Chapter 11, Problem 11.13P

(a)

To determine

The graph for the characteristic strength and Weibull modulus.

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,

  $22\text{Mpa,26}\text{Mpa,33}\text{Mpa,50}\text{Mpa},29\text{Mpa},56\text{Mpa},58\text{Mpa,41}\text{Mpa}$ and $43\text{Mpa}$ .

Formula Used:

Write the expression for the calculation of polymer specimen as:

  $PS=1-\frac{n}{N+1}$ …… (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.

Rank	$PS=1-\frac{n}{N+1}$	$1/PS$	$\ln \left(1/PS\right)$	$ln\ln \left(1/PS\right)$	Stress	$\ln \left(\text{stress}\right)$
$1$	$0.9$	$1.11111$	$0.105361$	$-2.25037$	$22$	$3.091042$
$2$	$0.8$	$1.25$	$0.223144$	$-1.49994$	$26$	$3.258097$
$3$	$0.7$	$1.428571$	$0.356675$	$-1.03093$	$29$	$3.367296$
$4$	$0.6$	$1.666667$	$0.510826$	$-0.67173$	$33$	$3.496508$
$5$	$0.5$	$2$	$0.693147$	$-0.36651$	$37$	$3.610918$
$6$	$0.4$	$2.5$	$0.916291$	$-0.08742$	$41$	$3.713572$
$7$	$0.3$	$3.333333$	$1.203973$	$0.185627$	$43$	$3.7612$
$8$	$0.2$	$5$	$1.609438$	$0.475885$	$49$	$3.89182$
$9$	$0.1$	$10$	$2.302585$	$0.834032$	$56$	$4.025352$

Table (1)

The graph between $ln\ln \left(1/PS\right)$ and $\ln \left(\text{stress}\right)$ as shown in figure (1)

  

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 $40.4\text{MPa}$ .

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.

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 $22\text{Mpa}$ .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 $22\text{Mpa}$ .