# One-dimensional systems of bars

A system of bars consists of a number of elastic bars, each with a constant cross-sectional area and Young’s modulus. The system is only loaded at the endpoints and joints between the elements. An example is shown in the figure below.

Acting on the system are the external forces $P_1$ och $P_2$ at the joint in the middle and to the right. The leftmost joint cannot move — the bar is said to be clamped. $p_1$ and $p_2$ are joint displacements and since the left joint is clamped, we have not introduced any displacement there. The displacements $p_1$ and $p_2$ are called structural degrees of freedom (and naturally dependent on the external forces).

## Method of joints — statically determinate systems

One method for analyzing the system above is the so-called method of joints which we will illustrate by analyzing the structure above. A first step is to construct a free body diagram around each joint (which has degrees of freedom) and consider force equilibrium:

$$\rightarrow: \; -N_1+N_2+P_1=0, \quad \quad \rightarrow: \; -N_2+P_2=0$$

We see that the number of equilibrium equations equals the number of structural degrees of freedom. In this case, we have as many unknown normal forces, $N_1$ och $N_2$, as we have equilibrium equations. This system is called statically determinate — the number of structural degrees of freedom equals the number of unknown normal forces. Consequently, the normal forces can be calculated from the equations above.

We can use the normal forces to determine the change in length for each bar by using Hooke’s law, combined with the expression for mean-strain. This gives (for each bar)

$$N= \sigma \ A = E \ \epsilon \ A = \frac{EA}{L} \delta$$

$$N_1=\frac{E \, A_1}{L_1} \, \delta_1, \quad \quad N_2=\frac{E \, A_2}{L_2} \, \delta_2$$

If we are interested in the structural degrees of freedom, $p_1$ och $p_2$, they can be determined from the so-called compatibility relations:

$$\delta_1=p_1, \quad \quad \delta_2=p_2-p_1$$

## Statically indeterminate systems

Systems of bars where the number of unknown bar forces (normal forces) is larger than the number of structural degrees of freedom are called statically indeterminate. To analyze such structures, one must generally combine three types of relationships: equilibrium, constitutive and compatibility. We illustrate the procedure by the following example.

Example

An elastic system of bars consists of two bar elements, with different lengths and geometry. The system is loaded by a point force $P$, at the joint between the two bars, according to the figure.

The left and right joints are clamped so there is only one structural degree of freedom $p$.

Equilibrium around the middle joint gives (same as in the previous example):
$$\rightarrow: \quad -N_1+N_2+P=0$$

We note that the number of unknown normal forces is two and the number of structural degrees of freedom is one. The system is said to be statically indeterminate of order one.

Constitutive relations for both of the bars are:
$$N_1=\frac{E \, A_1}{L_1} \, \delta_1, \quad \quad N_2=\frac{E \, A_2}{L_2} \, \delta_2$$

Finally, we have the compatibility relationships:
$$\delta_1=p, \quad \quad \delta_2=-p$$
Which means that if one bar elongates the other must shorten the same amount.

Summarizing, we see the number of equations are five (one equilibrium, two constitutive and two compatibility) and the number of unknowns (assuming we know the applied load $P$) is also five ($N_1$, $N_2$, $\delta_1$, $\delta_2$ och $p$). Hence, the system is solvable.