This is question 2.9 in my Foundation Design book. \"We wish to to design a shal

Question

This is question 2.9 in my Foundation Design book.

"We wish to to design a shallow foundation with a probability of failure of 10^-3. The footing supports a column carrying a dead load with a mean of 30k and COV of 0.05 and a live load with a mean of 10k and COV of 0.15. Based on the uncertainty of soil properties and our analysis method, we estimate the COV of the foundation capacity to be 0.2. For what mean capacity does the foundation need to be designed? Assume both loads and capacity are normally distributed.

Explanation / Answer

Given that foundation capacity(Y) has a probability of failure=0.001, then p=1-q=1-0.001=0.999. So it bases on binomial distribution with mean=np and variance=npq=0.2(given), then by simplifying mean=200.

Load capacity(X)=dead load(X1)+live load(X2)=>E(X)=E(X1)+E(X2)=30+10=40, V(X)=V(X1)+V(X2)[here cov=variance because it is of single variables]=0.05+0.15=0.2.

Then given X~N(40,0.2) & Y~N(200,0.2),then X+Y~N(40+200,0.2+0.2)=>X+Y~N(240,0.4).

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