## 46 Application to deflection problems

Generally, deflection problems are most readily solved by the complementary energy approach, although for linearly elastic systems there is no difference between the methods of complementary and potential energy since, as we have seen, complementary and strain energy then become completely interchangeable. We shall illustrate the method by reference to the deflections of frames and beams which may or may not possess linear elasticity.

Let us suppose that we require to find the deflection A2 of the load P2 in the simple pin-jointed framework consisting, say, of k members and supporting loads P\, P2, ■ . ■, P„, as shown in Fig. 4.7. From Eqs (4.14) the total complementary energy of the framework is given by k

vertical reaction at B is therefore zero, since both of their corresponding displacements are zero. If we examine Eq. (4.17) we note that A,- is the extension of the ith member of the framework due to the applied loads PUP2,..., P„. Therefore, the loads Fj in the substitution for A,- in Eq. (4.17) are those corresponding to the loads Py,P2, ■ • ■ yP„. The term dFjdP^ in Eq. (4.17) represents the rate of change of Ft with P2 and is calculated by applying the load P2 to the unloaded frame and determining the corresponding member loads in terms of P2. This procedure indicates a method for obtaining the displacement of either a point on the frame in a direction not coincident with the line of action of a load or, in fact, a point such as C which carries no load at all. We place at the point and in the required direction a fictitious or dummy load, say P{, the original loads being removed. The loads in the members due to P[ are then calculated and dF/8P{ obtained for each member. Substitution in Eq. (4.17) produces the required deflection.

It must be pointed out that it is not absolutely necessary to remove the actual loads during the application of P{. The force in each member would then be calculated in terms of the actual loading and P{. Ft follows by substituting Pf = 0 and dFj/dPf is found by differentiation with respect to P{. Obviously the two approaches yield the same expressions for F, and dFjdPf, although the latter is arithmetically clumsier.

### Example 4.2

Calculate the vertical deflection of the point B and the horizontal movement of D in the pin-jointed framework shown in Fig. 4.8(a). All members of the framework are

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