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 mm² ; 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

Y₃ = 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 Y₃ from curve (2) can be read as 1.87.)

∴