Chapter 1, Problem 19P

A round cold-drawn 1045 steel rod has a mean strength $\overline{S_y}\text{ = 95}\text{.5}$ kpsi with a standard deviation of ${\hat{\sigma }}_{S_y}\text{ = 6}\text{.59}$ kpsi. The rod is to be subjected to a mean static axial load of $\bar{P}$ = 65 kip with a standard deviation of ${\hat{\sigma }}_P$ = 5.0 kip. Assuming the strength and load have normal distributions. determine the reliabilities corresponding to the design factors of (a) 1.2. (b) 1.5. Also, determine the diameter corresponding to each case.

To determine

The reliability corresponding to $1.2$ design factor for cold-drawn 1045 steel rod.

The diameter of cold-drawn 1045 steel rod for $1.2$ design factor.

Answer to Problem 19P

The reliability corresponding to $1.2$ design factor for cold-drawn 1045 steel rod is $94.7\%$.

The diameter of cold-drawn 1045 steel rod for $1.2$ design factor is $1.020\text{in}$.

Explanation of Solution

Write the expression for coefficient of variance for strength.

$C_s=\frac{{\widehat{\sigma }}_{S_{xy}}}{{\overline{S}}_{sy}}$ (I)

Here, the coefficient of variance for strength is $C_s$, the standard deviation for shear yield strength is ${\widehat{\sigma }}_{S_{xy}}$ and shear yield strength is ${\overline{S}}_{sy}$.

Write the expression for coefficient of variance for stress.

$C_{\sigma }=\frac{{\widehat{\sigma }}_P}{\overline{P}}$ (II)

Here, the coefficient of variance for stress is $C_{\sigma }$, the standard deviation for axial load is ${\widehat{\sigma }}_P$ and axial load is $\overline{P}$.

Write the expression for transform variable.

$z=-\frac{{\overline{n}}_d-1}{\sqrt{{\overline{n}}^2{}_d{C}_S^2+C_{\sigma }^2}}$ (III)

Here, the design factor is ${\overline{n}}_d$ and transform variable is $z$.

Write the expression for axial stress.

$\begin{array}{c}\overline{\sigma }=\frac{\overline{P}}{\left(\frac{\pi d^2}{4}\right)}\\ =\frac{4\overline{P}}{\pi d^2}\end{array}$ (IV)

Here, the axial stress is $\overline{\sigma }$, the axial load is $\overline{P}$ and the diameter of steel rod is $d$.

Calculate the reliability for steel rod.

$R=1-\Phi \left(z\right)$ (V)

Here, the reliability for steel is $R$ and the transform function is $\Phi \left(z\right)$.

Calculate the diameter of steel rod.

${\overline{n}}_d=\frac{{\overline{S}}_y}{\overline{\sigma }}$ (VI)

Substitute $\frac{4\overline{P}}{\pi d^2}$ for $\overline{\sigma }$ in Equation (VI).

$\begin{array}{c}{\overline{n}}_d=\frac{{\overline{S}}_y}{\frac{4\overline{P}}{\pi d^2}}\\ {\overline{n}}_d=\frac{\pi d^2{\overline{S}}_y}{4\overline{P}}\\ d=\sqrt{\frac{4\overline{P}{\overline{n}}_d}{\pi {\overline{S}}_y}}\end{array}$ (VII)

Conclusion:

Substitute $6.59\text{kpsi}$ for ${\widehat{\sigma }}_{S_{xy}}$ and $95.5\text{kpsi}$ for ${\overline{S}}_y$ in Equation (I).

$\begin{array}{c}{C}_s=\frac{\left(6.59\text{kpsi}\right)}{\left(95.5\text{kpsi}\right)}\\ =0.06901\end{array}$

Substitute $5\text{kip}$ for ${\widehat{\sigma }}_P$ and $65\text{kip}$ for $\overline{P}$ in Equation (II).

$\begin{array}{c}{C}_{\sigma }=\frac{\left(5\text{kip}\right)}{\left(65\text{kip}\right)}\\ =0.077\end{array}$

Substitute $1.2$ for ${\overline{n}}_d$, $0.06901$ for $C_s$ and $0.077$ for $C_{\sigma }$ in Equation (III).

$\begin{array}{c}z=-\frac{\left(\left(1.2\right)-1\right)}{\sqrt{{\left(1.2\right)}^2{\left(0.06901\right)}^2+{\left(0.077\right)}^2}}\\ =-\frac{\left(0.2\right)}{\sqrt{0.01278}}\\ =-\frac{\left(0.2\right)}{0.113048}\\ \approx -1.77\end{array}$

Write the formula of interpolation.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (VIII)

Here, the variables denote by $x$ and $y$ are the value of $z$ and $\Phi \left(z\right)$.

Refer to table A-10 “Cumulative distribution function of Normal distribution” and obtain the value given in Table below.

$x$	y
$1.61$	$0.0537$
$1.6127$	$y_2$
$1.62$	$0.0526$

Table (1)

Substitute $1.61$ for $x_1$, $1.6127$ for $x_2$, $1.62$ for $x_3$, $0.0537$ for $y_1$ and $0.0526$ for $y_3$ in Equation (VIII).

$\begin{array}{c}{y}_2=\frac{\left(1.6127-1.61\right)\left(0.0526-0.0537\right)}{\left(1.62-1.61\right)}+0.0537\\ =\frac{\left(-0.00000297\right)}{\left(0.01\right)}+0.0537\\ =0.053403\end{array}$

Substitute $-1.77$ for $\Phi \left(z\right)$ in Equation (V).

$\begin{array}{c}R=\left(1-0.053403\right)\times 100\%\\ =\left(0.9465\right)\times 100\%\\ \approx 94.7\%\end{array}$

Thus, the reliability of steel rod is $94.7\%$.

Substitute $65\text{kip}$ for $\overline{P}$, $1.2$ for ${\overline{n}}_d$ and $95.5\text{kpsi}$ for ${\overline{S}}_y$ in Equation (VII).

$\begin{array}{c}d=\sqrt{\frac{4\left(65\text{kip}\right)\left(1.2\right)}{\pi \left(95.5\text{kpsi}\right)}}\\ =\sqrt{\frac{\left(312\text{kip}\right)}{\pi \left(95.5\text{kpsi}\right)}}\\ =\sqrt{\left(1.0399{\text{in}}^{\text{2}}\right)}\\ \approx 1.020\text{in}\end{array}$

Thus, the diameter of steel rod is $1.020\text{in}$.

To determine

The reliability corresponding to $1.5$ design factor for cold-drawn 1045 steel rod.

The diameter of cold-drawn 1045 steel rod for $1.5$ design factor.

Answer to Problem 19P

The reliability corresponding to $1.5$ design factor for cold-drawn 1045 steel rod is $99.98\%$.

The diameter of cold-drawn 1045 steel rod for $1.5$ design factor is $1.140\text{in}$.

Explanation of Solution

Write the expression for coefficient of variance for strength.

$C_s=\frac{{\widehat{\sigma }}_{S_{xy}}}{{\overline{S}}_{sy}}$ (IX)

Write the expression for coefficient of variance for stress.

$C_{\sigma }=\frac{{\widehat{\sigma }}_P}{\overline{P}}$ (X)

Write the expression for transform variable.

$z=-\frac{{\overline{n}}_d-1}{\sqrt{{\overline{n}}^2{}_d{C}_S^2+C_{\sigma }^2}}$ (XI)

Write the expression for axial stress.

$\begin{array}{c}\overline{\sigma }=\frac{\overline{P}}{\left(\frac{\pi d^2}{4}\right)}\\ =\frac{4\overline{P}}{\pi d^2}\end{array}$ (XII)

Calculate the reliability for steel rod.

$R=\left(1-\Phi \left(z\right)\right)\times 100\%$ (XIII)

Calculate the diameter of steel rod.

${\overline{n}}_d=\frac{{\overline{S}}_y}{\overline{\sigma }}$ (XIV)

Substitute $\frac{4\overline{P}}{\pi d^2}$ for $\overline{\sigma }$ from Equation (XII) in Equation (XIV).

$\begin{array}{c}{\overline{n}}_d=\frac{{\overline{S}}_y}{\frac{4\overline{P}}{\pi d^2}}\\ {\overline{n}}_d=\frac{\pi d^2{\overline{S}}_y}{4\overline{P}}\\ d=\sqrt{\frac{4\overline{P}{\overline{n}}_d}{\pi {\overline{S}}_y}}\end{array}$ (XV)

Conclusion:

Substitute $6.59\text{kpsi}$ for ${\widehat{\sigma }}_{S_{xy}}$ and $95.5\text{kpsi}$ for ${\overline{S}}_y$ in Equation (IX)

$\begin{array}{c}{C}_s=\frac{6.59\text{kpsi}}{95.5\text{kpsi}}\\ =0.06901\end{array}$

Substitute $5\text{kip}$ for ${\widehat{\sigma }}_P$ and $65\text{kip}$ for $\overline{P}$ in Equation (X).

$\begin{array}{c}{C}_{\sigma }=\frac{5\text{kip}}{65\text{kip}}\\ =0.09231\end{array}$

Substitute $1.5$ for ${\overline{n}}_d$, $0.06901$ for $C_s$ and $0.09231$ for $C_{\sigma }$ in Equation (XI).

$\begin{array}{c}z=-\frac{\left(\left(1.5\right)-1\right)}{\sqrt{{\left(1.5\right)}^2{\left(0.06901\right)}^2+{\left(0.09231\right)}^2}}\\ =-\frac{0.5}{0.13869}\\ \approx -3.6051\end{array}$

Write the formula of interpolation.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (XVI)

Here, the variables denote by $x$ and $y$ are the value of $z$ and $\Phi \left(z\right)$ respectively.

Refer to table A-10 “Cumulative distribution function of Normal distribution” and obtain the value given in Table below.

$x$	y
$3.6$	$0.000159$
$3.605$	$y_2$
$3.7$	$0.000108$

Table (2)

Substitute $3.6$ for $x_1$, $3.605$ for $x_2$, $3.7$ for $x_3$, $0.000159$ for $y_1$ and $0.000108$ for $y_3$ in Equation (XVI).

$\begin{array}{c}{y}_2=-\frac{\left(3.605-3.6\right)\left(0.000108-0.000159\right)}{\left(3.7-3.6\right)}+\mathrm{0.0.000159}\\ =\left(-0.00000255\right)+0.000159\\ =0.00015645\end{array}$

Substitute $0.00015645$ for $\Phi \left(z\right)$ in Equation (XIII).

$\begin{array}{c}R=\left(1-\left(0.00015645\right)\right)\times 100\%\\ =\left(0.99984355\right)\times 100\%\\ \approx 99.98\%\end{array}$

Thus, the reliability of steel rod is $99.98\%$.

Substitute $65\text{kip}$ for $\overline{P}$, $1.5$ for ${\overline{n}}_d$ and $95.5\text{kpsi}$ for ${\overline{S}}_y$ in Equation (XV).

$\begin{array}{c}d=\sqrt{\frac{4\left(65\text{kip}\right)\left(1.5\right)}{\pi \left(95.5\text{kpsi}\right)}}\\ =\sqrt{\frac{\left(390\right)\text{kip}}{\pi \left(95.5\right)\text{kpsi}}}\\ =\sqrt{1.29990{\text{in}}^2}\\ \approx 1.140\text{in}\end{array}$

Thus, the diameter of steel rod is $1.140\text{in}$.