Question 2.8.5: Draw the shearing force and bending moment diagrams for the ...
Draw the shearing force and bending moment diagrams for the bent member ABC shown in Fig. 2.8-13(a).
Since the given structure is not a horizontal beam, the simplified sign convention in Fig. 2.8-4 is not suitable ; its use in this case is confusing since phrases such as ‘forces to the left of the section act downwards ‘ or ‘ sagging moment’ are ambiguous. Hence we should refer to Fig. 2.8-3 for the sign convention.
Suppose coordinate axes for member AB and those for member BC are chosen as shown in Fig. 2.8-13(a). Consider a section a-a (Fig. 2.8-13(b)). It is easy to verify that the directions of the shear force S and the bending moment M are as indicated.
Since the outward normal to the face of the section is in the positive x-direction, it is seen that S is negative (because the shear force vector is in the negative y-direction) and that M is positive (because the bending moment vector is in the positive z-direction, i.e. perpendicular to the plane of the paper and pointing towards the reader). Also, from equilibrium consideration, the magnitudes of S and M are
S = P \sin α (negative, according to sign convention)
M= Px\sin α (positive, according to sign convention)
M therefore varies linearly from 0 at C to PL_{2} \sin α at B, while S remains constant at — P \sin α throughout length CB.
At a section b-b in member AB at distance x from A the directions of S and M can be shown to be as in Fig. 2.8-13(c). In this case the outward normal to the section is in the negative x-direction while the shear force vector and the bending moment vector are in the positive y- and z-directions respectively. Hence the sign convention makes both S and M negative. For equilibrium of the free body in Fig. 2.8-13(c),
S = P (negative)
M = P(L_{2} \sin α + L_{1} – x) (negative)
M therefore varies linearly from — PL_{2} \sin α at B to —P(L_{2} \sin α + L_{1}) at A.
The shear force diagram and the bending moment diagram are shown in Figs. 2.8-13 (d) and (e) respectively.
Suppose, instead of choosing member coordinate axes as shown in Fig. 2.8-13(a), we choose axes as shown in Fig. 2.8-14(a). The reader should verify that the shear force diagram and the bending moment diagram become as shown in Figs. 2.8-14(b) and (c) respectively.
This example has demonstrated that the sign of the shear force or bending moment at a section depends on the system of coordinate axes chosen for the member. The necessity to specify a system of coordinate axes for each member appears cumbersome, but actually it is a highly efficient and systematic way of handling structures, and is in fact a standard procedure in matrix computer structural analysis (see Chapter 7).
