Simple bolted connection between structural members

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Reference no: EM131444433

Consider a simple bolted connection between structural members. In the structural system at hand, all bolts always carry equal loads. If there are n bolts in the connection, and the total load being carried by the connection is L [kN], then each bolt carries L/n.

For the critical failure mode, the load capacity of individual bolts can be modeled as a Normal random variable with mean μ = 100 [kN] and standard deviation σ = 2 [kN].

For a two-bolt connection, if both bolts were exactly average, the load capacity would be 2μ = 200 [kN]. However, the connection fails if either of the two bolts fail. Thus, if one of the bolts was weaker than average with an individual load capacity less than μ, say 95 [kN], then the load capacity of the connection would be 2 · 95 = 190 [kN], which is less than 2μ.

The load capacity of an n-bolt connection is n times the strength of the weakest bolt. Thus, the strength of the weakest bolt determines the load capacity of the connection.

  1. What is the average and standard deviation of the load capacity for one-bolt connections? (Yes, this is the obvious answer.)
  2. What is the average and standard deviation of the load capacity for two-bolt connec- tions? (No, there is no simple analytic formula for this question.)
  3. Estimate and plot and/or tabulate the averages and standard deviations of the load capacity of n-bolt connections, for n = 1, 2, . . . , 10. (Yes, you will want to use the computer on this problem.)
  4. Plot and/or tabulate the number of bolts in a connection "n" (on the abscissa: hor- izontal axis), versus the probability that the load capacity of a connection is greater than or equal to nμ (on the ordinate: vertical axis). The plot and/or table should include the range n = 1, 2, . . . 10. Hint: this is not a direct extension of the previous parts. This question can be answered using a relatively simple analytic formula.

Reference no: EM131444433

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