Question

Qu.2 Chapter 8: Failure (2096) crack of length of 1.0 mm was detocted on the surface of the shaft of wind turbine (see Figure 3), made from 4340 steel. The diameter of the shaft of wind turbinc is cqual to 30 mm. Assuming subjected to a maximum load of 20,000 N during operation. 10.2 mm and Y ?! the radius of the curvature of the crack is 1x What is the maximum stress(o.) experienced by the shaft during operations? (a) (b) (c) What is the maximum stress (o) experienced by the crack tip during operations? What is the stress concentration factor, K,? (d) Given that the shaft is made of an alloy with a Kic of 87 4 MPav/m and using an answer you calculated in (b) as the operating stress, should you immediately stop the wind turbine from operating to prevent a potential catastrophic failure? Explain your answer. 2) Assuming minimum load of 2,000 N determine i) the mean stress experienced by the shaft of wind turbine ii) the stress amplitude ii) maximum stress amplitude it operates in order to not fail prematurely if his service life is 265,000 cycles ( Figure 4). 700 4340 se 400 G5 steel 300 200 2014-T6 AU aloy 100 0 10 10" 10 10 Figure 3: Wind turbine Figure 4: Maximum Stress Amplitude vs Cycles to failure

Answer 1

Given data c = 1mm y = 1.1 d = 30mm P_max = 20, 000N P_t = 1 times 10^-2mm 4340 steel that means k_10 = 99 MPa squareroot m a. Maximum stess 60 = P/A = 20, 000 times 4/Pi times 0.03^2 = 28.294 Mpa b. 6_max1 = 60 [1 + 2 (c/e_t) ^1/2] = 28.294 [1 + 2 times (1/10^-2) ^1/2] 6_max1 = 594.174 MPa 6_c lessthenorequalto K_1c/y squareroot Pia = 99/1.1 squareroot Pi times 0.001 = 1605.71 MPa As 6_c > 6_max, shapt fail catastrophically P_min = 2500N 6_min = 2000 times 4/Pi times 0.03^2 = 2.829 MPa 6_mean = 6_max + 6_min/2 = 594.174 + 2.829/2 = 898.50MPa 6_a = 6_max - 6_min/2 = 594.174 - 2.829/2 = 295.67 MPa from fig N = 265, 000 cycle for 1045 steel curve stress = 380 MPa