Constraint reactions: statics exercises

End of September 2026: in Year 3 of the Italian technical institutes (technology sector), on the Mechanics, Mechatronics and Energy track, the subject Mechanics, Machines and Energy almost always opens with statics (the same block appears in Design, Construction and Systems on the CAT track, construction and land management). It is the first topic of the year, before kinematics and dynamics, and the first test typically falls four to seven weeks from now — so between mid-October and mid-November. It is worth arriving with a solid method: the statics of beams is also the direct foundation of structural mechanics (Scienza delle Costruzioni), one of the most feared exams of the first years of engineering.

Constraint reactions: what they are and how to calculate them

A beam resting on a support does not “stay still by magic”: the support exerts a force (or a moment) that prevents movement in one or more directions. That force is called a constraint reaction, and it is the unknown that almost every statics exercise asks you to find. The method for calculating it is always the same, in three steps: work out how many supports the structure has and what they prevent, draw all the forces acting (the free-body diagram), and write the three equations of statics.

Review: force and moment

First a quick review, because everything else is built on it: the moment of a force about a point (the pole) is the product of the magnitude of the force and the perpendicular distance between the line of action of the force and the pole:

M = F \cdot b

where $b$ is the lever arm. By convention, a moment that tends to turn anticlockwise is positive and one that tends to turn clockwise is negative (or vice versa, as long as you are consistent throughout the exercise). A couple is a pair of equal and opposite forces that are not aligned: its moment does not depend on the pole chosen, but this is the only case where that is true — for a single force the moment changes with the pole, and this is exactly what we will use shortly.

Supports in the plane

In the statics of beams, in a planar problem, supports are classified by how many reactions they introduce:

Support	Symbol	Reactions	What it prevents
Roller (simple support)	triangle on wheels	1	Only translation perpendicular to the support
Pin (hinge)	solid triangle	2	Horizontal and vertical translation, not rotation
Fixed support (clamp)	solid hatching	2 + 1 moment	Horizontal and vertical translation, and rotation

The roller gives a single reaction because it leaves the beam free to slide along the support and to rotate; the pin gives two because it blocks movement in both directions but lets the beam rotate; the fixed support blocks everything, so besides the two force reactions it also needs a reaction moment that prevents rotation.

Degrees of freedom and static determinacy

A beam in the plane has 3 degrees of freedom (two translations and one rotation). For it to be in equilibrium without being either unstable or over-constrained, the total number of constraint reactions must be exactly 3:

fewer than 3 reactions $\Rightarrow$ unstable structure (it moves; it is not a statics problem solvable with the equilibrium equations alone);
exactly 3 reactions, arranged so that they are not all parallel or all concurrent at one point $\Rightarrow$ statically determinate structure (the case you solve with the three equilibrium equations, the one in this post);
more than 3 reactions $\Rightarrow$ statically indeterminate structure (the equilibrium equations are no longer enough: you need structural mechanics to add the compatibility equations — out of reach in Year 3).

A beam supported by a pin and a roller has $2 + 1 = 3$ reactions: statically determinate. A fixed cantilever has $2 + 1 = 3$ reactions: also statically determinate, with a single support.

The free-body diagram

Before writing any equation, draw the beam on its own, replacing each support with the reactions it introduces, and mark all the external loads (weights, applied forces, distributed loads) with their direction and point of application. This is the step most people skip — and it is indeed where most mistakes begin.

The equilibrium equations of statics

With the free-body diagram ready, equilibrium is written with three equations:

\sum F_{x} = 0 \sum F_{y} = 0 \sum M_{P} = 0

where $P$ is a pole chosen by you, not necessarily a point on the beam. The choice of pole is the “clever” part of the method: if you choose the pole at a support, the moment of that reaction about itself is zero, and the unknown disappears from the equation. With two supports (a pin and a roller, for example), choosing the pole at one of them gives an equation with a single unknown — the reaction of the other support — which you solve straight away without systems.

Distributed loads

A uniform distributed load of intensity $q$ (in $N/m$ ) over a stretch of beam of length $L$ is replaced, for the purpose of calculating the reactions, by its resultant: a concentrated force of magnitude $q \cdot L$ , applied at the midpoint of the loaded stretch. A load at an angle, on the other hand, must be resolved into its horizontal and vertical components before entering the equations — exactly as for vectors.

The typical mistake

The most frequent: forgetting the reaction moment of the fixed support and treating it as if it were a pin with only two reactions — the beam “looks” balanced in the force equations, but the rotation is not balanced, and the mistake only shows up at the end, if it shows up at all. The second, almost as common: choosing a pole that eliminates no unknowns (for example the centre of mass of the beam, when there are two unknown supports), which leads to a system of two equations in two unknowns instead of two independent equations already solved one at a time. The third: losing the sign of a moment halfway through the calculation — it is best to write “anticlockwise positive” (or the opposite) explicitly at the start of the exercise and never change it.

Three worked exercises

Exercise 1 — beam on a pin and a roller, two concentrated loads

Beam of length $L = 6 m$ , pin at A (on the left), roller at B (on the right). Two downward vertical loads: $P_{1} = 1000 N$ at $2 m$ from A, $P_{2} = 1500 N$ at $4 m$ from A.

I choose the pole at A, so the horizontal and vertical reactions of the pin do not appear in the moment:

\sum M_{A} = 0 : R_{B} \cdot 6 - P_{1} \cdot 2 - P_{2} \cdot 4 = 0

R_{B} \cdot 6 = 1000 \cdot 2 + 1500 \cdot 4 = 2000 + 6000 = 8000 N m

R_{B} = \frac{8000}{6} = 1333.3 N

Now vertical equilibrium (there are no horizontal forces, so the horizontal component at the pin is zero):

\sum F_{y} = 0 : R_{A} + R_{B} = P_{1} + P_{2} = 2500 N

R_{A} = 2500 - 1333.3 = 1166.7 N

Check with a different pole, B:

\sum M_{B} = 0 : R_{A} \cdot 6 - P_{1} \cdot 4 - P_{2} \cdot 2 = 0

R_{A} = \frac{1000 \cdot 4 + 1500 \cdot 2}{6} = \frac{4000 + 3000}{6} = \frac{7000}{6} = 1166.7 N

It matches: the reactions are correct.

Exercise 2 — same beam, uniform distributed load

The same beam, pin and roller, $L = 6 m$ , but now with a uniform distributed load $q = 500 N/m$ over the whole length. I replace the load with its resultant: $R = q \cdot L = 500 \cdot 6 = 3000 N$ , applied at the midpoint, $3 m$ from A.

\sum M_{A} = 0 : R_{B} \cdot 6 - 3000 \cdot 3 = 0 \Rightarrow R_{B} = \frac{9000}{6} = 1500 N

\sum F_{y} = 0 : R_{A} = 3000 - 1500 = 1500 N

Check with the pole at B: $R_{A} \cdot 6 - 3000 \cdot 3 = 0 \Rightarrow R_{A} = 1500 N$ . Consistent — and it is also the result you expect from symmetry, since the load is centred: the two reactions are equal.

Exercise 3 — fixed cantilever, inclined concentrated load

Cantilever beam (fixed only at A, free at B), length $L = 2 m$ , horizontal. At the free end B a force $P = 1000 N$ acts, inclined at $3 0^{\circ}$ downwards relative to the axis of the beam.

First I resolve the force into its two components, along the beam and perpendicular to it:

P_{x} = P cos 3 0^{\circ} = 1000 \cdot 0.866 = 866.0 N

P_{y} = P sin 3 0^{\circ} = 1000 \cdot 0.5 = 500.0 N

The fixed support reacts with two forces and a moment. Horizontal and vertical equilibrium:

\sum F_{x} = 0 : H_{A} = P_{x} = 866.0 N

\sum F_{y} = 0 : V_{A} = P_{y} = 500.0 N

For the moment, I choose the pole at A (the fixed support): the component $P_{x}$ has zero lever arm about A because it acts along the axis of the beam, so it does not contribute to the moment. Only $P_{y}$ contributes, with lever arm $L$ :

\sum M_{A} = 0 : M_{A} - P_{y} \cdot L = 0 \Rightarrow M_{A} = 500 \cdot 2 = 1000 N m

Check with the pole at B, the free end: there $P$ has zero lever arm, leaving the vertical reaction $V_{A}$ with lever arm $L$ and the fixed-support moment. $M_{A} - V_{A} \cdot L = 1000 - 500 \cdot 2 = 0$ : it works. If you had got a non-zero number, the mistake would be in the resolution or in the sign of $M_{A}$ .

On this specific topic — the beams of the technical track — there is not yet a dedicated practice test on the site. The closest one in terms of prerequisites, moment and equilibrium of the rigid body, is at /simulazioni/verifiche-liceo/prima/corpo-rigido/: useful for reviewing the definition of moment and the equilibrium conditions before tackling constrained beams.

If in class you started from vectors and have not yet consolidated them, you will find them covered again in Vectors and forces in Year 3: resolving without trigonometry and in Resolving forces: vectors in Year 3 without trigonometry. The next step of mechanics on the technical track, after statics, is dynamics: I talk about it in Dynamics and the inclined plane in Year 2. And if this road leads you to engineering, the natural sequel to statics at university is in Analytical mechanics and dynamics: the Lagrange exam without panic. The full picture of the physics subjects is at /materie/fisica/.

If you have a specific statics exercise that does not add up, you can ask for a free 30-minute introductory call on /contatti/.

Constraint reactions: worked statics exercises