From a design point of view, it is often necessary to know the maximum and minimum internal loads in a structure and where they are located.

If the shear force and bending moment are calculated and graphed, then the maximum and minimum of each are easily identified and located.

For the simply-supported beam with only one point force load, the reactions can be found as

     A_y = F/3       B_y = 2F/3

The section of the beam to the left of the applied load will have an expression for the shear force and bending moment that will differ from the section to the right of the applied load. Therefore, the sections must be evaluated separately. In other words, there is not a single function that will model shear or moment from A to B.

Section 1 Cut and Analysis

Cutting the beam at an arbitrary location x in section 1 and forming free-body diagrams for the piece to the left and to the right of the cut will result in the diagrams shown.

Either side of the cut beam can be used to solve for the internal shear force and internal bending moment. Generally, the simpler side is used. In this case, if the left side is used, internal shear load is

     ΣF_y = 0
     A_y - V₁ = 0
     V₁ = A_y = F/3

Summing the moments about the right edge give,

     ΣM_cut = 0
     M₁ - x (F/3) = 0
     M₁ = xF/3

For any location x between x = 0 and x = 2/3 L, the shear and moment are given by

     V₁ = F/3          0 ≤ x ≤ 2/3 L                 (1)

     M₁ = xF/3         0 ≤ x ≤ 2/3 L                (2)

Section 2 Cut and Analysis

This process can be repeated for for the section to the right of the load. The free-body diagrams for the two pieces are shown at the left. For this section, it is easier to solve for the piece to the right of the cut.

     ΣF_y = 0
     2F/3 + V₂ = 0
     V₂ = -2F/3

The moments about the left edge can be used to determine the bending moment.

     ΣM_cut = 0
     (2F/3) (L - x) - M₂ = 0
     M₂ = 2F(L - x)/3

Thus, for any location x between x = 2/3 L and x = L, the shear and moment are given by

     V₂ = -2F/3               2/3 L ≤ x ≤ L               (3)

     M₂ = 2F(L - x)/3       2/3 L ≤ x ≤ L               (4)

If the shear equations (Eqs. 1 and 3) are graphed on one single axis and the moment equations (Eqs. 2 and 4) on another single axis, the shear and moment diagrams are obtained as shown on the left.

The location for maximum and minimum shear force and bending moment are easily found and evaluated.

The number of sections required for complete shear and moment diagrams depends on the beam configuration and loading. The beam must be sectioned at each point that loading or support conditions change. A cut must be made in each section of the beam so that the shear and moment can be evaluated over that section.

Multiple sections are required since shear and moment is not a continuous function over changing supports and loads.

For complex beams with more than a couple loads, determining moment and shear diagrams is very difficult. Thus, most structural engineers will use a beam analysis tool to calculate moments and shear. A basic one is avaible for mobile devices. The tool can generate the moment and shear diagram, and give support reactions.

For Android mobile devices, "Beam HPC" calculator, can be downloaded for free at Google Play.

For iOS (Apple) mobile devices, "Beam HPC" calculator, can be downloaded for free at iTunes.