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Chapter 4: Q90E (page 192)

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

Find the following percentiles for the standard normal distribution. Interpolate where appropriate.

\(\begin{array}{*{20}{l}}{{\rm{a}}{\rm{. 91st}}}\\\begin{array}{l}{\rm{b}}{\rm{. 9th }}\\{\rm{c}}{\rm{. 75th }}\\{\rm{d}}{\rm{. 25th }}\\{\rm{e}}{\rm{. }}{{\rm{6}}^{{\rm{th}}}}\end{array}\end{array}\)

Let \({\rm{X}}\) denote the voltage at the output of a microphone, and suppose that \({\rm{X}}\) has a uniform distribution on the interval from \({\rm{ - 1}}\) to \({\rm{1}}\). The voltage is processed by a 鈥渉ard limiter鈥� with cut-off values \({\rm{ - }}{\rm{.5}}\) and \({\rm{.5}}\), so the limiter output is a random variable \({\rm{Y}}\) related to \({\rm{X}}\) by \({\rm{Y = X}}\) if \({\rm{|X|}} \le {\rm{.5,Y = }}{\rm{.5}}\) if \({\rm{X > }}{\rm{.5}}\), and \({\rm{Y = - }}{\rm{.5}}\) if \({\rm{X < - }}{\rm{.5}}\). a. What is \({\rm{P(Y = }}{\rm{.5)}}\)? b. Obtain the cumulative distribution function of \({\rm{Y}}\) and graph it.

The article "Error Distribution in Navigation"\({\rm{(J}}\). of the Institute of Navigation, \({\rm{1971: 429 442}}\)) suggests that the frequency distribution of positive errors (magnitudes of errors) is well approximated by an exponential distribution. Let \({\rm{X = }}\) the lateral position error (nautical miles), which can be either negative or positive. Suppose the pdf of \({\rm{X}}\) is

\(f(x) = (.1){e^{ - .2k1}} - 楼 < x < 楼\)

a. Sketch a graph of \({\rm{f(x)}}\) and verify that \({\rm{f(x)}}\) is a legitimate pdf (show that it integrates to\({\rm{1}}\)).

b. Obtain the cdf of \({\rm{X}}\) and sketch it.

c. Compute \(P(X拢0),P(X拢2),P( - 1拢X拢2)\), and the probability that an error of more than \({\rm{2}}\) miles is made.

The article 鈥淎 Model of Pedestrians鈥� Waiting Times for Street Crossings at Signalized Intersections鈥� (Transportation Research, \({\rm{2013: 17--28}}\)) suggested that under some circumstances the distribution of waiting time X could be modelled with the following pdf:

\({\rm{f(x;\theta ,\tau ) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{\theta }}}{{\rm{\tau }}}{{{\rm{(1 - x/\tau )}}}^{{\rm{\theta - 1}}}}}&{{\rm{0}} \le {\rm{x < \tau }}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

a. Graph \({\rm{f(x;\theta ,80)}}\) for the three cases \({\rm{\theta = 4,1}}\) and \({\rm{.5}}\) (these graphs appear in the cited article) and comment on their shapes. b. Obtain the cumulative distribution function of X. c. Obtain an expression for the median of the waiting time distribution. d. For the case \({\rm{\theta = 4,\tau = 80}}\) calculate \({\rm{P(50}} \le {\rm{X}} \le {\rm{70)}}\) without at this point doing any additional integration.

Suppose the reaction temperature \({\rm{X}}\) (in \(^{\rm{o}}{\rm{C}}\)) in a certain chemical process has a uniform distribution with \({\rm{A = - 5}}\) and \({\rm{B = 5}}\).

a. Compute \({\rm{P(X < 0)}}\).

b. Compute \({\rm{P( - 2}}{\rm{.5 < X < 2}}{\rm{.5)}}\).

c. Compute \({\rm{P( - 2}} \le {\rm{X}} \le {\rm{3)}}\).

d. For \({\rm{k}}\) satisfying \({\rm{ - 5 < k < k + 4 < 5}}\), compute \({\rm{P(k < X < k + 4)}}\).
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