In our previous article we study the Beams and types of beams based on their geometry. if you missed the topic so you can click here to clear the pervious concept. Now let’s look at another important aspect of beams: types of beams based on equilibrium conditions.

Â When categorizing beams based on equilibrium conditions, we can classify them into two fundamental types:

## Letâ€™s Understand the Types of Beams based on the equilibrium

- statically determinate beams
- statically indeterminate beam

## 1. Statically determinant = 0

â€œA Structure is Said to be determinate; If conditions at static Equilibrium are sufficient to analyse the structure

or

A type of structured system in which all forces and reactions can be calculated using the principal of Equilibrium and equation of static

Sum of all forces and Sum of moments is Equal to Zero(= 0)

## Here are some examples of statically determinate beams

## 2. statically indeterminate beam > 0

Is a beam that has more unknown reaction forces than the number of equilibrium equations available. This means that the reactions cannot be determined solely by applying the laws of statics. In order to determine the reactions it is necessary to make additional assumptions or to use more advanced methods of analysis.

## Statically indeterminate beams can be analyzed using a variety of methods, including

- Force method
- Moment distribution method:
- Stiffness method

## The choice of method depends on the complexity of the beam and the accuracy required.

- Continuous beam with three supports
- Beam with one end fixed and the other end supported on a roller
- Beam with one end fixed and the other end restrained from rotating

## Now Lets look ibto the other important definations

## Equilibrium Principle

In structural engineering, engineers rely on the principles of equilibrium, which state that for any structure to be in a state of balance, the following conditions must be met:

- The sum of all forces acting in the horizontal direction (Î£Fx) must equal zero.
- The sum of all forces acting in the vertical direction (Î£Fy) must equal zero.
- The sum of all moments (rotational forces) about any point must equal zero (Î£M = 0).

This means that the number of unknown reaction forces and moments can be determined using these equations.

## Degree of Determinacy

The degree of determinacy of a structure is a measure of how many unknown forces can be determined by applying the laws of statics. A structure is said to be statically determinate if the number of unknown forces is equal to the number of equilibrium equations. If the number of unknown forces is greater than the number of equilibrium equations, then the structure is said to be statically indeterminate.

The degree of determinacy is denoted by the letter **D**. For a two-dimensional structure, the degree of determinacy can be determined using the following for

**D = 3m – 2j – r**

- m is the number of members
- j is the number of joints
- r is the number of support reactions

## Here are some examples of the degree of determinacy of different types of structures -

- Simply supported beam: Statically determinate (D = 1)
- Fixed beam: Statically indeterminate (D = 2)
- Continuous beam: Statically indeterminate (D > 2)
- Truss: Statically determinate (D = 2j – 3)
- Frame: Statically determinate or indeterminate (D = 3m – 2j – r)

## Stabilities in 2-d structure

## For a 2-D structures we need to make a note of the below some point,

- Maximum no of externally independent support reaction should be available for 2D structure, is R=3
- All Reaction should be non-Parallel
- All Reactions should be Nin-concurrent
- Reaction should be non-Trivial
- There Should be no condition of mechanism l.e no three colinear hinges

## Procedure analysis or calculation steps for the beam

Letâ€™s discuss it one by one, how to calculates the forces and reaction in a statically determinate structures follow are the important Steps

## 1. Identify Supports and Joints

The first step is to Determine and identify the type of Support**, **whether it is Pin joint, fixed or Roller support. And then locate the joint where member meet.

## 2. Draw Free Body Diagrams (FBDs)

Isolate each component of the structure (beams, trusses, etc.) and draw a free body diagram for each one. This involves representing the structure with all forces acting on it and applying equilibrium equations (Î£F = 0 and Î£M = 0) to find unknown forces and reactions.

## 3. Apply Equilibrium Equations

For each free body diagram, apply the equations of equilibrium

- Î£Fx (sum of horizontal forces) = 0
- Î£Fy (sum of vertical forces) = 0
- Î£M (sum of moments about a point) = 0

## 4. Solve for Unknowns forces

Use the equilibrium equations to solve for unknown forces (e.g., internal member forces, support reactions). In statically determinate structures, you will have as many equations as unknowns, making it possible to solve for everything

Now Letâ€™s understand the above steps with an Examples, With the help of this calculation we can Draw a SFD and BMD for any Beam,

## Example: Consider a simply supported beam with a point load 2000N applied at its centre.

Given data

- Point load (P) = 2000N
- Length of the beam (L) = 4mÂ

## Step 1: Identify Supports and Joints

In this example, we have a simply supported beam, which means it is supported at both ends by roller supports shown in Fig:1 (R1 and R2).

Locate the joint where the beam meets the support. In this case, we have two joints: one at the left end and one at the right end.

## Step 2: Draw Free Body Diagrams (FBDs)

Isolate each component of the structure. In this case, we have the entire beam as one component.

Draw a free body diagram for the beam. Include all forces acting on it.

## Step 3: Apply Equilibrium Equations

- To calculate the reaction forces, we can use the following equations:
- Î£Fy = 0
- Î£MA = 0

where:

- Î£Fy is the sum of all forces in the y-direction
- Î£MA (R1) is the sum of all moments about point A

## Step 4 : Solve for Unknowns forces

**In this case, point A is the left support.**

Substituting the forces and moments into the equations, we get:

- R1 + R2 = 2000 N
- -R1 * L/2 + 2000 N * L/2 = 0

Solving these equations, we get:

- R1 = 1000 N
- R2 = 1000 N

## Conclusion

In the field of structural engineering understanding the degree of determination is essential to assess the solvability and stability of structures. Statically determinate structures are well-balanced and can be solved using standard equilibrium equations. Statically indeterminate structures, on the other hand, present challenges due to the excess of unknowns, necessitating advanced methods for analysis. Over determinate structures, while rare, indicate an unstable condition that requires careful intervention to achieve stability

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