Question 15.6: Calculate the vertical deflection of the joint B and the hor...

The virtual force systems, i.e. unit loads, required to determine the vertical deflection of B and the horizontal deflection of D are shown in Fig. 15.12(b) and (c), respectively. Therefore, if the actual vertical deflection at \mathrm{B} is \delta_{\mathrm{B}, \mathrm{v}} and the horizontal deflection at \mathrm{D} is \delta_{\mathrm{D}, \mathrm{h}} the external virtual work done by the unit loads is 1 \delta_{\mathrm{B}, \mathrm{v}} and 1 \delta_{\mathrm{D}, \mathrm{h}}, respectively. The internal actual and virtual force systems comprise axial forces in all the members.

These axial forces are constant along the length of each member so that for a truss comprising n members, Eq. (15.12)

w_{i,N}=\sum\int_{L}{\frac{N_{\mathrm{A}}N_{\mathrm{v}}}{EA}}\,\mathrm{d}x       (15.12)

reduces to

W_{\mathrm{i}, N}=\sum\limits_{j=1}^{n} \frac{F_{\mathrm{A}, j} F_{\mathrm{v}, j} L_{j}}{E_{j} A_{j}}    (i)

in which F_{\mathrm{A}, j} and F_{\mathrm{v}, j} are the actual and virtual forces in the j th member which has a length L_{j}, an area of cross-section A_{j} and a Young’s modulus E_{j}.

Since the forces F_{\mathrm{v}, j} are due to a unit load, we shall write Eq. (i) in the form

W_{\mathrm{i}, N}=\sum\limits_{j=1}^{n} \frac{F_{\mathrm{A}, j} F_{1, j} L_{j}}{E_{j} A_{j}}       (ii)

Also, in this particular example, the area of cross section, A, and Young’s modulus, E, are the same for all members so that it is sufficient to calculate \sum_{j=1}^{n} F_{\mathrm{A}, j} F_{1, j} L_{j} and then divide by E A to obtain W_{\mathrm{i}, N}.

The forces in the members, whether actual or virtual, may be calculated by the method of joints (Section 4.6). Note that the support reactions corresponding to the three sets of applied loads (one actual and two virtual) must be calculated before the internal force systems can be determined. However, in Fig. 15.12(c), it is clear from inspection that F_{1, \mathrm{AB}}=F_{1, \mathrm{BC}}=F_{1, \mathrm{CD}}=+1 while the forces in all other members are zero. The calculations are presented in Table 15.1; note that positive signs indicate tension and negative signs compression.

Thus equating internal and external virtual work done (Eq. (15.23))

W_{\mathrm{e}}=W_{\mathrm{i}}      (15.23)

we have

1 \delta_{\mathrm{B}, \mathrm{v}}=\frac{1263.6 \times 10^{6}}{200000 \times 1800}

whence

\delta_{\mathrm{B}, \mathrm{v}}=3.51 \mathrm{~mm}

and

1 \delta_{\mathrm{D}, \mathrm{h}}=\frac{880 \times 10^{6}}{200000 \times 1800}

which gives

\delta_{\mathrm{D}, \mathrm{h}}=2.44 \mathrm{~mm}

Both deflections are positive which indicates that the deflections are in the directions of the applied unit loads. Note that in the above it is unnecessary to specify units for the unit load since the unit load appears, in effect, on both sides of the virtual work equation (the internal F_{1} forces are directly proportional to the unit load).