## 5 Local Strengthening through FRP Jacketing
#### Beam-Column Joint Strengthening  
External beam-column joints are not laterally supported on four sides by beams, hence require horizontal shear reinforcement to prevent deterioration due to cracking induced by the high shear stress demand. This demand is the result of the steep moment gradients as they facilitate reversal of moment from one face of the member to the other. In old structures, r.c. beam-column joints rarely had in their body horizontal stirrups. So, it is deemed necessary to retrofit such a critical element of the structure. As it is observed in the plan layout of Figure 1a, T- and L-shaped (knee or corner) joints exist. The strengthening design procedure is presented in detail for the L-shaped joint B1-C1-B7 of the ground storey; apparently all other L-shaped joints of the ground floor follow the same requirements. It is noted that in Chapter 8 of fib Bulletin 90 [4] two design approaches are presented. In this example, Approach 1 is implemented aiming to determine the required amount of horizontal FRP reinforcement through its function as added shear reinforcement in the joint panel. Approach 1 generally leads to significant amount of FRP reinforcement because, in the interest of conservatism, neglects any confining contribution, which could in turn lead to a retrofit inferior to expectations. The first step is to calculate the sums of yield moments in the beams and in the columns framing into the joint. Hence, \(\Sigma\mathrm{M}_{\mathrm{yb}}\) is the sum of yield moments of the beams that frame into the joint and \(\Sigma\mathrm{M}_{\mathrm{yc}}\) is the sum of yield moments of the columns that frame into the joint:  
x-x direction--weak axis (C1-B1): Beam B1-left has yield moment 134 kNm. The yield moments of the column below and above the joint are 53 kNm (N\({}_{\mathrm{Ed}}\) = 181.3 kN) and 48 kNm (N\({}_{\mathrm{Ed}}\) = 120.8 kN), respectively (the cross-section analysis was performed using Response2000 [11]). Hence, \(\Sigma\mathrm{M}_{\mathrm{yc}}\) = 53 + 48 = 101 kNm \(<\Sigma\mathrm{M}_{\mathrm{yb}}\) = 134 kNm. By being columns weaker than beams, they define the magnitude of shear stress in the joint. Hence, the vertical shear force V\({}_{\mathrm{j}\
u}\) is derived by following the Greek Code for Assessment and Retrofit [36], from Equation (13) as:  
\[\mathrm{V}_{\mathrm{j}\
u}=\Sigma\mathrm{M}_{\mathrm{yc}}(1/\mathrm{jd}_{ \mathrm{c}}-1/\mathrm{L}_{\mathrm{b,n}}\cdot\mathrm{H}_{\mathrm{n}}/\mathrm{H })+1/2\,|\,(\mathrm{V}_{\mathrm{g}\leftrightarrow\mathrm{b}\mathrm{d}}\mathrm{ b})_{1}-(\mathrm{V}_{\mathrm{g}\leftrightarrow\mathrm{b}\mathrm{d}}\mathrm{ b})_{\mathrm{r}}\,| \tag{13}\]  
where \(\mathrm{j}\mathrm{d}_{\mathrm{c}}\) is the internal lever arm of the column section, \(\mathrm{H}_{\mathrm{n}}\) and \(\mathrm{H}\) are the theoretical and clear storey heights, and \(\mathrm{L}_{\mathrm{b,n}}\) is the theoretical half span of the beam. The beams are considered as fully fixed at their ends. The load from the slabs is assumed to be transferred to beam B1. Conservatively, it is assumed that the loads transferred from the slab to B1 correspond to a quarter of the area of the slab. Hence, the total linear load for the seismic combination is estimated as \(\mathrm{g}\) = 9.75 kNm, \(\mathrm{q}\) = 4.38 kNm and \(\mathrm{g}\) + 0.3q = 11.06 kNm, hence \((\mathrm{V}_{\mathrm{g}\leftrightarrow\mathrm{b}\mathrm{d}\mathrm{b}}\mathrm{ b})_{\mathrm{l}}\) = \((\mathrm{g}+0.3\mathrm{q})\mathrm{l}/2\) = 33.19 kN. Then, by implementing Equation (13) results to \(\mathrm{V}_{\mathrm{j}\
u}\) = 510.8 kN.