Question 16.7: Determine the support reactions in the three-span continuous...

It is clear from inspection that the degree of statical indeterminacy of the beam is two. Therefore, if we choose the supports at \mathrm{B} and \mathrm{C} as the releases, the primary structure is that shown in Fig. 16.17(b). We therefore require the vertical displacements, v_{\mathrm{B}, 0} and v_{\mathrm{C}, 0}, at the points \mathrm{B} and \mathrm{C}. These may readily be found using any of the methods previously described (unit load method, moment-area method, Macauley’s method (Section 13.2)) and are

v_{\mathrm{B}, 0}=\frac{8.88}{E I} \quad v_{\mathrm{C}, 0}=\frac{9.08}{E I} \quad(\text { downwards) }

We now require the flexibility coefficients, a_{11}, a_{12}, a_{22} and a_{21}. The coefficients a_{11} and a_{21} are found by placing a unit load at B (point 1) as shown in Fig. 16.17(c) and then determining the displacements at \mathrm{B} and \mathrm{C} (point 2). Similarly, the coefficients a_{22} and a_{12} are found by placing a unit load at \mathrm{C} and calculating the displacements at \mathrm{C} and \mathrm{B}; again, any of the methods listed above may be used. However, from the reciprocal theorem (Section 15.4) a_{12}=a_{21} and from symmetry a_{11}=a_{22}. Therefore it is only necessary to calculate the displacements a_{11} and a_{21} from Fig. 16.17(c). These are

a_{11}=a_{22}=\frac{0.45}{E I} \quad a_{21}=a_{12}=\frac{0.39}{E I} \quad \text { (downwards) }

The total displacements at the support points B and C are zero so that

\begin{aligned}& v_{\mathrm{B}, 0}-a_{11} R_{1}-a_{12} R_{2}=0 & (\text{i})\\& v_{\mathrm{C}, 0}-a_{21} R_{1}-a_{22} R_{2}=0 & (\text{ii})\end{aligned}

or, substituting the calculated values of v_{\mathrm{B}, 0}, a_{11}, etc., in Eqs (i) and (ii), and multiplying through by E I

\begin{aligned} & 8.88-0.45 R_{1}-0.39 R_{2}=0 & (\text{iii})\\ & 9.08-0.39 R_{1}-0.45 R_{2}=0 & (\text{iv}) \end{aligned}

Note that the negative signs in the terms involving R_{1} and R_{2} in Eqs (i) and (ii) are due to the fact that the unit loads were applied in the opposite directions to R_{1} and R_{2}. Solving Eqs (iii) and (iv) we obtain

R_{1}\left(=R_{\mathrm{B}}\right)=8.7 \mathrm{kN} \quad R_{2}\left(=R_{\mathrm{C}}\right)=12.68  \mathrm{kN}

The remaining reactions are determined by considering the statical equilibrium of the beam and are

R_{\mathrm{A}}=1.97 \mathrm{kN} \quad R_{\mathrm{B}}=4.65  \mathrm{kN}