Question 15.13: An elastic member is pinned to a drawing board at its ends A...
An elastic member is pinned to a drawing board at its ends \mathrm{A} and \mathrm{B}. When a moment, M, is applied at \mathrm{A}, A rotates by \theta_{\mathrm{A}}, \mathrm{B} rotates by \theta_{\mathrm{B}} and the centre deflects by \delta_{1}. The same moment, M, applied at \mathrm{B} rotates \mathrm{B} by \theta_{\mathrm{C}} and deflects the centre through \delta_{2}. Find the moment induced at \mathrm{A} when a load, W, is applied to the centre in the direction of the measured deflections, and A and B are restrained against rotation.
The three load conditions and the relevant displacements are shown in Fig. 15.25. Thus, from Fig. 15.25(a) and (b) the rotation at A due to M at \mathrm{B} is, from the reciprocal theorem, equal to the rotation at \mathrm{B} due to M at \mathrm{A}.
Thus
\theta_{\mathrm{A}(\mathrm{b})}=\theta_{\mathrm{B}} (i)
It follows that the rotation at \mathrm{A} due to M_{\mathrm{B}} at \mathrm{B} is
\theta_{\mathrm{A}(\mathrm{c}), 1}=\frac{M_{\mathrm{B}}}{M} \theta_{\mathrm{B}}
where (b) and (c) refer to (b) and (c) in Fig. 15.25.
Also, the rotation at \mathrm{A} due to a unit load at \mathrm{C} is equal to the deflection at \mathrm{C} due to a unit moment at A. Therefore
\frac{\theta_{\mathrm{A}(\mathrm{c}), 2}}{W}=\frac{\delta_{1}}{M}
or
\theta_{\mathrm{A}(\mathrm{c}), 2}=\frac{W}{M} \delta_{1} (ii)
in which \theta_{\mathrm{A}(\mathrm{c}), 2} is the rotation at \mathrm{A} due to W at \mathrm{C}. Finally the rotation at \mathrm{A} due to M_{\mathrm{A}} at \mathrm{A} is, from Fig. 15.25(a) and (c)
\theta_{\mathrm{A}(\mathrm{c}), 3}=\frac{M_{\mathrm{A}}}{M} \theta_{\mathrm{A}} (iii)
The total rotation at A produced by M_{\mathrm{A}} at \mathrm{A}, W at \mathrm{C} and M_{\mathrm{B}} at \mathrm{B} is, from Eqs (i), (ii) and (iii)
\theta_{\mathrm{A}(\mathrm{c}), 1}+\theta_{\mathrm{A}(\mathrm{c}), 2}+\theta_{\mathrm{A}(\mathrm{c}), 3}=\frac{M_{\mathrm{B}}}{M} \theta_{\mathrm{B}}+\frac{W}{M} \delta_{1}+\frac{M_{\mathrm{A}}}{M} \theta_{\mathrm{A}}=0 (iv)
since the end A is restrained against rotation. In a similar manner the rotation at B is given by
\frac{M_{\mathrm{B}}}{M} \theta_{\mathrm{C}}+\frac{W}{M} \delta_{2}+\frac{M_{\mathrm{A}}}{M} \theta_{\mathrm{B}}=0 (v)
Solving Eqs (iv) and (v) for M_{\mathrm{A}} gives
M_{\mathrm{A}}=W\left(\frac{\delta_{2} \theta_{\mathrm{B}}-\delta_{1} \theta_{\mathrm{C}}}{\theta_{\mathrm{A}} \theta_{\mathrm{C}}-\theta_{\mathrm{B}}^{2}}\right)
The fact that the arbitrary moment, M, does not appear in the expression for the restraining moment at \mathrm{A} (similarly it does not appear in M_{\mathrm{B}} ) produced by the load W indicates an extremely useful application of the reciprocal theorem, namely the model analysis of statically indeterminate structures. For example, the fixed beam of Fig. 15.25(c) could possibly be a full-scale bridge girder. It is then only necessary to construct a model, say, of perspex, having the same flexural rigidity, E I, as the full-scale beam and measure rotations and displacements produced by an arbitrary moment, M, to obtain the fixed-end moments in the full-scale beam supporting a full-scale load.
