Calculate sif of crack - beam:

An edge cracked beam carries crack in its central plane whose lengths is 5 mm. A load of 1000 N is applied opposite to crack so that crack would tend to open in bending. Calculate SIF of crack if the beam has following dimensions :

W = 25 mm, B = 10 mm, S = 100 mm, i.e.  S /W =4

Solution

Z = Modulus of section = (1 /6 )BW ² =1/6 x 10 x(25)₂ =1041.67 mm²

M = Bending moment at central section

=PS/4 =(1000x100) /4= 25000 N-mm

s = Gross stress =  M /Z = 25000 /1041.67 = 24 N/mm²

Y₁ is calculated from Eq. (7)

a /W=  5 /25 = 1/5 ;(a/W)²  = 1/25 ; (a/W)³ = 1/125 ;  (a/W)⁴ = 1/625

∴          Y₁ = 1.93 - 3.07 x(1/ 5)+ 14.53 x(1/ 25 )- 25.11 x(1/ 125) + 25.8 x(1/ 625)

                    =1.93 - 0.614 + 0.5812 - 0.2 + 0.04 =1.737

∴          K _I  = 24x1.737 x √5=  95.13 N/ mm²

 (Note that from Figure 5.7 for a 3 pt. bend specimen at  a /W= 5 is read 1.73).