5.2. Determining the support reactions

Let us consider some examples.

Example 1. Determine the support reactions of the cantilever beam (Fig. 5.3 a).

Solving. Represent the reaction of the patching up in the form of two forces and directed as indicated in Fig. 5.3 a and the reaction moment .

a) b)

Fig. 5.3.

Compose the beam equilibrium equation.

1. Equal the projection sum of all forces on the axis Z acting on the beam to zero: . We obtain . The horizontal component is equal to zero without any horizontal load.

2. The same is for The resultant force replaces the uniform load q as applied in the middle of the region

from where we have

.

The vertical component reaction of the cantilever beam is equal to the force sum applied to the beam.

3. Compose the third equilibrium equation. Equal the moment sum of all forces about the some point to zero, for example, about the point A:

from which

The minus sign indicates that the accepted previously direction of the reaction moment has to be changed in the opposite direction. So, the reaction moment is equal to the moments sum of the external forces about the patching up.

Example 2. Determine the support reaction of the simple beam (Fig. 5.3 b).

Solving. Since the horizontal load is lacking, then Az = 0.

1. .

2.

.

5.3. Determining the internal stresses in bending

Two internal stresses (internal force factors) – the bending moment M and the shearing force Q arise in the plane ordinary bending at the beam cross sections. To determine them the method of sections is applied. Mentally cut the beam, for example, a distance Z from the left support (Fig. 5.4 a). Remove one of the beam portion, for example, the right one, and consider the equilibrium of the left portion. Replace the interaction of the beam portions by the internal stresses: the bending moment M and the shearing force Q (Fig. 5.4 b).

To determine M and Q two equilibrium equations are used:

1.

2.

a)

b)

Fig. 5.4.

Hereby:

1. The shearing force Q at the beam cross section is equal to the algebraic sum of all external force projection on the section plane acting to one side of the section.

2. The bending moment M at the beam cross-section is equal to the algebraic sum (calculated about the section centroid) of the external forces acting to one side of the given section.

5.4. The sign rule for the bending moments and the shearing forces

The shearing force at the beam section m – n (Fig. 5.5 a) is considered as positive, if the resultant of the external forces to the left from the section is directed upward and to the right from it – downward and in the opposite direction is negative (Fig. 5.5 b).

b)

a)

Fig. 5.5.

The bending moment at the beam section, for example at the section m–n (Fig. 5.6 a) is considered as positive if the resultant moment of the external forces to the left from the section is directed clockwise and to the right from it - counterclockwise, otherwise it is negative (Fig. 5.6 b). The moments represented in Fig. 5.6 a bend the beam so that it is concave upward and the moments represented in Fig. 5.6 b bend the beam so that it is concave downward. It can be easily checked by bending a ruler.

a)

b)

Fig. 5.6.

From this it follows that a more convenient sign rule of the bending moment is to be remembered. The bending moment is considered positive if it bends the beam so that it is concave upward.