Structural mechanics

One-dimensional systems of bars

Let us study an elastic bar element where we have introduced the local displacements $p^1_1$ and $p^1_2$ as well as the associated forces $P^1_1$ and $P^1_2$ according to the figure below.

For this bar element, we can write the following stiffness relationship:

$$
\begin{bmatrix}
P^1_1 \\ P^1_2
\end{bmatrix} = \frac{E \, A_1}{L_1}
\begin{bmatrix}
1 & -1 \\
-1 & 1
\end{bmatrix}
\begin{bmatrix}
p^1_1 \\ p^1_2
\end{bmatrix}
$$

 

$$\begin{bmatrix}P^1_1 \ P^1_2\end{bmatrix}=
\begin{bmatrix}-N(0) \\ N(L)\end{bmatrix}\begin{bmatrix}
p^1_1 \\ p^1_2
\end{bmatrix}=\begin{bmatrix}u(0) \\ u(L)\end{bmatrix}$$

A systematic methodology for analyzing a system of bars is to introduce global degrees of freedom, establish elements with local stiffnesses and then assemble these into the global system. This methodology is the same used in the finite element method.

Example

Determine the displacement $p_2$ for the system below.

Introduce global displacements $p_1$-$p_3$ and forces $P_1$-$P_3$ at the nodes.

In this case, we have two bar elements whose stiffness relations describes relations between local displacements and forces.

The following locals stiffness relations holds:
$$
\begin{bmatrix}
P^1_1 \\\ P^1_2 \\\
\end{bmatrix} =\frac{E \, A_1}{L_1}
\begin{bmatrix}
1 & -1 \\\
-1 & 1
\end{bmatrix}
\begin{bmatrix}
p^1_1 \\\ p^1_2
\end{bmatrix}, \\\
\begin{bmatrix}
P^2_1 \\\ P^2_2
\end{bmatrix} =\frac{E \, A_2}{L_2}
\begin{bmatrix}
1 & -1 \\\
-1 & 1
\end{bmatrix}
\begin{bmatrix}
p^2_1 \\\
p^2_2
\end{bmatrix}
$$

We identify relations between the local and global degrees of freedom (compatibility):
$$
p^1_1=p_1,\\\
p^1_2=p_2, \\\
p^2_1=p_2, \\\
p^2_2=p_3
$$

then the local stiffness relations can be written as

$$
\begin{bmatrix}
P^1_1 \\\ P^1_2
\end{bmatrix} =\frac{E \, A_1}{L_1}
\begin{bmatrix}
1 & -1 \\\
-1 & 1
\end{bmatrix}
\begin{bmatrix}
p_1 \\\ p_2
\end{bmatrix},
\\\
\begin{bmatrix}
P^2_1 \\\ P^2_2
\end{bmatrix} =
\frac{E \, A_2}{L_2}
\begin{bmatrix}
1 & -1 \\\
-1 & 1
\end{bmatrix}
\begin{bmatrix}
p_2 \\\ p_3
\end{bmatrix}
$$

Now study equilibrium for each joint/node

which gives:
$$
\begin{bmatrix}
P_1 \\\
P_2 \\\
P_3
\end{bmatrix}=
\begin{bmatrix}
P^1_1 \\\
P^1_2+P^2_1 \\\
P^2_2
\end{bmatrix}
$$
We can now assemble (sum) the local stiffness realtions to the global system:
$$
\begin{bmatrix}
P_1 \\\ P_2 \\\ P_3
\end{bmatrix}=
\frac{EA_1}{L_1}
\begin{bmatrix}
1 & -1 &0 \\\
-1 & 1 & 0\\\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
p_1 \\\ p_2 \\\ p_3
\end{bmatrix} \\\ +
\frac{EA_2}{L_2}
\begin{bmatrix}
0 & 0 & 0 \\\
0 & 1 & -1 \\\
0 & -1 & 1
\end{bmatrix}
\begin{bmatrix}
p_1 \\\ p_2 \\\ p_3
\end{bmatrix}
$$

In the example, we know that $p_1=p_3=0$ (boundary conditions). If this information is used, the following is obtained:
$$
\begin{bmatrix}
P_1 \\\ P_2 \\\ P_3
\end{bmatrix}=\frac{EA_1}{L_1}
\begin{bmatrix}
-p_2 \\\ p_2 \\\ 0
\end{bmatrix}+
\frac{EA_2}{L_2}
\begin{bmatrix}
0 \\\ p_2 \\\ -p_2
\end{bmatrix}
$$
alternatively, each equation written separately
$$
P_1 = -\frac{EA_1}{L_1} p_2 \\\
P_2 = \left(\frac{EA_1}{L_1} +\frac{EA_2}{L_2} \right) p_2 \\\
P_3 = -\frac{EA_2}{L_2} p_2
$$

For a given load $P_2$, the displacement $p_2$ is determined as
$$
p_2 = \frac{P_2}{EA_1/L_1+EA_2/L_2}
$$
After $p_2$ has been determined, $P_1$ and $P_3$ can be calculated.

Two-dimensional systems of bar

An elastic 2D bar element is studied where we have introduced the local element displacements $p_{1x}, \, p_{1y},\, p_{2x}, \, p_{2y}$ as well as the associated forces $P_{1x}, \, P_{1y},\, P_{2x}, \, P_{2y}$.

For this bar element, we can establish the following stiffness relation (no proof provided in this text).

 

$$\begin{bmatrix}
P_{1x} \\ P_{1y} \\ P_{2x} \\ P_{2y}
\end{bmatrix} =
\frac{EA}{L}
\begin{bmatrix}
\cos^2(\alpha) & \cos(\alpha)\sin(\alpha) & -\cos^2(\alpha) &
\cos(\alpha)\sin(\alpha) \\
%
\cos(\alpha)\sin(\alpha) & \sin^2(\alpha) & -\cos(\alpha)\sin(\alpha) &
-\sin^2(\alpha) \\
%
-\cos^2(\alpha) & -\cos(\alpha)\sin(\alpha) & \cos^2(\alpha) &
\cos(\alpha)\sin(\alpha) \\
%
-\cos(\alpha)\sin(\alpha) & -\sin^2(\alpha) & \cos(\alpha)\sin(\alpha) &
\sin^2(\alpha) \\
\end{bmatrix}
\begin{bmatrix}
p_{1x} \\
p_{1y} \\
p_{2x} \\
p_{2y}
\end{bmatrix}$$

The element forces $P_{1x}$, $P_{1y}$, $P_{2x}$ och $P_{2y}$ can be related to the normal forces (along the bar) at $\xi=0$ and $\xi=L$:

$$
\begin{bmatrix}
P_{1x} \\ P_{1y} \\
\end{bmatrix}=-N(0) \,
\begin{bmatrix}
\cos(\alpha) \\ sin(\alpha)
\end{bmatrix}, \quad
\begin{bmatrix}
P_{2x} \\ P_{2y} \\
\end{bmatrix}=N(L) \,
\begin{bmatrix}
\cos(\alpha) \\
\sin(\alpha)
\end{bmatrix}
$$

and in the same manner for the displacements:
$$
\begin{bmatrix}
p_{1x} \\ p_{1y} \\
\end{bmatrix}=u(0) \,
\begin{bmatrix}
\cos(\alpha) \\
\sin(\alpha)
\end{bmatrix}, \quad
\begin{bmatrix}
p_{2x} \\ p_{2y} \\
\end{bmatrix}=u(L) \,
\begin{bmatrix}
\cos(\alpha) \\
\sin(\alpha)
\end{bmatrix}
$$

Example

Element stiffnesses for each bar (for simplicity we will directly use information about the boundary conditions)

Bar element 1

The angle $\alpha=0$ $\rightarrow$ $\sin(\alpha)=0$, $\cos(\alpha)=1$ as well as $p^1_{1x}=p^1_{1y}=0$ and $p^1_{2x}=p_{1}$, $p^1_{2y}=p_{2}$:
$$
\begin{bmatrix}
P^1_{2x} \\\ P^1_{2y}
\end{bmatrix}=\frac{EA}{4 \, L}
\begin{bmatrix}
1 & 0 \\\
0 & 0
\end{bmatrix}
\begin{bmatrix}
p^1_{2x} \\\ p^1_{2y}
\end{bmatrix}=\frac{EA}{4 \, L}
\begin{bmatrix}
1 & 0 \\\
0 & 0
\end{bmatrix}
\begin{bmatrix}
p_{1} \\\ p_{2}
\end{bmatrix}
$$

Bar element 2

$\sin(\alpha)=-3/5$, $\cos(\alpha)=4/5$ as well as $p^2_{1x}=p^2_{1y}=0$ and $p^2_{2x}=p_{1}$, $p^2_{2y}=p_{2}$:

$$
\begin{bmatrix}
P^2_{2x} \\\ P^2_{2y}
\end{bmatrix}
=\frac{EA}{5L}
\begin{bmatrix}
16/25 & -12/25 \\\ -12/25 & 9/25
\end{bmatrix}
\begin{bmatrix}
p^2_{2x} \\\ p^2_{2y}
\end{bmatrix} \\\
=\frac{EA}{5L}
\begin{bmatrix}
16/25 & -12/25 \\\
-12/25 & 9/25
\end{bmatrix}
\begin{bmatrix}
p_{1} \\\ p_2
\end{bmatrix}
$$

Bar element 3

The angle $\alpha=\pi/2$ $\rightarrow$ $\sin(\alpha)=1$, $\cos(\alpha)=0$ as well as $p^3_{2x}=p^3_{2y}=0$ and $p^3_{1x}=p_{1}$, $p^3_{1y}=p_{2}$:

$$
\begin{bmatrix}
P^3_{1x} \\\ P^3_{1y}
\end{bmatrix}
=\frac{EA}{3 L}
\begin{bmatrix}
0 & 0 \\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
p^3_{1x} \\\
p^3_{1y}
\end{bmatrix}
=\frac{EA}{3 L}
\begin{bmatrix}
0 & 0 \\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
p_{1} \\\ p_{2}
\end{bmatrix}
$$

If we establish equilibrium for the node

then we obtain

$$
\begin{bmatrix}
P_1 \\\ P_2
\end{bmatrix}=
\begin{bmatrix}
P^1_{2x} \\\ P^1_{2y}
\end{bmatrix} +
\begin{bmatrix}
P^2_{2x} \\\ P^2_{2y}
\end{bmatrix} +
\begin{bmatrix}
P^3_{1x} \\ P^3_{1y}
\end{bmatrix}
$$
Thereby, the global stiffness relation is obtained by assembling (summing) the local stiffness relations:

$$
\begin{align}
\begin{bmatrix}
P_1 \\\ P_2
\end{bmatrix}
&= \frac{EA}{L}
\begin{bmatrix}
1/4+16/125 & -12/125\\\
-12/125 & 9/125+ 1/3
\end{bmatrix}
\begin{bmatrix}
p_1 \\\ p_2
\end{bmatrix} \\\
&\approx \dfrac{EA}{L}
\begin{bmatrix}
0.378 & -0.096 \\\ -0.096 & 0.405
\end{bmatrix}
\begin{bmatrix}
p_1 \\\ p_2
\end{bmatrix}
\end{align}
$$

Leave a Reply