Question 16.22: Obtain the bending moment diagram for the portal frame shown...

First we shall determine the end moments in the members assuming that the frame does not sway. The corresponding FEMs are found using the results in Table 16.6 and are as follows:

\begin{aligned} & M_{\mathrm{AB}}^{\mathrm{F}}=M_{\mathrm{BA}}^{\mathrm{F}}=0 \quad M_{\mathrm{CD}}^{\mathrm{F}}=M_{\mathrm{DC}}^{\mathrm{F}}=0 \\ & M_{\mathrm{BC}}^{\mathrm{F}}=-\frac{4 \times 5 \times 10^{2}}{15^{2}}=-8.89 \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{CB}}^{\mathrm{F}}=+\frac{4 \times 10 \times 5^{2}}{15^{2}}=+4.44  \mathrm{kNm} \end{aligned}

The DFs are

\mathrm{DF}_{\mathrm{BA}}=\frac{K_{\mathrm{BA}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}}=\frac{4 E I / 10}{4 E I / 10+4 \times 2 E I / 15}=0.43

Hence

\mathrm{DF}_{\mathrm{BC}}=1-0.43=0.57

From the symmetry of the frame, \mathrm{DF}_{\mathrm{CB}}=0.57 and \mathrm{DF}_{\mathrm{CD}}=0.43.

The no-sway moments are determined in the table overleaf. We now assume that the frame sways by an arbitrary amount, \delta, as shown in Fig. 16.47(b). Since we are ignoring the effect of axial strains, the horizontal movements of B and C are both \delta. The FEMs corresponding to this sway are then (see Table 16.6)

\begin{aligned} & M_{\mathrm{AB}}^{\mathrm{F}}=M_{\mathrm{BA}}^{\mathrm{F}}=-\frac{6 E I \delta}{10^{2}}=M_{\mathrm{DC}}^{\mathrm{F}}=M_{\mathrm{CD}}^{\mathrm{F}} \\ & M_{\mathrm{BC}}^{\mathrm{F}}=M_{\mathrm{CB}}^{\mathrm{F}}=0 \end{aligned}

Suppose that \delta=100 \times 10^{2} / 6 E I. Then

M_{\mathrm{AB}}^{\mathrm{F}}=M_{\mathrm{BA}}^{\mathrm{F}}=M_{\mathrm{DC}}^{\mathrm{F}}=M_{\mathrm{CD}}^{\mathrm{F}}=-100  \mathrm{kNm} \quad \text { (a convenient value) }

The DFs for the members are the same as those in the no-sway case since they are functions of the member stiffness. We now obtain the member end moments corresponding to the arbitrary sway.

No-sway case (fig.1)

Sway case (fig.2)

Comparing the frames shown in Figs 16.47 and 16.46 we see that they are virtually identical. We may therefore use 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)

directly. Thus, substituting for the no-sway and arbitrary-sway end moments we have

2.36+4.75-3.25-1.63+k(-82.9-66.8-66.8-82.9)+2 \times 10=0

which gives

k=0.074

The actual sway moments are then

M_{\mathrm{AB}}^{\mathrm{S}}=k M_{\mathrm{AB}}^{\mathrm{AS}}=0.074 \times(-82.9)=-6.14  \mathrm{kN} \mathrm{m}

Similarly

Thus the final end moments are

M_{\mathrm{AB}}=M_{\mathrm{AB}}^{\mathrm{NS}}+M_{\mathrm{AB}}^{\mathrm{S}}=2.36-6.14=-3.78  \mathrm{kN} \mathrm{m}

Similarly

M_{\mathrm{BA}}=-0.19  \mathrm{kNm} \quad M_{\mathrm{BC}}=0.19  \mathrm{kNm} \quad M_{\mathrm{CB}}=8.19  \mathrm{kNm}

M_{\mathrm{CD}}=-8.19  \mathrm{kNm} \quad M_{\mathrm{DC}}=-7.77  \mathrm{kNm}

The bending moment diagram is shown in Fig. 16.48 and is drawn on the tension side of the members.