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 mm^{2}
M = Bending moment at central section
=PS/4 =(1000x100) /4= 25000 N-mm
s = Gross stress = M /Z = 25000 /1041.67 = 24 N/mm^{2}
Y_{1} 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
∴ Y_{1} = 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/ mm^{2 }
(Note that from Figure 5.7 for a 3 pt. bend specimen at a /W= 5 is read 1.73).