In the previous two sections, methods where presented to construct shear and moment diagrams. However, there are some important relationships between loading, shear and moment which can aid in the construction of the diagrams.
Beam Element with Distributed Load
Example of Loading Function "q" equaling the Slope of the Shear Diagram
Example of Change in the Shear Force
Relationship between Shear Force and Distributed Loads
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
Relationship between Bending Moment and Shear Force
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
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
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
Principle of Superposition
Principle of Superposition
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.