Question 16.23: Calculate the end moments in the members of the frame shown ...

Initially, the FEMs produced by the applied loads are calculated. Thus, from Table 16.6

\begin{aligned} & M_{\mathrm{BA}}^{\mathrm{F}}=-M_{\mathrm{BA}}^{\mathrm{F}}=-\frac{40 \times 6}{8}=-30  \mathrm{kNm} \\ & M_{\mathrm{BC}}^{\mathrm{F}}=-M_{\mathrm{CB}}^{\mathrm{F}}=-\frac{20 \times 6^{2}}{12}=-60  \mathrm{kNm} \\ & M_{\mathrm{CD}}^{\mathrm{F}}=M_{\mathrm{DC}}^{\mathrm{F}}=0 \end{aligned}

The DFs are calculated as before. Note that the length of the member CD = \sqrt{6^{2}+4.5^{2}}=7.5 \mathrm{~m}.

\mathrm{DF}_{\mathrm{BA}}=\frac{K_{\mathrm{BA}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}}=\frac{4 E I / 6}{4 E I / 6+4 E I / 6}=0.5

Hence

No-sway case (fig.1)

Unlike the frame in Ex. 16.22 the frame itself in this case is unsymmetrical. Therefore the geometry of the frame, after an imposed arbitrary sway, will not have the simple form shown in Fig. 16.47(b). Furthermore, since the member CD is inclined, an arbitrary sway will cause a displacement of the joint \mathrm{C} relative to the joint \mathrm{B}. This also means that in the application of the principle of virtual work a virtual rotation of the member \mathrm{AB} will result in a rotation of the member \mathrm{BC}, so that the end moments M_{\mathrm{BC}} and M_{\mathrm{CB}} will do work; Eq. (16.46)

M_{\mathrm{AB}}^{\mathrm{NS}}+M_{\mathrm{BA}}^{\mathrm{NS}}+M_{\mathrm{CD}}^{\mathrm{NS}}+M_{\mathrm{DC}}^{\mathrm{NS}}+k\left(M_{\mathrm{AB}}^{\mathrm{AS}}+M_{\mathrm{BA}}^{\mathrm{AS}}+M_{\mathrm{CD}}^{\mathrm{AS}}+M_{\mathrm{DC}}^{\mathrm{AS}}\right)+P h=0       (16.46)

cannot, therefore, be used in its existing form. In this situation we can make use of the geometry of the frame after an arbitrary virtual displacement to deduce the relative displacements of the joints produced by an imposed arbitrary sway; the FEMs due to the arbitrary sway may then be calculated.

Figure 16.50 shows the displaced shape of the frame after a rotation, \theta, of the member AB. This diagram will serve, as stated above, to deduce the FEMs due to sway and also to establish a virtual work equation similar to Eq. (16.46). It is helpful, when calculating the rotations of the different members, to employ an instantaneous centre, I. This is the point about which the triangle IBC rotates as a rigid body to I B^{\prime} \mathrm{C}^{\prime}; thus all sides of the triangle rotate through the same angle which, since \mathrm{BI}=8 \mathrm{~m} (obtained from similar triangles AID and BIC), is 3 \theta / 4. The relative displacements of the joints are then as shown.

The FEMs due to the arbitrary sway are, from Table 16.6 and Fig. 16.50

\begin{gathered} M_{\mathrm{AB}}^{\mathrm{F}}=M_{\mathrm{BA}}^{\mathrm{F}}=-\frac{6 E I(6 \theta)}{6^{2}}=-E I \theta \\ M_{\mathrm{BC}}^{\mathrm{F}}=M_{\mathrm{CB}}^{\mathrm{F}}=+\frac{6 E I(4.5 \theta)}{6^{2}}=+0.75 E I \theta \\ M_{\mathrm{CD}}^{\mathrm{F}}=-\frac{3 E I(7.5 \theta)}{7.5^{2}}=-0.4 E I \theta \end{gathered}

If we impose an arbitrary sway such that E I \theta=100 we have

M_{\mathrm{AB}}^{\mathrm{F}}=M_{\mathrm{BA}}^{\mathrm{F}}=-100  \mathrm{kNm} \quad M_{\mathrm{BC}}^{\mathrm{F}}=M_{\mathrm{CB}}^{\mathrm{F}}=+75  \mathrm{kNm} \quad M_{\mathrm{CD}}^{\mathrm{F}}=-40  \mathrm{kNm}

Now using the principle of virtual work and referring to Fig. 16.50 we have

Sway case (fg.2)

Hence

4\left(M_{\mathrm{AB}}+M_{\mathrm{BA}}+M_{\mathrm{CD}}\right)-3\left(M_{\mathrm{BC}}+M_{\mathrm{CB}}\right)-600=0     (i)

Now replacing M_{\mathrm{AB}}, etc., by M_{\mathrm{AB}}^{\mathrm{NS}}+k M_{\mathrm{AB}}^{\mathrm{AS}}, etc., Eq (i) becomes

Substituting the values of M_{\mathrm{AB}}^{\mathrm{NS}} and M_{\mathrm{AB}}^{\mathrm{AS}}, etc., we have

from which k=-0.352. The final end moments are calculated from M_{\mathrm{AB}}= M_{\mathrm{AB}}^{\mathrm{NS}}-0.352 M_{\mathrm{AB}}^{A S}, etc., and are given below.