Calculation of Support Reactions

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When designing buildings, frames, beams or even trusses, it is important to know the magnitude and direction of the load that resides within a support. Using different methods and techniques it is possible to calculate these support reactions. However, most techniques always boil down to using the same method. It is always necessary to identify the unknown forces using a free-body diagram followed by using the equations of equilibrium to solve for the unknown reactions[1].

Support Reactions

When designing buildings, there are several different types of supports that are used to connect members to each other as well as fixed bodies. Each type of support has different combinations of reactions associated with it. When considering support reactions in two dimensions, there are a maximum of three different types of reactions that can occur to any given support, these included vertical forces, horizontal forces and moments. These reactions can be solved using the Equations of Equilibrium so long as the structure or beam is statically determinate. If the number of support reactions is greater than the number of equations of equilibrium, then the structure is to be considered statically indeterminate. For example, if a structure were to have two fixed end connections with no internal supports, there would be a total of six unknown reaction forces, two in the x direction, two in the y direction and two moments. Since there are only three possible equations of equilibrium, the problem would need to be solved using a method of analysis for indeterminate structures.

Types of Supports

Table 1: Types of Supports, adapted from Kassimali[2]

There are 5 main types of supports. These supports are adapted from Kassimali[2] and consist of the following:

  • Fixed end - This support prevents rotation, horizontal and vertical translation. By doing so it can be determined that the support will have a moment, and a horizontal and vertical force component.
  • Pinned - The reaction forces for this support consist of a horizontal force and vertical force. This support does not have a moment because the support does not resist rotation.
  • Roller - This support only has one reaction force which is a vertical force. This support only has one reaction force because it does not resist rotation or movement in the horizontal direction.
  • Link - This support has a resultant support reaction which can consist of components vertically and horizontally.
  • Rocker - Like the roller, this support also has a single reaction force which is a vertical force because is does not resist rotation and horizontal translation.

See Table 1 for examples of types of supports and their reaction forces.

Internal Reactions

In some cases structures may have internal supports. If there are internal supports the structure maybe be separated at these supports. By separating the structure at these supports into members is may be easier to solve for the external reaction forces of the structure (see Problem 2). Internal supports are the same has external supports in the sense that they both have different types of reaction combinations for different types of supports. Since internal supports also have support reactions they will have to be considered or solved when solving for a structure with internal supports. The method to solving the structure using internal supports is the same has solving for the support reactions [3]. If the structure is indeterminate with no internal support reactions it may be possible to solve for the reaction forces using different types of methods such has the Force Method for One Degree of Indeterminacy.

Types of Internal Supports

Like external supports there are different types of internal supports adapted from Hibbeler[3] . Some of these supports are:

  • Hinge - A hinge resists horizontal and vertical translation but allows rotation. Therefore a hinge consists of horizontal and vertical support reaction (see Problem 2).
  • Internal Roller - This is the same has a roller that is used has an external support since it allows rotation and horizontal translation. Therefore it will have a vertical support reaction.

Free-Body Diagrams

Before calculating for the support reactions on a body, a free-body diagram should be drawn. A free-body diagram is a representation of a body that is free from any support or external object[3]; however, the forces and moments from the supports attached to the body are to be considered when drawing a free-body diagram, along with any other external forces or moments. If there is a force that acts in both the x and y direction (i.e. the force is on an angle), it must be broken up into its respective x and y components. Once all the known and unknown forces are drawn, the free-body diagram is complete. From here, all the forces that need to be solved can be clearly seen, and their respective directions clearly labelled, therefore all forces and moments are able to be separated into their respective equations of equilibrium. All forces in the x-direction are to be placed into the x-direction equation of equilibrium, the same procedure is used for the y-direction as well as all the moments that act on the body. There are many different types of loads that can be applied to a body, for explanations on this see Loads and Load combinations.

Method of Analysis

This method is adapted from Kassimali[2].

1. Check Stability and Determinacy of the structure. If indeterminate or unstable use a method of analysis for indeterminate structures.

2. Draw the Free-Body Diagram including all applied forces and support reactions.

3. Using the Equations of Equilibrium, solve reactions if possible.

4. For more complicated structures, it may be cut into sections. Sections should be cut between every time the loading condition changes. It is also useful to cut at hinges where no moments occur.

5. Continue using the Equilibrium equations to solve the structure.

6. Redraw the Free-Body Diagram with forces and reactions in the correct directions. A negative answer shows that the direction drawn on the initial Free-Body Diagram was assumed incorrectly.

Equations of Equilibrium

The equations of equilibrium are as follows:

1. $ \sum F_{x} = 0 $ , Equation 1 represents the sum of all the forces acting on a body in the x direction, and states that this sum must be equal to zero.[3]

2. $ \sum F_{y} = 0 $ , Equation 2 represents the sum of all the forces acting on a body in the y direction, and states that this sum must be equal to zero.[3]

3. $ \sum M_{o} = 0 $ , Equation 3 represents the sum of all the couple moments and the moments caused by all force components acting on the member around the z axis, which is perpendicular to both the x and y axis and passes through an arbitrary point o.[3]

Example Problems

Problem 1

Solve the following support reactions.


Step 1: Check for determinacy.

$ i=(3m+r)-(3j+e_{c}) $
$ i=(3*2+3)-(3*3+0) $
$ i=0 $

The beam is determinate and has no collapse mechanism, therefore the analysis continues.

Step 2: Using Table 1 to determine the appropriate reactions, draw the free-body diagram. Include all reactions and external loads.


Step 3: Solve the reactions using the Equations of Equilibrium.

$ \sum M_{A} = 0 $
$ -100 kN * 5 m + C_{y} * 10 m = 0 $
$ C_{y} = 50 kN $

$ \sum F_{y} = 0 $
$ A_{y} + C_{y} - 100 kN = 0 $
$ A_{y} = 50 kN $

$ \sum F_{x} = 0 $
$ A_{x} = 0 $

Steps 4 and 5 can be skipped as all reactions have been solved.

Step 6: Redraw FBD.


Picture above is wrong. Shold be 50 kN for the reaction forces

Problem 2

Solve the following support reactions.


Step 1: Check for determinacy.

$ i=(3m+r)-(3j+e_{c}) $
$ i=(3*3+4)-(3*4+1) $
$ i=0 $

The beam is determinate and has no collapse mechanisms, therefore the analysis continues.

Step 2: Using Table 1 to determine the appropriate reactions, draw the free-body diagram. Include all reactions and external loads.


Step 3: Split at the hinge. The hinge has no moments.


Step 4: Starting with the right side as it is simpler, solve for reactions.

$ \sum M_{B} = 0 $
$ C_{y} * 1 m - 30 kN * 3 m = 0 $
$ C_{y} = 90 kN $

$ \sum F_{y} = 0 $
$ -V_{B} + C_{y} - 30 kN = 0 $
$ V_{B} = 60 kN $

$ \sum F_{x} = 0 $
$ N_{B} = 0 $

Step 5: Transfer the internal reactions to the other side of the hinge. Solve for remaining unknowns.

$ \sum M_{A} = 0 $
$ -M_{A} - 20 kN * 2 m + 60 kN * 3 m = 0 $
$ M_{A} = 140 kNm $

$ \sum F_{y} = 0 $
$ A_{y} + V_{B} - 20 kN = 0 $
$ A_{y} = -40 kN $

$ \sum F_{x} = 0 $
$ A_{x} = 0 $

Step 6: Redraw FBD.


Step 7: Check the answers by checking the equilibrium of the entire system.

$ \sum M_{A} = 0 $
$ -140kNm - 20kN*2m + 90kN*4m - 30kN*6m = 0 $
$ 0 = 0 $

$ \sum F_{y} = 0 $
$ -40kN - 20kN + 90kN - 30kN = 0 $
$ 0 = 0 $

$ \sum F_{x} = 0 $
$ 0 = 0 $


  1. Hibbeler, R. C. (2011). Mechanics of Materials (8th ed.). Upper Saddle River, NJ: Pearson Prentice Hall.
  2. 2.0 2.1 2.2 Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning.
  3. 3.0 3.1 3.2 3.3 3.4 3.5 Hibbeler, R. C. (2010). Engineering Mechanics: Statics and Dynamics (12th ed.). Upper Saddle River, NJ: Pearson Prentice Hall.