Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
Question
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Chapter 11, Problem 11.13P

(a)

To determine

The graph for the characteristic strength and Weibull modulus.

(a)

Expert Solution
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Answer to Problem 11.13P

The graph for the characteristic strength and Weibull modulus is shown in figure (1)

Explanation of Solution

Given:

The following characteristic strengths are given as,

  22Mpa,26Mpa,33Mpa,50Mpa,29Mpa,56Mpa,58Mpa,41Mpa and 43Mpa .

Formula Used:

Write the expression for the calculation of polymer specimen as:

  PS=1nN+1 …… (I)

Here, N is the total number of specimen and n is the number of specimen.

Calculation:

The coordinates for the characteristic strength and Weibull modulus is calculated in table below.

    RankPS=1nN+11/PSln(1/PS)lnln(1/PS)Stressln(stress)
    10.91.111110.1053612.25037223.091042
    20.81.250.2231441.49994263.258097
    30.71.4285710.3566751.03093293.367296
    40.61.6666670.5108260.67173333.496508
    50.520.6931470.36651373.610918
    60.42.50.9162910.08742413.713572
    70.33.3333331.2039730.185627433.7612
    80.251.6094380.475885493.89182
    90.1102.3025850.834032564.025352

Table (1)

The graph between lnln(1/PS) and ln(stress) as shown in figure (1)

  Materials Science And Engineering Properties, Chapter 11, Problem 11.13P

Figure (1)

From Figure (1) it is found that the natural log of characteristics strength on the horizontal axis is 3.7 and at this point, the vertical axis is equal to zero providing a strength value of 40.4MPa .

Conclusion:

Thus, the graph for the characteristic strength and Weibull modulus is shown in figure (1)

(b)

To determine

The stress for the 90% probability of the survival for the silicon carbide material.

(b)

Expert Solution
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Explanation of Solution

Introduction:

At very low stresses, such as 0 to 20% of the yield stress, some structural materials are truly elastic, and these materials will not fail from fatigue at certain very low stresses unless a crack or some other major flaws is already present in the material. At higher stresses, such as 20 to 50% of the yield stress in a tensile test, there is a plastic strain.

From Table (1), it is found that the stress for 90% probability of survival is 22Mpa .At low stress, some structural materials are truly elastic, and these materials will not fail from fatigue at certain very low stresses unless a crack or some other major flaws is already present in the material whereas at higher stresses in a tensile test there is a production of plastic strain.

Conclusion:

Therefore, the stress for 90% probability of survival is 22Mpa .

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