In the previous article, I presented the procedure for calculating support reactions in the case of straight (continuous) beams. If you haven’t read it yet, I encourage you to catch up on the material, and then proceed with calculating beams online using the EquiBeam application.
Now it’s time to address the topic of calculations for a hinged beam:
Calculating Beams – Detaching the System from Supports
Gerber Beam Static Determinacy Condition
Calculating Gerber beam
We will now focus on the calculations of a hinged beam, as shown in the static diagram below:
These types of beams are called Gerber beams (Gerber beams) due to the presence of a hinge at point B allowing the relative rotation of the two parts of the beam.
The left part of the beam is supported by a fixed support at point A (a clamped support – restricting 3 degrees of freedom). The right part of the beam is supported by a roller support.
This beam is subjected to
- a concentrated moment K = 2qL² on the left side of point B (remember – a moment cannot be applied directly at the hinge; it can only be applied to one side or the other),
- a vertical concentrated force P₁ = qL at point C and
- a continuous load of intensity q between points B and D.
Calculating beams – detaching the system from supports
As in the first example, the first step in beam calculations is detaching the system from its supports – replacing the supports with their support reactions.
The diagram of the beam with reactions is shown below:.
Notice that in this example, the beam is not divided into two parts, and the hinge is replaced with reactions – there is also a second approach to the calculations, which you will see in EquiBeam if you select the option to calculate hinge forces.
Either way – in the clamped support, we can record 3 support reactions because it restricts 3 degrees of freedom: in the direction of the x-axis (HA), the y-axis (VA), and rotation around the z-axis creating a clamping moment (MA). In the movable support, we record one reaction in the direction of the y-axis (VD).
Z tego powodu w podporze stałej możemy zapisać dwie reakcje podporowe – na kierunku osi x (HB) i osi y (VB). Tak naprawdę w tej podporze występuje jedna siła, ale jej kierunek jest dowolny, uzależniony od obciążeń, więc rozkładamy ją na dwie składowe. W podporze ruchomej zapisujemy jedną reakcję na kierunku osi y (VD).
Gerber beam static determinacy condition
The static determinacy condition for beams with hinges can be expressed as (since each hinge restricts 2 relative degrees of freedom):
3t-r-2p=0
Static determinacy condition of the system:
3t-r-2p=0→3⋅2-4-2⋅1=0-ok!
The static determinacy condition of the system is satisfied – the beam CAN BE statically determinate.
We perform kinematic analysis
Our beam has 6 degrees of freedom because it consists of two rigid parts.
All three degrees of freedom of the left part of the beam are restricted by the clamped support. This means the left part is immediately geometrically stable.
The right side of the beam is connected to the left side by a hinge, which restricts 2 out of 3 degrees of freedom. In such a situation, the right side can freely rotate around the hinge point.
This rotation is eliminated by the movable support, whose direction does not pass through the rotation point of the hinge. Ultimately, the entire beam is geometrically stable and statically determinate.
Static equilibrium equations
We now need to write the equilibrium equations.
In the case of Gerber beams, in addition to the three basic equilibrium equations, we also have additional equations relative to the hinge (∑ MiBl i ∑ MiBp). This is due to the fact that the hinge, by allowing the free rotation of the beam, does not transfer a bending moment.
As a result, we have 5 equations available to determine 4 unknown reactions. Thanks to this, one equation (usually the moment equation for the entire beam) can be omitted, simplifying the calculations
Let’s write the global equations, starting with the sum of forces along the x-axis
∑F_{ix} =H_A=0Apart from the HA reaction, there are no horizontal forces in the beam – meaning the value of this reaction is 0. The equation for forces along the y-axis:
∑F_{iy} =V_A-q⋅3L+qL+V_D=0 → V_A-2qL+V_D=0→(1)We have two vertical reactions here, as well as force components from the continuous load q and the concentrated force P. The signs in the equation, of course, depend on the direction of the forces relative to the assumed y-axis direction
∑M_{iA }=M_A+2qL^2-q⋅3L⋅2.5L+qL⋅2L+V_D⋅4L=0 → M_A+4V_D L-3.5qL^2=0In the moment equation relative to point A, there are no surprises either: the VA and HA reactions are not present (their directions pass through point A) – we have concentrated moments from the reactions (MA) and loads (K), moments from concentrated forces – VD at a distance of 4L and force P at a distance of 2L, as well as the moment from the continuous load, where the distance of the concentrated force from point A is 2.5L.
The last equations to write relate to a specific side of the hinge. Here, it is important to remember that when writing the equation on one side, we completely disregard what is happening on the other side – and, of course, vice versa, as shown in the diagram below. The base (or pole) of the moment is, of course, the hinge:
∑M_{iBl} =M_A+2qL^2-V_A⋅L=0 →(2)∑M_{iBp} =-q⋅3L⋅1.5L+qL⋅L+V_D⋅3L=0 → 3V_D L-3.5qL^2=0→(3)Support reaction calculations
Finally, using the equilibrium equations, we calculate the values of the unknown support reactions:
- from equation (3), we calculate the reaction VD,
- substitute it into equation (1) to calculate VA ,
- substitute into equation (2) to calculate the reaction MA.
H_B=0,(3)→V_D=1 \frac{1}{6} qL,(1)→V_A=\frac{5}{6} qL,(2)→M_A=-1 \frac{1}{6} qL^2In the case of complex Gerber beams, it is worth performing a verification – for example, by writing the moment equation relative to another point in the system and checking whether the moment correctly cancels out.
If it does, the reactions have been calculated correctly. In this case, we can use the moment equation for the entire system, which was not used to solve the system of equations:
-1\frac{1}{6} qL^2+4⋅(1 \frac{1}{6} qL)⋅L-3.5qL^2=0-ok!With that, we conclude the second calculation example.
Summary
Kinematic analysis, static analysis, and the calculation of support reactions are the first steps in the static assessment of beams.
The examples discussed in the articles show that understanding the basic principles of statics and correctly applying the equilibrium equations allow for quick and efficient calculation of reactions, even in complex systems.
Go through the presented solutions. Solve your own examples in the EquiBeam app. You will have a solid foundation to tackle more advanced beam calculations.
