Two parts are to be assembled in a way that if one part fails, the entire assembly fails. Each of the parts have undergone exten
sive individual analysis and you are provided with statistics on the stresses that are expected. Specifically, you are provided with the mean and standard deviation of the stress, and the yield stress of the material for each part. Failure of a part will be defined as a part experiencing stress exceeding the yield stress. What is the probability that the assembly fails? (Hint: Use Table A-10 in the textbook to find the probability of failure for each part.)
Part #1: σnom = 260 MP a, U (σ) = 25 MP a, σyield = 310 MPa
Part #2: σnom = 475 MP a, U (σ) = 5 MP a, σyield = 490 MPa
Part #1: σnom = 260 MP a, U (σ) = 25 MP a, σyield = 310 MPa
Part #2: σnom = 475 MP a, U (σ) = 5 MP a, σyield = 490 MPa
1 answer:
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Answer:
radius = 9.1 × m
Explanation:
given data
applied load = 5560 N
flexural strength = 105 MPa
separation between the support = 45 mm
solution
we apply here minimum radius formula that is
radius = .................1
here F is applied load and is length
put here value and we get
radius =
solve it we get
radius = 9.1 × m