2: Determinate Beams and Frames

2.3: Free Body Diagrams

The construction of a proper free body diagram (FBD) is an essential first step in a structural analysis. A free body diagram is a sketch of the outline of a structure, roughly to scale and isolated from its surroundings. It shows key features of the geometry, all applied forces (loads), and all relevant reactive forces at the points of contact between the portion shown and its surroundings.

To determine the character of the reactive forces (i.e. their types and directions), you determine the displacement contraints, and show the force corresponding to each.

Beam Model and FBD

Fig 3-1: Beam model and FBD of complete structure

Fig. 3-1 shows the structural model of a beam in the top portion and a free body diagram of the entire beam in the bottom portion. Note that the figure shows a beam that is completely fixed at the left end, has a roller support 2m from the right end, and an internal hinge at point b, 6m to the left of the roller support. Two applied loads, a uniformly distributed load of 28kN/m over the entire beam and a concentrated load of 65kN are shown.

At two locations, a and d, the beam is in contact with its surroundings; the surroundings are not shown. At these locations, then, we must determine the displacement constraints, and show the corresponding forces.

At point a, the symbol on the model shows that that point is constrained against rotation, against vertical displacement and against horizontal displacement. Therefore, at point a on the FBD we show a moment corresponding to the rotational constraint, a vertical force corresponding to the vertical displacement constraint, and a horizontal force corresponding to the horizontal displacement constraint. These are unknown forces, and so we label them with convenient symbols: $M_a$, $V_a$, and $H_a$, respectively.

At point d, there is only the constraint against vertical displacement, and so we show only a vertical force, $V_d$.

The location of the internal hinge is not particularly relevant on this FBD, so it is not shown.

Free body diagrams should not show uniformly distributed loads replaced by their statically equivalent concentrated loads. That does not make things simpler, but it does increase chances for confusion and error later when you draw shear force and bending moment diagrams.
After a little experience, you can tell by 'inspection' that the horizontal force at a, $H_a$, is 0. However, you use an equation of equilibrium to know that, and it is important not to fool yourself into thinking you have fewer unknowns than you do. All forces that could develop under other loading conditions should be shown, even if you can trivially determine they are zero under the given loading condition.

FBD of portion b-c-d

Fig 3-2: Beam - FBD of portion b-c-d

We will see later that it is often necessary to draw a FBD of a portion of the structure, particularly when we can ‘break’, or ‘cut’, the structure at an internal release condition such as the hinge at b.

Fig. 3-2 shows a FBD of portion b-c-d of one portion of the structure. Of course, we show only those applied loads that act on the portion of the structure that we show. The 65kN concentrated force is as before, but the length of the distributed load has been reduced to 8m to match the length of the portion.

At point b, there are relative displacement constraints that require that both the vertical and horizontal displacements of the beam on either side of the hinge must each be equal. Therefore we must show forces corresponding to that portion of the beam that is in contact but not shown. These are internal forces.

There is no constraint against relative rotation of the beam to either side of the hinge – after all, that is what a hinge allows – and so we do not show an internal moment at b.

Recognizing points in a structure at which internal forces are known to be zero is crucial to analyzing many complex structures.

FBD of portion a-b

Fig 3-3: Beam - FBD of portion a-b

Occasionally it is important to also draw the FBD of adjacent portions of a structure. Fig. 3-3 shows one such FBD for the portion a-b which is immediately adjacent to the FBD portion shown in Fig. 3-2. It is vital to realize that the forces shown at b are internal forces and that internal forces always occur in equal and opposite pairs. A single FBD of a structure cut at a point only shows one of the forces at each pair. The other FBD must then show the other forces of the pair.

Once you have established the assumed direction of forces on one FBD, you must show those forces as equal and opposite on the adjacent FBD. You establish equality by using the same label, and you establish ‘oppositeness’ by showning the forces in the opposite direction.

The FBD of Fig. 1-3 shows the forces at b, $H_b$ and $V_b$ being equal and opposite to those shown in Fig. 3-2.

Unknown internal forces shown on one FBD of a portion of a structure must be shown as equal and opposite at the same point on the adjacent FBD.

References