Chapter 3, Problem 14SE

a.

To determine

Find the estimate of strength of the cable using the ductile wire method.

Find the uncertainty in the strength of the cable using the ductile wire method.

Answer to Problem 14SE

The estimate of strength of the cable using the ductile wire method is $\underset{\_}{T=71,600\pm 108lb}$.

The uncertainty in the strength of the cable using the ductile wire method is $\underset{\_}{{\sigma }_T=108lb}$.

Explanation of Solution

Given info:

The strength of first four wires is measured to be $6,000\pm 20\text{lb}$, strength of second four wires is measured to be $5,700\pm 30\text{lb}$ and the strength of third four wires is measured to be $6,200\pm 40\text{lb}$. In the ductile wire method, the strength of the cable is estimated to be the sum of the strengths of the 12 wires and in the brittle wire method, the strength of the cable is estimated to be strength of the weakest wire multiplied by the number of wires.

Calculation:

Let the first four wires be denoted by $X_1,X_2,X_3\text{and}{X}_4$. The second four wires be denoted by $Y_1,Y_2,Y_3\text{and}{Y}_4$ and the third four wires be denoted by $Z_1,Z_2,Z_3\text{and}{Z}_4$.

Let the strength of the cable is denoted by T.

The strength of the cable using the ductile wire method is sum of the strengths of the 12 wires.

That is, $T=X_1+X_2+X_3+X_4+Y_1+Y_2+Y_3+Y_4+Z_1+Z_2+Z_3+Z_4$

The form of the measurements of a process is,

$\text{Measured}\text{value}\left(\mu \right)\pm \text{Standard deviation}\left(\sigma \right)$.

Here, for a random sample the measured value will be sample mean and the population standard deviation will be sample standard deviation.

The form of the measurements of strength of first four wires is $6,000\pm 20\text{lb}$.

Here, the measured value or mean of the strength of first four wires is $X_1=X_2=X_3=X_4=6,000$ and the uncertainty in the strength of first four wires is ${\sigma }_{X_1}={\sigma }_{X_2}={\sigma }_{X_3}={\sigma }_{X_4}=20$.

The form of the measurements of strength of second four wires is $5,700\pm 30\text{lb}$.

Here, the measured value or mean of the strength of second four wires is $Y_1=Y_2=Y_3=Y_4=5,700$ and the uncertainty in the strength of second four wires is ${\sigma }_{Y_1}={\sigma }_{Y_2}={\sigma }_{Y_3}={\sigma }_{Y_4}=30$.

The form of the measurements of strength of third four wires is $6,200\pm 40\text{lb}$.

Here, the measured value or mean of the strength of third four wires is $Z_1=Z_2=Z_3=Z_4=6,200$ and the uncertainty in the strength of third four wires is ${\sigma }_{Z_1}={\sigma }_{Z_2}={\sigma }_{Z_3}={\sigma }_{Z_4}=40$.

Measured value of the strength of the cable using the ductile wire method:

The formula for the strength of the cable using the ductile wire method is $T=X_1+X_2+X_3+X_4+Y_1+Y_2+Y_3+Y_4+Z_1+Z_2+Z_3+Z_4$.

Here, $X_1=X_2=X_3=X_4=6,000$, $Y_1=Y_2=Y_3=Y_4=5,700$ and $Z_1=Z_2=Z_3=Z_4=6,200$.

The measured value of strength of the cable using the ductile wire method is obtained as follows:

$\begin{array}{c}T=X_1+X_2+X_3+X_4+Y_1+Y_2+Y_3+Y_4+Z_1+Z_2+Z_3+Z_4\\ =\left(4\times 6,000\right)+\left(4\times 5,700\right)+\left(4\times 6,200\right)\\ =71,600\end{array}$

Thus, the measured value of strength of the cable using the ductile wire method is $\underset{\_}{T=71,600lb}$.

Uncertainty:

The uncertainty of a process is determined by the standard deviation of the measurements. In other words it can be said that, measure of variability of a process is known as uncertainty of the process.

Therefore, it can be said that uncertainty is simply $\left(\sigma \right)$.

Standard deviation:

The standard deviation is based on how much each observation deviates from a central point represented by the mean. In general, the greater the distances between the individual observations and the mean, the greater the variability of the data set.

The general formula for standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum x_i{}^2-\frac{{\left({\displaystyle \sum x_i}\right)}^2}{n}}}{n-1}}$.

From the properties of uncertainties for functions of one measurement it is known that,

• If $X_1,X_2,\mathrm{...},X_n$ are independent measurements with uncertainties ${\sigma }_{X_1},{\sigma }_{X_2},\mathrm{...},{\sigma }_{X_n}$ and if $U=U\left(X_1,X_2,\mathrm{..},X_n\right)$ is a function of $X_1,X_2,\mathrm{...},X_n$ then the uncertainty in the variable U is ${\sigma }_U=\sqrt{{\left(\frac{\partial U}{\partial X_1}\right)}^2{\sigma }_{X_1}^2+{\left(\frac{\partial U}{\partial X_2}\right)}^2{\sigma }_{X_2}^2+\mathrm{....}+{\left(\frac{\partial U}{\partial X_n}\right)}^2{\sigma }_{X_n}^2}$.

Here, ${\sigma }_{X_1}={\sigma }_{X_2}={\sigma }_{X_3}={\sigma }_{X_4}=20$, ${\sigma }_{Y_1}={\sigma }_{Y_2}={\sigma }_{Y_3}={\sigma }_{Y_4}=30$ and ${\sigma }_{Z_1}={\sigma }_{Z_2}={\sigma }_{Z_3}={\sigma }_{Z_4}=40$.

Uncertainty in the strength of the cable using the ductile wire method:

The uncertainty in the strength of the cable using the ductile wire method is,

$\begin{array}{c}{\sigma }_{T=\left(\begin{array}{l}{X}_1+X_2+X_3+X_4+\\ Y_1+Y_2+Y_3+Y_4+\\ Z_1+Z_2+Z_3+Z_4\end{array}\right)}=\sqrt{\left(\begin{array}{l}{\left(\frac{\partial T}{\partial X_1}\right)}^2{\sigma }_{X_1}^2+{\left(\frac{\partial T}{\partial X_2}\right)}^2{\sigma }_{X_2}^2+{\left(\frac{\partial T}{\partial X_3}\right)}^2{\sigma }_{X_3}^2+{\left(\frac{\partial T}{\partial X_4}\right)}^2{\sigma }_{X_4}^2\\ +{\left(\frac{\partial T}{\partial Y_1}\right)}^2{\sigma }_{Y_1}^2+{\left(\frac{\partial T}{\partial Y_2}\right)}^2{\sigma }_{Y_2}^2+{\left(\frac{\partial T}{\partial Y_3}\right)}^2{\sigma }_{Y_3}^2+{\left(\frac{\partial T}{\partial Y_4}\right)}^2{\sigma }_{Y_4}^2\\ {\left(\frac{\partial T}{\partial Z_1}\right)}^2{\sigma }_{Z_1}^2+{\left(\frac{\partial T}{\partial Z_2}\right)}^2{\sigma }_{Z_2}^2+{\left(\frac{\partial T}{\partial Z_3}\right)}^2{\sigma }_{Z_3}^2+{\left(\frac{\partial T}{\partial Z_4}\right)}^2{\sigma }_{Z_4}^2\end{array}\right)}\\ =\sqrt{\begin{array}{l}{\sigma }_{X_1}^2+{\sigma }_{X_2}^2+{\sigma }_{X_3}^2+{\sigma }_{X_4}^2+{\sigma }_{Y_1}^2+{\sigma }_{Y_2}^2\\ +{\sigma }_{Y_3}^2+{\sigma }_{Y_4}^2+{\sigma }_{Z_1}^2+{\sigma }_{Z_2}^2+{\sigma }_{Z_3}^2+{\sigma }_{Z_4}^2\end{array}}\\ =\sqrt{\left(4\times {\left(20\right)}^2\right)+\left(4\times {\left(30\right)}^2\right)+\left(4\times {\left(40\right)}^2\right)}\\ =108\end{array}$

Thus, the uncertainty in the strength of the cable using the ductile wire method is $\underset{\_}{{\sigma }_T=108lb}$.

Estimate of the strength of the cable using the ductile wire method:

The estimate of the measurement of a process is,

$\text{Measured}\text{value}\left(\mu \right)\pm \text{Standard deviation}\left(\sigma \right)$.

Here, for a random sample the measured value will be sample mean and the population standard deviation will be sample standard deviation.

The estimate of strength of the cable using the ductile wire method is,

$\begin{array}{c}T=\text{Measured value of }T\pm {\sigma }_T\\ =71,600\pm 108lb\end{array}$

Thus, the estimate of the strength of the cable using the ductile wire method is $\underset{\_}{T=71,600\pm 108lb}$.

b.

To determine

Find the estimate of strength of the cable using the brittle wire method.

Find the uncertainty in the strength of the cable using the brittle wire method.

Answer to Problem 14SE

The estimate of strength of the cable using the brittle wire method is $\underset{\_}{T=68,400\pm 360lb}$.

The uncertainty in the strength of the cable using the brittle wire method is $\underset{\_}{{\sigma }_T=360lb}$.

Explanation of Solution

Calculation:

The strength of the cable using the brittle wire method is the strength of the weakest wire multiplied by the number of wires.

That is, $T=12\times \text{strength of weakest wire}$.

Here, the second four wires are the wires with weakest strength.

Hence, the strength of the cable is $T=12Y_i$.

From part (a), $X_1=X_2=X_3=X_4=6,000$, ${\sigma }_{X_1}={\sigma }_{X_2}={\sigma }_{X_3}={\sigma }_{X_4}=20$, $Y_1=Y_2=Y_3=Y_4=5,700$, ${\sigma }_{Y_1}={\sigma }_{Y_2}={\sigma }_{Y_3}={\sigma }_{Y_4}=30$ and $Z_1=Z_2=Z_3=Z_4=6,200$, ${\sigma }_{Z_1}={\sigma }_{Z_2}={\sigma }_{Z_3}={\sigma }_{Z_4}=40$.

Measured value of the strength of the cable using the brittle wire method:

The formula for the strength of the cable using the brittle wire method is $T=12Y_i$.

Here,$Y_1=Y_2=Y_3=Y_4=5,700$.

The measured value of strength of the cable using the brittle wire method is obtained as follows:

$\begin{array}{c}T=12\left(Y_1+Y_2+Y_3+Y_4\right)\\ =12\left(4\times 5,700\right)\\ =68,400\end{array}$

Thus, the measured value of strength of the cable using the brittle wire method is $\underset{\_}{T=68,400lb}$.

Uncertainty in the strength of the cable using the brittle wire method:

The uncertainty in the strength of the cable using the brittle wire method is,

$\begin{array}{c}{\sigma }_{T=12\left(Y_1+Y_2+Y_3+Y_4\right)}=\sqrt{{\left(\frac{\partial T}{\partial Y_1}\right)}^2{\sigma }_{Y_1}^2+{\left(\frac{\partial T}{\partial Y_2}\right)}^2{\sigma }_{Y_2}^2+{\left(\frac{\partial T}{\partial Y_3}\right)}^2{\sigma }_{Y_3}^2+{\left(\frac{\partial T}{\partial Y_4}\right)}^2{\sigma }_{Y_4}^2}\\ =\sqrt{\left(\begin{array}{l}{\left(\frac{\partial 12\left(Y_1+Y_2+Y_3+Y_4\right)}{\partial Y_1}\right)}^2{\sigma }_{Y_1}^2+{\left(\frac{\partial 12\left(Y_1+Y_2+Y_3+Y_4\right)}{\partial Y_2}\right)}^2{\sigma }_{Y_2}^2\\ +{\left(\frac{\partial 12\left(Y_1+Y_2+Y_3+Y_4\right)}{\partial Y_3}\right)}^2{\sigma }_{Y_3}^2+{\left(\frac{\partial 12\left(Y_1+Y_2+Y_3+Y_4\right)}{\partial Y_4}\right)}^2{\sigma }_{Y_4}^2\end{array}\right)}\\ =12\times \sqrt{{\sigma }_{Y_1}^2+{\sigma }_{Y_2}^2+{\sigma }_{Y_3}^2+{\sigma }_{Y_4}^2}\\ =12\times \sqrt{\left(4\times {\left(30\right)}^2\right)}\\ =360\end{array}$

Thus, the uncertainty in the strength of the cable using the brittle wire method is $\underset{\_}{{\sigma }_T=360lb}$.

Estimate of the strength of the cable using the brittle wire method:

The estimate of the measurement of a process is,

$\text{Measured}\text{value}\left(\mu \right)\pm \text{Standard deviation}\left(\sigma \right)$.

Here, for a random sample the measured value will be sample mean and the population standard deviation will be sample standard deviation.

The estimate of strength of the cable using the brittle wire method is,

$\begin{array}{c}T=\text{Measured value of }T\pm {\sigma }_T\\ =68,400\pm 360lb\end{array}$

Thus, the estimate of the strength of the cable using the brittle wire method is $\underset{\_}{T=68,400\pm 360lb}$.