Consider a simple bolted connection between structural members. In the structura
ID: 3181887 • Letter: C
Question
Consider a simple bolted connection between structural members. In the structural system at hand^6, 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 mu = 100 [kN] and standard deviation sigma = 2 [kN]. For a two-bolt connection, if both bolts were exactly average, the load capacity would be 2mu = 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 mu, say 95 [kN], then the load capacity of the connection would be 2 middot 95 = 190 [kN], which is less than 2mu. 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. What is the average and standard deviation of the load capacity for one-bolt connections? (Yes, this is the obvious answer.) What is the average and standard deviation of the load capacity for two-bolt connections? (No, there is no simple analytic formula for this question.) 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.) Plot and/or tabulate the number of bolts in a connection "n" (on the abscissa: horizontal axis), versus the probability that the load capacity of a connection is greater than or equal to n mu (on the ordinate: vertical axis). The plot and/or table should include the range n = 1,2,... 10.Explanation / Answer
Part (a)
Load capacity of individual bolts is Normal with µ = 100 and = 2, average and standard deviation of load capacity of a one-bolt connection are 100 and 2 respectively. ANSWER
Part (b)
If X = Load capacity of individual bolts and Yn = load capacity of an n-bolt connection, we are given that Y = nX. So, average load capacity of an 2-bolt connection = E(Y2) = E(2X) = 2E(X) = 2x100 = 200. Standard deviation of load capacity of an 2-bolt connection = SD(Y2) = SD(2X) = 2SD(X) = 2x2 = 4. ANSWER
Part (c)
Using the very rule as in Part (b), the following table gives the answer.
Number of bolts in the connection (n)
1
2
3
4
5
6
7
8
9
10
Average Load Capacity
100
200
300
400
500
600
700
800
900
1000
SD of Load c
Capacity
2
4
6
8
10
12
14
16
18
20
Part (d)
We need to apply the given rule that the load capacity of an n-bolt connection = n times the load capacity of the weakest bolt. Thus, if the load capacity of an n-bolt connection is greater than nµ, load of each bolt must be greater than µ. µ being the mean of a Normal Distribution, probability the load capacity of a bolt exceeds µ = ½ and hence probability that the load capacity of an n-bolt connection is greater than nµ = (½)n
Now, the values of n (number of bolts in the connection) and P (probability that the load capacity of the connection is greater than nµ) can be easily tabulated.
n
1
2
3
4
5
6
7
8
9
10
P
½
¼
1/8
1/16
1/32
1/64
1/128
1/256
1/512
1/1024
DONE
Number of bolts in the connection (n)
1
2
3
4
5
6
7
8
9
10
Average Load Capacity
100
200
300
400
500
600
700
800
900
1000
SD of Load c
Capacity
2
4
6
8
10
12
14
16
18
20