## 4 An Example  
The proposed method was applied on a test structure consisting in a three storey-four columns frame with interstorey height equal to 3 m and beams length equal to 4 m. The 400 mm \(\times\) 400 mm square columns are longitudinally reinforced with 4\(\phi\)20 bars in the corners and 8\(\phi\)16 bars along the sides and transversely reinforced with \(\phi\)10 stirrups 100 mm-spaced. The 300 mm-deep rectangular beams are 500 mm wide and reinforced with 4+4\(\phi\)16 bars and \(\phi\)10 stirrups 100 mm-spaced. Yield stress and Young's modulus of the reinforcement steel are 500 MPa and 200 GPa, respectively; the concrete is ascribed to C25/30 class according to Eurocode 2 [38]. The frame is subjected to a 10 kN/m live load;  
Figure 7: A sketch of the plastic hinges that originates when the column is removed is sketched in (**a**). The mechanism can be described with a unique coordinate, for example, the rotation \(\varphi\). The forces applied on the structure onto which the virtual work principle (VWP) is applied is depicted in (**b**). beam and column linear weights are 3.60 kN/m and 3.84 kN/m, respectively. Figure 8 sketches the frame and identifies the suddenly removed middle column at the bottom floor. The top end of the removed element is set as the control node and its vertical displacement as the control displacement. The steps of the simplified method are proposed in reference to the enumerated list reported in Section 3. 1. The analytical model proposed by [PERSON] and [PERSON] [39] was adopted for the evaluation of the bending moments and the corresponding curvatures. It results a bending moment at yielding equal to \(M_{y}=94.5\) kNm and a curvature equal to \(\chi_{y}=1.18\times 10^{-2}\) m\({}^{-1}\). Referring to the ultimate capacity, it follows that at the ultimate concrete compressive strain, \(\varepsilon_{cu}=0.0035\), the compression steel is not yielding. The ultimate bending moment and curvature are \(M_{u}=102.8\) kNm and \(\chi_{u}=8.16\times 10^{-2}\) m\({}^{-1}\), respectively. 2. The axial force in the selected column was evaluated considering the tributary area of the column, that is, half length of the beam. The force results from the following expression \[\text{N}=n_{f}\left(\text{DL}+\text{LL}\right)\ell+g_{b}\left(\ell-w_{c}\right) n_{f}+g_{c}\left(n_{f}-1\right)h,\] (9) where DL and LL are the dead and the live loads, respectively, \(g_{b}\) and \(g_{c}\) are the linear weights of beam and column, respectively, \(\ell\) is the length of the beam, \(h\) is the interstorey height, \(n_{f}\) is the number of floors and \(w_{c}\) is the width of the column. For the present example, the axial force is \(\text{N}=181.9\) kN. 3. The upper bound theorem of the plastic analysis was applied and the load multiplier that defines the collapse load \(\lambda\left(\text{m}^{*}\right)\) is determined as \[\lambda\left(\text{m}^{*}\right)=\frac{1}{\ell}\left\{4n_{f}\text{m}^{*}+N \ell-n_{f}\left[\left(\text{DL}+\text{LL}\right)\ell+g_{b}\left(\ell-w_{c} \right)\right]\ell-g_{c}\left(n_{f}-1\right)h\ell\right\}.\] (10) Substituting the values of the yielding and the ultimate bending moments it results \(\lambda\left(M_{y}\right)=283.5\) and \(\lambda\left(M_{u}\right)=308.5\), respectively. 4. The chord rotation of beams end were computed through the analytical models proposed in the Italian Building Code [40; 41] for an equivalent cantilever: the elastic rotation is due to an elastic curvature increasing from zero (at the free end of the cantilever) to \(\chi_{y}\) (at the opposite end); the plastic curvature, that is, \(\chi_{u}-\chi_{y}\), is kept constant on the plastic hinge length [42]. According to  
Figure 8: A sketch of the test frame onto which the simplified method is applied. The removed middle column is reported in red, the removed side column in violet. The corresponding control nodes are marked with letters A\({}_{1}\) and A\({}_{2}\), respectively. The geometry and the reinforcement layouts of beams and columns are depicted on the right-hand side.