PSI - Issue 72

Albena Doicheva / Procedia Structural Integrity 72 (2025) 243–251

248

Table 1. Comparison of the parameters of the three support reactions determined by Equations (5)-(7) for the two values of E 1       1 1 1 1700 4000 100% 1700 H E H E H E     Section / h b 0 g m  1,0 g m  2,0 g m  3,0 g m  1 H 2.0 -43.46 -5.23 -5.20 10.91 3.8 -42.73 -5.16 -4.88 12.54 50 -41.99 -5.06 -4.45 14.21 2 H 2.0 12.82 9.31 12.00 10.21 3.8 13.29 9.95 13.89 11.19 50 13.75 10.62 16.11 12.19 3 H 2.0 11.85 8.16 8.69 8.91 3.8 12.28 8.63 9.55 9.89 50 12.72 9.10 10.38 10.91 Table 1 shows the differences between the values of the three support reactions for the two values of the elastic moduli 2 1 1700kN/ cm E  and 2 1 4000kN/ cm E  and for values of 0 g m  , 1,0 g m  , 2,0 g m  , 3,0 g m  .

h=25 cm, g=0 to 3,5m

h=25 cm, g=0 to 3,5m

H1/P_g=0m H1/P_g=1,0m H1/P_g=2,0m H1/P_g=3,0m H1/P_g=3,5m

H2/P_g=0m H2/P_g=1,0m H2/P_g=2,0m H2/P_g=3,0m H2/P_g=3,5m

-0.30 -0.10 0.10 0.30 0.50

4.50

2.50

H1/P

H2/P

0.50

2.0

2.6

3.8 h/b

7.1

-1.50

50.0

2.0

2.6

3.8 h/b

7.1

50.0

a)

b)

2 1 1700kN/cm E  a) 1 H ; b) 2 H , calculated by

Figure 4. The parameters of the support reactions for a beam with symmetrical cross-section and

Equations (5) – (7).

2 andH H , calculated by Equations (5) – (7),

Figure 4 shows the variation in the parameters of the support reactions 1 while the crack between the beam and the column grows, and 2 1 1700kN/cm E  .

0 g m  to

2,0 g m  . At

3,0 g m  it decreases to negative values, which

1 H increases from

The support reaction

/ 2 3,5 g L m   it becomes zero.

indicates that it becomes tensile, and at

0 g m  . As the moments move and g increases,

2 H decreases.

2 H has the largest value at

The support reaction

3,0 g m  , the support reaction

/ 2 3,5 g L m   . The support

2 H has increased again and reaches a value of zero at

At

3 H has values and behavior similar to that of

2 H .

reaction

6. Shear force The magnitudes of the forces H 3 , H' 2 and H' 1 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)