Page 122 - Aircraft Stuctures for Engineering Student

106 Energy methods of structural analysis

Example 4.10
An elastic member is pinned to a drawing board at its ends A and B. When a moment
M is applied at A, A rotates 0.4, B rotates and the centre deflects S1. The same
moment M applied to B rotates B, Oc and deflects the centre through 62. Find the
moment induced at A when a load W is applied to the centre in the direction of
the measured deflections, both A and B being restrained against rotation.

Fig. 4.26 Model analysis of a fixed beam.

The three load conditions and the relevant displacements are shown in Fig. 4.26.
Thus from Fig. 4.26(a) and (b) the rotation at A due to M at B is, from the reciprocal
theorem, equal to the rotation at B due to M at A. Hence

eA(b) = OB
It follows that the rotation at A due to MB at B is
MB
eA(c),l = MeB
Also the rotation at A due to unit load at C is equal to the deflection at C due to unit
moment at A. Therefore
eA(c)
I-_
2
61
-
W M
or
W
eA(c),2 = z61 (ii)
where 6A(c),2 is the rotation at A due to W at C. Finally the rotation at A due to MA at
A is, from Fig. 4.26(a) and (c)

(iii)
The total rotation at A produced by MA at A, W at C and MB at B is, from Eqs (i), (ii)
and (iii)