When you make a cut in an object, similar to a fixed reaction, we describe what is happening at that point using one horizontal force (called normal force), one vertical force (called shear force), and a bending moment.

6.1.1 Types of Internal Forces

There are 3 types of internal forces (& moments):

• normal force (N) – the horizontal force we calculated in trusses in the last chapter
• shear force (V) – the vertical force that changes based on the applied loads
• bending moment (M) – changes based on the applied loads and applied moments

Normal force is represented by ‘N’. Shear force, the vertical force is represented with ‘V’. Bending moment is ‘M’.  Normal and shear have units of N or lb and bending moment has units of Nm or ft-lb. The following table summarizes information on internal forces (and moments).

Note that for a vertical column, the normal force would be vertical. For this reason, the normal force is often called ‘axial’ as in: along the axis. The shear force for a column would be horizontal and is sometimes called ‘transverse’.

This is for a 2d analysis of the beam assuming there is negligible loading in the third dimension.

In 3 dimensions, there are:

• 1 normal force (N)
• 2 shear forces (V₁ & V₂), and
• 3 bending moments (M₁, M₂, & T – torsion).

6.1.2 Sign Convention

So that there is a standard within the industry, a sign convention is necessary so we agree on what is positive and what is negative. On the right for shear – up is positive. Notice that both of the following figures show the identical sign convention.

When you look at the beam as a whole (in the figure below), positive shear is right side down. When you cut into beam, for it to be in static equilibrium, the positive shear must then be up on the right to be equal and opposite of the overall motion.

6.1.3 Calculating the Internal Forces

To solve the internal forces at a certain point along the beam,

1. Find the external & reaction forces
2. Make a cut.
3. In a FBD of one side of the cut, add the internal forces (and moments) using the positive sign convention.
4. Use the equilibrium equations to solve for the unknown internal forces and moments.

Example: For the following distributed load, a) what are reaction forces? b) what are the internal forces at the midpoint b) between reaction forces?

1. Solve external forces:

[latex]\sum F_{X}=A_{x}=0[/latex]
[latex]\sum F_{y}=A_{y}+C-\omega L=0[/latex]
[latex]\sum M_{A}=-(\omega L)\left(\frac{L}{2}\right)+d_{A C} C=0[/latex]
$$C = \left(\frac{\omega L^2}{2d_{A C}}\right) = \frac{(100 \frac{lb}{ft} )*(7ft)^2}{2 * (4ft)} = 612.5 lb \text{ (+j direction)} $$
$$A_y = \omega*L- C = (100 \frac{lb}{ft})*(7 ft) – 612.5 lb = 87.5 lb \text{ (+j direction) }$$
$$\underline{A_x = 0 \qquad A_y = 87.5 \text{ (+j )} \qquad C = 612.5 lb \text{ (+j )} }$$

2. Make a cut at B.

3. In a FBD of one side of the cut, add the internal forces (and moments) using the positive sign convention.

4. Use the equilibrium equations to solve for the unknown internal forces and moments.

For just this portion, the force from intensity is: F_w = ( 100 lb/ft ) * ( 2 ft) = 200 lb and acts 1 ft from the left, so the moment due to intensity is: M_w = w * 2 ft * 1 ft = F_w * 1 ft = ( 100 lb/ft ) * ( 2 ft) * (1 ft)  =  200 ft-lb

$$\sum F_y = 87.5 lb – 200 lb – V = 0 \\ V = -112.5 lb \text{ (- indicates going up not down)} $$

$$ \sum M_A = – (w * 2 ft) * (1 ft) – V * (2 ft)  + M = 0 \\ M = (100 \frac{lb}{ft}) * 2 ft^2 + (-112.5 lb) * (2 ft) \\ M = 200 ft \cdot lb – 225 ft \cdot lb \\ M = -25 ft \cdot lb \text{ (- indicates going reverse direction)} $$

$$\underline{N = 0 \qquad V = -112.5 lb \text{ (+j )} \qquad M = -25 ft \cdot lb \text{ (clockwise)} }$$