2: Forces in Statically Determinate Beams and Frames

2.8: Shear, Moment and Normal Force Diagrams

2.8.1: Relationship of Transverse Loads to Shears and Moments

Fig. 2.8-1: Loads, Shears and Moments

Fig. 2.8-1 shows a beam with a varying distributed load, acting upwards on the beam. We are interested in developing a relationship between the external loads and the internal forces.

The bottom portion of Fig. 2.8-1 shows a free body diagram of an element of the beam of length $dx$. All forces are shown in a direction that is positive by the beam sign convention:

lateral loads are +ive acting in the upward (+ive y) direction.
shear forces are +ive if they tend to rotate the element clockwise.
bending moments are +ive if they cause compression in the top fibers of the beam.

We now write an equilibrium equation for the sum of forces in the vertical direction.

\[\sum F_y = 0~~~~~~~(+\uparrow)\] \[V - (V + dV) + w dx + (dw)(dx){1\over 2} = 0\]

Neglecting the second-order term and rearranging, we get:

\[dV = w dx\]

That can be interpreted two different ways.

As $dV = w dx$, it says that the change in shear between any two points is equal to the area under the distributed load diagram between those two points.
As ${dV/dx} = w$, it says that the slope of the shear diagram at any point is equal to the intensity of the distributed load at that point.

Now, summing moments about point o on the right face, we have:

\[\sum M_o = 0~~~~~~(+{\mathrm{CCW}})\] \[-M + (M+dM) - V dx - w(dx){dx\over2} - (dw)(dx){1\over2}{dx\over3} = 0\]

Neglecting the second- and third-order terms and rearranging gives us:

\[dM = V dx\]

Again, that can be interpreted two different ways:

As $dM = V dx$, it says that the change in moment between any two points is equal to the area under the shear diagram between those two points.
As ${dM/dx} = V$, it says that the slope of the moment diagram at any point is equal to the intensity of the shear at that point.

Because internal forces always occur in equal and opposite pairs, when we plot shear and moment diagrams we must associate a sign with a *pair* of forces. Thus, beam sign convention.

Because the original differential equation was developed assuming a distributed force, the resulting relationships work for the case of distributed loads.

The relationships also work at locations of concentrated loads, recognizing that the load intensity is infinite, and the shear change across a concentrated force is equal to the value of the force.

The relationships do not work directly for concentrated moments. Shear does not change acroos a concentrated moment, and bending moment changes by the value of the applied moment.

2.8.2: Relationship of Longitudinal Loads to Normal Force

Figure 2.8-2: Longitudinal loads and normal forces.

All loads that action on beams and members in frames can be resolved into two components, one perpendicular to the axis of the member and one parallel to the axis. The perpendicular components are the transverse loads, and they affect shear forces and bending moments as outlined above. The parallel components are the longitudinal loads, $w_p$, and they have direct effect only on the normal forces. We also assume that the beam depth has no significance and that all longitudinal forces are coincident with the longitudinal axis.

By inspection from Fig. 2.8-2: we can derive the relationship:

The change in normal force between two points along the axis of a beam is equal to the total longitudinal load between those two points.