PSI - Issue 71

A. Kumar et al. / Procedia Structural Integrity 71 (2025) 453–460

455

3 U Fw F AE   = =  3

(2)

3

where A is the RCE's cross-sectional area

3   3

E

=

(3)

3

where s A is the RCE's projected area

3 A E E A =

(4)

s

(3 4 ) 4 sin ( A t h l A l = + =

(5)

cos ) 

h l  −

Where

s

4 3    +       h l 

t E E   =  

(6)

3

h

l  





4sin

cos  −    l 



Out-of-Plane Shear modulus (G 13 )

1.2.

The shear force 13 F is loaded in direction- 1 . There is no shearing in the cell wall's perpendicular direction, shear stress 13  is transmitted to the inclined cell walls.

Fig. 3. (a) Shear force 13 F applied on the whole structure in direction -1; (b) Shear force 13 F acting on an RCE.

RCE force 13 F is evenly distributed to the four inclined walls. So,

13 4sin F

(7)

F

=

AB



Where AB F is the produced shear force in inclined cell wall AB since shear force 13 F ( Fig. 3(b)). The RCE's shear strain energy from the four inclined cell walls reflects the total strain energy from applied shear force 13 F . 2 AB

0 2 w F

4 = 

d

(8)

U

z

AG

Where A lt =

(9)