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The article "A Probabilistic Model of Fracture in Concrete and Size Effects on Fracture Toughness" (Magazine of Concrete Res., \({\rm{1996: 311 - 320}}\)) gives arguments for why fracture toughness in concrete specimens should have a Weibull distribution and presents several histograms of data that appear well fit by superimposed Weibull curves. Consider the following sample of size \({\rm{n = 18}}\) observations on toughness for high strength concrete (consistent with one of the histograms); values of \({{\rm{p}}_{\rm{i}}}{\rm{ = (i - }}{\rm{.5)/18}}\) are also given.

\(\begin{array}{*{20}{c}}{{\rm{ Observation }}}&{{\rm{.47}}}&{{\rm{.58}}}&{{\rm{.65}}}&{{\rm{.69}}}&{{\rm{.72}}}&{{\rm{.74}}}\\{{{\rm{p}}_{\rm{i}}}}&{{\rm{.0278}}}&{{\rm{.0833}}}&{{\rm{.1389}}}&{{\rm{.1944}}}&{{\rm{.2500}}}&{{\rm{.3056}}}\\{{\rm{ Observation }}}&{{\rm{.77}}}&{{\rm{.79}}}&{{\rm{.80}}}&{{\rm{.81}}}&{{\rm{.82}}}&{{\rm{.84}}}\\{{{\rm{p}}_{\rm{i}}}}&{{\rm{.3611}}}&{{\rm{.4167}}}&{{\rm{.4722}}}&{{\rm{.5278}}}&{{\rm{.5833}}}&{{\rm{.6389}}}\\{{\rm{ Observation }}}&{{\rm{.86}}}&{{\rm{.89}}}&{{\rm{.91}}}&{{\rm{.95}}}&{{\rm{1}}{\rm{.01}}}&{{\rm{1}}{\rm{.04}}}\\{{{\rm{p}}_{\rm{i}}}}&{{\rm{.6944}}}&{{\rm{.7500}}}&{{\rm{.8056}}}&{{\rm{.8611}}}&{{\rm{.9167}}}&{{\rm{.9722}}}\end{array}\)

Construct a Weibull probability plot and comment.

Short Answer

Expert verified

A Weibull distribution may be used to simulate the distribution of fracture toughness in concrete sample.

Step by step solution

01

Definition

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

02

Construct a Weibull probability plot and comment

If we denote observations by\({\rm{x}}\). Then the Weibull plot uses \({\rm{ln(x)}}\) on vertical axis and \({{\rm{(100p)}}^{{\rm{th }}}}\)percentile of the \({{\rm{p}}_{\rm{i}}}\) values given.

The value of the \({{\rm{(100p)}}^{{\rm{th}}}}\)percentile for given \({\rm{p}}\)is written as:

\({\rm{\eta (p) = ln( - ln(1 - p))}}\)

The table given below calculates \({\rm{ln(x)}}\)and \({\rm{\eta (p)}}\)for the given \({\rm{x}}\)and \({\rm{p}}\)values respectively.

The probability plot of Weibull is shown below. It appears to be straightforward enough to support the notion that the distribution of fracture toughness in concrete specimens might be properly described by a Weibull distribution.

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Most popular questions from this chapter

Suppose that \({\rm{10\% }}\) of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that X is

a. At most 30?

b. Less than 30 ?

c. Between 15 and 25 (inclusive)?

Let \({\rm{X}}\) denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The article 鈥淏lade Fatigue Life Assessment with Application to VAWTS鈥 (J. of Solar Energy Engr., \({\rm{1982: 107 - 111}}\)) proposes the Rayleigh distribution, with pdf

\({\rm{f(x;\theta ) = \{ }}\begin{array}{*{20}{c}}{\frac{{\rm{x}}}{{{{\rm{\theta }}^{\rm{2}}}}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2}}{{\rm{\theta }}^{\rm{2}}}{\rm{)}}}}}&{{\rm{x > 0}}}\\{\rm{0}}&{{\rm{otherwise}}}\end{array}\)

otherwise as a model for the \({\rm{X}}\) distribution.

a. Verify that \({\rm{f(x;\theta )}}\) is a legitimate pdf.

b. Suppose \({\rm{\theta = 100}}\) (a value suggested by a graph in the article). What is the probability that \({\rm{X}}\) is at most \({\rm{200}}\)? Less than \({\rm{200}}\)? At least \({\rm{200}}\)?

c. What is the probability that \({\rm{X}}\) is between \({\rm{100}}\) and \({\rm{200}}\) (again assuming \({\rm{\theta = 100}}\))?

d. Give an expression for \({\rm{P(X}} \le {\rm{x)}}\).

The reaction time (in seconds) to a certain stimulus is a continuous random variable with pdf

\(f(x)= \left\{ {\begin{array}{*{20}{c}}{\frac{3}{2} \times \frac{1}{{{x^2}}}}&{1拢x拢3} \\0&{{\text{ }}otherwise{\text{ }}}\end{array}} \right.\)

a. Obtain the cdf.

b. What is the probability that reaction time is at most\({\rm{2}}{\rm{.5sec}}\)? Between \({\rm{1}}{\rm{.5}}\) and\({\rm{2}}{\rm{.5sec}}\)?

c. Compute the expected reaction time.

d. Compute the standard deviation of reaction time.

e. If an individual takes more than \({\rm{1}}{\rm{.5sec}}\) to react, a light comes on and stays on either until one further second has elapsed or until the person reacts (whichever happens first). Determine the expected amount of time that the light remains lit. (Hint: Let \({\rm{h(X) = }}\) the time that the light is on as a function of reaction time\({\rm{X}}\).)

a. Show that if X has a normal distribution with parameters \({\rm{\mu }}\) and\({\rm{\sigma }}\), then \({\rm{Y = aX + b}}\) (a linear function of X ) also has a normal distribution. What are the parameters of the distribution of Y (i.e., E(Y) and V(Y)) ? (Hint: Write the cdf of\({\rm{Y,P(Y}} \le {\rm{y)}}\), as an integral involving the pdf of X, and then differentiate with respect to y to get the pdf of Y.)

b. If, when measured in\(^{\rm{^\circ }}{\rm{C}}\), temperature is normally distributed with mean 115 and standard deviation 2 , what can be said about the distribution of temperature measured in\(^{\rm{^\circ }}{\rm{F}}\)?

The article 鈥淪econd Moment Reliability Evaluation vs. Monte Carlo Simulations for Weld Fatigue Strength鈥 (Quality and Reliability Engr. Intl., \({\rm{2012: 887 - 896}}\)) considered the use of a uniform distribution with \({\rm{A = }}{\rm{.20}}\) and \({\rm{B = 4}}{\rm{.25}}\) for the diameter \({\rm{X}}\) of a certain type of weld (mm).

a. Determine the pdf of \({\rm{X}}\) and graph it.

b. What is the probability that diameter exceeds \({\rm{3 mm}}\)?

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d. For any value \({\rm{a}}\) satisfying \({\rm{.20 < a < a + 1 < 4}}{\rm{.25}}\), what is \({\rm{P(a < X < a + 1)}}\)?

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