In this entry:
- How to solve statically non-determinable tasks
- Example of a solution to a concentrated force task
- Example solution of a continuous load task
The normal stress induced by tensile and compressive forces in a prismatic bar was discussed in the entry Tension and compression. In this post, we dealt with statically determinable examples, that is, examples where we had only one support reaction and were able to determine it from the equilibrium condition.
Statically indeterminate tasks
Now we will deal with examples that are a little more difficult, that is, statically non-determinant. These are tasks where the bar is fixed at both its ends (you can say that it is placed between two non-sliding walls). In this case, we have two unknown reaction forces of these restraints and only one equilibrium equation, so that's why we talk about statically non-equivocal examples.
Equilibrium equations
Geometric condition
In this case, we use an additional geometric condition. This condition states that the total elongation of the rod must be equal to zero. Since both ends of the bar cannot move (they are restrained), so the total deformation of our system is equal to "0".
You may encounter tasks in which the bar is set vertically. It does not matter when it comes to the method of solution we proceed exactly the same.
When it comes to the number of fragments into which we divide our rod, two factors matter:
- change of load - additional force or continuous load
- Change in the cross-sectional area or stiffness of the material ( Young's modulus)
Both of these factors affect the amount of deformation of our rod.
Tension compression statically indeterminate - tasks
Example of a solution to a concentrated force task
As a first example, we will solve a task with a rod loaded with a single concentrated force. The bar will consist of two parts with different cross-sectional areas. The task will be solved on symbols without numerical data, this is common in this subject of tasks.
| All examples used in this post were created and calculated in my Stretch Calculator. I invite you to try the application, with its help you will determine the normal forces, normal stress and elongation or shortening of the bar. |
The figure above shows the example we will solve. Let's start by determining the reactions in Ra and Rb in the supports. As a reminder, the expression of the reactions is arbitrary and we decide how to take them.
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. Next, we work out the formulas for the elongations of the segment L1 and L2. After substituting the known quantities for the forces N1 and N2 in the corresponding intervals, and the product of E and A and assimilating the total to zero, we are able to calculate the reaction Ra. Then, after substituting the reaction Ra into the equilibrium condition, we get Rb.
In the next step, already knowing the value of support reactions, we can determine the value of normal (axial) forces, normal stresses and strain for each compartment. This stage of calculation is already described in the entry Tension and compression.
Knowing all the quantities, we can proceed to draw graphs showing the change in these quantities for each interval. The graphs are included in the figure below.
Example solution of a continuous load task
The next example we will analyze is a task with a rod with a continuous load of q=4 kN/m. The bar will consist of two parts with different cross-sectional areas. This time the task will be solved on numerical data.
The task will also be solved with Stretch Calculator for solving this type of task.
The figure above shows the example we will solve. Let's start by determining the reactions in Ra and Rb in the supports. As you can see, the bar is positioned vertically to show you how to solve such a task and how to draw graphs.
In the next step, we write an 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 that point to point B.
We then work out the formulas for the elongations of the L1 and L2 segments. After substituting the known quantities for the forces N1 and N2 in the corresponding intervals. As you can see in the second interval, where there is a continuous load, we use the integral of the quotient of the normal force by the product of the Young's modulus and the cross-sectional area to determine the elongation.
After solving this expression, we get the reaction Ra. Then, after substituting the Ra reaction into the equilibrium condition, we get Rb.
In the next step, already knowing the value of support reactions, we can determine the value of normal (axial) forces, normal stresses and strain for each compartment. This stage of calculation is already described in the entry Tension and compression.
Knowing all the quantities, we can proceed to draw graphs showing the change in these quantities for each interval. The graphs are included in the figure below.
As you can see, the elongation at the end of the bar is zero, which confirms that we have solved the task well. This concludes the entry Extension compression statically indeterminate - tasks.