Beam Element with Distributed Load

Example of Loading Function "q" equaling
the Slope of the Shear Diagram

Example of Change in the Shear Force

For a distributed load, q, the forces can be summed in the y direction to give,

     ΣF_y = V - (V + ΔV) - q Δx = 0

This can be rearrange to give,

     ΔV/Δx = -q

If the limit of both sides is taken as Δx goes to zero, then the relationship between shear force and distributed loads becomes

     dV/dx = -q

This can also be written in integral form as

Graphical examples of both these equations are shown at the left.

Beam Element with Distributed Load

Example of Shear Function "V" equaling
the Slope of the Moment Diagram

Example of Change in the Bending Moment

Similar to the shear force in the previous paragraphs, the moments can be summed for a beam element with a distributed load of "q", giving,

     ΣM = -M + (M+ ΔM)
                  - (V + ΔV)Δx - q Δx (Δx/2) = 0

This equation can be rearranged to give,

     ΔM/Δx = V + ΔV + q (Δx/2)

Using the previous relationship for ΔV gives

     ΔM/Δx = V - q Δx + q (Δx/2)
                = V - q (Δx/2)

Now, take the limit of both sides as Δx goes to zero. Last term vanishes and a direct relationship be between bending moment and shear force is identified as,

     dM/dx = V

This can also be written as an integral relationship,

Two examples using the above equations are shown at the left.

Beam Element with Point Load

Effect of a Point Load

For a beam element subjected to a point load, the forces can be summed in the vertical to give,

     ΣF_y = V - (V + ΔV) - F = 0

     ΔV = -F

This represents that point loading will cause a jump in the shear diagram equal to the point load. The moment will not jump with a point load.

Beam Element with Point Moment

Effect of a Point Moment

For a beam element subjected to a point moment as shown, the shear force at that point is unaffected; but if the moments are summed about the left edge, then

     ΣM = ΔM - (V + ΔV)Δx - T = 0

If the limit as Δx goes to zero is taken, then it becomes

     ΔM = -T

For beams with several different loads, it is often easier to solve for the shear and moment diagrams for the individual loads and then combine them to find the total shear and moment diagrams. This is known as the method, or principle, of superposition.

For the cantilevered beam shown, shear and moment diagrams can be constructed for the two point loads and for the distributed load, separately; they can then be combined to form the full shear and moment diagrams.