A grade 30 gray cast iron is subjected to a load F applied t

A grade 30 gray cast iron is subjected to a load F applied to a 1 by 2/3 - in cross-section line with a 1/4, in diameter hole drilled in the center as depicted in Fig. 6-31a. The surfaces are machined. In the neighborhood of the hole, what is the factor of safety guarding against failure under the following conditions: The load F = 1000 lbf tensile, steady. The load is 1000 lbf repeatedly applied. The load fluctuates between - 1000 lbf and 300 lbf without column action. Use the Smith-Dolan fatigue locus. The grade 30 cast-iron par in social fatigue with (a) its geometry displayed and 9b) is designer\'s fatigue diagram for the circumstances of Ex.6 - 13. Mechanical Engineering Design Some preparatory work is needed. From Table A-24. S_ur = 31 kpsi. S_ac = 109 kpsi, k_a k_b S_e\' = 14 kpsi. Since k_c for axial loading is 0.9, then s_c = (k_a k_b S_e\')k_c = 14(0.9) = 12.6 kpsi. From Table A-15-1, A = r(w - d) = 0.375 (1 - 0.25) = 0.281 in^2, d/w = 0.25/1 = 0.25, and K_1 = 2.45. The notch sensitivity for cast iron is 0.20 (sec p. 304), so K_f = 1 + q(K_1 - 1) = 1 + 0.20(2.45 - 1) = 1.29 Since the load is steady, sigma_a = 0, the load is static. Based on the discussion of cast iron is Sec. 5 - 2, K_t, and consequently K_f, need not be applied. Thus, sigma_m = F_m/A = 1000 (10^- 1)/0.281 = 3.56 kpsi, and n = S/sigma = 31.0/3.56 = 8.71 F_d = F_ = F/2 = 1000/2 = 500 Ibf sigma = sigma = K_f F_/A = 1.29(500)/0.281 (10^- 3) = 2.30 kpsi r = sigma _/sigma_ = 1 From Eq. (6 - 52). S_a = (1)31 + 12.6/2 [- 1 + squareroot 1 + 4(1) 31(12.6)/[(1)31 + 12.6]^2] = 7.63 kpsi n = S_e/sigma_a = 7.63/2.30 = 3.32 F_a = 1/2|300 - (- 1000)| = 650 lbf sigma_a = 1.29(650) - 0.281 (10^- 3) = 2.98 kpsi F_infinity = 1/2[300 + (- 1000)] = - 350 lbf sigma_infinity = 1.29(- 350)/0.281 (10^- 1) = - 1.61 kpsi r = sigma_a/sigma_infinity = 3.0/- 1.61 = - 1.86 From Eq. (6 - 53). S_a = S_c + (S_c/S_infinity - 1) S_infinity = S_infinity/r. It follows that S_a = S_d/1 - 1/r (S_/S_ - 1) = 12.6/1 - 1/- 1.86 (12.6/31 - 1) = 18.5 kpsi n = S_a/sigma_a = 18.5/2.98 = 6.20 Figure 6 - 31 b shows the portion of the designer\'s fatigue diagram that was constructed.

Solution

As the load is static in first part the factor of safety that is required is very low but in the above solution the factor of safety is quite high considering the static load condition,so even a low factor of safety could have been suitable for this condition.

B. In second part as the load is repeatidely applied ,so there is more chance of failure of the system ,so a high factor of safety is required in this situation but in solution above the factor of safety is less considering the repeatedly applied load.

C. In the third case the load applied is fluctuating load and this kind of load cause a much higher failure of system due to fatigue ,and so the factor of safety required should be highest in the above three cases.