Question 16.16: Determine the end moments in the members of the portal frame...
Determine the end moments in the members of the portal frame shown in Fig. 16.36; the second moment of area of the vertical members is 2.5 I while that of the horizontal members is I.
In this particular problem the approach is very similar to that for the continuous beam of Ex. 16.15. However, due to the unsymmetrical geometry of the frame and also to the application of the 10 \mathrm{kN} load, the frame will sway such that there will be horizontal displacements, v_{\mathrm{B}} and v_{\mathrm{C}}, at \mathrm{B} and \mathrm{C} in the members \mathrm{BA} and \mathrm{CD}. Since we are ignoring displacements produced by axial forces then v_{\mathrm{B}}=v_{\mathrm{C}}=v_{1}, say. We would, in fact, have a similar situation in a continuous beam if one or more of the supports experienced settlement. Also we note that the rotation, \theta_{\mathrm{A}}, at A must be zero since the end \mathrm{A} of the member \mathrm{AB} is fixed.
Initially, as in Ex. 16.15, we calculate the FEMs in the members of the frame, again using the results of Exs 13.20 and 13.22. The effect of the cantilever CE may be included by replacing it by its end moment, thereby reducing the number of equations to be solved. Thus, from Fig. 16.36 we have
\begin{aligned} & M_{\mathrm{CE}}^{\mathrm{F}}=-\frac{3 \times 6^{2}}{2}=-54 \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{AB}}^{\mathrm{F}}=-M_{\mathrm{BA}}^{\mathrm{F}}=-\frac{10 \times 10}{8}=-12.5 \mathrm{kNm} \\ & M_{\mathrm{BC}}^{\mathrm{F}}=-M_{\mathrm{CB}}^{\mathrm{F}}=-\frac{3 \times 20^{2}}{12}=-100 \mathrm{kNm} \quad M_{\mathrm{CD}}^{\mathrm{F}}=M_{\mathrm{DC}}^{\mathrm{F}}=0 \end{aligned}
Now, from Eqs (16.28) and (16.30)
M_{\mathrm{AB}} =-\frac{2 E I}{L}\left[2 \theta_{\mathrm{A}}+\theta_{\mathrm{B}}+\frac{3}{L}\left(v_{\mathrm{A}}-v_{\mathrm{B}}\right)\right]+M_{\mathrm{AB}}^{\mathrm{F}} \qquad (16.28)
M_{\mathrm{BA}} =-\frac{2 E I}{L}\left[2 \theta_{\mathrm{B}}+\theta_{\mathrm{A}}+\frac{3}{L}\left(v_{\mathrm{A}}-v_{\mathrm{B}}\right)\right]+M_{\mathrm{BA}}^{\mathrm{F}} \qquad (16.30)
\begin{aligned} & M_{\mathrm{AB}}=-\frac{2 \times 2.5 E I}{10}\left(\theta_{\mathrm{B}}-\frac{3}{10} v_{1}\right)-12.5 & (\text{i})\\ & M_{\mathrm{BA}}=-\frac{2 \times 2.5 E I}{10}\left(2 \theta_{\mathrm{B}}-\frac{3}{10} v_{1}\right)+12.5 & (\text{ii}) \end{aligned}
In Eqs (i) and (ii) we are assuming that the displacement, v_{1}, is to the right. Furthermore
M_{\mathrm{BC}}=-\frac{2 E I}{20}\left(2 \theta_{\mathrm{B}}+\theta_{\mathrm{C}}\right)-100 (iii)
\begin{aligned} & M_{\mathrm{CB}}=-\frac{2 E I}{20}\left(2 \theta_{\mathrm{C}}+\theta_{\mathrm{B}}\right)+100 & (\text{iv}) \\ & M_{\mathrm{CD}}=-\frac{2 \times 2.5 E I}{10}\left(2 \theta_{\mathrm{C}}+\theta_{\mathrm{D}}+\frac{3}{10} v_{1}\right) & (\text{v})\\ & M_{\mathrm{DC}}=-\frac{2 \times 2.5 E I}{10}\left(2 \theta_{\mathrm{D}}+\theta_{\mathrm{C}}+\frac{3}{10} v_{1}\right) & (\text{vi}) \end{aligned}
From the equilibrium of the member end moments at the joints
M_{\mathrm{BA}}+M_{\mathrm{BC}}=0 \quad M_{\mathrm{CB}}+M_{\mathrm{CD}}-54=0 \quad M_{\mathrm{DC}}=0
Substituting in the equilibrium equations for M_{\mathrm{BA}}, M_{\mathrm{BC}}, etc., from Eqs (i)-(vi) we obtain
\begin{aligned} 1.25 E I \theta_{\mathrm{B}}+0.1 E I \theta_{\mathrm{C}}-0.15 E I v_{1}+87.5 & =0 & (\text{vii}) \\ 1.2 E I \theta_{\mathrm{C}}+0.1 E I \theta_{\mathrm{B}}+0.5 E I \theta_{\mathrm{C}}+0.15 E I v_{1}-46 & =0 & (\text{viii})\\ E I \theta_{\mathrm{D}}+0.5 E I \theta_{\mathrm{C}}+0.15 E I v_{1} & =0 & (\text{ix}) \end{aligned}
Since there are four unknown displacements we require a further equation for a solution. This may be obtained by considering the overall horizontal equilibrium of the frame. Thus
S_{\mathrm{AB}}+S_{\mathrm{DC}}-10=0
in which, from Eq. (16.29)
S_{\mathrm{AB}} =\frac{6 E I}{L^2}\left[\theta_{\mathrm{A}}+\theta_{\mathrm{B}}+\frac{2}{L}\left(v_{\mathrm{A}}-v_{\mathrm{B}}\right)\right]+S_{\mathrm{AB}}^{\mathrm{F}} \qquad (16.29)
S_{\mathrm{AB}}=\frac{6 \times 2.5 E I}{10^{2}} \theta_{\mathrm{B}}-\frac{12 \times 2.5 E I}{10^{3}} v_{1}+5
where the last term on the right-hand side is S_{\mathrm{AB}}^{\mathrm{F}}(=+5 \mathrm{kN}), the contribution of the 10 \mathrm{kN} horizontal load to S_{\mathrm{AB}}. Also
S_{\mathrm{DC}}=\frac{6 \times 2.5 E I}{10^{2}}\left(\theta_{\mathrm{D}}+\theta_{\mathrm{C}}\right)-\frac{12 \times 2.5 E I}{10^{3}} v_{1}
Hence, substituting for S_{\mathrm{AB}} and S_{\mathrm{DC}} in the equilibrium equations, we have
E I \theta_{\mathrm{B}}+E I \theta_{\mathrm{D}}+E I \theta_{\mathrm{C}}-0.4 E I v_{1}-33.3=0 (x)
Solving Eqs (vii)-(x) we obtain
E I \theta_{\mathrm{B}}=-101.5 \quad E I \theta_{\mathrm{C}}=+73.2 \quad E I \theta_{\mathrm{D}}=-9.8 \quad E I v_{1}=-178.6
Substituting these values in Eqs (i)-(vi) yields
\begin{aligned} & M_{\mathrm{AB}}=11.5 \mathrm{kN} \mathrm{m} \quad M_{\mathrm{BA}}=87.2 \mathrm{kN} \mathrm{m} \quad M_{\mathrm{BC}}=-87.2 \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{CB}}=95.5 \mathrm{kN} \mathrm{m} \quad M_{\mathrm{CD}}=-41.5 \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{DC}}=0 \quad M_{\mathrm{CE}}=-54 \mathrm{kN} \mathrm{m} \end{aligned}