture notes, laboratory experiments, and homework suggestions for every day. Each day the student reads the necessary book pages, and then performs the activities.
 * If this works out, there are probably about a dozen other subjects that I can start donating some of my personal teaching materials for. I've taught several classes, and I have lecture notes and laboratory assignments for all of them. --Whiteknight 18:24, 12 July 2007 (UTC)

Re: Welcome back
Is your thesis online? --JWSchmidt 21:58, 8 July 2007 (UTC)
 * Not yet, still have a bit more corrections to do.--Rayc 01:48, 16 July 2007 (UTC)

sorry, but...
...I'm reverting you on Cite sources. There was no argument given by the one supporting voter either, and while there are agruments on the talk page, there was never a quorum. It's not for you to decide this John: if you want to get some action on it, bring it up on the colloquium. You unilaterally decided to move this from proposed to official, 10 months after the last "vote" was cast. You know I love you and respect you, but that's a pretty heavy-handed move.

Way too much time was spent arguing over policies at the beginning (including near flame wars over rounded corners), and I'm sure a few folks were scared off (I would have been, if I weren't so distracted by the possibilities of what we can achieve here). Making a policy after that kind of discussion and feedback is a bit scary dude.

There are all sorts of proposed policies that deserve to be revisited... you guys wrote some good ones back in the day, but it all fell apart because writing policies isn't a way to make a project grow (it's about content and learning, silly!). I think we can start moving those through now (maybe a "policy of the week" thread on the colloquium), but you saying it's law because it makes sense to you is a really bad idea. Please think about this... and yes, I'm reverting you. --SB_Johnny | PA! 01:09, 18 July 2007 (UTC)

 * Just an afterthought: I'd vote for that policy. It's the procedure that's upsetting me here. --SB_Johnny | PA! 01:14, 18 July 2007 (UTC)
 * Just an FYI that I support Johnny's revert on the grounds he has given. We shouldn't be so cavalier about these things.  The Jade Knight 02:23, 18 July 2007 (UTC)

 * Citing sources is a great thing when there are indeed sources to cite. The problem with the Cite, Verifiability, and Reliable Sources triad is that they focus on secondary sources. Primary sources are permitted on Wikibooks, and my understanding of the OR allowance here is that for research projects Wikiversity can itself be the primary source.
 * I understand that it's nicer to have these as enforced policies through their links on the welcome template, but citation, sources, and verifiability could just as easily be addressed as help pages. If they're going to be policy pages, they should be discussed again, rather than switched from proposed to enforced months later and without any recent discussion or announcement.
 * It's not that we shouldn't have a policy that "people should cite sources", but rather that we need to figure a lot of stuff out before we enact a policy saying that anything unsourced will be removed. Sugar cubes can be just as effective as cattle prods :). --SB_Johnny | PA! 12:39, 18 July 2007 (UTC)

Thanks for the contact
So you can access my website now! I fixed the slow response to a tolerable degree. I know you have little spare time, but you're welcome to create an account and comment on whatever interests you - there is no substitute for experience and I don't have any. I'd rather not publish my url until the wiki is secure. PeterMG 05:16, 29 July 2007 (UTC)
 * I just visited your site because of what you "said" on IRC.<- Yes thanks, I wanted to make sure it was publically accessible. I'm still setting up the schools and topics, so there are many dead links, but I think there is enough for you to see what I'm proposing. If you see any 'gotchas' I haven't, I'd like to know sooner rather than later. PeterMG 06:01, 29 July 2007 (UTC)

Thank you for the resource
Hello JW,

Thank you very much for your kind welcome.

At the moment I am trying to become a middle school science teacher, and I have to take tests. The tests are not that difficult, though the breadth of knowledge is wide.

I have been experimenting with various wikis attempting to create a collaborative project, including zoho and pbwiki -- without success.

Most teachers take the praxis exams; the expectations are well documented in the Praxis Inc web site. I am using their framework to create what is thought of as "required learning" for teaching -- what is on the often hated standardized exams. Since much of the testing frame work is still embedded in my text, it is not yet ready for Wiki.

I have a very attractive page under construction still on my laptop; it includes a hilarious picture of a mad scientist I found here in the wiki resources. A scratch version is installed on Zoho wiki service if you are curious to my approach: Middle school science learning

Since I am very interested in project-based learning, which is also called group science, I will attempt to wrap the required learnings within more conceptual learning, what the constructivists call "constructing a community of knowledge."

For now I have to work memorizing all of the required factual information, so as to do well on the tests, but I have a lot of material developed on group science.

I also have contributed to the Wikipedia "Project-Based Learning" article; there was talk on the discussion page about it being meaningless "cruft," so I made hurried changes to it in order to save it from deletion. I would very much like to extend it.

Thinmanis my home page, and this is my star achievement, which enjoys a high search engine rating: Empathy

Thanks for the encouragement & a word about "project"
Hi, you are right, the way to build solid knowledge neurons is by doing, for certain. But the word project is loaded.

I think I can put this in bullet form:
 * kids develop a relationship with the world as they become more aware of it
 * as they begin to socialize and interrelate, this relationship develops into a shared conception
 * when they get to school, and teachers try to straighten them on certain misconceptions,
 * they think the teachers are crazy

Kids will tell you what you want to hear, but they think they are really being taught patience with elders. As soon as they back with their friends, the misconceptions are restored, and all is well from their perspective.

The project, or group, science, strategy is to allow them to discover the realities of science on their own by providing them with much the same equipment that our early scientists had, such as prisms, telescopes, microscopes. Their misconceptions are not actually wrong so much as "intelligently wrong," and often the entire community holds the same misconceptions. Here is were the true value is; as the students develop scientific technique and learn how things really are, they go back home and tell their parents what they did. In this way the entire community gets educated -- now that's value.

Science today is seeing a rise in the social science; as we apply scientific technique to social situations, we are far better able to get from these situations the kinds of results we are looking for. When kids do this, they can better experiment with life and conclude earlier what it is that they are good at, and what they like to do. This gives them factors better chances of getting the highest potential out of life, of self-actualizing.

Wow, thats great writing, I think I might use it myself.

John van v 01:47, 2 August 2007 (UTC)John

Gizmo
I am very tempted to get a nokia 770 to experiment w/ Gizmo and its underlying SIP protocol.

What is your experience? I read that it works over dial-up, but I read that the sound can be garbled.

Middle_school_science
Are you saying that I should just go ahead and create this page? And throw up an outline of my proposed format.

I have put my personal page online: http://thinman.com/middle_school_science

So far, the effort is for looks, and the content is designed to make readers feel comfortable.

My goal for the midl schl sci wiki is to create a framework that, as I mentioned, wraps the required learning as it appears in required and standardized testing.

I have a prototype for this kind of framework from my computer experience. I created a simplified view of all computers I called the CMDN categories: [http://docs.google.com/Doc?id=ddk32zv4_274dx95sh "When working with common computer systems, observations fall into four distinct categories, CPU, Memory, Disk and Network." (JvV) ]

I would like to create a similar "high level" simplifications for each of the "disciplines:" Physics, Chemistry, Earth and Space, and so forth, each of which would consist of a limited number of categories.

Then from these categories I want to create yet a higher "wrapper" that resembles the Russian matrioshka nesting dolls, which is becoming a common analogous symbol for evolution. Childhood growth through inquiry is of cousre an example of evolution, but genetic.

This will create a hierarchial structure for the information, or constructed knowledge, that will then be turned into a true concept map through the linking of concepts within -- creating a vast concept map. To build this map, however, I am hoping I can prevail in insisting that proprietary software not be used. Concept mapping MUST, I feel, be a fully open architecture in the truest sense. And it must be a simple and easily maintained teachnology.

almost :P
Actually, it's the school I'm worried about... has a record of blockings on wikipedia, and it would be a bit problematic if I were showing people how to set up accounts and found that we, um, couldn't do that ;-). --SB_Johnny | PA! 09:33, 4 August 2007 (UTC)

Wikiversity beta
I am a user in Wikiversity beta I asked to change my name from Diaza to ZaDiak but they told to ask it from you could you please do that?
 * I'm looking. --JWSchmidt 16:57, 5 August 2007 (UTC)

Content Does Not Seem to Match Page Names.
Greetings my Friend from Dionysios. Your Servant is wondering if the following Pages are correctly named:

 * 1) User talk:JWSchmidt/Early 2006 discussions Jan Feb March
 * 2) User talk:JWSchmidt/More 2006 discussions April May June

They appear to contain Archived Discussions for 2007, not 2006.

Best regards,

(s) Dionysios  (talk), a Participant in the Wikiversity School of Advanced General Studies, Date: 2007-08-06 (August 6, 2007) Time:  1621 UTC

Latest Thoughts, Discoveries, and Writings of Dionysios
Greetings, My Friend ,

The Latest Thoughts, Discoveries, and Writings of Your Servant  are found at "Wikisphere".

And what among your many Thoughts, Discoveries, and Writings are you finding most Compelling just Now?

Best regards,

(s) Dionysios  (talk), 2007-08-06 (August 6, 2007) Time:  2149 UTC

Wikiversity Video Will Result in Much Progress
Your Servant is so pleased that you are working on Collaborative Video Creation.  Video inside Wikiversity Pages will certainly carry its Guarantee to bring much Progress in the Wikiendeavor in which we find ourselves Engaged.  Support at Wikiversity for Video Content is a tremendous Wikiendeavor.  Thank you for your many Excellent Contributions.

DAZ Sudio Animation looks like a fine Tool. Thank you for calling it to the attention of  Your Servant.   It promises to be Valuable.

(s) Dionysios  (talk), Date: 2007-08-07 (August 7, 2007) Time:  1445 UTC

Screen shots with things circled...
Hi John. While going through the colloquium archives (slowly but surely!), I noticed that you seem to be a wiz at making screenshots making circles and arrows. Would you be willing to do a bunch of these for the bloom clock? I ask because I was showing a non-wiki person how to do it yesterday, and a graphical introduction to how the clock works and what all the links mean would be a big help (including simple things like "what is 'commons'?").

Speaking of commons, do we have a good help page for image uploads? If so, does it encourage people to use commons rather than Wikiversity? --SB_Johnny | PA! 15:41, 7 August 2007 (UTC)

 * Ah, thankyou-thankyou-thankyou :). I'll tell you what I'm looking for in a bit, going out to take some photos now.
 * No-one knows "the rules" on commons these days, it seems :). If you want to start a page on it, I'll gladly give some pointers on the categorization/linking part. The only absolutes are that it fits in with the scope (i.e., the image is useful to at least one wikimedia project), and that it's under a free license (personally I prefer GFDL). --SB_Johnny | PA! 16:47, 7 August 2007 (UTC)

hi man!how are u?
i have a question for u!well, can i founded a new department becouse today i founded a new department on Biology School and his name's Department of tropical diseases!I'll wait your answer

thanks very much!
well i'll wait ur ideas and ur advices.i'll write there smt links to university where u can study this and to hospitals, and i'll write also general thinks about tropical diseases but i want to help me someone to show me what i must to do!

Wiki Templates
Hi, I have reached a first milestone with my Middle School Science page; the content for the earth and space science is now fully original, qualifying it to be a wikiversity page.

For career reasons, I will have to finish all the other content for the other topics (phys, chem, math, bio) so I can actually take the tests.

The structure is based on the Praxis exams, and I am not sure of the Praxis licensing so I have ommited all their text.

The GACE exam (the teacher exam I will be taking), has a somewhat open license that makes the text in it available for educators -- which is what we are. It is more generalized than the Praxis framework, and without detail, but I will include it anyway to help the structure of middle school science page.

I really do need collaborators, and I am not sure how to find them. I have written an invitation for teachers to join; here is the link: http://docs.google.com/Doc?id=ddk32zv4_420dvkwqw

Middle School Science Progress
Hi, I have reached a first milestone with my Middle School Science page; the content for the earth and space science is now fully original, qualifying it to be a wikiversity page.

For career reasons, I will have to finish all the other content for the other topics (phys, chem, math, bio) so I can actually take the tests.

The structure is based on the Praxis exams, and I am not sure of the Praxis licensing so I have ommited all their text.

The GACE exam (the teacher exam I will be taking), has a somewhat open license that makes the text in it available for educators -- which is what we are. It is more generalized than the Praxis framework, and without detail, but I will include it anyway to help the structure of middle school science page.

I really do need collaborators, and I am not sure how to find them. I have written an invitation for teachers to join; here is the link: http://docs.google.com/Doc?id=ddk32zv4_420dvkwqw

Thank you
Thank you for the warm welcome Yourdeadin 17:49, 18 August 2007 (UTC)yourdeadin

Image Redirects prohibited?
Hi JW. I see that you've undone the image redirects we put on to the Technical writing course.

Is there a reason for your prohibition? Is there a page somewhere that details all such "rules"?

Thank you,

80.95.102.226 10:07, 23 August 2007 (UTC)

I've also been undoing some of these redirects. Whether or not it is against policy, I wonder what you're trying to achieve? Your redirects simply prevent the images being used or viewed. Perhaps you could let us know what you are trying to do? McCormack 10:14, 23 August 2007 (UTC)

 * Please see the discussion at User talk:AlistairReece and User talk:AlistairReece. All pages in the "Image:" namespace should contain information about the origin of the media file and how it is licensed for re-use....holding this information is the only purpose for these pages. Community members frequently look at this information about media files. It is disruptive to redirect such pages to pages that do not have the author/copyright/license information. --JWSchmidt 14:04, 23 August 2007 (UTC)

namespaces
Hey there, I have made a suggestion on wiki 101 regarding namespaces and would really appreciate your feedback. Cheers AlistairReece 14:51, 23 August 2007 (UTC)

What is AD?
It's ad, as in advertising. I wasn't able to access stuff on the ad-supported site, I think. I don't have the links right now for the (wikimania) sessions. Historybuff 18:24, 25 August 2007 (UTC)

editing
how come i can edit your profile page.
 * Because this is a wiki (changes can also be undone, see here). Also, if you edit others user pages this is seen as not polite. And please do not forget to sign your posts, so you will get faster responses. Erkan Yilmaz (evaluate me!, discussion ) 08:41, 26 August 2007 (UTC)
 * In my case, I invite others to edit my user pages. --JWSchmidt 15:20, 26 August 2007 (UTC)

Modern Greek
Hi I think I can help some interested to learn modern greek I am a native speaker tell me what I can do to help.--ZaDiak 19:55, 26 August 2007 (UTC)

I like your idea I know what to do to make use of it but the users must know a bit Greek and then we should make use of your idea. Bellow I tell you how we can make use of your idea. The greek wikiversity is under construction maybe they can learn and help a bit also the whole plan by translating from English to Greek.--ZaDiak

 * I like the idea of having a page here at this Wikiversity with suggested "trips to places where Greek is spoken". Students could be invited to help with translation projects, they could be sent on "trips to Greek websites" and asked to write 'travel reports" about what they find, they might be directed to chat channels where they can interact with Greek speakers. --JWSchmidt 21:11, 26 August 2007 (UTC)

Yeah very good idea I think it would be very good.

Linux Programming
Yes and no. It would focus on languages, toolkits, APIs, etc. that are predominantly used on Linux (e.g. GTK+, Qt) instead of languages, toolkits, APIs, etc. that are not just used on Linux, but other OSes as well (e.g. OpenGL). Cybiko123 02:19, 31 August 2007 (UTC)


text: Category:Tools

title: Category:Tools


text:
Nonlinear finite elements/Homework 6/Solutions

 * Problem 1: Continuum-based Nonlinear Beam Bending
 * Problem 1: Part 1
 * Problem 1: Part 2
 * Problem 1: Part 3
 * Problem 1: Part 4
 * Problem 1: Part 5
 * Problem 1: Part 6
 * Problem 1: Part 7
 * Problem 1: Part 8
 * Problem 1: Part 9
 * Problem 1: Part 10
 * Problem 1: Part 11
 * Problem 1: Part 12
 * Problem 1: Part 13
 * Problem 1: Part 14
 * Problem 1: Part 15
 * Problem 2: Exploring commercial software


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 1

Problem 1: Part 1
What is the continuum-based approach for the finite element analysis of beams? Draw figures to elucidate.

In the continuum-based approach, instead of starting off with simplified governing equations, a beam (or shell) element is developed directly from a continuum element by imposing structural assumptions.

The kinematic assumptions (for example, normals to midsurface remain normal) are imposed on the discretized equations derived from the weak form. A plane stress assumption in the direction normal to the beam is also enforced.

In the continuum-based approach for beams (in two dimensions), we start with a parent element which is a six-noded square (see Figure 1).

The parent element maps to the actual beam element in the reference and current configurations. The reference line of the beam element maps to the $$\eta = 0$$ line in the parent element. Master nodes are placed at the intersection of the fibers connecting the slave nodes with the reference line.

The momentum equation is solved on the master nodes and the resulting velocities are projected to the slave nodes. A set of quadrature points is set up along the fiber and the stresses are computed at these points. The plane stress assumption is imposed and the internal forces are computed at the slave nodes. The internal forces are interpolated to the master nodes and the process is repeated for each load step.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 2

Problem 1: Part 2
Why is the continuum-based approach preferred in nonlinear finite element analysis?

The continuum-based approach is preferred because the governing equations for nonlinear beams and shells are very complex and features such as variations in thickness, junctions, and stiffeners are difficult to incorporate. It is also easier to model complex constitutive models (plasticity, anisotropy) with the continuum-based approach.

Also, codes that use this approach are generally quite robust and accurate.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 3

Problem 1: Part 3
The rate of deformation is defined as

\boldsymbol{D} = \text{sym}(\boldsymbol{\nabla} \mathbf{v}) = \frac{1}{2}\left[\boldsymbol{\nabla} \mathbf{v} + (\boldsymbol{\nabla} \mathbf{v})^T\right] $$ where $$\mathbf{v}$$ is the velocity. In index notation, we write

D_{ij} = \text{sym}(v_{i,j}) = \frac{1}{2}\left(v_{i,j} + v_{j,i}\right) = \frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right), \qquad i,j=1,2,3~. $$ Given the above definition, derive equations (9.2.1) through (9.2.7) of the book chapter.

The motion of a point $$P$$ on the beam with respect to a point on the reference line $$C$$ is shown in Figure 2.

Since the normal ($$\mathbf{n}$$) rotates as a rigid body, the velocity of point $$P$$ with respect to $$C$$ is given by

\mathbf{v}_{\text{PC}} = \boldsymbol{\omega}\times\mathbf{r} $$ where $$\boldsymbol{\omega}$$ is the angular velocity of the normal, and $$\mathbf{r}$$ is the vector from $$C$$ to $$P$$.

Expressed in terms of the local basis vectors $$\mathbf{e}_x$$, $$\mathbf{e}_y$$, and $$\mathbf{e}_z$$, the angular velocity and the radial vector are
 * $$\begin{align}

\boldsymbol{\omega} & = 0~\mathbf{e}_x + 0~\mathbf{e}_y + \omega~\mathbf{e}_z = \omega~\mathbf{e}_z \\ \mathbf{r} & = 0~\mathbf{e}_x + y~\mathbf{e}_y + 0~\mathbf{e}_z = y~\mathbf{e}_y~. \end{align}$$ Therefore,

\mathbf{v}_{\text{PC}} = (\omega~\mathbf{e}_z)\times(y~\mathbf{e}_y) = y~\omega~\mathbf{e}_z\times\mathbf{e}_y = -y~\omega~\mathbf{e}_x ~. $$

Let $$\mathbf{v}^M(\mathbf{x},t)$$ be the velocity of the point $$C$$ at time $$t$$. Then the actual velocity of point $$P$$ is

\mathbf{v} = \mathbf{v}^M + \mathbf{v}_{\text{PC}} ~. $$

Now, in terms of the local basis vectors

\mathbf{v}^M = v^M_x~\mathbf{e}_x + v^M_y~\mathbf{e}_y + 0~\mathbf{e}_z~. $$

Therefore,

\mathbf{v} =v^M_x~\mathbf{e}_x + v^M_y~\mathbf{e}_y- y~\omega~\mathbf{e}_x =(v^M_x - y~\omega)\mathbf{e}_x + v^M_y \mathbf{e}_y ~. $$

Therefore, the velocity of any point $$P$$ in terms of the local basis at its orthogonal projection at the reference line is
 * $$\begin{align}

\mathbf{v}(x,y,z,t) & = v_x~\mathbf{e}_x & + & v_y~\mathbf{e}_y & + &v_z~\mathbf{e}_z \\ & = [v^M_x(x,t) - y~\omega(x,t)]~\mathbf{e}_x & + &v^M_y(x,t)~\mathbf{e}_y & + &0~\mathbf{e}_z~. \end{align}$$

The components of the rate of deformation tensor are

D_{ij} = \frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right), \qquad i,j=1,2,3~. $$

In terms of the local basis, these components are

\begin{align} D_{xx} & = \frac{\partial v_x}{\partial x} = \frac{\partial v^M_x}{\partial x} - y\frac{\partial \omega}{\partial x} \\ D_{yy} & = \frac{\partial v_y}{\partial y} = 0 \\ D_{xy} & = \frac{1}{2}\left(\frac{\partial v_x}{\partial y} + \frac{\partial v_y}{\partial x}\right) = \frac{1}{2}\left(-\omega + \frac{\partial v^M_y}{\partial x}\right)\\ D_{zz} & = \frac{\partial v_z}{\partial z} = 0 \\ D_{yz} & = \frac{1}{2}\left(\frac{\partial v_y}{\partial z} + \frac{\partial v_z}{\partial y}\right) = 0 \\ D_{zx} & = \frac{1}{2}\left(\frac{\partial v_z}{\partial x} + \frac{\partial v_x}{\partial z}\right) = 0 \end{align} $$

For the Euler-Bernoulli beam theory, the normals remain normal to the reference line. Let $$\theta$$ be the rotation of the normal. Then, the rotation is given by (see Figure 3)

\theta = \frac{\partial u^M_y}{\partial x} $$ where $$u^M_y$$ is the displacement in the local $$y$$-direction at a point on the reference line.

The angular velocity of the normal is given by

\omega = \frac{\partial \theta}{\partial t} = \frac{\partial v^M_y}{\partial x}~. $$

Hence,

\begin{align} D_{xx} & = \frac{\partial v^M_x}{\partial x} - y\frac{\partial \omega}{\partial x} = \frac{\partial v^M_x}{\partial x} - y\frac{\partial^2 v^M_y}{\partial x^2} \\ D_{yy} & = 0 \\ D_{xy} & = \frac{1}{2}\left(-\omega + \frac{\partial v^M_y}{\partial x}\right) = 0 \\ D_{zz} & = 0 \\ D_{yz} & = 0 \\ D_{zx} & = 0 \end{align} $$


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 4

Problem 1: Part 4
Why is the Euler-Bernoulli beam theory $C^1$? Why is the Timoshenko beam theory $C^0$?

In the Euler-Bernoulli theory, the rate of deformation in the $$xx$$ direction is given by

D_{xx} = \frac{\partial v^M_x}{\partial x} - y\frac{\partial^2 v^M_y}{\partial x^2} ~. $$

Since second-derivatives of $$v^M_y$$ are involved, any approximate solution needs to be at least twice differentiable (or $$C^1$$) to be acceptable. This is why this theory is called $$C^1$$.

On the other hand, in the Timoshenko theory, only first derivatives of the primary dependent variables appear. Therefore, once differentiable (or $$C^0$$) shape functions are acceptable and the theory is called $$C^0$$.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 5

Part 5
Why is a $C^0$ beam theory preferable?

A $$C^1$$ approximation is difficult to construct in multi-dimensions (for example, for beams that can deform in the $$z$$-direction too). That is why $$C^0$$ theories are preferred.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 6

Problem 1: Part 6
What are the major assumptions in the Timoshenko beam formulation described in the book chapter (the authors call these the modified Mindlin-Reissner assumptions)?

The assumptions are:
 * 1) The fibers (or directors) remain straight.
 * 2) The transverse normal stress is negligible (this is also called the plane stress condition.)
 * 3) The fibers are inextensible.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 7

Problem 1: Part 7
Derive the equation (9.3.13) of the book chapter starting from equation (9.3.3).

Figure 4 shows the notation used to represent the motion of the master and slave nodes and the directors at the nodes.

Since the fibers remain straight and do not change length, we have
 * $$ \text{(5)} \qquad

\begin{align} \mathbf{x}_{i-}(t) & = \mathbf{x}_i(t) - \frac{1}{2} h_i^0~\mathbf{p}_i(t) \\ \mathbf{x}_{i+}(t) & = \mathbf{x}_i(t) + \frac{1}{2} h_i^0~\mathbf{p}_i(t) \end{align} $$ where $$\mathbf{x}_i$$ is the location of master node $$i$$, $$\mathbf{p}_i$$ is the unit director vector at master node $$i$$, and $$h_i^0$$ is the initial thickness of the beam.

Taking the material time derivatives of equations (5), we get
 * $$ \text{(6)} \qquad

\begin{align} \mathbf{v}_{i-}(t) & = \mathbf{v}_i(t) - \frac{1}{2} \frac{\partial }{\partial t}[h_i^0~\mathbf{p}_i(t)] = \mathbf{v}_i(t) - \frac{1}{2} h_i^0~\boldsymbol{\omega}_i(t)\times\mathbf{p}_i(t) \\ \mathbf{v}_{i+}(t) & = \mathbf{v}_i(t) + \frac{1}{2} \frac{\partial }{\partial t}[h_i^0~\mathbf{p}_i(t)] = \mathbf{v}_i(t) + \frac{1}{2} h_i^0~\boldsymbol{\omega}_i(t)\times\mathbf{p}_i(t) \end{align} $$ where $$\boldsymbol{\omega}_i$$ is the angular velocity of the director.

From equations (5) we have
 * $$ \text{(7)} \qquad

\begin{align} \mathbf{p}_i(t) & = -\cfrac{2}{h_i^0}[\mathbf{x}_{i-}(t) - \mathbf{x}_i(t)] = -\cfrac{2}{h_i^0}\left[(x_{i-} - x_i)~\mathbf{e}_x + (y_{i-} - y_i)~\mathbf{e}_y\right] \\ \mathbf{p}_i(t) & = \cfrac{2}{h_i^0}[\mathbf{x}_{i+}(t) - \mathbf{x}_i(t)] = \cfrac{2}{h_i^0}\left[(x_{i+} - x_i)~\mathbf{e}_x + (y_{i+} - y_i)~\mathbf{e}_y\right] \end{align} $$ where $$(\mathbf{e}_x,\mathbf{e}_y,\mathbf{e}_z)$$ is the global basis.

In terms of the global basis, the angular velocity is given by

\boldsymbol{\omega}_i= \omega_i~\mathbf{e}_z ~. $$

Therefore,
 * $$\text{(8)} \qquad

\begin{align} \boldsymbol{\omega}_i\times\mathbf{p}_i & = -\cfrac{2}{h_i^0} \begin{vmatrix} \mathbf{e}_x & \mathbf{e}_y & \mathbf{e}_z \\ 0& 0 & \omega_i \\ x_{i-}-x_i& y_{i-}-y_i& 0 \end{vmatrix} = -\cfrac{2}{h_i^0} \left[ - \omega_i (y_{i-}-y_i)~\mathbf{e}_x + \omega_i (x_{i-}-x_i)~\mathbf{e}_y \right] \\ \boldsymbol{\omega}_i\times\mathbf{p}_i & = \cfrac{2}{h_i^0} \begin{vmatrix} \mathbf{e}_x & \mathbf{e}_y & \mathbf{e}_z \\ 0& 0 & \omega_i \\ x_{i+}-x_i& y_{i+}-y_i& 0 \end{vmatrix} = \cfrac{2}{h_i^0} \left[ - \omega_i (y_{i+}-y_i)~\mathbf{e}_x + \omega_i (x_{i+}-x_i)~\mathbf{e}_y \right] ~. \end{align} $$

Substituting equation (8) into equations (6), we get
 * $$ \text{(9)} \qquad

\begin{align} \mathbf{v}_{i-} & = \mathbf{v}_i - \omega_i (y_{i-}-y_i)~\mathbf{e}_x + \omega_i (x_{i-}-x_i)~\mathbf{e}_y \\ \mathbf{v}_{i+} & = \mathbf{v}_i - \omega_i (y_{i+}-y_i)~\mathbf{e}_x + \omega_i (x_{i+}-x_i)~\mathbf{e}_y ~. \end{align} $$

Let the velocity vectors be expressed in terms of the global basis as

\begin{align} \mathbf{v}_i & = v_i^x~\mathbf{e}_x + v_i^y~\mathbf{e}_y \\ \mathbf{v}_{i-} & = v_{i-}^x~\mathbf{e}_x + v_{i-}^y~\mathbf{e}_y \\ \mathbf{v}_{i+} & = v_{i+}^x~\mathbf{e}_x + v_{i+}^y~\mathbf{e}_y ~. \end{align} $$

Then equations (9) can be written as
 * $$ \text{(10)} \qquad

\begin{align} v_{i-}^x~\mathbf{e}_x + v_{i-}^y~\mathbf{e}_y & = [v_i^x - \omega_i (y_{i-}-y_i)]~\mathbf{e}_x + [v_i^y + \omega_i (x_{i-}-x_i)]~\mathbf{e}_y \\ v_{i+}^x~\mathbf{e}_x + v_{i+}^y~\mathbf{e}_y & = [v_i^x - \omega_i (y_{i+}-y_i)]~\mathbf{e}_x + [v_i^y + \omega_i (x_{i+}-x_i)]~\mathbf{e}_y ~. \end{align} $$

Therefore, the components of the velocity vectors are
 * $$ \text{(11)} \qquad

\begin{align} v_{i-}^x & = v_i^x - \omega_i (y_{i-}-y_i) \\ v_{i-}^y & = v_i^y + \omega_i (x_{i-}-x_i) \\ v_{i+}^x & = v_i^x - \omega_i (y_{i+}-y_i) \\ v_{i+}^y & = v_i^y + \omega_i (x_{i+}-x_i) ~. \end{align} $$

Then, the matrix form of equations (11) is

\begin{bmatrix} v^x_{i-} \\ v^y_{i-} \\ v^x_{i+} \\ v^y_{i+} \end{bmatrix} = \begin{bmatrix} 1 & 0 & - (y_{i-}-y_i) \\ 0 & 1 &(x_{i-}-x_i)\\ 1 & 0 & - (y_{i+}-y_i) \\ 0 & 1 &(x_{i+}-x_i) \end{bmatrix} \begin{bmatrix} v_i^x \\ v_i^y \\ \omega_i \end{bmatrix} $$ or

\begin{bmatrix} \mathbf{v}_{i-} \\ \mathbf{v}_{i+} \end{bmatrix}^{\text{slave}} = \mathbf{T}_i~\dot{\mathbf{d}}_i^{\text{master}} ~. $$


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 8

Problem 1: Part 8
Consider the curved composite beam shown in Figure 12.

Assume that the beam has been shaped into an arc of a circle. The material of the beam is a transversely isotropic fiber composite material with the fibers running along the length of the beam. The rate constitutive relation of the material is given by

\cfrac{D}{Dt} \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \end{bmatrix} = \begin{bmatrix} C_{11} & C_{12} & C_{13} & 0 & 0 & 0 \\ C_{12} & C_{11} & C_{13} & 0 & 0 & 0 \\ C_{13} & C_{13} & C_{33} & 0 & 0 & 0 \\ 0 & 0 & 0 & C_{44} & 0 & 0 \\ 0 & 0 & 0 & 0 & C_{44} & 0 \\ 0 & 0 & 0 & 0 & 0 & (C_{11}-C_{12})/2 \end{bmatrix} \begin{bmatrix} D_{11} \\ D_{22} \\ D_{33} \\ D_{23} \\ D_{31} \\ D_{12} \end{bmatrix}~. $$

The problem becomes easier to solve if we consider numerical values of the parameters. Let the local nodes numbers of element $$5$$ be $$1$$ for node $$5$$, and $$2$$ for node $$6$$.

Let us assume that the beam is divided into six equal sectors. Then,

\theta = \cfrac{\pi}{4}~; \theta_1 = \cfrac{\pi}{3}~; \theta_2 = \cfrac{\pi}{6} ~. $$ Let $$r_1 = 1$$ and $$r_2 = 1.2$$. Since the blue point is midway between the two, $$r = 1.1$$.

Also, let the components of the stiffness matrix of the composite be

C_{11} = 10; C_{33} = 100; C_{12} = 6; C_{13} = 60; C_{44} = 30~. $$

Let the velocities for nodes $$1$$ and $$2$$ of the element be

\mathbf{v}_1 = \begin{bmatrix} v_1^x \\ v_1^y \\ \omega_1 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} ~; \mathbf{v}_2 = \begin{bmatrix} v_2^x \\ v_2^y \\ \omega_2 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} $$

The Maple code for these initializations is shown below


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 9

Problem 1: Part 9
Construct a laminar coordinate system at the blue point. (Use equations 9.3.16 and 9.3.17 from the book chapter). Assume that the blue point is at the center of element 5.

The $$x$$ and $$y$$ coordinates of the master and slave nodes are

\begin{bmatrix} x_1 \\ y_1 \\ x_2 \\ y_2 \end{bmatrix} = \begin{bmatrix} r\cos\theta_1 \\ r\sin\theta_1 \\ r\cos\theta_2 \\ r\sin\theta_2 \end{bmatrix} = \begin{bmatrix} 0.55000 \\ 0.95263 \\ 0.95263 \\ 0.55000 \end{bmatrix} $$

\begin{bmatrix} x_{1-} \\ y_{1-} \\ x_{2-} \\ y_{2-} \end{bmatrix} = \begin{bmatrix} r_1\cos\theta_1 \\ r_1\sin\theta_1 \\ r_1\cos\theta_2 \\ r_1\sin\theta_2 \end{bmatrix} = \begin{bmatrix} 0.50000 \\ 0.86603 \\ 0.86603 \\ 0.50000 \end{bmatrix} $$

\begin{bmatrix} x_{1+} \\ y_{1+} \\ x_{2+} \\ y_{2+} \end{bmatrix} = \begin{bmatrix} r_2\cos\theta_1 \\ r_2\sin\theta_1 \\ r_2\cos\theta_2 \\ r_2\sin\theta_2 \end{bmatrix} = \begin{bmatrix} 0.60000 \\ 1.0392 \\ 1.0392 \\ 0.60000 \end{bmatrix} $$

Since there are two master nodes in an element, the parent element is a four-noded quad with shape functions
 * $$\begin{align}

N_{1-}(\xi,\eta) & = \cfrac{1}{4}(1-\xi)(1-\eta) & N_{2-}(\xi,\eta) & = \cfrac{1}{4}(1+\xi)(1-\eta) \\ N_{1+}(\xi,\eta) & = \cfrac{1}{4}(1-\xi)(1+\eta) & N_{2+}(\xi,\eta) & = \cfrac{1}{4}(1+\xi)(1+\eta) ~. \end{align}$$

The isoparametric map is
 * $$\begin{align}

x(\xi, \eta) & = x_{1-}~N_{1^-}(\xi,\eta) + x_{2-}~N_{2^-}(\xi,\eta) + x_{1+}~N_{1^+}(\xi,\eta) + x_{2+}~N_{2^+}(\xi,\eta)\\ y(\xi, \eta) & = y_{1-}~N_{1^-}(\xi,\eta) + y_{2-}~N_{2^-}(\xi,\eta) + y_{1+}~N_{1^+}(\xi,\eta) + y_{2+}~N_{2^+}(\xi,\eta)~. \end{align}$$

Plugging in the numbers, we get
 * $$\begin{align}

x &= 0.12500 (1 - \xi) (1 - \eta) + 0.21651 (1 + \xi) (1 - \eta) \\ & + 0.15000 (1 - \xi) (1 + \eta) + 0.25981 (1 + \xi) (1 + \eta) \\ y &= 0.21651 (1 - \xi) (1 - \eta) + 0.12500 (1 + \xi) (1 - \eta) \\ & + 0.25981 (1 - \xi) (1 + \eta) + 0.15000 (1 + \xi) (1 + \eta)~. \end{align}$$

Therefore, the derivatives with respect to $$\xi$$ are
 * $$\begin{align}

x_{,\xi} = \frac{\partial x}{\partial \xi} &= 0.20131 + 0.018301 \eta \\ y_{,\xi} = \frac{\partial y}{\partial \xi} &= -0.20131 - 0.018301 \eta~. \end{align}$$

Since the blue point is at the center of the element, the values of $$\xi$$ and $$\eta$$ at that point are zero. Therefore, the values of the derivatives at that point are
 * $$\begin{align}

x_{,\xi} = \frac{\partial x}{\partial \xi} &= 0.20131 \\ y_{,\xi} = \frac{\partial y}{\partial \xi} &= -0.20131 ~. \end{align}$$

Therefore,

\mathbf{x}_{,\xi} = 0.20131 \mathbf{e}_x - 0.20131 \mathbf{e}_y, \lVert\mathbf{x}_{,\xi}\rVert_{}= \sqrt{0.20131^2 + 0.20131^2} = 0.28470 ~. $$

The local laminar basis vector $$\widehat{\mathbf{e}}_x$$ is given by

\widehat{\mathbf{e}}_x = \cfrac{\mathbf{x}_{,\xi}}{\lVert\mathbf{x}_{,\xi}\rVert_{}} = 0.70711 \mathbf{e}_x - 0.70711 \mathbf{e}_y ~. $$

The laminar basis vector $$\widehat{\mathbf{e}}_y$$ is given by

\widehat{\mathbf{e}}_y = \mathbf{e}_z\times\widehat{\mathbf{e}_x} = 0.70711 \mathbf{e}_x + 0.70711 \mathbf{e}_y ~. $$

A plot of the vectors is shown in Figure 13.

The Maple code for this calculation is shown below.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 10

Problem 1: Part 10
Assume that the velocities at nodes 5 and 6 are $\mathbf{v}_5$ and $\mathbf{v}_6$, respectively. Compute the velocity gradient at the blue point.

To compute the velocity gradient, we have to find the velocities at the slave nodes using the relation

\begin{bmatrix} v^x_{i-} \\ v^y_{i-} \\ v^x_{i+} \\ v^y_{i+} \end{bmatrix} = \begin{bmatrix} 1 & 0 & - (y_{i-}-y_i) \\ 0 & 1 &(x_{i-}-x_i)\\ 1 & 0 & - (y_{i+}-y_i) \\ 0 & 1 &(x_{i+}-x_i) \end{bmatrix} \begin{bmatrix} v_i^x \\ v_i^y \\ \omega_i \end{bmatrix} $$

For node 1 of the element (global node 5), the $$\mathbf{T}$$ matrix is

\mathbf{T}_1 = \begin{bmatrix} 1 & 0 & 0.0866 \\ 0 & 1 & -0.05 \\ 1 & 0 & -0.0866 \\ 0 & 1 &0.05 \end{bmatrix} ~. $$

Therefore, the velocities of the slave nodes are

\begin{bmatrix} v^x_{1-} \\ v^y_{1-} \\ v^x_{1+} \\ v^y_{1+} \end{bmatrix} = \begin{bmatrix} 1.0866 \\ 1.95 \\ 0.9134 \\ 2.05 \end{bmatrix} ~. $$

For node 2 of the element (global node 6), the $$\mathbf{T}$$ matrix is

\mathbf{T}_2 = \begin{bmatrix} 1 & 0 & 0.05 \\ 0 & 1 & -0.0866 \\ 1 & 0 & -0.05 \\ 0 & 1 & 0.0866 \end{bmatrix} ~. $$

Therefore, the velocities of the slave nodes are

\begin{bmatrix} v^x_{2-} \\ v^y_{2-} \\ v^x_{2+} \\ v^y_{2+} \end{bmatrix} = \begin{bmatrix} 2.05 \\ 0.9134 \\ 1.95 \\ 1.0866 \end{bmatrix} ~. $$

The interpolated velocity within the element is given by

\begin{align} \mathbf{v}(\boldsymbol{\xi}, t) & = \cfrac{D}{Dt}[\mathbf{x}(\boldsymbol{\xi},t)] \\ & = \sum_{i- = 1}^2 \frac{\partial }{\partial t}[\mathbf{x}_{i-}(t)]~N_{i^-}(\xi,\eta) + \sum_{i+ = 1}^2 \frac{\partial }{\partial t}[\mathbf{x}_{i+}(t)]~N_{i^+}(\xi,\eta)\\ & = \sum_{i- = 1}^2 \mathbf{v}_{i-}(t)~N_{i-}(\xi,\eta) + \sum_{i+ = 1}^2 \mathbf{v}_{i+}(t)~N_{i+}(\xi,\eta) ~. \end{align} $$

Plugging in the values of the nodal velocities and the shape functions, we get
 * $$\begin{align}

v_x & = 0.27165 (1 - \xi) (1 - \eta) + 0.51250 (1 + \xi) (1 - \eta)\\ & + 0.22835 (1 - \xi) (1 + \eta) + 0.48750 (1 + \xi) (1 + \eta)\\ v_y & = 0.48750 (1 - \xi) (1 - \eta) + 0.22835 (1 + \xi) (1 - \eta)\\ & + 0.51250 (1 - \xi) (1 + \eta) + 0.27165 (1 + \xi) (1 + \eta)~. \end{align}$$

The velocity gradient is given by

\boldsymbol{\nabla} \mathbf{v} = v_{i,j} = \frac{\partial v_i}{\partial x_j} ~. $$

The velocities are given in terms of the parent element coordinates ($$\xi,\eta$$). We have to convert them to the ($$x,y$$) system in order to compute the velocity gradient. To do that we recall that
 * $$\begin{align}

\frac{\partial v_x}{\partial \xi} & = \frac{\partial v_x}{\partial x}\frac{\partial x}{\partial \xi} + \frac{\partial v_x}{\partial y}\frac{\partial y}{\partial \xi} \\ \frac{\partial v_x}{\partial \eta} & = \frac{\partial v_x}{\partial x}\frac{\partial x}{\partial \eta} + \frac{\partial v_x}{\partial y}\frac{\partial y}{\partial \eta} \\ \frac{\partial v_y}{\partial \xi} & = \frac{\partial v_y}{\partial x}\frac{\partial x}{\partial \xi} + \frac{\partial v_y}{\partial y}\frac{\partial y}{\partial \xi} \\ \frac{\partial v_y}{\partial \eta} & = \frac{\partial v_y}{\partial x}\frac{\partial x}{\partial \eta} + \frac{\partial v_y}{\partial y}\frac{\partial y}{\partial \eta} ~. \end{align}$$

In matrix form

\begin{bmatrix} \frac{\partial v_x}{\partial \xi} \\ \frac{\partial v_x}{\partial \eta} \end{bmatrix} = \begin{bmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi} \\ \frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta} \\ \end{bmatrix} \begin{bmatrix} \frac{\partial v_x}{\partial x} \\ \frac{\partial v_x}{\partial y} \end{bmatrix} = \mathbf{J} \begin{bmatrix} \frac{\partial v_x}{\partial x} \\ \frac{\partial v_x}{\partial y} \end{bmatrix} $$ and

\begin{bmatrix} \frac{\partial v_y}{\partial \xi} \\ \frac{\partial v_y}{\partial \eta} \end{bmatrix} = \begin{bmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi} \\ \frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta} \\ \end{bmatrix} \begin{bmatrix} \frac{\partial v_y}{\partial x} \\ \frac{\partial v_y}{\partial y} \end{bmatrix} = \mathbf{J} \begin{bmatrix} \frac{\partial v_y}{\partial x} \\ \frac{\partial v_y}{\partial y} \end{bmatrix}~. $$

Therefore,

\begin{bmatrix} \frac{\partial v_x}{\partial x} \\ \frac{\partial v_x}{\partial y} \end{bmatrix} = \mathbf{J}^{-1} \begin{bmatrix} \frac{\partial v_x}{\partial \xi} \\ \frac{\partial v_x}{\partial \eta} \end{bmatrix} \qquad \text{and} \qquad \begin{bmatrix} \frac{\partial v_y}{\partial x} \\ \frac{\partial v_y}{\partial y} \end{bmatrix} = \mathbf{J}^{-1} \begin{bmatrix} \frac{\partial v_y}{\partial \xi} \\ \frac{\partial v_y}{\partial \eta} \end{bmatrix}~. $$

The isoparametric map is
 * $$\begin{align}

x &= 0.12500 (1 - \xi) (1 - \eta) + 0.21651 (1 + \xi) (1 - \eta) \\ & + 0.15000 (1 - \xi) (1 + \eta) + 0.25981 (1 + \xi) (1 + \eta) \\ y &= 0.21651 (1 - \xi) (1 - \eta) + 0.12500 (1 + \xi) (1 - \eta) \\ & + 0.25981 (1 - \xi) (1 + \eta) + 0.15000 (1 + \xi) (1 + \eta)~. \end{align}$$

The derivatives with respect to $$\xi$$ and $$\eta$$ are
 * $$\begin{align}

\frac{\partial x}{\partial \xi}&= 0.20131 + 0.018301 \eta & \frac{\partial y}{\partial \xi}&= -0.20131 - 0.018301 \eta \\ \frac{\partial x}{\partial \eta} &= 0.0683 + 0.018301 \xi & \frac{\partial y}{\partial \eta} &= 0.0683 - 0.018301 \xi \end{align}$$

Therefore,

\mathbf{J} = \begin{bmatrix} 0.20131 + 0.018301 \eta & -0.20131 - 0.018301 \eta \\ 0.0683 + 0.018301 \xi & 0.0683 - 0.018301 \xi \end{bmatrix}~. $$

Since we are only interested in the point ($$\xi,\eta$$) = (0,0), we can compute $$\mathbf{J}$$ at this point

\mathbf{J} = \begin{bmatrix} 0.20131 & -0.20131 \\ 0.0683&0.0683 \end{bmatrix}~. $$

The inverse is

\mathbf{J}^{-1} = \begin{bmatrix} 2.4837 & 7.3205 \\ -2.4837 & 7.3205 \end{bmatrix}~. $$

The derivatives of $$\mathbf{v}$$ with respect to $$\xi$$ and $$\eta$$ are
 * $$\begin{align}

\frac{\partial v_x}{\partial \xi}&= 0.5 + 0.018301 \eta & \frac{\partial v_x}{\partial \eta} &= -0.0683 + 0.018301 \xi \\ \frac{\partial v_y}{\partial \xi}&= -0.5 + 0.018301 \eta & \frac{\partial v_y}{\partial \eta} &= 0.0683 + 0.018301 \xi \end{align}$$

At ($$\xi,\eta$$) = (0,0), we get
 * $$\begin{align}

\frac{\partial v_x}{\partial \xi}&= 0.5 & \frac{\partial v_x}{\partial \eta} &= -0.0683 \\ \frac{\partial v_y}{\partial \xi}&= -0.5 & \frac{\partial v_y}{\partial \eta} &= 0.0683 \end{align}$$

Therefore,

\begin{bmatrix} \frac{\partial v_x}{\partial x} \\ \frac{\partial v_x}{\partial y} \end{bmatrix} = \begin{bmatrix} 0.74184 \\ -1.7418 \end{bmatrix}; \qquad \begin{bmatrix} \frac{\partial v_y}{\partial x} \\ \frac{\partial v_y}{\partial y} \end{bmatrix} = \begin{bmatrix} -0.74184 \\ 1.7418 \end{bmatrix}~. $$

Therefore, the velocity gradient at the blue point is

\boldsymbol{\nabla} \mathbf{v} = \begin{bmatrix} 0.74184 & -1.7418 \\ -0.74184 &1.7418 \end{bmatrix} $$

The Maple code for doing the above calculations is shown below.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 11

Problem 1: Part 11
Compute the rate of deformation at the blue point.

The rate of deformation is given by

\boldsymbol{D} = \frac{1}{2}[\boldsymbol{\nabla} \mathbf{v} + (\boldsymbol{\nabla} \mathbf{v})^T] ~. $$

Using the velocity gradient from the previous question, the rate of deformation at the blue point is given by

\mathbf{D} = \begin{bmatrix} 0.74184 & -1.2418 \\ -1.2418&1.7418 \end{bmatrix} $$

The Maple code is given below.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 12

Problem 1: Part 12
Rotate the rate of deformation so that its components are with respect to the laminar coordinate system.

The global base vectors are

\mathbf{e}_x = \begin{bmatrix} 1 \\ 0 \end{bmatrix}; ~\qquad \mathbf{e}_y = \begin{bmatrix} 0 \\ 1 \end{bmatrix}~. $$

Therefore, the rotation matrix is

\mathbf{R}_{\text{lam}} = \begin{bmatrix} \mathbf{e}_x\bullet\widehat{\mathbf{e}_x} & \mathbf{e}_x\bullet\widehat{\mathbf{e}_y} \\ \mathbf{e}_y\bullet\widehat{\mathbf{e}_x} & \mathbf{e}_y\bullet\widehat{\mathbf{e}_y} \end{bmatrix} = \begin{bmatrix} 0.7071 &0.7071 \\ -0.7071 &0.7071 \end{bmatrix} ~. $$

Therefore, the components of the rate of deformation tensor with respect to the laminar coordinate system are

\mathbf{D}_{\text{lam}} = \mathbf{R}^T_{\text{lam}}\mathbf{D}\mathbf{R}_{\text{lam}} = \begin{bmatrix} 2.4837 & -0.5 \\ -0.5 &0 \end{bmatrix} ~. $$ The Maple script used to compute the above is shown below.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 13

Problem 1: Part 13
Compute the stress rate at the blue point. This stress rate is expressed in terms of the laminar coordinate system.

The rate constitutive relation of the material is given by

\cfrac{D}{Dt} \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \end{bmatrix} = \begin{bmatrix} C_{11} & C_{12} & C_{13} & 0 & 0 & 0 \\ C_{12} & C_{11} & C_{13} & 0 & 0 & 0 \\ C_{13} & C_{13} & C_{33} & 0 & 0 & 0 \\ 0 & 0 & 0 & C_{44} & 0 & 0 \\ 0 & 0 & 0 & 0 & C_{44} & 0 \\ 0 & 0 & 0 & 0 & 0 & (C_{11}-C_{12})/2 \end{bmatrix} \begin{bmatrix} D_{11} \\ D_{22} \\ D_{33} \\ D_{23} \\ D_{31} \\ D_{12} \end{bmatrix}~. $$

The map between the notations for a composite beam are shown in Figure 14.

Since the problem is a 2-D one, based on the figure, the reduced constitutive equation is

\cfrac{D}{Dt} \begin{bmatrix} \sigma_{11} \\ \sigma_{33} \\ \sigma_{31} \end{bmatrix} = \begin{bmatrix} C_{11} & C_{13} & 0 \\ C_{13} & C_{33} & 0\\ 0& 0& C_{44} \end{bmatrix} \begin{bmatrix} D_{11} \\ D_{33} \\ D_{31} \end{bmatrix}~. $$

The laminar $$x$$-direction maps to the composite $$3$$-direction and the laminar $$y$$-directions maps to the composite $$1$$-direction. Hence the constitutive equation can be written as

\cfrac{D}{Dt} \begin{bmatrix} \sigma_{yy} \\ \sigma_{xx} \\ \sigma_{xy} \end{bmatrix} = \begin{bmatrix} C_{11} & C_{13} & 0 \\ C_{13} & C_{33} & 0\\ 0& 0& C_{44} \end{bmatrix} \begin{bmatrix} D_{yy} \\ D_{xx} \\ D_{xy} \end{bmatrix}~. $$

Rearranging,

\cfrac{D}{Dt} \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{xy} \end{bmatrix} = \begin{bmatrix} C_{33} & C_{13} & 0\\ C_{13} & C_{11} & 0 \\ 0& 0& C_{44} \end{bmatrix} \begin{bmatrix} D_{xx} \\ D_{yy} \\ D_{xy} \end{bmatrix}~. $$

Plugging in the values, we get

\cfrac{D}{Dt} \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{xy} \end{bmatrix} = \begin{bmatrix} 100 & 60 & 0\\ 60 & 10 & 0 \\ 0 & 0& 30 \end{bmatrix} \begin{bmatrix} 2.4837 \\ 0 \\ -0.5 \end{bmatrix} = \begin{bmatrix} 248.37 \\ 149.02 \\ -15 \end{bmatrix} ~. $$

The Maple script for this calculation is shown below.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 14

Problem 1: Part 14
Apply the plane stress condition and solve for $D_{yy}$.

The plane stress condition requires that $$\sigma_{yy} = 0$$ in the laminar coordinate system. We assume that the rate of $$\sigma_{yy}$$ is also zero. In that case, we get

0 = \cfrac{D}{Dt}(\sigma_{yy}) = C_{13}~D_{xx} + C_{11}~D{yy} $$ or,

D_{yy} = - \cfrac{C_{13}}{C_{11}}~D_{xx} ~. $$

Plugging in the values we get

D_{yy} = - \left(\cfrac{60}{10}\right)~(2.4837) = -14.902 ~. $$


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 15

Problem 1: Part 15
Express the stress rate and the modified rate of deformation in the global coordinate system.

The modified laminar rate of deformation is

\begin{bmatrix} D_{xx} \\ D_{yy} \\ D_{xy} \end{bmatrix} = \begin{bmatrix} 2.4837 \\ -14.902 \\ -0.5 \end{bmatrix} $$

Alternatively, we can write

\mathbf{D}_{\text{lam}} = \begin{bmatrix} 2.4837 & -0.5 \\ -0.5 & -14.902 \end{bmatrix} ~. $$

The modified laminar stress rate is

\cfrac{D}{Dt} \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{xy} \end{bmatrix} = \begin{bmatrix} 100 & 60 & 0\\ 60 & 10 & 0 \\ 0 & 0& 30 \end{bmatrix} \begin{bmatrix} 2.4837 \\ -14.902 \\ -0.5 \end{bmatrix} = \begin{bmatrix} -645.76 \\ 0 \\ -15 \end{bmatrix} ~. $$

Alternatively, we can write

\cfrac{D}{Dt}(\boldsymbol{\sigma}_{\text{lam}}) = \begin{bmatrix} -645.76 & -15 \\ -15 & 0 \end{bmatrix} ~. $$

To get the global stress rate and rate of deformation, we have to rotate the components to the global basis using

\cfrac{D}{Dt}(\boldsymbol{\sigma}) = \mathbf{R}_{\text{lam}} \cfrac{D}{Dt}(\boldsymbol{\sigma}_{\text{lam}}) \mathbf{R}^T_{\text{lam}} ~; \qquad \mathbf{D} = \mathbf{R}_{\text{lam}} \mathbf{D}_{\text{lam}} \mathbf{R}^T_{\text{lam}} ~. $$

Computing these quantities gives us

\cfrac{D}{Dt}(\boldsymbol{\sigma}) = \begin{bmatrix} -337.88 & 322.88 \\ 322.88 & -307.88 \end{bmatrix} ~. $$ and

\mathbf{D} = \begin{bmatrix} -6.7092 & -8.6929 \\ -8.6829 & -5.7092 \end{bmatrix} ~. $$

The Maple code for the above computations is given below.


text:
Nonlinear finite elements/Homework 6/Solutions/Problem 2

Problem 2: Exploring commercial software
Track down the different types of beam and shell elements that ANSYS and LS-DYNA provide. For each element that you find, answer the following
 * 1) What is the name of the element?
 * 2) How many nodes does the element have?
 * 3) How many displacement type (displacements, rotations) degrees of freedom are there at each node?List them.
 * 4) How many force type (forces, moments) boundary conditions can be applied at each node?List them.
 * 5) What is the form of the shape functions used by the element? (Alternatively, you may write down the approximate solutions assumed by the element.)
 * 6) Is the beam element a continuum-based one or one derived from the equations of beam theory?
 * 7) Does the element allow for full and reduced integration?


text:
Category:Audio media

__NOGALLERY__

title: Category:Audio media


text:
User:Emesee/Archives

 * Anthropomorphic robotics
 * Wiki business plan
 * Open design
 * Decentralizing Wikipedia and its sister projects
 * Free top level domains
 * Unmanned aerial systems
 * Investigating Ripple
 * Distributed P2P search engine
 * Novel economic mechanisms
 * Morp
 * Developments in medicine
 * Efficiency in computing project
 * University news
 * Martian Boom Town

title: User:Emesee/Archives


text:
User:Remi/Archives/Mission and slogan

Motto and slogan contests completed
After nearly a year of discussion, the motto and slogan contests have concluded.

 * The motto portion of the contest ended August 5, 2007. The slogan selected is "open learning community".
 * The slogan portion of the contest ended May 2, 2007. The slogan selected is "set learning free".

The following reproduces the finalist section of the motto contest page. Please feel free to add comments at the end of this section or on the contest talk page.

Winning Motto: "Open learning community"
The winning motto is '"open learning community"''.

Summary of final round: From the goals of the contest, the Wikiversity motto is both a phrase that conveys the general mission of Wikiversity (or part of the mission) and a "short description" that briefly describes Wikimedia project. The motto will be listed with our logo.

The winning motto, "open learning community," had 61% of the positive support statements in round 6. There were 27 support statements for "open learning community," and many included substantive points in favor of the option. "Open education" had 3 in support and "free learning community" had 14 in support. The objections are few in comparison to supportive statements (except for the least popular "open education" option). The winning "open" motto had 3 comments against, the "education" motto had four against, and the "free" motto had two against. The objections do not seem to invalidate the value or meaning of the free/open learning community motto options.

As for the choice of an article to possibly precede "open learning community, an important criticism of using the "the" was made on the talk page (Open University already exists). Twelve supporters of this motto did not favor using an article or did not mention the issue. The definite article had twelve supporters as well. The indefinite article had three supporters. This is a difficult call. Because of the substantive criticism made on the talk page, the winning motto is presented without a suggested article. A merit of using no article is that the motto is shorter and will fit better under the logo. This can be an issue for later discussion.

See the "summary of contest process issues and possible further steps" section (of the motto contest page) for other possible discussion points.

Winning Slogan: "Set learning free"
The winning slogan is '"set learning free"''.

Summary of final round: The Wikiversity slogan is a phrase for listing at top of main page with "Welcome to Wikiversity". "Set learning free" had 68% of the positive support statements in round 6. There were 30 support statements for "set learning free". "Knowledge is free" had 4 in support and "Because knowledge should be free" had 10 in support. The latter 2 options had 3 comment each against use. "Set learning free" had no statements against. As for the wording of "set learning free" vs "learning set free," a few more people preferred "set learning free".

Archive of comments: Motto_contest/Round_6

Thank you for your interest and participation! --Reswik 02:18, 7 August 2007 (UTC)

title: User:Remi/Archives/Mission and slogan


text:
Category:Video media

__NOGALLERY__

title: Category:Video media