PSI - Issue 44

Francesca Barbagallo et al. / Procedia Structural Integrity 44 (2023) 363–370 Francesca Barbagallo et al. / Structural Integrity Procedia 00 (2022) 000 – 000

366

4

 h from the relations between tensile and compressive strains and then  sh ; from this it is possible to evaluate  sv from equilibrium conditions and then once again  v to start a new iteration. This iterative procedure is easily implementable, but it doesn’t allow a close formulation to solve the problem. 3. Critical examination and alternative proposal to Fardis and Eurocode 8 formulations According to Fardis formulation, for each value of  the design (horizontal) shear resistance is given by the lowest value between those obtained by horizontal and vertical equilibrium. Fig. 2(a) plots V Rdj ( eq.h ) and V Rdj ( eq.v ) as a function of  , while Fig. 2(b) plots V Rdj (  )=MIN( V Rdj ( eq.h ) ; V Rdj ( eq.v ) ). V Rdj is the maximum value read in this last plot. The trend of both curves shows discontinuities in value and slope due to the variation of  sh and  sv , which differ from f yd in the central part of the plot. If we assume  sh =  sv = f yd , the trend of the curves becomes regular, as shown in Fig. 3(a). Nevertheless, in this example, the value of V Rdj remains unchanged, as shown in Fig. 3(b). To consider the actual value of  sh and  sv in evaluating the shear strength of a beam-column joint is not much complex, requiring just an iterative procedure, but it makes less easy the proposal of design criteria. For this reason, the authors suggest assuming  sh =  sv = f yd . To check if this assumption may have significant influence on prediction of shear strength of the joint, a large set of beam-column joints, with different geometry, axial force and horizontal or vertical joint reinforcement, is analyzed. The features of the analyzed joints are resumed in Table 1.

3000

3000

 = 

 = 

(a)

(b)

VRdj,h VRdj,v V Rdj(eq.h) V R j(eq.v)

VRdj,v Rdj (  )

2000

2000

V [kN]

V [kN]

1000

1000

V Rdj

0

0

0.00

0.31

0.63

0.94

1.26

1.57

0.00

0.31

0.63

0.94

1.26

1.57

 [rad]

 [rad]

Fig. 2. Graphical application of Fardis model: Beams 30  60, columns 40  70, A sh = A sv =9 cm

2 ,  =0.16: (a) V

Rdj ( eq.h ) and V Rdj ( eq.v ) ; (b) V Rdj (  ).

3000

3000

 = 

 = 

(a)

(b)

VRdj,h VRdj,v V Rdj(eq.h) V Rdj(eq.v)

VRdj,v Rdj (  )

2000

2000

V [kN]

V [kN]

1000

1000

V Rdj

0

0

0.00

0.31

0.63

0.94

1.26

1.57

0.00

0.31

0.63

0.94

1.26

1.57

 [rad]

 [rad]

Fig. 3. Graphical application of the proposed model: Beams 30  60, columns 40  70, A sh = A sv =9 cm

2 ,  =0.16: (a) V

Rdj ( eq.h ) and V Rdj ( eq.v ) ; (b) V Rdj (  ).

Table 1. Data set of beam-column joints.

Geometry (beams and columns) Beams 30  50, column 30  60 Beams 30  60, column 40  70 Beams 30  70, column 40  70 Beams 60  28, column 30  70 Beams 30  50, column 30  40 Beams 30  60, column 30  40

Axial load ratio  = N Ed / A c f cd

Horizontal A sh and vertical A sv joint reinforcement

0.00 0.16 0.32 0.48

A sv = A sh =0.5% A beam

A sv =0.42% A column + A sh =0, 0.5 A sv , A sv , 1.5 A sv A sh =0.5% A beam + A sv =0, 0.5 A sh , A sh , 1.5 A sh

A sv =0.42% A column + A sh =3 A sv A sh =0.5% A beam + A sv =3 A sh