Issue 55

M. M. Konieczny et alii, Frattura ed Integrità Strutturale, 55 (2021) 241-257; DOI: 10.3221/IGF-ESIS.55.18

  2 :

b) in the applied layer of the plate (titanium)

3 m

     2 2 h z h a

 2

  

z

, dla

(13)



    3 3

 

2

C h a h





2

2

The equivalent von Mises stress was determined from the relationship: a) for the steel layer:

          1 1 1 1 1 2 2 red r r

(14)

b) for the titanium layer:

          2 2 2 2 2 2 2 red r r

(15)

In the formulas for determining the intensity of bending moments in the radial direction (2) and circumferential direction (3) as well as in the formulae applied for determining the stresses in the radial direction (10) and (11) and circumferential directions (12) and (13) h 1 and h 2 determine the position of the layer inert sections of the plate (Fig. 3):

2

2

 2 E H E a E H E a   1 2



h

(16)



2

1

2

  1 2 h H h

(17)

where: h 1 , h 2 – values determining the state of the inactive layer in the cross section of the steel plate; E 1 , E 2 – Young's modules

for the steel and titanium parts; H, a – thickness of steel and titanium plate, respectively. Inactive layer of the plate – it is a layer in which the deformations are equal to zero [15, 16].

Bimetallic perforated plate freely supported on the perimeter and loaded with a concentrated force P acting perpendicular to the surface In the first case, the analysis the state of stress in a bimetallic perforated plate loaded with a concentrated force applied perpendicularly to the surface with the value P = 10 kN was presented below. The geometry of the plate is presented in Fig. 2. The concentrated force P was applied centrally to the pressure stamp with a diameter of b 1 = 14 mm. A free support of the plate on the outer perimeter was assumed. The method of support and loading the plate is shown in Fig. 4. The boundary conditions are assumed to take the following form (Fig. 4):  0 r m . In this case, the coefficients k 1 , k 2 , k 3 inside the function of the lateral force intensity t(r) (6) are equal to, respectively: a) for the radius r – ½ b = 6 mm – plate deflection angle φ = 0; b) for the radius r = ½ D = 150 mm – radial moment strength

P ; k

k 1 = 0; k 2 =

3 = 0

 2

The roots of the Eqn. (8) 1 λ and 2 λ are:     1 λ 1.127 ;      2 λ 1.065

246