After reviewing the theory on calculating support reactions, it’s time for some practice.
In this article, I will analyze two typical examples to help you better understand how to “calculate beams” in practice – i.e., perform static and kinematic analysis and determine support reactions in statically loaded beams.
I will show you step by step how to use the known equilibrium equations to correctly solve tasks related to reactions. In the article, we will discuss beams with different types of supports and loads, as well as highlight the most common mistakes.
Our agenda:
Detaching the system from constraints
Static determinacy of the beam
Calculation of the reaction forces
To learn how to calculate beams, it’s worth reviewing the presented solutions and then practicing on your own using the EquiBeam app. By using it, you’ll quickly master calculating beam reactions, which is a crucial step in the static analysis of structures.
Calculating a Simple Beam
In this example, we’ll deal with the calculations of a beam presented in the static diagram below:
Beams of this type are called two-support beams because of the pin support at point B (which restricts two degrees of freedom, in vertical and horizontal directions) and the roller support at point D (which restricts one degree of freedom in the direction perpendicular to its base).
This beam is loaded with a concentrated moment K = 2qL2 at point A, a vertical concentrated force P1 = 2qL at point C, a horizontal concentrated force P2=qL at point D, and a continuous load of intensity qqq between points A and C.
Calculating beams – detaching the system from constraints
The first step in calculating beams is usually to detach the system from its constraints – replacing the supports with their support reactions. The diagram of the beam with reactions is shown here:
As stated in the context of degrees of freedom restricted by the supports – we similarly define the support reactions present in the system. In other words, if a support blocks movement in, for example, the y-axis direction, a reaction force will appear in that direction.
For this reason, we can write two support reactions at the fixed support – in the x-axis direction (HB ) and y-axis direction (VB ). Technically, there is one force in this support, but its direction is dependent on the loads, so we split it into two components. At the roller support, we write one reaction in the y-axis direction (VD ).
Condition for static determinacy of the beam
Now that we know the reactions in the system, we can proceed to determine whether the beam is statically determinate. The condition for static determinacy for beams without hinges can be written as:
3t-r=0
where t is the number of panels (rigid elements of the structure), and r tis the number of reactions. In our case t=1, r=3:
3t-r=0→3⋅1-3=0-ok!
The condition for static determinacy of the system is met – which means that the beam CAN BE statically determinate. More specifically, if the beam is geometrically stable, the system will be statically determinate!!! Therefore, we still need to check geometric stability, meaning we need to perform kinematic analysis.
Here, kinematic analysis is very simple – our beam has 3 degrees of freedom. Two of them are restricted by the fixed support. Analyzing this support alone, the beam can only rotate around it. However, the system also includes a roller support, whose reaction direction (which is very important) does not pass through the fixed support. As a result, rotation is also restricted, and the beam is geometrically immutable and statically determinate. Only now can we proceed to analyze the equilibrium equations and calculate the support reaction values.
Static equilibrium equations
To do this, we analyze the projections of all forces onto the x and y axes, as well as the moments relative to one of the points in the system. The first equation will be the sum of the projections of forces on the x-axis:
∑F_{ix} =H_B+qL=0→(1)In the equation, there will be the reaction HB from the fixed support and the force P2 . Both of these forces act in the same direction as the assumed x-axis, so we record them with a “+” sign. The equation ends with “= 0” because only then can the system be in static equilibrium.
The second equation we write will be the sum of the projections of forces on the y-axis:
∑F_{iy} =V_B-q⋅3L-2qL+V_D=0 → V_B-5qL+V_D=0→(2)In the equation, we include the reactions VB and VD, recorded with a “+” sign (as their directions align with the y-axis). The concentrated force P1 is recorded with a “-” sign.
Also, a concentrated force arising from the continuous load is included. This force is written at the center of gravity of the continuous load, and its value is the product of the load intensity and the length over which it acts – in this case, q·3L. This force is directed according to the continuous load, so we record it with a “-” sign in the equation.
The final equation will be the sum of moments relative to any point in the system.
Interestingly, this point doesn’t even need to be on the beam, but a good practice is to choose it at a location that simplifies the calculations – for example, at the location of a support where reaction forces occur. Such a point reduces the number of unknowns in the equilibrium equation. Let’s choose point B:
∑M_{iB} =-2qL^2-q⋅3L⋅0.5L-2qL⋅2L+V_D⋅3L=0 → 3V_D L-7.5qL^2=0→(3)In the moment equations, we include all loads and reactions that generate a moment relative to the selected point. Remember that
a moment is a force acting at some distance.
If the force’s direction passes through the point, it doesn’t generate a moment relative to that point.
Therefore, reactions HB and VB, won’t appear in the equation, but reaction VD, will, as its direction doesn’t pass through point B, and the distance between this point and the direction of VB is 3L. This force rotates the beam around point B counterclockwise, so we record it with a “+” sign (this would be the moment direction determined by the right-hand rule relative to the z-axis!!!).
Now for the moments from the loads. Force P2 doesn’t generate a moment because its direction intersects point B. Force P1 generates a moment with a moment arm of 2L , rotating clockwise, so we record it with a “-” sign. Then we have the concentrated moment M – which is already a moment by itself, so it doesn’t need a moment arm, we just add it directly to the equation, noting the sign – in this case, “-“. And here’s an important note:
a common mistake is multiplying a concentrated moment by a moment arm in the moment equilibrium equation!!
The last point is the moment generated by the continuous load.
We already mentioned that this moment can be recorded as a concentrated force acting at the center of gravity of the load. Therefore, we write the moment as we would for any other force – taking into account the value of the concentrated force (q·3L). In this case, the distance of this substitute force from point B is 0.5L.
And here’s another common mistake – students often assume that the moment arm of the continuous load is always half the length of the load’s action! From analyzing this example, we can see that this isn’t always true, although sometimes such a case may happen!
Calculating support reactions
Now that we have correctly written the equilibrium equations, we can treat them purely mathematically – as a system of 3 equations with 3 unknowns. In this situation, from equation (1), we calculate reaction HBH_BHB, from equation (3), reaction VD and substituting into equation (2), we calculate reaction VB
(1)→H_B=-qL, (3)→V_D=2.5qL,(2)→V_B=2.5qL
In this way, we have “calculated” the beam – from system analysis, through static and kinematic analysis, to writing the equilibrium equations and calculating the support reactions.
In the next article, I will present the calculations for a hinged (Gerber) beam.
