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Simply supported beam calculator
Deflections and slopes of simply supported beam
Simply supported beam diagrams
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Support reactions of a simply supported beam

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Introduction

The simply supported beam is one of the most simple structures. It features only two supports, one at each end. One is a pinned support and the other is a roller support. With this configuration, the beam is inhibited from any vertical movement at both ends whereas it is allowed to rotate freely. Due to the roller support it is also allowed to expand or contract axially, though free horizontal movement is prevented by the other support.

Supports of a simply supported beam

Removing any of the supports inserting an internal hinge, would render the simply supported beam to a mechanism, that is body the moves without restriction in one or more directions. Obviously this is unwanted for a load carrying structure. Therefore, the simply supported beam offers no redundancy in terms of supports, and if a local failure occurs the whole structure would collapse. These type of structures that offer no redundancy are called critical or determinant structures. To the contrary, a structure that features more supports than required to restrict its free movements is called redundant or indeterminate structure. 

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Static analysis - how to find reactions

Since a simply supported beam is a determinate structure, it is possible to obtain its static response using just equilibrium equations. These equations enforce that the sum of all forces and moments, acting upon the structure, in any direction, including both applied loads and support reactions, must be zero. For a plane structure, with in plane loading, the equilibrium equations are:

\sum F_x = 0

\sum F_y = 0

\sum M^{any point} = 0

The first two equations, enforce force equilibrium in directions x and y, (a Cartesian system of axes that can be arbitrarily defined, depending on the case). The third equation enforces the equilibrium of all moments around a certain point, which can be any point of the plane.

Example 1 : support reactions of a simply supported beam with a point load

Determine the support reactions of a centrally loaded simply supported beam, with a point force P , at the middle

Support reactions of a simply supported beam with a point load (force) at center

Assigning the unknown support reactions to variables R_A , H_A and R_B , as shown in the figure, the three equilibrium equations are defined this way:

  • Along direction x, there is no imposed force applied to the structure. There is only the unknown support reaction H_A . Thus the first equilibrium equation is:

\sum F_x = 0\Rightarrow H_A=0

  • Along direction y, the imposed force P is applied to the center of the beam, as well as support reactions R_A and R_B . Thus the first equilibrium equation is:

\sum F_y = 0\Rightarrow R_A+R_B-P=0

  • For the third equation we have to choose one point around which the moments are calculated. It is often more convenient to select a point through which some of the forces are directed (such as point A in our example), because, the resulting moments for these forces would be zero. So, around point A, support reactions H_A and R_A have no lever arm, imposed force P have a lever arm equal to half the beam length and support reaction R_B have a lever arm equal to the beam length. Assuming counter-clockwise positive rotation, the third equation becomes:

\ \circlearrowleft\sum M^{A}= 0 \Rightarrow H_A 0+R_A 0-P{L\over2}+R_B L=0

There are three unknowns, and we have three equations, therefore it is possible to solve the system of equations and obtain the unknown support reactions. H_A is directly found from the first equation equal to zero. Unless there is an imposed load along the beam longitudinal axis, this reaction will always be zero. From the third equation we can directly obtain R_B :

-P{L\over2}+R_B L=0\Rightarrow

R_B={P\over2}

And finally, substituting R_B to the second equation, R_A should be found too:

R_A+{P\over2}-P=0\Rightarrow

R_A={P\over2}

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When the imposed load is a distributed one, the procedure remains the same, provided that you replace it with an equivalent point force, having:

  • same direction
  • magnitude equal to the total force of the distributed load
  • application point at the centroid of the distributed load
Example 2 : support reactions of a simply supported beam with distributed load

Find the reactions of the following simply supported beam, with a uniform distributed load applied to its half span.

Simply supported beam with a uniform distributed load in half span

Before proceeding to the equilibrium equations, we will replace the distributed load with an equivalent point force W . This force should have the same downward direction and a magnitude equal to the total load, that is W=wL/2 . Its application point should be located at the centroid of the distributed load it replaces, in this case one quarter length from the right end of the beam. The following figure illustrates the beam, with the equivalent load we are going to use, for the needs of equilibrium.

Equivalent point force for the half span distributed load

Next we assign the unknown support reactions to variables R_A , H_A and R_B , and we define the x, y axes as shown in the figure below.

Support reactions for the simply supported beam with a half span distributed load

Because there there is no horizontal component of imposed loads, reaction H_A should be zero, by virtue of the equilibrium equation \sum F_x = 0 . The remaining two non-trivial equilibrium equations, are written below, taken into account the positive directions, indicated by our axes system:

\sum F_y = 0\Rightarrow R_A+R_B-{wL\over 2}=0

\ \circlearrowleft\sum M^{A}= 0 \Rightarrow R_A 0-{wL\over 2}{3L\over4}+R_B L=0

From the last equation we can find R_B :

-{3wL^2\over4}+R_B L=0\Rightarrow

R_B = {3wL\over8}

Substituting R_B to the first equation, we can find the remaining unknown reaction R_A :

R_A+ {3wL\over8}-{wL\over 2}=0\Rightarrow

R_A -{wL\over8}=0\Rightarrow

R_A ={wL\over8}

Related pages

Simply supported beam calculator
Deflections and slopes of simply supported beam
Simply supported beam diagrams
Centroids Table

See also
Simply supported beam calculator
Deflections and slopes of simply supported beam
Simply supported beam diagrams
Centroids Table