PSI - Issue 79

Albena Doicheva et al. / Procedia Structural Integrity 79 (2026) 370–378

375

Fig. 4 shows that as the modulus of elasticity of concrete increases, the support response H 1 increases while H 2 decreases. The support response H 3 is larger for higher concrete modulus until the crack opening around b/h=2.1 to 2.3, where there is an extremum and then decreases rapidly to values smaller than those before the crack opening. 6. Shear force The magnitudes of the forces 3 2 1 , and H H H are already known. Equation (1) becomes:

3 C c VTCCVTTV HHHV      2 1 j S C C

(10)

Equation (5.22) of Eurocode (2004) gives us the magnitude of the shear force:   1 2 jhd Rd s s yd c V A A f V    

(11)

where yd f [kN/cm

2 ] — design value of the yield strength of steel;

Rd  should not be taken less than 1.2. The comparison of Equations (10) with (1) and (11) is expressed in the comparison shown in Equations (12).



M M

?

?





;

(12)

H H H

3 s yd H H H A A f      2 1 1 2 Rd s

b    

b



3

2

1

j

j

b

b

2 / 6

b M qL 

is the moment in the beam on the face of the column.

where

4kN/cm' q  — for all numerical results on section 6. The spring coefficients 1 k , 2 k and 3 k of the three supports are reduced to the tension/compression stiffness of the beam by the 1  , 2  and 3  multipliers. When we consider the rigid support between structural elements using static schemes, we assume that the connections between them do not allow the sections to move and to rotate. This was implied in the equation of equilibrium - the sum of moments about the support is zero. For this reason, multipliers 1  , 2  and 3  will be assumed equal to 1. 2 37,5kN/ cm yd f  — design value of the yield strength of steel, Doicheva (2023c).

h=75 cm

h=75 cm

1,75 2,25 2,75 3,25 3,75 4,25 4,75

1,75 2,25 2,75 3,25 3,75 4,25 4,75

(H1+H2+H3)/qL/2 2×(Mb/jb)/qL/2 Eurocode/qL/2

(H1+H2+H3)/qL/2 2×(Mb/jb)/qL/2 Eurocode/qL/2

(H1+H2+H3)/qL/2

(H1+H2+H3)/qL/2

h/b

h/b

1,9

2,1

2,4

2,7

a) 3,2

3,8

4,8

6,4

9,8

1,9

2,1

2,4

2,7

3,2

b) 3,8

4,8

6,4

9,8

20,7

20,7

Figure 5. Comparison of the parameters on Equation (12) calculated by Equations (5) – (7) : ( a ) 2 1 4200kN/ cm E  . Figure 5 shows the variation in the sum with respect to the parameters of the three support reactions, /2 /2 H qL H qL H qL   , calculated by Equations (5) – (7), while the crack between the beam and the column grows. The comparison is made by Equation (12). The graphs show that the proposed new model for calculating shear force gives us not only its most unfavorable value but also makes the shear force traceable throughout the crack development process. 2 1 1700kN/ cm E  ; ( b ) 1 2 3 /2