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Chapter 3: Problem 89

The article "Reliability-Based Service-Life Assessment of Aging Concrete Structures" (J. Structural Engr., 1993: 1600-1621) suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the mean time between occurrences of loads is . 5 year. a. How many loads can be expected to occur during a 2-year period? b. What is the probability that more than five loads occur during a 2 -year period? c. How long must a time period be so that the probability of no loads occurring during that period is at most .1?

Short Answer

Expert verified
a. Expect 4 loads in 2 years. b. Probability of more than 5 loads is calculated via cumulative Poisson probabilities. c. Time period for probability 鈮� 0.1 is approximately 1.15 years.

Step by step solution

01

Identify Parameters for the Poisson Process

The Poisson process is characterized by a rate \( \lambda \), which is the mean number of occurrences per time unit. Since the mean time between occurrences is 0.5 years, the rate \( \lambda \) is \( \frac{1}{0.5} = 2 \) loads per year.
02

Calculate Expected Loads for 2-Year Period

To find the expected number of loads in a 2-year period, multiply the rate \( \lambda = 2 \) by the duration 2 years: \( \text{Expected loads} = 2 \times 2 = 4 \). Hence, 4 loads can be expected during a 2-year period.
03

Probability of More Than Five Loads

To find the probability of more than five loads occurring in a 2-year period, calculate the cumulative probability of 0 to 5 loads and subtract from 1. The Poisson distribution formula is \( P(X = k) = \frac{e^{-\lambda t} (\lambda t)^k}{k!} \) where \( \lambda t = 4 \): - \( P(X \leq 5) = \sum_{k=0}^{5} \frac{e^{-4} 4^k}{k!} \)- Compute \( P(X \leq 5) \) and subtract from 1 to find \( P(X > 5) = 1 - P(X \leq 5) \).
04

Determine Time Period for No Loads with Probability 0.1

To find the period \( t \) where the probability of no loads is at most 0.1, use \( P(X=0) = e^{-\lambda t} \). - Set \( e^{-2t} = 0.1 \) and solve for \( t \): - \( -2t = \ln(0.1) \), hence \( t = \frac{\ln(0.1)}{-2} \).- Calculate \( t \) to find the duration in years.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Structural Loads in Poisson Process
Structural loads refer to forces, weights, or other external factors that concrete structures like bridges or buildings must withstand. In engineering, it is crucial to understanding when these loads might occur and how often. A Poisson process is a statistical method used to model the random occurrence of such events over time.
This method assumes that events happen independently and at a constant average rate. In our problem, structural loads happening over time on aging concrete is modeled using this process to predict future occurrences.
Understanding Probability with Poisson Distribution
In a Poisson process, probability helps us determine the likelihood of a certain number of events occurring within a given interval. Probability is a measure of chance, ranging from 0 (impossible event) to 1 (certain event).
For a Poisson distribution, the probability of observing exactly \( k \) events in an interval is calculated using the formula:
\[ P(X = k) = \frac{e^{-\lambda t} (\lambda t)^k}{k!} \]
Here, \( \lambda t \) represents the mean number of events in the period, \( e \) is the base of the natural logarithm, and \( k! \) denotes the factorial of \( k \). This formula helps in figuring out the probability of actions such as more than five loads occurring.
Rate Parameter (\( \lambda \)) in Poisson Process
The rate parameter, represented by \( \lambda \), is a key component in a Poisson process. It defines the average number of times an event occurs in a fixed interval of time or space.
In our case, \( \lambda = 2 \) gives us the mean occurrence rate鈥攊ndicating loads typically occur twice a year on the structure. Knowing \( \lambda \) allows us to predict and plan for the expected frequency of load occurrences.
This is essential for the reliability and maintenance of structures, ensuring they remain safe over time by understanding the likely frequency of stress they endure.
Expected Number of Occurrences Over Time
The expected number of occurrences provides an average prediction of how often events like structural loads might happen over a specific period. In Poisson processes, it's simply calculated by multiplying the rate parameter \( \lambda \) by the length of time under consideration.
For our example, the expected number during a 2-year period is \( \lambda \times 2 = 4 \) loads. This means we can expect, on average, four load occurrences in the given duration. Understanding this helps engineers prepare and act appropriately to manage such loads, maintaining the structure's integrity over time.

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Most popular questions from this chapter

A concrete beam may fail either by shear \((S)\) or flexure \((F)\). Suppose that three failed beams are randomly selected and the type of failure is determined for each one. Let \(X=\) the number of beams among the three selected that failed by shear. List each outcome in the sample space along with the associated value of \(X\).

A certain brand of upright freezer is available in three different rated capacities: \(16 \mathrm{ft}^{3}, 18 \mathrm{ft}^{3}\), and \(20 \mathrm{ft}^{3}\). Let \(X=\) the rated capacity of a freezer of this brand sold at a certain store. Suppose that \(X\) has pmf \begin{tabular}{l|ccc} \(x\) & 16 & 18 & 20 \\ \hline\(p(x)\) & \(.2\) & \(.5\) & \(.3\) \end{tabular} a. Compute \(E(X), E\left(X^{2}\right)\), and \(V(X)\). b. If the price of a freezer having capacity \(X\) is \(70 X-650\), what is the expected price paid by the next customer to buy a freezer? c. What is the variance of the price paid by the next customer? d. Suppose that although the rated capacity of a freezer is \(X\), the actual capacity is \(h(X)=X-.008 X^{2}\). What is the expected actual capacity of the freezer purchased by the next customer?

If \(a \leq X \leq b\), show that \(a \leq E(X) \leq b\).

A student who is trying to write a paper for a course has a choice of two topics, \(A\) and \(B\). If topic \(A\) is chosen, the student will order two books through interlibrary loan, whereas if topic B is chosen, the student will order four books. The student believes that a good paper necessitates receiving and using at least half the books ordered for either topic chosen. If the probability that a book ordered through interlibrary loan actually arrives in time is \(.9\) and books arrive independently of one another, which topic should the student choose to maximize the probability of writing a good paper? What if the arrival probability is only \(.5\) instead of \(.9\) ?

Possible values of \(X\), the number of components in a system submitted for repair that must be replaced, are 1 , 2,3 , and 4 with corresponding probabilities .15, .35,.35, and \(.15\), respectively. a. Calculate \(E(X)\) and then \(E(5-X)\). b. Would the repair facility be better off charging a flat fee of \(\$ 75\) or else the amount \(\$[150 /(5-X)]\) ? [Note: It is not generally true that \(E(c / Y)=c / E(Y)\).]
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