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Question 8.1: Find the stress at mid-height of the axially loaded bar show...

Find the stress at mid-height of the axially loaded bar shown in Figure 8.8a.

Take Young’s modulus of the bar material as E = 70 GPa.

8.8
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The bar is discretized using three elements as shown in Figure 8.8b.

Element stiffness matrices are calculated as follows (Equation 8.94):

[K^{(e)}]=\frac{AE}{l} \begin{bmatrix} 1&-1\\-1&1 \end{bmatrix} \quad\quad\quad\quad\quad 8.94\\ [K^{(1)}]=\frac{900\times 70,000}{70}\begin{bmatrix} 1&-1\\-1&1 \end{bmatrix}=\begin{bmatrix} 9&-9\\-9&9 \end{bmatrix} \times 10^5 \\ [K^{(2)}]=\frac{400\times 70,000}{70}\begin{bmatrix} 1&-1\\-1&1 \end{bmatrix} = \begin{bmatrix} 4&-4\\-4&4 \end{bmatrix}\times 10^5 \\ [K^{(2)}]=\frac{100\times 70,000}{70}\begin{bmatrix} 1&-1\\-1&1 \end{bmatrix} =\begin{bmatrix} 1&-1\\-1&1 \end{bmatrix} \times 10^5

The assembly of the element stiffness matrices, before imposition of boundary conditions, is done as follows:

[\bar{K} ]=\begin{bmatrix} 9&-9&0&0\\-9&9+4&-4&0\\0&-4&4+1&-1\\0&0&-1&1 \end{bmatrix} \times 10^5=\begin{bmatrix} 9&-9&0&0\\-9&13&-4&0\\0&-4&5&-1\\0&0&-1&1 \end{bmatrix} \times 10^5

Similarly, the nodal displacement vector and the nodal load vector, before the imposition of boundary conditions, are

\left\{\bar{\phi_i}\right\} =\begin{Bmatrix} \phi _1\\\phi _2\\\phi _3\\\phi _4 \end{Bmatrix}

and

\left\{\bar{P}\right\} =\begin{Bmatrix} R_1\\0\\0\\-100 \end{Bmatrix}

The equilibrium equations before the imposition of boundary conditions are given by

\begin{bmatrix} 9&-9&0&0\\-9&13&-4&0\\0&-4&5&-1\\0&0&-1&1 \end{bmatrix} \begin{Bmatrix} 0\\\phi _2\\\phi _3\\\phi _4 \end{Bmatrix} \times 10^5=\begin{Bmatrix} R_1\\0\\0\\-100 \end{Bmatrix}

The known boundary condition is \phi _1=0. Using Equations 8.66 through 8.68, the above equation is reframed as follows:

\begin{bmatrix} K&K_1\\K_2&K_3 \end{bmatrix} \begin{Bmatrix} \phi _i\\\phi _1 \end{Bmatrix} =\begin{Bmatrix} P_1\\P_2 \end{Bmatrix} \quad\quad\quad\quad\quad 8.66 \\ [K_2]\left\{\phi _i\right\} +[K_3]\left\{\phi _1\right\} =\left\{P_2\right\} \quad\quad\quad\quad\quad 8.68 \\ \begin{bmatrix} \begin{bmatrix}13&-4&0\\-4&5&-1\\0&-1&1 \end{bmatrix}&\begin{bmatrix} -9\\0\\0 \end{bmatrix} \\ \begin{bmatrix} -9&0&0 \end{bmatrix} & \begin{bmatrix} 9 \end{bmatrix} \end{bmatrix} \begin{Bmatrix} \begin{Bmatrix} \phi _2\\\phi _3\\\phi _4 \end{Bmatrix} \\ \begin{Bmatrix} 0 \end{Bmatrix} \end{Bmatrix} \times 10^5=\begin{Bmatrix} \begin{Bmatrix} 0\\0\\-100 \end{Bmatrix} \\ \begin{Bmatrix} R_1 \end{Bmatrix} \end{Bmatrix}

We see that the final global stiffness matrix, nodal displacement vector, and load vector are

[K]=\begin{bmatrix} 13&-4&0\\-4&5&-1\\0&-1&1 \end{bmatrix} \times 10^5 \\ \left\{\phi _i\right\} =\begin{Bmatrix} \phi _2\\\phi _3\\\phi _4 \end{Bmatrix} \\ \left\{P\right\} =\begin{Bmatrix} 0\\0\\-100 \end{Bmatrix}

Thus, the final equilibrium equations are given by

[K]=\begin{bmatrix} 13&-4&0\\-4&5&-1\\0&-1&1 \end{bmatrix} \begin{Bmatrix} \phi _2\\\phi _3\\\phi _4 \end{Bmatrix}\times 10^5=\begin{Bmatrix} 0\\0\\-100 \end{Bmatrix}

and

\begin{bmatrix} -9&0&0 \end{bmatrix} \begin{Bmatrix} \phi _2\\\phi _3\\\phi _4 \end{Bmatrix} \times 10^5= \left\{R_1\right\}

Solving the above equations, we get the nodal displacements and support reaction as follows:

\begin{Bmatrix} \phi _2\\\phi _3\\\phi _4 \end{Bmatrix} =-\begin{Bmatrix} 1.11\\3.61\\13.61 \end{Bmatrix} \times 10^{-4} \ mm

and

R_1=-100 \ N

For Element 2, matrix [B] is given by (Equation 8.92)

[B]=\begin{bmatrix} \frac{\delta}{\delta x} \end{bmatrix} [N]=\begin{bmatrix} \frac{\delta}{\delta x} \end{bmatrix}\begin{bmatrix} \frac{l-x}{l} &\frac{x}{l} \end{bmatrix} =\begin{bmatrix} -\frac{1}{l} & \frac{1}{l} \end{bmatrix} \quad\quad\quad\quad\quad 8.92 \\ [B]=\begin{bmatrix} -\frac{1}{l}&\frac{1}{l}\end{bmatrix} =\begin{bmatrix} -\frac{1}{70}&\frac{1}{70} \end{bmatrix}

And, the strain vector is given by

\left\{\varepsilon \right\} =[B]\left\{\phi _i^{(2)}\right\} =-\begin{bmatrix} -\frac{1}{70}&\frac{1}{70} \end{bmatrix} \begin{Bmatrix} 1.11\\3.61 \end{Bmatrix}\times 10^{-4}=-3.57\times 10^{-6}

The corresponding stress is given by

\left\{\sigma \right\} =[C]\left\{\varepsilon \right\} =-(70\times 10^3)\times (3.57\times 10^{-6})=-0.25 \ MPa