Engineering Statics

Find the external reactions.

Begin by drawing a free-body diagram of the entire beam, simplified by replacing the distributed load \(w\) with an equivalent concentrated load at the centroid of the rectangle. The magnitude of the equivalent load \(W\) is equal to the “area” under the rectangular loading curve.

Then apply and simplify the equations of equilibrium to find the external reactions at \(A\) and \(B\text{.}\)

Cut the beam.

Cut the beam at the point of interest and separate the beam into two sections.

For this example cut is taken at the midpoint.

Notice that as the beam is cut in two, the distributed load \(w\) is cut as well.

Each of these distributed load halves will support equivalent point loads of \(wL/2\) acting through the centroid of each cut half.

At each cut, a shear force, a normal force, and a bending moment will be exposed, and these need to be included on the free-body diagram.

At this point, we don't know the actual directions of the internal loads, but we do know that they act in opposite directions. We will assume that they act in the positive sense as defined by the List 8.2.1.

The next steps will add the internal loads one at a time.

Add Normal Forces.

Axial forces are positive in tension and act in opposite directions on the two halves of the cut beam.

Add Shear Forces.

Positive shear forces act down when looking at the cut from the right, and up when looking at the cut from the left.

An alternate definition of positive shears is that the positive shears cause clockwise rotation. This definition is useful if you are dealing with a vertical column instead of a horizontal beam.

Add Bending Moment.

Bending moments are positive when the moment tends to bend the beam into a smiling U-shape. Negative moments bend the beam into a frowning shape.

For vertical columns, positive bending moments bend a beam into a C shape and negative into a backward C-shape.

Solve for the internal loads.

You may uses either FBD to find the internal loads using the techniques you have already learned. So, with a standard \(xy\) coordinate system, forces to the left or up are positive when summing forces and counterclockwise moments are positive when summing moments.

Determine the reactions.

Draw an FBD of the entire, un-cut beam and determine the reactions.

Notice that only the applied loads and support reactions are included on this un-cut beam FBD. The internal loads are only exposed and shown on a FBD after the beam is cut.

Use this free-body diagram and the equations of equilibrium to determine the external reaction forces.

Section the beam.

Take a cut at the point of interest and draw a FBD of either or both parts.

The free-body diagrams of both portions have been drawn with the internal loads indicated in the positive direction defined by the standard sign convention.

The axial force is shown in tension on both parts. This force has been named \(\vec{N}\) so its name doesn't conflict with the forces at point \(A\text{.}\)

The shear force \(\vec{V}\) is positive when the shear is down on the right face of the cut and up on the left face.

The bending moment \(\vec{M}\) is positive if the bending direction would tend to bend the beam into a concave upward curve.

Always assume that the unknown internal loads act in the positive direction as defined by the standard sign convention.

Solve for the internal loads.

Solve for the three unknown internal loads.

Once you have found the reactions and drawn a free-body diagram of the simpler portion with the normal force, shear force, and bending moment assumed positive, you then solve for the unknown values and signs just like any other equilibrium problem.