Question

The bearing capacity, ?, of soil under a square foundation of size 9 ft^2 is determined to be a random variable with a mean of 3 ksf and a standard deviation of 0.3 ksf. The applied axial load, ?, acting on the foundation is also a random variable with a mean of 16 kip and a standard deviation of 2 kip. Assume ? and ? are statistically independent normal variables. Using the limit state function of the form ?() = 9?– ?, use the normal format approach to calculate the reliability index and the corresponding probability of failure of the foundation.

Answer 1

First of all lets look at some definations:

Reliability: In simple statistical terms, reliability refers to a function defining probability of failure. Here failure refers to the failure of system. Every system has a capacity (Resistance [R]) and is subjected to certain amount of demand (Load [S[). When the demand exceed capacity, system failure is reached.

Limit State function: State function is the difference between capacity (R) and load (S). Limit State Function returns a negetive value in case pf system failure and positive value when system is stable.

MVFOSM: Mean-value First-Order Second-moment Method is used to calculate reliability index when we have knowledge of only second moment (i.e. mean, standard deviation, correlation coefficient etc.). In MVFOSM, reliablity index is calculated with the equation:

Where, and are the mean and variance of limit state function, and is the reliability index. Here we can see that is inversly proportional to variance indicating high variance implies lower reliability index and vice versa. If the variance is high, failure domain is away from the average failure region ( because variance measures how far a set of number is spread from the mean, so if variance is high, values are far away from the failure region). Thus if the is high, probability of failure is small and vice versa.

Now lets calculate the reliablity index:

Given value:

= 3ksf, = 16kip, = (0.3ksf)², = (2kip)²

g() = 9C-P

= = 9*3 - 16 = 27-16 = 11

= 9²*0.3²+2² = 81*0.09+4 = 7.29+4 = 11.29

Hence the reliability index is : = 0.97431

For 2nd question I am not sure about the answer. Please repost the 2nd question. Thank you for asking. Please rate..