User:EML5526.S11.Team1.HS/HW3

=HW3 Draft=

2.4
Given a three bar structure subjected to the prescribed load at point C equal to $$ 10^3 ~N $$ as shown in Figure 2.19. The Young's modulus is $$ E = 10^{11} ~Pa $$, the cross-sectional area of the bar BC is $$ 2 \times 10^{-2} ~m^2 $$ and that of BD and BF is $$ 10^{-2} ~m^2 $$. Note that point D is free to move in the x-direction. Coordinates of joints are given in meters.

a. Construct the global stiffness matrix and load matrix.

b. Partition the matrices and solve for the unknown displacements at Point B and displacement in the x-direction at point D.

c. Find stresses in the three bars.

d. Find the reactions at nodes C,D,F.

a)
For Element 3, $$ \phi = 45^o,I = 3,J =4 $$

$$ K^{\left(3\right)} = \frac{10^{-2} \times 10^{11}}{\sqrt{2}} \begin{bmatrix} \frac{1}{2} &\frac{1}{2} &-\frac{1}{2} &-\frac{1}{2}\\ \frac{1}{2} &\frac{1}{2} &-\frac{1}{2} &-\frac{1}{2}\\ -\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} &\frac{1}{2}\\ -\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} &\frac{1}{2} \end{bmatrix} $$

For Element 2, $$ \phi = 90^o,I = 2,J = 4 $$

$$ K^{\left(2\right)} = \frac{2 \times 10^{-2} \times 10^{11}}{1} \begin{bmatrix} 0 &0  &0  &0\\  0  &1  &0 &-1\\  0  &0  &0 &0 \\  0  &-1  &0 &1 \end{bmatrix} $$

For Element 1, $$ \phi = 135^o,I = 1,J =4 $$

$$ K^{\left(1\right)} = \frac{10^{-2} \times 10^{11}}{\sqrt{2}} \begin{bmatrix} \frac{1}{2} &-\frac{1}{2} &-\frac{1}{2} &\frac{1}{2}\\ -\frac{1}{2} &\frac{1}{2} &\frac{1}{2} &-\frac{1}{2}\\ -\frac{1}{2} &\frac{1}{2} &\frac{1}{2} &-\frac{1}{2}\\ \frac{1}{2} &-\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} \end{bmatrix} $$

Global Stiffness Matrix

$$ K = 10^{9} \begin{bmatrix} \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &0  &0  &0  &-\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0  &0  &0  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0  &0  &0  &0  &0 \\ 0 &0  &0  &2  &0  &0  &0  &-2 \\ 0 &0  &0  &0  &\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0  &\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &\frac{1}{\sqrt{2}}  &0 \\ \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &-2  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &0  &\frac{1}{\sqrt{2}}+2 \end{bmatrix} $$

$$ d = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ u_{3x}\\ 0\\ u_{4x}\\ u_{4y} \end{bmatrix} f = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 10^3\\ 0 \end{bmatrix} r = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ r_{3y}\\ 0\\ 0 \end{bmatrix} $$

Therefore

$$ load~matrix = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix} $$

b)
Global system of equations:

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

Partitions

$$ \bar{d}_{E} = \begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{2x}\\ u_{2y} \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix},

d_{F}= \begin{bmatrix} u_{3x}\\ 0\\ u_{4x}\\ u_{4y} \end{bmatrix},

r_{E} = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y} \end{bmatrix} ,

f_{F} = \begin{bmatrix} 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix} $$

$$ K_{EF} = \begin{bmatrix} 0 &0 &-\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}} \\ 0 &0 &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0 \\ 0 &0  &0  &-2\ \end{bmatrix},

K_{EF}^{T} = \begin{bmatrix} 0 &0 &0  &0 \\ 0 &0  &0  &0 \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}}  &0  &0 \\ \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &-2 \end{bmatrix},

K_{F} = \begin{bmatrix} \frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &\frac{1}{\sqrt{2}}  &0 \\ -\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &\frac{1}{\sqrt{2}}+2 \end{bmatrix},

K_{E} = \begin{bmatrix} \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &0 \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0 \\ 0 &0 &0  &0 \\ 0 &0  &0  &2 \end{bmatrix} $$

Unknown displacement at B is solved by

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

Subtracting the first term from both sides of the above equation and premultiply by $$ K_F^{-1} $$, we obtain

$$ d_F = K_F^{-1} \left( f_F - K_{EF}^T \bar{d}_E \right) $$

Solving for the displacements we get: $$ d_{F}= 10^{-6} \begin{bmatrix} 1+2\sqrt{2}\\ 0\\ \frac{1}{2}+2\sqrt{2}\\ \frac{1}{2} \end{bmatrix} ~m $$

c)
Element 3 Stress

$$ \sigma = 0 ~Pa $$

Element 2 Stress

$$ \sigma = \frac{E}{l} \begin{bmatrix} 0 &-1 &0  &1 \end{bmatrix} 10^{-6} \begin{bmatrix} \frac{1}{2} + 2\sqrt{2}\\ \frac{1}{2}\\ 0\\ 0 \end{bmatrix} = -5 \times 10^4 ~Pa $$

Element 1 Stress

$$ \sigma = \frac{E}{l} \begin{bmatrix} \frac{\sqrt{2}}{2} &-\frac{\sqrt{2}}{2} &\frac{\sqrt{2}}{2}  &\frac{\sqrt{2}}{2} \end{bmatrix} 10^{-6} \begin{bmatrix} \frac{1}{2} + 2\sqrt{2}\\ \frac{1}{2}\\ 0\\ 0 \end{bmatrix} = \sqrt{2} \times 10^5 ~Pa $$

d)
Reactions can be solved using this equation and the global matrix

$$ r_E = K_E \bar{d}_E + K_{EF} d_F $$

$$ load~matrix = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix}

= \begin{bmatrix} -10^3\\ 10^3\\ 0\\ -10^3\\ 0\\ 0\\ 10^3\\ 0 \end{bmatrix} ~ N $$

3.3
Consider a trial (candidate) solution of the form $$ u\left ( x \right ) = \alpha_{0} + \alpha_{1}\left ( x-3 \right ) $$ and a weight function of the same form. Obtain a solution to the weak form in Problem 3.1. Check the equilibrium equation in the strong form in Problem 3.1; is it satisfied?

Check the natural boundary condition; is it satisfied?

$$ u\left ( x \right ) = \alpha_{0} + \alpha_{1}\left ( x-3 \right ) $$

$$ w\left ( x \right ) = \beta_{0} + \beta_{1}\left ( x-3 \right ) $$

Using boundary conditions from Problem 3.9

$$ w \left(3 \right) = 0 \therefore  \beta_{0} = 0 $$

$$ u \left(3 \right) = \alpha_{0} = 0.001 $$

Only one unknown parameter and one arbitrary parameter remain

$$ u\left ( x \right ) = 0.001 + \alpha_{1}\left ( x-3 \right ) \frac{du}{dx} = \alpha_{1} $$

$$ w\left( x \right) = \beta_{1} \left(x - 3 \right) \frac{dw}{dx} = \beta_{1} $$

Now substitute into Problem 3.1 weak form

$$ \int_{1}^{3} \frac{\mathrm{d} w }{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx =  -0.1(wA)_{x=1} + \int_{1}^{3} 2xwdx$$

$$ \int_{1}^{3} \beta_{1}\alpha_{1}EAdx + 0.1(\beta_{1} \left(x-3 \right)A)_{x=1} - \int_{1}^{3} 2\beta_{1}x \left(x-3 \right)dx = 0 $$

Integrating and factoring out $$ \beta_1 $$ we get:

$$ \beta_{1} \left( 2 \alpha_{1} EA - 0.2A + \frac{20}{3} \right) = 0 $$

The $$ \beta_1 $$ goes to zero and $$ \alpha_1 $$ can be solved:

$$ \alpha_{1} = \frac{0.1A - \frac{10}{3}}{EA} $$

Therefore the solution to the weak form is

$$ u \left( x \right) = 0.001 + \frac{0.1A - \frac{10}{3}}{EA} \left( x-3 \right) $$

Checking for equilibrium in the strong form, values are substituted into

$$ \frac{d}{dx} \left( AE \frac{du}{dx} \right) + 2x = 0 $$

This yields $$ 2x = 0 $$ which does not satisfy.

Checking the natural boundary condition, this equation is used

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

Substituting in $$ \alpha_1 $$ we get

$$ 0.1 - \frac{\frac{10}{3}}{A} = 0.1 $$

Thus, this boundary condition will satisfy only for large values of $$ A $$