Shearing force and bending moments

4. As stated in para. 3, for equilibrium of a beam, the forces to the left of any section such as X in Figure 36.5, must balance the forces to the right. Also the moment about X of the forces to the left must balance the moment about X of the forces to the right.

    Although for equilibrium the forces and moments cancel, the magnitude and nature of these forces and moments are important as they determine both the stresses at X and the beam curvature and deflection. The resultant force to the left of X and the resultant force to the right of X (forces or components of forces transverse to the beam), constitute a pair of forces tending to shear the beam at this section. Shearing force is defined as the force transverse to the beam at a given section tending to cause it to shear at that section.

    By convention, if the tendency is to shear as shown in Figure 36.6(a), the shearing force is regarded as positive, i.e. + F; if the tendency to shear is as shown in Figure 36.6(b), it is regarded as negative, i.e. − F.

The values of shearing force and bending moment will usually vary along a beam. Diagrams showing the shearing force and bending moment for all sections of a beam are called shearing force and bending moment diagrams respectively.

Shearing forces and shearing force diagrams are less important than bending moments, but can be very useful in giving pointers to the more important bending moment diagrams. For example, wherever the shearing force is zero, the bending moment will be a maximum or a minimum. The shearing force and bending moment diagrams for the beam shown in Figure 36.7 are obtained as follows:

It is first necessary to calculate the reactions at A and B. The beam is simply-supported at A and B which means that it rests on supports at these points giving vertical reactions. The general conditions for equilibrium require that the resultant moment about any point must be zero, and total upward force must equal total downward force. Therefore, taking moments about A, the moment of R_B must balance the moment of the load at C:

$\begin{array}{l}{R}_B\text{×8m=24}\text{kN×5m=120}\text{kNm}\\ R_B=\frac{\text{120}\text{kNm}}{\text{8m}}\text{=15}\text{kN,}\text{and}\\ R_A=\text{24}\text{kN-15}\text{kN=9}\text{kN}\text{.}\end{array}$

Immediately to the right of A the shearing force is due to R_A and is therefore 9 kN. As this force to the left of the section considered, is upwards, the shearing force is positive. The shearing force is the same for all points between A and C as no other forces come on the beam between these points.

When a point to the right of C is considered, the load at C as well as R_A must be considered, or alternatively, R_B on its own. The shearing force is 15 kN, either obtained from R_B = 15 kN, or from load at C - R_A = 15 kN. For any point between C and B the force to the right is upwards and the shearing force is therefore negative. It should be noted that the shearing force changes suddenly at C.

The bending moment at A is zero, as there are no forces to its left. At a point 1 m to the right of A the moment of the only force R_A to the left of the point is R_A × 1 m = 9 kNm. As this moment to the left is clockwise the bending moment is positive, i.e. it is + 9 kNm. At points 2 m, 3 m, 4 m and 5 m to the right of A the bending moments are respectively:

R_A × 2 m = 9 kN × 2 m = 18 kNm

R_A × 3 m = 9 kN × 3 m = 27 kNm

R_A × 4 m = 9 kN × 4 m = 36 kNm

R_A × 5 m = 9 kN × 5 m = 45 kNm

All are positive bending moments.

For points to the right of C, the load at C as well as R_A must be considered or, more simply, R_B alone can be used. At points 5 m, 6 m and 7 m from A the bending moments are respectively:

R_B × 3 m = 15 kN × 3 m = 45 kNm

R_B × 2 m = 15 kN × 2 m = 30 kNm

R_B × 1 m = 15 kN × 1 m = 15 kNm

As these moments to the right of the points considered are anticlockwise they are all positive bending moments. At B the bending moment is zero as there is no force to its right. The results are summarised in the table below.

Making use of the above values, the diagrams are as shown (in the usual manner) in Figure 36.8. A stepped shearing force diagram, with horizontal and vertical lines only, is always obtained when the beam carries concentrated loads only. A sudden change in shearing force occurs where the concentrated loads, including the reactions at supports, occur. For this type of simple loading the bending moment diagram always consists of straight lines, usually sloping. Sudden changes of bending moment cannot occur except in the unusual circumstances of a moment being applied to a beam as distinct from a load.

Bending stress

6. 

$\frac{\sigma }{y}=\frac{M}{I}\left(=\frac{E}{R}\right)$

where

    

σ = stress due to bending at distance y from the neutral axis;

M = bending moment;

I = second moment of area of section of beam about its neutral axis;

E = modulus of elasticity;

R = radius of curvature.

Section modulus Z = I/y_max.

The second moments of area of the beam sections most commonly met with are (about the central axis XX):

(a) Solid rectangle (Figure 36.9).

$I=\frac{B{D}^3}{12}$

(b) Symmetrical hollow rectangle or I section (Figure 36.10)

$I=\frac{B{D}^3-b{d}^3}{12}$

(c) Solid rod (Figure 36.11)

$I=\frac{\pi D^4}{64}$

(d) Tube (Figure 36.12)

$I=\frac{\pi \left(D^4-d^4\right)}{64}$

The neutral axis of any section, where bending produces no strain and therefore no stress, always passes through the centroid of the section. For the symmetrical sections listed above this means that for vertical loading the neutral axis is the horizontal axis of symmetry.

For example, let the maximum bending moment on a beam be 120 Nm. If the beam section is rectangular 18 mm wide and 36 mm deep, the maximum bending stress is calculated as follows:

Second moment of area of section about the neutral axis,

$I=\frac{b{d}^3}{12}=\frac{\left(18\right){\left(36\right)}^3}{12}=6.9984\times 1{\text{0}}^{\text{4}}{\text{mm}}^{\text{4}}$

Maximum distance from neutral axis,

$y=\frac{36}{2}=18\text{mm}\text{.}$

Since σ/y = M/I then the maximum bending stress σ will occur where M and y have their maximum values, i.e.

$\begin{array}{l}\sigma =\frac{My}{I}=\frac{\text{120}\text{Nm×18}\text{mm}}{\text{6}{\text{.9984×10}}^{\text{4}}{\text{×10}}^{\text{−12}}{\text{mm}}^{\text{4}}}\\ \text{=}\frac{\text{120}{\text{Nm×18×10}}^{\text{−3}}\text{m}}{\text{6}{\text{.9984×10}}^{\text{4}}{\text{×10}}^{\text{−12}}{\text{m}}^{\text{4}}}\\ \text{=30}\text{.86}{\text{MN/m}}_{\text{2}}\\ \text{=30}\text{.86}{\text{MP}}_{\text{a}}\end{array}$