Question 4.6: Determine the force in the member AB of the truss shown in F...

We are required to calculate the force in the member AB, so that again we need to relate this internal force to the externally applied loads without involving the internal forces in the remaining members of the truss. We therefore impose a virtual extension, \Delta_{v, B }, at B in the member AB, such that B moves to B ^{\prime}. If we assume that the remaining members are rigid, the forces in them do no work. Further, the triangle BCD rotates as a rigid body about D to B ^{\prime} C ^{\prime} D, as shown in Fig. 4.12(b). The horizontal displacement of C , \Delta_{ C }, is then given by

while

Hence,

\Delta_{ C }=\frac{4 \Delta_{v, B }}{3}  (i)

Equating the external virtual work done by the 30 kN load to the internal virtual work done by the force, F_{ BA }, in the member, AB, we have (see Eq. (4.23) and Fig. 4.9)

W_{e}=W_{i}  (4.23)

30 \Delta_{ C }=F_{ BA } \Delta_{V, B }  (ii)

Substituting for \Delta_{ C } from Eq. (i) in Eq. (ii),

from which

In the preceding, we are, in effect, assigning a positive (that is, F_{ BA } is tensile) sign to F_{ BA } by imposing a virtual extension on the member AB. The actual sign of F_{ BA } is then governed by the sign of the external virtual work. Thus, if the 30 kN load were in the opposite direction to \Delta_{ C }, the external work done would have been negative, so that F_{ BA } would be negative and therefore compressive. This situation can be verified by inspection. Alternatively, for the loading shown in Fig. 4.12(a), a contraction in AB implies that F_{ BA } is compressive. In this case, DC would have rotated in a counterclockwise sense, \Delta_{ C } would have been in the opposite direction to the 30 kN load so that the external virtual work done would be negative, resulting in a negative value for the compressive force F_{ BA } ; F_{ BA } would therefore be tensile, as before. Note also that the 10 kN load at D does no work, since D remains undisplaced.