HW 6 - Flight 2120 failure analysis

When design for a cyclic load, if in one case, the designed reliability is 0.99, what is the approximate reliability factor which should be used according to the textbook?     1.0     0.9     0.81     0.75

A beginner reliability engineer did not realize that the failures of the system should be grouped by type instead of having them in one group. The system was observed to fail because of two types of failures: electrical (E) and mechanical (M). The failure data for E are 316, 138, 87, 923, 921, 1113, 1152, 577, 480, 1401 The data for M are 746, 1281, 1304, 1576, 1386, 671, 2106, 660, 1149, 425 The true data for E comes from an exponential distribution with mean = 1000 h, and the data for M comes from Weibull with γ = 2 and θ = 1000. a. What is the reliability expression for the true distribution? b. What is the reliability expression for the combined failures? c. Is the analysis of the engineer correct? Why?

True or False a) “If the most conservative fatigue criterion were used for infinite life design, the first cycle yield check is not needed.”  b) If an accurate value for the notch sensitivity is hard to find, I should consider the fatigue stress concentration factor higher than the stress concentration factor under static loading condition in design.  c) It is easier to find the asymptotic value of fatigue strength as the endurance limit on the S-N curve for a brass alloy (Cu70Zn30) than that of endurance limit on the S-N curve for an AISI 1045 steel.

A new electronic component for aircraft is tested to withstand temperatures as low as -50°F. One engineer suggests that the device cannot actually stand temperatures that low. What are the Type I and Type II errors? Type I: The device CAN NOT withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN.Type II: The device CAN withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN NOT.   Type I: The device CAN withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN NOT.Type II: The device CAN NOT withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN.    Is the answer the first or second bullet point?

Presented below are the quality ratings with regard to soil moisture availability. Erosion rating Moisture rating Erosion Rating MC (%) Rating Slope Slope rating Rating hazard class Low 1 0-5 4 0-5 1 Moderate 2 5-15 3 5-10 2 High 3 15-20 2 10-25 3 Very high 4 >20 1 >25 4 Quality ratings of three land units are tabulated below. Land quality rating Land unit Nutrient availability Trafficability 1 2 1 2 3 1 3 2 2 Additional characteristics of the land units areas follows: Land Permeability Land use M.C. Slope Clay V. Silt OM Structure % % Fine % % unit sand % 1 2 Slow Moderate Rangeland Arable 10 9 40% 10 30 4 blocky 18 8 30% 20 20 2 Coarse granular 3 Moderate to Settled 22 12 20% 15 50 3 Fine rapid granular Suitability Class Nutrient The suitability rating for land qualities for tomatoes and cabbage production is also indicated below; Suitability classes for Tomato production Trafficability Moisture Land Qualities Slope Erosion availability hazard S1 1 2 2 2 2 S2 2 3 3 2 S3 3 4 3 3 4 S4…

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List 3 of the main reliability design analysis techniques

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(a) A very important component of a system has a mean time between failures of 100 hr and a mean time to repair of 10 hr. Determine the inherent availability of the system.           (b) If the equipment in Q7 (c i) above is required to meet an inherent availability requirement of 0.985 and a mean time between failures of 100 hr. Calculate the permissible mean time to repair and if the system as a mean of the time intervals between corrective maintenance actions of 900 hrs and the sum of the mean corrective maintenance time intervals including supply downtime and administrative down time of 10 hrs Determine the operational availability of the system

1- Select the optimum choice for: high speed air craft skins material. Table 1: The properties of metallic materials E, GPa Max. use lemp, 'C | Density, Mg/cm' ay, MPa | Relative cost per unit sheet area Material 2.8 300 2 Aluminum alloy 200 69 200 7,8 350 350 700 Carbon steel Stainless steel 193 7.8 650 6 8.9 600 6 Nickel alloy Titanium alloy 207 1000 116 600 4.5 900 10 Table 2: Weighting factor Specific yield strength: Max, use temperature: Youngs modulus: 4 2