Question 2.8.1: Determine the shear force and bending moment at any section ...
Determine the shear force and bending moment at any section of the beam AB in Fig. 2.8-5(a).
By inspection (or successive application of the condition \sum{M=0} to B and A):
R_{A}=\frac{4}{6} P=0.67 P kN; R_{B}=\frac{2}{6} P=0.33 P kN
Consider a section a-a at distance x from support A.
For 0 < x < 2 m, the only force to the left of section a-a is R_{A} (0.67 P kN) and it acts upwards. Therefore according to the sign convention in Fig. 2.8-4(a)(which is a particular case of the general sign convention in Fig. 2.8-3) the shear force at section a-a is -0.67P kN.
For 2 m < x < 6 m, the resultant of the forces to the left of section a-a acts downwards and has a magnitude of (P — 0.67P) kN = 0.33P kN. Hence the shear force at section a-a is + 0.33P kN.
The shear force at any section of the beam is as shown in the shear force diagram (often referred to as the SF diagram or the SFD) in Fig. 2.8-5(b). Note that, according to the above calculations, the shear forces at sections at A, C, and B (where x exactly equals 0, 2 m and 6 m respectively) are indeterminate. This is because of the theoretical assumption that the forces R_{A}, P, and R_{C} are each concentrated at one point. Such concentrated forces cannot occur in practice because they would immediately cut through the beam, since a concentrated force produces an infinitely large pressure at its point of application. In fact, so called concentrated forces are distributed over a small length of the beam, so that at section C, for example, the shear-force distribution would appear somewhat as shown by the dotted line in Fig. 2.8-5(b).
Next consider the bending moment at section a-a at x from A:
For 0 < x < 2 m, the moment is a ‘sagging’ moment of magnitude 0.67 Px kN m produced by the reaction R_{A}. Hence the bending moment at x from A is +0.67 Px kN m, i.e. the bending moment varies linearly from 0 for x = 0 at A to +1.33P kN m for x = 2 m at C.
For 2 m < x < 6 m, the resultant moment at the section is made up of a ‘sagging’ moment of 0.67 Px kN m due to R_{A} and a ‘hogging’ moment of P(x — 2) kN m,
i.e. the resultant sagging moment is
0.67Px- P(x – 2) = (2 – 0.33x)P kN
i.e. the bending moment varies linearly from +1.33P kN m for x = 2 m at C to 0 for x = 6 m at B. The bending moment diagram (often referred to as BM diagram or the BMD) is as shown in Fig. 2.8-5(c).
The reader should verify the shear force and bending moment diagrams in Figs 2.8-6 to Fig. 2.8-9 inclusive. The relationships shown in Fig. 4.18-1 in Chapter 4 will be found exceptionally valuable in facilitating both calculation and sketching.
