In this entry:

An example solution to a problem with continuous load

The normal stress caused by tensile and compressive forces in a prismatic rod is discussed in the entry __Tension and compression__. In this entry, we dealt with statically determinate examples, i.e. those where we had only one support reaction and we were able to determine it from the equilibrium condition.

### Statically indeterminate tasks

Now we will deal with slightly more difficult examples, i.e. statically indeterminate ones. These are tasks where the rod is fixed at both ends (you can say that it is placed between two non-movable walls). In such a case, we have two unknown reaction forces of these restraints and only one equilibrium equation, so that is why we are talking about statically indeterminate examples.

In this case, we use the additional **Equation of compatibility**. This condition states that the total extension of the rod must be zero. Since both ends of the bar cannot move (they are fixed), the total deformation of our system is equal to "0".

You may encounter tasks in which the rod is positioned vertically. It doesn't matter when it comes to the solution, we proceed exactly the same.

When it comes to the number of fragments into which we divide our rod, two factors are important:

load change - additional force or continuous load

change in cross-sectional area or material stiffness (Young's modulus)

Both of these factors influence the amount of deformation of our rod.

## Tension and compression statically indeterminate - examples

### An example solution to a problem with point load

As a first example, we will solve a problem with a rod loaded with one point load. The rod will consist of two parts with different cross-sectional areas. The task will be solved using symbols without numerical data, which is common in this topic.

All examples used in this post were created and calculated in my |

The figure above shows the example we will solve. Let's start by labeling the reactions in Ra and Rb in the supports. As a reminder, the expression of the reaction is arbitrary and it is up to us to decide how to take it.

In the next step, we write the equilibrium equation for forces in the horizontal direction and the geometric condition. In our case, we will have two intervals from A to the application of force F and from the point of application of force F to point B. Then we write formulas for the elongations of the segments L1 and L2. After substituting the known values for the forces N1 and N2 in the appropriate ranges and the product of E and A and equating the whole to zero, we are able to calculate the reaction Ra. Then, after substituting the Ra reaction into the equilibrium condition, we obtain Rb.

In the next step, knowing the value of the support reactions, we can determine the value of normal (axial) forces, normal stresses and deformations for each compartment. This stage of calculations is already described in the __Tension and compression__ entry __.__

Knowing all the quantities, we can start drawing graphs showing the change in these quantities for each interval. The charts are shown in the figure below.

### An example solution to a problem with continuous load

The next example we will analyze is a problem with a bar with a continuous load of q = 4 kN/m. The rod will consist of two parts with different cross-sectional areas. This time the task will be solved using numerical data.

The task will also be solved using a __calculator__ for solving this type of tasks.

The figure above shows the example we will solve. Let's start by labeling the reactions in Ra and Rb in the supports. As you can see, the rod is positioned vertically to show you how to solve such a problem and how to draw graphs.

In the next step, we write the equilibrium equation and add a geometric condition. In our case, we will have two intervals from A to the start of the load q and from this point to point B.

Then we write formulas for the extensions of sections L1 and L2. After substituting the known quantities for the forces N1 and N2 in the appropriate ranges. As you can see, in the second interval, where there is a continuous load, to determine the elongation we use the integral of the quotient of the normal force by the product of Young's modulus and the cross-sectional area.

After solving this expression, we obtain the reaction Ra. Then, after substituting the Ra reaction into the equilibrium condition, we obtain Rb.

In the next step, knowing the value of the support reactions, we can determine the value of normal (axial) forces, normal stresses and deformations for each compartment. This stage of calculations is already described in the __Tension and compression__ entry __.__

Knowing all the quantities, we can start drawing graphs showing the change in these quantities for each interval. The charts are shown in the figure below.

As you can see, the elongation at the end of the rod is zero, which confirms that we have solved the problem correctly. This concludes the post Tension and compression statically indeterminate - examples.

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