In Subsection 3.3.3 you were introduced to axial internal loadings, loadings which were either tension or compression, or possibly zero. This section will explain two other internal loads found in two-dimensional systems, the internal shear and internal bending moment.

Internal loads are present at every point within a rigid body, but they always occur in equal-and-opposite pairs which cancel each other out, so they’re not obvious. They’re there however, and when an object is cut (in your imagination) into two parts the internal loads become visible and can be determined.

You are familiar with straight, two-force members which only exist in equilibrium if equal and opposite forces act on either end. Now imagine that we cut the member at some point along its length. To maintain equilibrium forces must exist at the cut, equal and opposite to the external forces. These forces are internal forces.

Now let’s examine the two-force member shown in Figure 8.1.2. This time, the member is L shaped, not straight, but the external forces must still share the same line of action to maintain equilibrium. If you cut across the object, you will obtain two rigid bodies which must also be in equilibrium. However, adding an equal and opposite horizontal force at the cut won't produce static equilibrium because the two forces form a couple which causes the piece to rotate. This means that something is missing!

Two-dimensional rigid bodies have three degrees-of-freedom and require three equilibrium equations to satisfy for static equilibrium.

\begin{align*} \Sigma F_x \amp = 0 \amp \amp \text{ prevents translation in the} x \text{ direction,} \\ \Sigma F_y \amp = 0 \amp \amp \text{ prevents translation in the} y \text{ direction, and} \\ \Sigma M \amp = 0 \amp\amp \text{ prevents rotation.} \end{align*}

Assuming the material is rigid, the connection between the two halves must resist both translation and rotation, so we can model this connection as a fixed support and replace the removed half of the link with a force reaction and a couple-moment reaction as shown in the free-body diagrams of Figure 8.1.3. This internal loading is actually a simplification of a more complex loading distributed across the section plane. The couple $$\vec{M}$$ represents the net rotational effect of the force system on the surface of the cut.

The horizontal force can also be resolved into orthogonal components parallel and perpendicular to the cut. These components have special names in the context of internal loads.

The internal force component perpendicular to the cut is called the normal force. This is the same internal tension or compression force that we assumed to be the only significant internal load for trusses. If the object has an axis, and the cut is perpendicular to it, the the normal force may also be properly called an axial force.

The internal force component parallel to the cut is called the shear force. The word shear refers to the shearing between that occurs between adjacent planes due to this force. You can get a feel for shearing adjacent planes by sliding two pieces of paper together.

The internal couple-moment is called the bending moment because it tends to bend the material by rotating the cut surface.

The shear force is often simply referred to as shear, and the bending moment as moment; together with the normal or axial force the three are referred to as the “internal loading”. The symbol $$\vec{V}$$ is commonly chosen for the shear force, and $$\vec{A}\text{,}$$ $$\vec{P}$$ or $$\vec{N}$$ for the normal force and $$\vec{M}$$ for the bending moment.

The controlling design parameter for most engineering systems is deformation. Thankfully, due to a property called elasticity, most materials will bend, stretch, and compress, long before they ultimately break. For example, when designing the floor in a new building, the floor is often limited to deflecting less than the length of the span in inches, divided by 360. Any more deformation than this would be considered disconcerting to the building residents and also start damaging surface materials like drywall. For example, for a $$\ft{20}$$ span, the deflection would need to be less than

\begin{equation*} \delta = \dfrac{\ft{20} \cdot \dfrac{\inch{12}}{\ft{1}}}{360}=\inch{0.667}\text{.} \end{equation*}

To meet this deformation limit, we need to consider the magnitude and location of applied loads, the size and shape of the floor beams, and the material the floor beams are made from. As deflection is an internal property of the flooring materials, the first step is to determine the internal loadings that arise from the externally applied loads, using the methods of this chapter.