Determination of Support Reactions- Analysis of Structures-Part-3

Determination of support reactions is first step in analysing a structure. In this article I will be answering the questions like how to determine the reactions at supports, what are the state of equilibrium of a structures, what is the statically indeterminate structure, etc. Answer will also be provided to a question- What is the difference between determinate and indeterminate structures? I will explain it in such a practical way that the people with limited knowledge can also get benefit from this article. Engineering students and those who want to learn structural engineering will get benefit from this article.

In the state of equilibrium a structure is stable, i.e., loads (actions) and reactions are equal in magnitude but opposite in direction. In the example below total action, P is 1000 kg and total reaction is 2x500=1000 kg. Action or load is acting downward but reactions are upward. Magnitude of load (action) is equal however, the direction of reactions are opposite. Therefore, this structure is in equilibrium state and stable.

Simply supported beam

In order to be in equilibrium a structure must meet the following six conditions, we also call them equation of equilibrium.

Summation of forces and moments must be zero, i.e.,
Summation of Fx=0, Summation of Fy=0, Summation of Fz=0,
Summation of Mx=0, Summation of My=0, Summation of Mz=0

Useful notes:

1.Support reaction (force) develops along the opposite direction to the direction of load (action)

2.Support reaction (moment) develops in the opposite direction to the direction of moment. The moment could be a direct action or posed by a load.

Cantilever beam

In the example of a cantilever beam above, the load P acts downward but the support reacts with equal but upward reaction force, i.e., F=P

Load causes a clockwise moment at the support but the fixed support reacts with equal but anti- clockwise moments, i.e., M=P x Span

3.Supports should be arranged such a way that they are able to react to the action (loads/ moments). For example, if there is a longitudinal force a roller supports cannot carry it, there should be at least a hinged support in the beam. If there is a moment the support must be a fixed. Fixed supports can carry all kinds of actions.

4.Support reactions could be of two types, i.e.,

a) Forces acting in any three orthogonal directions (axes- x,y,z)

b) Moments about any three orthogonal directions (axes- x,y,z)

Direction of reactions develop to match the direction of load application

5.In order to find support reactions, we need to write as many equations as many reactions are in the structure. These equations should be written based on above mentioned six equilibrium equations. In certain cases we are unable to write equal number of equations to number of reactions. I will discuss about this case later.

In the example of a simply supported beam shown above the axis-y is in perpendicular (normal) to the monitor. Since this is an x-z plane structure, not a space structure (3D), the support is assumed as rigid in axis-y. Therefore, the moments about axis-x, and -z are not applicable. Hence, in a planar structure, the above mentioned Fy=0, Mx=0 and Mz=0 are not applicable.

Don't worry if you don't understand the general theory I needed to say above. After reading the following example you will understand most of it.👌😎

How to determine support reactions? In order to determine the support reactions we use the applicable equations of equilibrium.

Example-1, Simply supported beam

We have vertical load P =1000 kg acting downward. Since there are no other forces acting along other axes, other support reactions do not develop along other axes. Therefore, support reactions must be forces acting upward only. Let's say RA and RB are support reactions at support A and B respectively.

Applying the Fz=0, we have an equation, 1000-(RA+RB)=0--------Eq1

But we don't know the values of RA and RB. Therefore, RA and RB are unknowns. When writing equilibrium equations we need to use "+" and "-" signs depending on the direction to which the forces are acting. Similarly, the signs of moments are different depending on whether they are clockwise or anti-clockwise.

In order to find two unknowns we must have at least two equations. We can write one more equation using My=0. For this, we will write an equation of moments about axis- y either at A or B.

Let's write the moment equation at A. 1000x2.5-RBx(2x2.5)=0--------Eq2

Note, the perpendicular distance to axis-y for RA is 0. Therefore, the term RAx0 is not used in this equation.

Using a first equation, 1000=RA+RB which gives, RB=1000-RA. Substitute RB in the second equation then, we get 1000x2.5-(1000-RA)x(2x2.5)=0 or 2500-5000+5xRA=0 with this, 5xRA= 2500 from this we get RA=500. Use this value of RA in the first equation, we get 1000-(500+RB)=0 or 1000=500+RB which gives us RB=500.

Now let's check the equation, Sum Fz=0, i.e., 1000-2x500=0 => True

Check for the equation, Sum My=0, i.e.,1000x2.5-500x2x2.5=0 => True. Here, the 500 is reaction at B 

Therefore, the simply supported beam is in state of equilibrium.

Example2, Cantilever

If we need to carry out this exercise in a cantilever beam, process is the same. Because of the fixed support, the support reaction M needs to be included in the second equation. Thus the two equations would be as follow:

Fz=0,  P-F=0-----------Eq1       My=0,  PxSpan-M=0----------Eq2

Whenever we write a moment equation, the arm for the force (load) must be measured from centre of gravity of the force. Following examples show some basic cases:

Note: Centre of gravity (c.g.) of the triangularly varying force lies at 1/3 of its base from the perpendicular. In the cases -3 and -4 the "w" is the perpendicular from where the c.g. is at 1/3 of its base distance.

In order to write equation for the forces, write an algebraic sum of the bold figures, which also includes the support reactions. When we have an inclined forces we need to resolve them into two axes matching the axis of structure. Necessary calculations will be carried out using these components that represent the inclined forces. Following figure shows how an inclined force is resolved in two axes. In the Case-1 one more equation must be written to find the longitudinal reaction at the support A.

Horizontal and vertical components of inclined force P

To find it, use Fx=0, we can write P2xcos a-RAh=0. Here we assume support B is a roller support

Where, P2xcos a is horizontal component of force P2 and RAh is longitudinal reaction that will develop at support A due to horizontal component of P2. Refer to article Bending Moment-Shear Force-Axial Force to see how the inclined forces are dealt in the Structural Analysis.

Statically determinate or indeterminate structures: In the example above we had same number of equations as unknowns (support reactions RA and RB). Therefore, we were able to find the unknowns  using the equations of equilibrium. The structures in which support reactions can be determined using the equilibrium equations are called Statically Determinate Structures.

On the other hand, because of some reasons such as, in order to limit deflections in a beam, we restrain degrees of freedom of a support or use more supports than absolutely necessary to support the given loads. In this case we have more unknowns (support reactions) than the equilibrium equations we can write using the applied loads. In such cases we are unable to find the support reactions in the regular way we did earlier. The structures in which support reactions cannot be determined using only the equilibrium equations are called Statically Indeterminate Structures.

Difference between determinate and indeterminate structures

Degrees of static indeterminacy: Additional number of equations that is required to solve for the support reactions. In the above mentioned indeterminate structures the degree of static indeterminacy is one. This is also called a degree of redundancy. Thus reduntants in these examples are the reactions RC and F2.

Related reading: Loads, Structural Supports, Terminology of Structural Analysis, Arm

Next article will focus on calculation of Bending Moment and Shear forces     Previous