2002 Quiz 12-11 Solutions 2. rectangular block with dimensions labeled: b, w, and 25w Given: σy = 1400 MPa Kic = 100 MPa√m (est.) Estimated plastic zone size: rp = (Kic / σy)^2 = (100 / 1400)^2 ≈ 0.81 mm SSY: in-plane dimensions (a, w, d, etc) ≥ 15rp = 12.15 mm Plane Strain: B ≥ 15rp = 12.15 mm 2.2 Evidently, choose plate thickness B ~ 12.5 mm as a minimum. Similar CONS: a/w ≈ 1/2 ⇒ d ≈ w-d ⇒ W = 2d > ~12.5 mm W ≥ 25.4 mm = 1 in rectangular block with dimensions labeled: b, w, and 25w Assume SLOT WIDTH ≈ .1 to .2 W (Cross Area) ≈ (1.25 W)(1.2 + .1 to .2) W GROSS VOLUME ≈ B W² (1.25) = (12.5) (156) 2 23 min Recall that there remains uncertainty in the actual value of σmin in the cycle. Thus, there is also uncertainty in the actual value of Δσ to be used in the fatigue propagation analysis. Observe from the integrated, constant-Q and constant Δσ equation, for m ≠ 2, that with material properties, di, and af fixed, N ∝ (Δσ)^-m We can use this proportionality to infer an actual value for Δσ, based on our previous calculations and on the actual propagation life of N = 190,000 cycles. 4.6 [Enlarged] With this revised estimate of actual σmin, the observed fatigue crack propagation life of only 140,000 cycles is easily understood.