Chapter 5, Problem 5.2.2P

To determine

(a)

The distance $\overline{y}$ from the top of the shape to the horizontal plastic neutral axis.

Answer to Problem 5.2.2P

$\underset{\_}{\overline{y}=5.4166\text{in}\text{.}}$

Explanation of Solution

Given:

An unsymmetrical flexural member consists of a $\frac{1}{2}\times 12$ top flange, a $\frac{1}{2}\times 7$ bottom flange, and a $\frac{3}{8}\times 16$ web. Following is the sectional view of the given unsymmetrical member.

Concept Used:

As the given flexural member is unsymmetrical, therefore, the plastic neutral axis of the section won’t lie on the center of the member.

We will use the concept of equilibrium of forces, to calculate the distance $\overline{y},$ from the top of the flange of the neutral axis, we have

$F_y\times A{}_{top}=F_y\times A_{bottom}$

Calculation:

Now, as we have the following equation

$F_y\times A{}_{top}=F_y\times A_{bottom}$

Where, $F_y$ is the yield strength of steel,

$A{}_{top}$ is the area of the top half of the section of the member,

$A_{bottom}$ is the area of the bottom section of the member.

As the force is equal, we have

$\begin{array}{l}A{}_{\text{top}}=\frac{F_y\times A_{\text{bottom}}}{F_y}\\ A{}_{\text{top}}=A_{\text{bottom}}\end{array}$

Now, calculating the area of the top flange, as follows:

${\text{A}}_{\text{top}\text{flange}}\text{=}\text{height}\text{of}\text{the flange × breadth of the flange}$

Substituting the values, we have

$\begin{array}{l}{\text{A}}_{\text{top}\text{flange}}\text{=}{\left(\frac{\text{1}}{\text{2}}\right)}^{\text{' '}}\text{ × 12''}\\ {\text{A}}_{\text{top}\text{flange}}\text{=}{\text{ 6 in}}^{\text{2}}\text{.}\end{array}$

Calculating the area of the web above the neutral axis, as follows:

$\text{Area of the web}=\text{height}\text{of}\text{the}\text{web above the neutral axis }\times \text{ breadth of the web}$

Substituting the values, we have

$\text{Area of the web}={\left(\overline{y}-\frac{1}{2}\right)}^{\text{' '}}\times {\frac{\text{3}}{8}}^{\text{'}\text{'}}$

$\text{Area of the web}=\left(\frac{2\overline{y}-2}{2}\right)\times \frac{\text{3}}{8}{\text{in}}^{\text{2}}.$

$\text{Area of the web}=\left(\frac{6\overline{y}-6}{16}\right){\text{in}}^{\text{2}}.$

Now, calculating the area of component above the plastic neutral axis as follows:

$\begin{array}{l}{\text{A}}_{\text{top}}\text{=}{\text{A}}_{\text{top}\text{flange}}+\text{Area of the web}\\ A_{top}={\text{ 6 in}}^{\text{2}}+\left(\frac{6\overline{y}-6}{16}\right){\text{in}}^{\text{2}}.\\ A_{top}=\left(\text{ 6 }+\left(\frac{6\overline{y}-6}{16}\right)\right){\text{in}}^{\text{2}}.\\ A_{top}=\left(\frac{96+6\overline{y}-6}{16}\right){\text{in}}^{\text{2}}.\\ A_{top}=\left(\frac{90+6\overline{y}}{16}\right){\text{in}}^{\text{2}}.\end{array}$

Now, calculating the area of the bottom flange, we have

$A_{bottomflange}=\text{height of the bottom flange }\times \text{ breadth of the bottom flange}\text{.}$

Substitute the $\text{height of the bottom flange = }\frac{1}{2}\text{inches,}$

$\text{breadth of the bottom flange}\text{=}\text{7}\text{inches}\text{.}$

$\begin{array}{l}{\text{A}}_{\text{bottom}\text{flange}}\text{=}\text{height of the bottom flange × breadth of the bottom flange}\text{.}\\ {\text{A}}_{\text{bottom}\text{flange}}\text{= }\frac{\text{1}}{\text{2}}\text{inches × 7}\text{inches}\text{.}\\ {\text{A}}_{\text{bottom}\text{flange}}\text{= }\frac{\text{7}}{\text{2}}{\text{in}}^{\text{2}}\text{.}\end{array}$

Calculating the area of the web below the neutral axis, as follows:

$\text{Area of the web}=\text{height}\text{of}\text{the}\text{web below the neutral axis }\times \text{ breadth of the web}$

Substituting the values, we have

$\text{Area of the web}={\left(\frac{1}{2}+16-\overline{y}\right)}^{\text{' '}}\times {\frac{\text{3}}{8}}^{\text{'}\text{'}}$

$\text{Area of the web}=\left(\frac{1}{2}+16-\overline{y}\right)\times \frac{\text{3}}{8}{\text{in}}^{\text{2}}.$

$\text{Area of the web}=\left(\frac{\text{3}}{16}+6-\frac{\text{3}}{8}\overline{y}\right)\times {\text{in}}^{\text{2}}.$

Calculating the area of the component below the plastic neutral axis as follows:

$\begin{array}{l}{A}_{bottom}=A_{bottomflange}+\text{Area of the web}\\ A_{bottom}=\left(\frac{\text{3}}{16}+6-\frac{\text{3}}{8}\overline{y}\right)\times {\text{in}}^{\text{2}}\text{+ }\frac{7}{2}{\text{in}}^{\text{2}}\text{.}\\ A_{bottom}=\left(6.1875-\frac{\text{3}}{8}\overline{y}\right)\times {\text{in}}^{\text{2}}\text{+ }3.5{\text{in}}^{\text{2}}\text{.}\\ A_{bottom}=\left(6.1875-\frac{\text{3}}{8}\overline{y}\text{+ }3.5\right){\text{in}}^{\text{2}}\text{.}\end{array}$

Now for computing the neutral axis, use $A{}_{top}=A_{bottom}$

Substitute the values, we have

$A_{top}=\left(\frac{90+6\overline{y}}{16}\right){\text{in}}^{\text{2}}$ and $A_{bottom}=\left(6.1875-\frac{\text{3}}{8}\overline{y}\text{+ }3.5\right){\text{in}}^{\text{2}}\text{.}$

$A{}_{top}=A_{bottom}$

$\begin{array}{l}\left(\frac{90+6\overline{y}}{16}\right){\text{in}}^{\text{2}}=\left(6.1875-\frac{\text{3}}{8}\overline{y}\text{+ }3.5\right){\text{in}}^{\text{2}}\text{.}\\ \left(\text{5}\text{.625}\text{+}\text{0}\text{.375}\overline{y}\right){\text{in}}^{\text{2}}=\left(9.6875-0.375\overline{y}\right){\text{in}}^{\text{2}}\text{.}\\ \text{0}\text{.375}\overline{y}+\text{0}\text{.375}\overline{y}=9.6875-5.625\\ 0.750\overline{y}=4.0625\\ \overline{y}=\frac{4.0625}{0.750}\\ \overline{y}=5.4166\text{in}\text{.}\end{array}$

Conclusion:

Therefore, the distance $\overline{y}$ from the top of the shape to the horizontal plastic neutral axis is $\overline{y}=5.4166\text{in}\text{.}$

To determine

(b)

Plastic moment M_Pfor the horizontal plastic neutral axis.

Answer to Problem 5.2.2P

$M_p=410.77\text{in-kips}$

Explanation of Solution

Given:

An unsymmetrical flexural member consists of a $\frac{1}{2}\times 12$ top flange, a $\frac{1}{2}\times 7$ bottom flange, and a $\frac{3}{8}\times 16$ web.

Calculation:

We have the following formula for the plastic moment of the section $M_p=F_y\times Z_p$.

Where, $F_y$ is the yield strength of the steel.

$Z_p$ is the plastic section modulus.

And we have following formula for calculating the plastic section modulus

$Z_p=\frac{A}{2}\left({\overline{y}}_1+{\overline{y}}_2\right)$

We have, $A$ is the area of cross section of the total member as follows:

$\text{Total}\text{area}\text{of}\text{cross-section = area of upper flange + area of web + area of lower flange}\text{.}$

$\begin{array}{l}\text{Total}\text{area}\text{of}\text{cross-section = }\left(12\text{'}\text{'}\times \left(\frac{1}{2}\right)\text{'}\text{'}\right)\text{ + }\left(16\text{'}\text{'}\times \left(\frac{3}{8}\right)\text{'}\text{'}\right)\text{ + }\left(7\text{'}\text{'}\times \left(\frac{1}{2}\right)\text{'}\text{'}\right)\text{.}\\ \text{Total}\text{area}\text{of}\text{cross-section = }\left(6.00{\text{in}}^{\text{2}}\right)\text{ + }\left(6.00{\text{in}}^{\text{2}}\right)\text{ + }\left(3.50{\text{in}}^{\text{2}}\right)\text{.}\\ \text{Total}\text{area}\text{of}\text{cross-section = }\left(15.50{\text{in}}^{\text{2}}\right).\end{array}$

As we have calculated, the distance $\overline{y}=5.4166\text{in}\text{.}$ from the top of the shape to the horizontal plastic neutral axis.

We now have the following diagram to consider:

Calculate the area of the top flange as follows:

$\begin{array}{l}A{}_{tf}=12\text{'}\text{'}\times \left(\frac{1}{2}\right)\text{'}\text{'}\\ A{}_{tf}=6.00{\text{in}}^{\text{2}}.\end{array}$

Now, the area of the web portion that is left is as follows:

$\begin{array}{l}A{}_{tw}=\left(\frac{3}{8}\right)\text{'}\text{'}\times \left(5.42-\frac{1}{2}\right)\text{'}\text{'}\\ A{}_{tw}=\left(\frac{3}{8}\right)\text{'}\text{'}\times \left(4.92\right)\text{'}\text{'}\\ A{}_{tw}=1.845{\text{in}}^{\text{2}}.\end{array}$

Now, calculating the centroidal distance of the web as follows:

$\begin{array}{l}{y}_w=\left(\frac{\overline{y}-\frac{1}{2}\text{'}\text{'}}{2}\right)\\ y_w=\left(\frac{5.42\text{'}\text{'}-\frac{1}{2}\text{'}\text{'}}{2}\right)\\ y_w=\frac{4.92\text{'}\text{'}}{2}.\\ y_w=2.46\text{'}\text{'}\end{array}$

Now, calculating the centroidal distance of the top flange as follows:

$\begin{array}{l}{y}_f=5.42\text{'}\text{'}-\frac{\left(\frac{1}{2}\right)\text{'}\text{'}}{2}\\ y_f=5.42\text{'}\text{'}-\frac{0.5\text{'}\text{'}}{2}.\\ y_f=5.42\text{'}\text{'}-0.25\text{'}\text{'}.\\ y_f=5.170\text{'}\text{'}.\end{array}$

Following figure shows the centroidal distances that were found in the above steps.

Calculating ${\overline{y}}_1$ which is the distance of the centroid of the top section from the plastic neutral axis by tabulating the values as follows:

Member Component	$\text{Area}\left({\text{inches}}^3\right)$	$y\left(\text{inches}\right)$	$A\times y\left({\text{inches}}^3\right)$
Web	$A{}_{tw}=1.845{\text{in}}^{\text{2}}.$	$y_w=2.460\text{'}\text{'}.$	$4.5387{\text{in}}^{\text{3}}.$
Top Flange	$A{}_{tf}=6.00{\text{in}}^{\text{2}}.$	$y_f=5.170\text{'}\text{'}.$	$31.02{\text{in}}^3.$
Total	${\displaystyle \sum =}7.845{\text{in}}^{\text{2}}.$		${\displaystyle \sum =}35.56{\text{in}}^3.$

Now calculating the value of ${\overline{y}}_1$ by the principle of moments as follows:

$\begin{array}{l}{\overline{y}}_1=\frac{{\displaystyle \sum Ay}.}{{\displaystyle \sum A}}\\ {\overline{y}}_1=\frac{35.56{\text{in}}^{\text{3}}.}{7.845{\text{in}}^{\text{2}}.}\\ {\overline{y}}_1=4.533\text{in}.\end{array}$

Now, similarly find for the lower half section, we have the following figure

Calculate the area of the bottom flange as follows:

$\begin{array}{l}A{}_{bf}=7\text{'}\text{'}\times \left(\frac{1}{2}\right)\text{'}\text{'}\\ A{}_{bf}=3.50{\text{in}}^{\text{2}}.\end{array}$

Now, the area of the web portion that is left is as follows:

$\begin{array}{l}A{}_{tw}=\left(\frac{3}{8}\right)\text{'}\text{'}\times \left(11.58-\frac{1}{2}\right)\text{'}\text{'}\\ A{}_{tw}=\left(\frac{3}{8}\right)\text{'}\text{'}\times \left(11.08\right)\text{'}\text{'}\\ A{}_{tw}=0.375\text{'}\text{'}\times \left(11.08\right)\text{'}\text{'}\\ A{}_{tw}=4.155{\text{in}}^{\text{2}}.\end{array}$

Now, calculating the centroidal distance of the web as follows:

$\begin{array}{l}{y}_w=\left(\frac{\overline{y}}{2}\text{'}\text{'}\right)\\ y_w=\left(\frac{11.08\text{'}\text{'}}{2}\right)\\ y_w=5.540\text{'}\text{'}.\end{array}$

Now, calculating the centroidal distance of the bottom flange as follows:

$\begin{array}{l}{y}_{bf}=11.08\text{'}\text{'}+\frac{\left(\frac{1}{2}\right)\text{'}\text{'}}{2}\\ y_{bf}=11.08\text{'}\text{'}+\frac{0.5\text{'}\text{'}}{2}.\\ y_{bf}=11.08\text{'}\text{'}+0.25\text{'}\text{'}.\\ y_{bf}=11.33\text{'}\text{'}.\end{array}$

Calculating ${\overline{y}}_2$ which is the distance of the centroid of the bottom section from the plastic neutral axis by tabulating the values as follows:

Member Component	$\text{Area}\left({\text{inches}}^3\right)$	$y\left(\text{inches}\right)$	$A\times y\left({\text{inches}}^3\right)$
Web	$A{}_{tw}=4.155{\text{in}}^{\text{2}}.$	$y_w=5.540\text{'}\text{'}.$	$23.02{\text{in}}^{\text{3}}.$
Top Flange	$A{}_{bf}=3.50{\text{in}}^{\text{2}}.$	$y_{bf}=11.33\text{'}\text{'}.$	$39.66{\text{in}}^3.$
Total	${\displaystyle \sum =}7.655{\text{in}}^{\text{2}}.$		${\displaystyle \sum =}62.68{\text{in}}^3.$

Now calculating the value of ${\overline{y}}_2$ by the principle of moments as follows:

$\begin{array}{l}{\overline{y}}_2=\frac{{\displaystyle \sum Ay}.}{{\displaystyle \sum A}}\\ {\overline{y}}_2=\frac{62.68{\text{in}}^{\text{3}}.}{7.655{\text{in}}^{\text{2}}.}\\ {\overline{y}}_2=8.188\text{in}.\end{array}$

Now, calculating the plastic section modulus as follows:

$Z_p=\frac{A}{2}\left({\overline{y}}_1+{\overline{y}}_2\right)$

Substituting the values, we have

${\overline{y}}_1=4.533\text{in}.$

${\overline{y}}_2=8.188\text{in}.$

And $\text{A}\text{= }15.50{\text{in}}^{\text{2}}$

$\begin{array}{l}{Z}_p=\frac{15.50{\text{in}}^{\text{2}}}{2}\left(4.533\text{in}.+8.188\text{in}.\right)\\ Z_p=\frac{15.50{\text{in}}^{\text{2}}}{2}\left(12.721\text{in}.\right)\\ Z_p=\frac{197.2{\text{in}}^3}{2}.\\ Z_p=98.59{\text{in}}^3.\end{array}$

Now, for the plastic moment of the section, we have

$M_p=F_y\times Z_p$.

Substituting the values

$F_y=50\text{ksi}\text{.}$

$Z_p=98.59{\text{in}}^3.$

$\begin{array}{l}{M}_p=50\text{ksi}\times 98.59{\text{in}}^3.\\ M_p=4929.25\text{in-kips}\\ M_p=4929.25\text{in-kips}\times \frac{\text{ft}}{\text{12 in}}\\ M_p=410.77\text{in-kips}\text{.}\end{array}$

Conclusion:

Therefore, the plastic moment M_P for the horizontal plastic neutral axis is $M_p=410.77\text{in-kips}\text{.}$

To determine

(c)

The plastic section modulus Z with respect to the minor principal axis.

Answer to Problem 5.2.2P

$Z_y=49.375{\text{in}}^3.$

Explanation of Solution

Given:

An unsymmetrical flexural member consists of a $\frac{1}{2}\times 12$ top flange, a $\frac{1}{2}\times 7$ bottom flange, and a $\frac{3}{8}\times 16$ web.

Calculation:

To find thePlastic section modulus Z with respect to the minor principal axis, we need to know that the vertical line or axis is considered to be the minor principal axis, therefore as we know that the member is symmetrical along the y-axis, and the same axis is the minor principal axis which confirms that the plastic neutral axis is passing through its center.

We have the following formula for the plastic section modulus:

$Z_y=\frac{A}{2}\left({\overline{x}}_t+{\overline{x}}_c\right)$

Where, ${\overline{x}}_t$ is the distance between centroid of the area under the tension from the plastic neutral axis,

${\overline{x}}_c$ is the distance between centroid of the area under the compression from the plastic neutral axis,

and A is the total area of cross section.

Now, as we know that the plastic neutral axis is passing through the center of the minor principal axis, we conclude ${\overline{x}}_t={\overline{x}}_c$.

And know we need to find any one of them and let that be equal to $\overline{x}$.

Thus, we have

$\begin{array}{l}{Z}_y=\frac{A}{2}\left(\overline{x}+\overline{x}\right)\\ Z_y=\frac{A}{2}\left(2\overline{x}\right)\\ Z_y=A\overline{x}.\end{array}$

We can find the $\overline{x}$ by going through the following tabulation:

Member Component	$\text{Area}\left({\text{inches}}^3\right)$	$x\left(\text{inches}\right)$	$A\times x\left({\text{inches}}^3\right)$
Top Flange	$6.00{\text{in}}^{\text{2}}.$	$6.0\text{'}\text{'}.$	$36.00{\text{in}}^{\text{3}}.$
Web	$6.00{\text{in}}^{\text{2}}.$	$0.1875\text{'}\text{'}.$	$1.125{\text{in}}^3.$
Bottom Flange	$3.50{\text{in}}^{\text{2}}.$	$3.50\text{'}\text{'}.$	$12.25{\text{in}}^3.$
Total	${\displaystyle \sum =}15.50{\text{in}}^{\text{2}}.$		${\displaystyle \sum =}49.375{\text{in}}^3.$

Calculating the centroidal distance as follows:

$\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum Ax}.}{{\displaystyle \sum A}}\\ \overline{x}=\frac{49.375{\text{in}}^{\text{3}}.}{15.50{\text{in}}^{\text{2}}.}\\ \overline{x}=3.185\text{in}.\end{array}$

Calculation of plastic section modulus is as follows:

$Z_y=A\overline{x}.$

Substituting the values, we have

$A=15.50{\text{in}}^{\text{2}}$ and $\overline{x}=3.185\text{in}.$

$\begin{array}{l}{Z}_y=A\overline{x}.\\ Z_y=15.50{\text{in}}^{\text{2}}\times 3.185\text{in}.\\ Z_y=49.375{\text{in}}^3.\end{array}$

Conclusion:

Therefore, the value of plastic section modulus Z with respect to the minor principal axis is $Z_y=49.375{\text{in}}^3.$