## Appendix A Solved Example  
[PERSON] [12] tested \(1/3\)-scale reinforced concrete interior beam-column joint specimen (NC1). The specimen and reinforcement used are detailed in Figure 11. The length of each beam from the face of the column, \(l_{bi}\) (i.e., \(l_{b1}\) and \(l_{b2}\)) was 600 mm, and the total length of the beam was 1350 mm between the point of loading. The total length of the column, \(l_{c}\), was 800 mm. Four longitudinal steel bars of 8 mm diameter were used for both the beam and the column. Steel stirrups, 6 mm in diameter, spaced at 100 mm intervals, were used as shear reinforcement. The beams and the column had a rectangular cross-section of 100 \(\times\) 150 mm with \(h_{c}=h_{b}=150\) mm and \(b_{b}=b_{c}=100\) mm. The concrete cylinder's average compressive strength was 39 MPa. The yield strength of the tensile, shear, and compression reinforcing bars was about 500 MPa. All beams were subjected to cyclic reverse loading. The column was loaded with a steady axial load of 100 kN. The joint panel was deficient in transverse reinforcement, which resulted in a premature shear failure; as a result, the structural system's seismic performance was adversely affected. To make possible further ductile flexural failure and prevent shear failure, the proposed approach is used to determine the correct quantity of FRP fibers for the joint panel. By using the principal stress approach Equation (11), the shear stress capacity of the as-built joint, \(v_{jh}\), may be calculated as given below:  
\[p_{t,tot}^{exp}=\frac{-\sigma_{c}}{2}+\sqrt{\left(\frac{-\sigma_{c}}{2}\right) ^{2}+{v_{jh}}^{2}}\]  
\[v_{jh}=p_{t,tot}\cdot\sqrt{1+\left(\frac{\sigma_{c}}{P_{t,tot}}\right)}\]  
Considering \(p_{t,tot}\) is equal to \(p_{t,c}\), and \(p_{t,c}\) can be computed by Equation (12). \(p_{t,c}=0.29\) (\(\ell c\))\({}^{0.5}=0.29\) (\(39\))\({}^{0.5}=1.81\) MPa. Where \(\sigma_{c}=\)\(f_{a}\) = axial stress = Axial load/Acol = \(\frac{N}{bc\cdot bc}\)  
\[\sigma_{c}=\frac{\text{N}}{b_{c}\cdot h_{c}}=\frac{100,000}{(100)\cdot(150)}= 6.67\text{ MPa}\]  
\[v_{jh}^{d}=1.81+\sqrt{1+\left(\frac{6.67}{1.81}\right)}=3.97\text{ MPa}\]  
The design shear stress \(v_{jh}^{d}\) is the joint shear stress related to the flexural yielding of the weaker beam and column, representing the goal value for the design of the FRP strengthening. This conforms to an approach for capacity design. Here in the considered case, the initial yielding of the specimen corresponds to the beam yielding matching to the \(v_{jh}^{d}=7.53\) MPa joint shear stress calculated with Equation (8). Due to the shear strength (3.97 MPa) being 52.7% of the design value of the as-built joint, strengthening the joint panel with FRP is necessary (i.e., 47.3 percent increase is required in the actual shear capacity). In this example, the joint panel of uniaxial GFRP fabric with fibers inclined at 0\({}^{\circ}\) at the axis of the beam is supposed to be strengthened with one layer (\(n_{\text{t}}=1\)). The modulus of elasticity, \(E_{f}\), is taken as 19.2 GPa. Dry fiber thickness, \(tf\), in each direction is approximately 2 mm. Due to the absence of orthogonal beams to the load plane, shear strengthening of the joint panel can be carried on both sides (\(n_{s}=2\)). At maximum capacity, the inclination of the crack, \(\theta\), may be calculated by Equation (14) to determine the FRP system's equivalent area by Equation (16). \[\theta=const=atan\bigg{(}\frac{h_{b}}{h_{c}}\bigg{)}=atan\bigg{(}\frac{150}{1 50}\bigg{)}=45^{{}^{\circ}}\]  
\[A_{f,eq}=n_{\text{t}}\cdot n_{s}\cdot t_{f}\cdot h_{b}\cdot sin\theta\text{ for }\beta=0^{{}^{\circ}}\]  
\[A_{f,eq}=1\cdot 2\cdot 2\cdot 150\cdot\sin\bigg{(}45^{{}^{\circ}}\bigg{)}=424.26\text{ mm}^{2}\]  
The effective FRP strain may be estimated by Equation (22) after the equivalent FRP area is known without assuming the joint panel's starting damage (\(C_{LD.}=1\)). Bonding conditions of the FRP system at the ends (\(C_{MA. }=1\)) are not enhanced through mechanical anchoring.