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

369

7

1. Determine the value of  for which V Rdj ( eq.v ) equates the required strength. Fig. 5(b) shows that, in this example, it happens when  <  and using Eq. (3) we have

Ed N A

+ 

sv sv

, Rdj v V V =

V

+

=

(14)

,min

Rdj

Edj

cot



that gives  = 0.598 2. Using the above indicated value of  , determine the value of A sh for which V Rdj ( eq.h ) equates the required strength. Being  <  and using Eq. ( 2 ) we have

f



cot

cot  − 

cd

, Rdj h V V A = ,min Rdj

A

V

sh sh +  +

=

(15)

, j ef

Edj

1.6

2 + 

1 cot

that gives A sh = 10.11 cm

2 . The result is shown in Fig. 5(c).

1600 V [kN] 2000

1600 V [kN] 2000

1600 V [kN] 2000

VRd,h VRd,v VRd,h max VRd,v max VRd,x VEd teta=beta

VRd,h VRd,v VRd,h max VRd,v max VRd,x VEd teta=beta

VRd,h VRd,v VRd,h max VRd,v max VRd,x VEd teta=beta

V Rd,j

V Rd,j

V Rd,j

 = 

 = 

 = 

1200

1200

1200

800

800

800

(a)

(b)

(c)

400

400

400

0

0

0

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40







Fig. 5. Joint with beams 30  60, columns 40  70,  =0.16, V Edj = 1000 kN: (a) A sh = A sv =0; (b) A sh =0, A sv =8.04 cm 2 ; (c) A

sh =10.11 cm

2 , A

sv =8.04 cm

2 .

Furthermore, the procedure may be repeated by assigning a set of values for A sv and calculating the corresponding values of required A sh , obtaining the curve showed in Fig. 6(a), which allows the designer to choose the couple of values he believes preferable. It is also possible to draw more curves, each corresponding to a different value of V Edj , as shown in Fig. 6(b).

25

25

A sh [cm 2 ]

A sh [cm 2 ]

couple of values A sh -A sv necessary

(a)

(b)

couple of values A sh -A sv necessary selected values A sh -A sv

20

20

15

15

1100 kN V Edj = 1200 kN

10

10

1000 kN

900 kN

5

5

V Edj = 800 kN

0

0

25 A sv

25 A sv

0

5

10

15

20

0

5

10

15

20

Fig. 6. Couple of values A sh and A sv necessary for V Edj = 1000 kN (a) and for more values of V Edj (b)

5. Conclusions A capacity model for RC beam-column joint resistance, alternative to that developed by Fardis (2021) and incorporated in the new draft of Eurocode 8 (2020), is formulated in this paper. The model relies on the same resisting mechanisms proposed by Fardis, the diagonal strut with variable inclination and the two residual triangular parts of joint, to allow the joint to resist the horizontal shear force. As an additional assumption with respect to the original