Calculate the crack Stress Intensity Factor:
If on the beam specimen of last problem two loads, each equal to 500 N, are applied at distance of 25 mm from central plane, calculate the crack SIF.
Solution
L/2 = 25 mm; a/ W = 0.2; Z = 1041.67 mm^{2} ; S = 100 mm
M= P (S - L)/2 where, P is the load which is 500 N.
∴ M = 500 (100 - 50)/2 = 12500 N-mm
∴ s= M/ Z = 12500 / 1041.67 =12 N-mm
Y_{3} = 1.99 - 2.47(a/ W) + 12.97 (a/W)^{ 2} - 23.17 (a/W)^{ 3} + 24.8(a/ W)^{ 4}
= 1.99 - 2.47 (0.2) + 12.97 (0.2)^{2 }- 23.17 (0.2)^{3} + 24.8 (0.2)4
= 1.99 - 0.494 + 0.5188 - 0.1854 +0.04
= 1.87
(Check from Figure 7 that Y_{3 }from curve (2) can be read as 1.87.)
∴