Question 13.19: Figure 13.26(a) shows a fixed beam carrying a central concen...
Figure 13.26(a) shows a fixed beam carrying a central concentrated load, W. Determine the value of the fixed-end moments, M_A and M_B .
Since the ends A and B of the beam are prevented from rotating, moments M_A and M_B are induced in the supports; these are termed fixed-end moments. From symmetry we see that M_A = M_B and R_A = R_B = W/2 .
The beam AB in Fig. 13.26(a) may be regarded as a simply supported beam carrying a central concentrated load with moments M_A and M_B applied at the supports. The bending moment diagrams corresponding to these two loading cases are shown in Fig.13.26(b) and (c) and are known as the free bending moment diagram and the fixed-end moment diagram, respectively. Clearly the concentrated load produces sagging (positive) bending moments, while the fixed-end moments induce hogging (negative) bending moments. The resultant or final bending moment diagram is constructed by superimposing the free and fixed-end moment diagrams as shown in Fig. 13.26(d).
The moment-area method is now used to determine the fixed end moments, M_A and M_B . From Eq. (13.7) the change in slope between any two sections of a beam is equal to the area of the M/EI diagram between those sections. Therefore, the net area of the bending moment diagram of Fig. 13.26(d) must be zero since the change of slope between the ends of the beam is zero. It follows that the area of the free bending moment diagram is numerically equal to the area of the fixed-end moment diagram; thus
\left(\frac{dv}{dx} \right) _B – \left(\frac{dv}{dx} \right) _A = \int_{A}^{B}{\frac{M}{EI} } \ dx (13.7)
M_AL = \frac{1}{2} \frac{WL}{4} L
which gives
M_A = M_B = \frac{WL}{8}
and the resultant bending moment diagram has principal values as shown in Fig. 13.27. Note that the maximum positive bending moment is equal in magnitude to the maximum negative bending moment and that points of contraflexure (i.e. where the bending moment changes sign) occur at the quarter-span points.
Having determined the support reactions, the deflected shape of the beam may be found by any of the methods described in the previous part of this chapter.
