Question 15.3: Determine the force in the member AB in the truss shown in F...
Determine the force in the member AB in the truss shown in Fig. 15.9(a).
We are required to calculate the force in the member \mathrm{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_{\mathrm{v}, \mathrm{B}}, at \mathrm{B} in the member \mathrm{AB}, such that \mathrm{B} moves to \mathrm{B}^{\prime}. If we assume that the remaining members are rigid, the forces in them will do no work. Further, the triangle \mathrm{BCD} will rotate as a rigid body about \mathrm{D} to \mathrm{B}^{\prime} \mathrm{C}^{\prime} \mathrm{D} as shown in Fig. 15.9(b). The horizontal displacement of \mathrm{C}, \Delta_{\mathrm{C}}, is then given by
\Delta_{\mathrm{C}}=4 \alpha
while
\Delta_{\mathrm{v}, \mathrm{B}}=3 \alpha
Hence
\Delta_{\mathrm{C}}=\frac{4 \Delta_{\mathrm{v}, \mathrm{B}}}{3} (i)
Equating the external virtual work done by the 30 \mathrm{kN} load to the internal virtual work done by the force, F_{\mathrm{BA}}, in the member, \mathrm{AB}, we have (see Eq. (15.23)
W_{\mathrm{e}}=W_{\mathrm{i}} (15.23)
and Fig. 15.6)
30 \Delta_{\mathrm{C}}=F_{\mathrm{BA}} \Delta_{\mathrm{v}, \mathrm{B}} (ii)
Substituting for \Delta_{\mathrm{C}} from Eq. (i) in Eq. (ii),
30 \times \frac{4}{3} \Delta_{\mathrm{v}, \mathrm{B}}=F_{\mathrm{BA}} \Delta_{\mathrm{v}, \mathrm{B}}
Whence
F_{\mathrm{BA}}=+40 \mathrm{kN} \quad \text { (i.e. } F_{\mathrm{BA}} \text { is tensile) }
In the above we are, in effect, assigning a positive (i.e. tensile) sign to F_{\mathrm{BA}} by imposing a virtual extension on the member \mathrm{AB}.
The actual sign of F_{\mathrm{BA}} is then governed by the sign of the external virtual work. Thus, if the 30 \mathrm{kN} load had been in the opposite direction to \Delta_{\mathrm{C}} the external work done would have been negative, so that F_{\mathrm{BA}} would be negative and therefore compressive. This situation can be verified by inspection. Alternatively, for the loading as shown in Fig. 15.9(a), a contraction in \mathrm{AB} would have implied that F_{\mathrm{BA}} was compressive. In this case DC would have rotated in an anticlockwise sense, \Delta_{\mathrm{C}} would have been in the opposite direction to the 30 \mathrm{kN} load so that the external virtual work done would be negative, resulting in a negative value for the compressive force F_{\mathrm{BA}} ; F_{\mathrm{BA}} would therefore be tensile as before. Note also that the 10 \mathrm{kN} load at \mathrm{D} does no work since \mathrm{D} remains undisplaced.
We shall now consider problems involving the use of virtual forces. Generally we shall require the displacement of a particular point in a structure, so that if we apply a virtual force to the structure at the point and in the direction of the required displacement the external virtual work done will be the product of the virtual force and the actual displacement, which may then be equated to the internal virtual work produced by the internal virtual force system moving through actual displacements. Since the choice of the virtual force is arbitrary, we may give it any convenient value; the simplest type of virtual force is therefore a unit load and the method then becomes the unit load method.
