Number of tests, n = 330 and the results follow a normal distribution (Figure 1)

Question

Number of tests, n = 330 and the results follow a normal distribution (Figure 1) with: Mean tensile strength, 20.3 lbs/inch width Standard deviation of the tensile strength, = 7.1 lbs/inch Range of tensile strengths, 5 to 40 lbs/inch-width 0.06 Count 330 Mean 20.3 Lbs/in. Stdev 7.1 Lbs/in. C.O.V. 0.348 Max 40 Lbs/in Min 5 Lbs/in. 0.05 0.04 0.03 0.02 0.01 0.00 0 10 15 20 25 30 35 40 45 50 Tensile Strength of Kraft Paper (Lbs/inch width)) Figure Probability Density Function (Normal Distribution) for tensile strength of paper reinforcing used in the Geowall project (CE 3010 Spring 2014) a. Calculate the coefficient of variation (c.o.v.) for the tensile strength data. b. What value of tensile strength (lbs/inch) of the paper should be used if a 95 percent reliability with respect to tensile failure is desired, i.e., there would only be a5 percent probability of tensile failure of the Kraft paper reinforcing? (Tensile failure occurs when the actual strength of the paper is less than the design strength.) c. If the paper reinforcement is loaded to 10 lbs/inch, and assuming that 25 paper reinforcing strips are used in the construction of the geowall and that the wall will fail if two or more reinforcing strips fail, what is the probability of failure of the wall? (Binomial distribution) d. Redo part c assuming the distribution of the tensile strength to be Lognormal?

Explanation / Answer

(a) Here coefficient of variation = / = 7.1/20.3 = 0.3498

(b) Here if 95% reliability is required so if the required tensile failure is k lbs/inch

Pr(x < k; 20.3 ; 7.1) = 0.95

here the z value for the given probability is 1.96

(k - 20.3)/ 7.1 = 1.96

k = 20.3 + 7.1 * 1.96 = 34.22 Lbs/in.

(c) here the paper reinforcement load 10 lbs/inch

Number of paper reinforcing strips = 25

Here Pr(x < 10 lbs/inch) = PR(x < 10 lbs/inch ; 20.3 ; 7.1)

Z = (10 -20.3)/7.1 = -1.451

Pr(x < 10 lbs/inch) = PR(x < 10 lbs/inch ; 20.3 ; 7.1) = Pr (Z < -1.451) = 0.0734

Here as now we have to find the probability of failure of the wall. when two or more strips fail so here n = 25 and p = 0.0734

Pr(y >= 2) = 1 - BINOMIALCDF(y <2 ; 25; 0.0734) = 1 - 0.4432 = 0.5568

(d) If tensile strength is lognormal distribution

then Pr(x < 10 ; 20.3 ; 7.1)

Here

20.3 = exp ( + 2/2)

3.011 = + 2/2

= 3.011 - 2/2

Here

7.12 = [exp (2) - 1] exp [2 * 3.011]

[exp (2) - 1] = 0.122235

[exp (2) = 1.122235

2 ln = 0.115322

ln = 0.3396

ln   = 3.011 - 0.115322/2 = 2.9533

Z = (ln X - ln )/ ln (ln) = (ln 10 - 2.9533)/ 0.3396 = -1.92

Here Pr(x < 10 ; 20.3 ; 7.1) = Pr(Z < -1.92) = 0.0275

Here as now we have to find the probability of failure of the wall. when two or more strips fail so here n = 25 and p = 0.0275

Pr(y >= 2) = 1 - BINOMIALCDF(y <2 ; 25; 0.0275) = 1 - 0.8500 = 0.15

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