Question 5.3: Draw the shear and bending moment diagrams for the beam show...

Reactions See Fig. 5.5(b).

+\longrightarrow \sum{F_x}=0            A_x = 0

Shear Diagram To determine the equation for shear in segment AB of the beam, we pass a section aa at a distance x from support A, as shown in Fig. 5.5(b). Considering the free body to the left of this section, we obtain

S = 203.89 kN                        for  0 < x < 3 m

As this equation indicates, the shear is constant at 203.89 kN from an infinitesimal distance to the right of point A to an infinitesimal distance to the left of point B. At point A, the shear increases abruptly from 0 to 203.89 kN, so a vertical line is drawn from 0 to 203.89 on the shear diagram (Fig. 5.5(c)) at A to indicate this change. This is followed by a horizontal line from A to B to indicate that the shear remains constant in this segment.

Next, by using section bb (Fig. 5.5(b)), we determine the equation for shear in segment BC as

S = 203.89 – 265 = -61.11 kN                for  3 m < x ≤ 6 m

The abrupt change in shear from 203.89 kN at an infinitesimal distance to the left of B to -61.11 kN at an infinitesimal distance to the right of B is shown on the shear diagram (Fig. 5.5(c)) by a vertical line from +203.89 to -61.11. A horizontal line at -61.11 is then drawn from B to C to indicate that the shear remains constant at this value throughout this segment.

To determine the equations for shear in the right half of the beam, it is convenient to use another coordinate, x_1, directed to the left from the end E of the beam, as shown in Fig. 5.5(b). The equations for shear in segments ED and DC are obtained by considering the free bodies to the right of sections dd and cc, respectively. Thus,

S = 30x_1               for 0 ≤ x_1 < 3 m

and

S = 30x_1 – 241.11             for  3 m < x_1 ≤ 6 m

These equations indicate that the shear increases linearly from zero at E to +90 kN at an infinitesimal distance to the right of D; it then drops abruptly to -151.11 kN at an infinitesimal distance to the left of D; and from there it increases linearly to -61.11 kN at C. This information is plotted on the shear diagram, as shown in Fig. 5.5(c).

Bending Moment Diagram Using the same sections and coordinates employed previously for computing shear, we determine the following equations for bending moment in the four segments of the beam. For segment AB:

M = 203.89x            for 0 ≤ x ≤ 3 m

For segment BC:

M = 203.89x – 265(x – 3) = -61.11x + 795     for   3 m ≤ x < 6 m

For segment ED:

M = -30x_1(\frac{x_1}{2})= -15x^2_1        for   0 ≤ x_1 ≤ 3 m

For segment DC:

M = -15x^2_1 + 241.11(x_1 – 3) = -15x^2_1 + 241.11x_1 – 723.33         for   3 m ≤ x_1 < 6 m

The first two equations, for the left half of the beam, indicate that the bending moment increases linearly from 0 at A to 611.67 kN-m at B; it then decreases linearly to 428.34 kN-m at C, as shown on the bending moment diagram in Fig. 5.5(d). The last two equations for the right half of the beam are quadratic in x_1. The values of M computed from these equations are plotted on the bending moment diagram shown in Fig. 5.5(d). It can be seen that M decreases from 0 at E to -135 kN-m at D, and it then increases to +183.33 kN-m at an infinitesimal distance to the right of C. Note that at C, the bending moment drops abruptly by an amount 428.34 – 183.33 = 245 kN-m, which is equal to the magnitude of the moment of the counterclockwise external couple acting at this point.

A point at which the bending moment is zero is termed the point of inflection. To determine the location of the point of inflection F (Fig. 5.5(d)), we set M = 0 in the equation for bending moment in segment DC to obtain

from which x_1 = 3.99 m; that is, point F is located at a distance of 3.99 m from end E, or 12 – 3.99 = 8.01 m from support A of the beam, as shown in Fig. 5.5(d).