Answer:
[tex]K_{I}=5.21 MPa\sqrt{m}[/tex]
Explanation:
given data
Load P = 35 kN
Width of bar W = 50.8 mm
Breadth of bar B = 25 mm
Ratio of crack length to width α = a/W = 0.2
solution
we get here KI for a rectangular bar that is express as
[tex]K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}[/tex] ................................1
here Y is the geometrical function
so
Y = [tex]\frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}[/tex]
Y = [tex]\frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}[/tex]
Y = [tex]\frac{0.8792}{0.89}[/tex]
Y = 0.9878
so put here value in equation 1
[tex]K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}[/tex]
[tex]K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319[/tex]
[tex]K_{I}[/tex] = 5210.45 × 10³ [tex]Pa\sqrt{m}[/tex]
[tex]K_{I}[/tex] = 5.21 MPa [tex]\sqrt{m}[/tex]