Question

Calculate and sketch the bending moment and shearing force diagrams for the horizontal beam shown in Figure, which is simply supported at its ends.

Answer 1

Firstly, it will be necessary to calculate the magnitude of reactions R_A and R_B. 

Taking moments about B gives:

Clockwise moments about B = anti-clockwise moments about B 

i.e. R_A × 5 m = 6 kN × 2 m = 12 kN m

from which, R_A = 12/5 = 2.4 kN

Resolving forces vertically gives:

Upward forces = downward forces

i.e. R_A + R_B = 6 kN

i.e. 2.4 + R_B = 6

from which, R_B = 6 − 2.4 = 3.6 kN

As there is a discontinuity at point C in Figure(a), due to the concentrated load of 6 kN, it will be necessary to consider the length of the beam AC separately from the length of the beam CB. The reason for this is that the equations for bending moment and shearing force for span AC are different to the equations for the span CB; this is caused by the concentrated load of 6 kN.

For the present problem, to demonstrate the nature of bending moment and shearing force, these values will be calculated on both sides of the point of the beam under consideration. It should be noted that normally, the bending moment and shearing force at any point on the beam, are calculated only due to the resultant couples or forces, respectively, on one side of the beam.

Consider span AC 

Bending moment

Consider a section of the beam at a distance x from the left end A, where the value of x lies between A and C, as shown in Figure (b)

From Figure (b) it can be seen that the reaction RA causes a clockwise moment of magnitude R_A ×x = 2.4x on the left of this section and as shown in the lower diagram of Figure. It can also be seen from the upper diagram of Figure (b)  that the forces on the right of this section on the beam causes an anti-clockwise moment equal to RB × (5 − x) or 3.6(5 − x) and a clockwise moment of 6 × (3 − x), resulting in an anti-clockwise moment of:

Thus, the left side of the beam at this section is subjected to a clockwise moment of magnitude 2.4x and the right side of this section is subjected to an anti-clockwise moment of 2.4x, as shown by the lower diagram of Figure (b) As the two moments are of equal magnitude but of opposite direction, they cause the beam to be subjected to a bending moment M = 2.4x. As this bending moment causes the beam to sag between A and C, the bending moment is assumed to be positive, or at any distance x between A and C:

Bending moment = M = +2.4x  ........(6.1)

Shearing force

Here again, because there is a discontinuity at C, due to the concentrated load of 6 kN at C, we must consider a section of the beam at a distance x from the left end A, where x varies between A and C, as shown in Figure (c)

From Figure (c), it can be seen that the resultant of the vertical forces on the left of the section at x are 2.4 kN acting upwards. This force causes the left of the section at x to slide upwards, as shown in the lower diagram of Figure (c). Similarly, if the vertical forces on the right of the section at x are considered, it can be seen that the 6 kN acts downwards and that R_B = 3.6 kN acts upwards, giving a resultant of 2.4 kN acting downwards. The effect of the two shearing forces acting on the left and the right of the section at x, causes the shearing action shown in the lower diagram of Figure (c). As this shearing action causes the right side of the section to glide downwards, it is said to be a positive shearing force.

Summarising, at any distance x between A and C:

F = shearing force = +2.4 kN .....(6.2)

Consider span CB:

Bending moment 

At any distance x between C and B, the resultant moment caused by the forces on the left of x is given by:

M = RA × x − 6(x − 3) = 2.4x − 6(x − 3) 

= 2.4x − 6x + 18

i.e. M = 18 − 3.6x (clockwise)...... (6.3)

Rest part of the solution is given below

Answer 2

The effect of this resultant moment on the left of x is shown in the lower diagram of Figure (d)

Now from Figure (d), it can be seen that on the right side of x, there is an anti-clockwise moment of: M = R_B × (5 − x) = 3.6(5 − x) = 18 − 3.6x

i.e. M = 18 − 3.6x (anti-clockwise) .....(6.4)

The effect of the moment of equation (6.3) and that of the moment of equation (6.4), is to cause the beam to sag at this point as shown by the lower diagram of Figure (d), i.e. M is positive between C and B, and

M = +18 − 3.6x .......(6.5)

Shearing force

Consider a distance x between C and B, as shown in Figure.(e)

From Figure (e), it can be seen that at x, there are two vertical forces to the left of this section, namely the 6 kN load acting downwards and the 2.4 kN load acting upwards, resulting in a net value of 3.6 kN acting downwards, as shown by the lower diagram of Figure (e). Similarly, by considering the vertical forces acting on the beam to the right of x, it can be seen that there is one vertical force, namely the 3.6 kN load acting upwards, as shown by the lower diagram of Figure (e). Thus, as the right hand of the section is tending to slide upwards, the shearing force is said to be negative, i.e. between C and B,

F = −3.6 kN   ....... (6.6)

It should be noted that at C, there is a discontinuity in the value of the shearing force, where over an infinitesimal length the shearing force changes from +2.4 kN to −3.6 kN, from left to right.

Bending moment and shearing force diagrams

The bending moment and shearing force diagrams are simply diagrams representing the variation of bending moment and shearing force, respectively, along the length of the beam. In the bending moment and shearing force diagrams, the values of the bending moments and shearing forces are plotted vertically and the value of x is plotted horizontally, where x = 0 at the left end of the beam and x = the whole length of the beam at its right end. In the case of the beam of Figure (a), bending moment distribution between A and C is given by equation (6.1), i.e. M = 2.4x, where the value of x varies between A and C.

At A, x = 0, therefore M_A = 2.4 × 0 = 0 and at C, x = 3 m, therefore M_C = 2.4 × 3 = 7.2 kN.

Additionally, as the equation M = 2.4x is a straight line, the bending moment distribution between A and C will be as shown by the left side of Figure. f(a).

Similarly, the expression for the variation of bending moment between C and B is given by equation (6.3), i.e. M = 18−3.6x, where the value of x varies between C and B. The equation can be seen to be a straight line between C and B.

Therefore, plotting of the equation M = 18 − 3.6x between C and B results in the straight line on the right of Figure f(a), i.e. the bending moment diagram for this beam has now been drawn.

Figure Bending moment and shearing force diagrams

In the case of the beam of Figure (a), the shearing force distribution along its length from A to C is given by equation (6.2), i.e. F = 2.4 kN, i.e. F is the constant between A and C. Thus the shearing force diagram between A and C is given by the horizontal the line shown on the left of C in Figure f(b).

Similarly, the shearing force distribution to the right of C is given by equation (6.6), i.e. F = −3.6 kN, i.e. F is a constant between C and B, as shown by the horizontal line to the right of C in Figure f(b). At the point C, the shearing force is indeterminate and changes from +2.4 kN to −3.6 kN over an infinitesimal length.