Chapter 5, Problem 5.2.1P

To determine

(a)

Plastic section modulus $Z$ and the plastic moment $M_p$ with respect to the major principal axis.

Answer to Problem 5.2.1P

The plastic section modulus $Z=185.437i{n}^3$ and the plastic moment $M_p=772.65\text{ft}\text{.}\text{kips}$ with respect to the major principal axis.

Explanation of Solution

Given:

A flexural member is fabricated from two flange plates $\frac{1}{2}\times 7\frac{1}{2}$ and a web plate $\frac{3}{8}\times 17$. The stress of the steel is $50Ksi.$

Concept used:

The section is a symmetrical section which implies that the plastic neutral axis of the given section is same as the neutral of the given section. Therefore, calculating the lever arm and the centroid of the upper half of the given section, we can find the plastic section modulus.

We have the following figure that will define the terms that we have been given as per the question.

Calculation:

The following is the tabular measurement of every component required:

Elements	$h(inches)$	$b(inches)$	$A=h\times b{(inches)}^2$	$y(inches)$	$Ay{(inches)}^3$
Web	$\frac{3}{8}=0.375$	$\frac{17}{2}=8.5$	$8.5\times 0.375=3.1875$	$8.50$	$27.09$
Flange	$\frac{1}{2}=0.5$	$7\frac{1}{2}=7.5$	$0.5\times 7.5=3.75$	$17.50$	$65.625$
Sum			${\displaystyle \sum A=}6.94$	${\displaystyle \sum y=}26.00$	${\displaystyle \sum Ay=}92.72$

Calculating the centroid of the top half as :

$\overline{y}=\frac{{\displaystyle \sum Ay}}{{\displaystyle \sum A}}$

Substitute the values in the above equation.

$\begin{array}{l}\overline{y}=\frac{{\displaystyle \sum Ay=}92.72in}{{\displaystyle \sum A=}6.94in}\\ \overline{y}=13.36in.\end{array}$

Now, calculating the moment arm, we have the following formula :

$a=2\times \overline{y}$

Where, a is the moment arm of the section.

$\begin{array}{l}a=2\times \overline{y}\\ a=2\times 13.36in.\\ a=26.72in.\end{array}$

Now, the plastic section modulus can be calculated as follows:

$Z=\left(\frac{A}{2}\right)\times a$

Where, Z is plastic section modulus and A is area.

$\begin{array}{l}Z=\left(\frac{A}{2}\right)\times a\\ Z=6.94i{n}^2\times 26.72in.\\ Z=185.437i{n}^3.\end{array}$

Calculating the plastic moment as follows:

$M_p=F_y\times Z$

Substitute the value of $F_y$ and $Z$, we have

$\begin{array}{l}{M}_p=F_y\times Z\\ M_p=50Ksi\times 185.437i{n}^3.\\ M_p=9271.85\text{in}\text{.}\text{kips}\text{.}\\ M_p=772.65\text{ft}\text{.}\text{kips}\text{.}\end{array}$

Conclusion:

Therefore, the plastic section modulus $Z=185.437i{n}^3$ and the plastic moment $M_p=772.65\text{ft}\text{.}\text{kips}$ with respect to the major principal axis.

To determine

(b)

Elastic section modulus, $S$ and the yield moment, $M_y$ of the section with respect to the major principal axis.

Answer to Problem 5.2.1P

The elastic section modulus, $S=272.284i{n}^3$ and the yield moment of the section with respect to the major principal axis is $M_y=1134.52\text{ft}\text{.}\text{kips}$.

Explanation of Solution

Given:

A flexural member is fabricated from two flange plates $\frac{1}{2}\times 7\frac{1}{2}$ and a web plate $\frac{3}{8}\times 17$. The stress of the steel is $50Ksi.$

Concept used:

The section is a symmetrical section which implies that the elastic neutral axis of the given section is coinciding with the neutral of the given section. Therefore, calculating the moment of inertia at the major axis using parallel axis theorem, we can find the elastic section modulus.

We have the following figure that will define the terms that we have been given as per the question.

Calculation:

The following is the tabular measurement of every component required:

Elements	$\overline{I}{(inches)}^4$	$A=h\times b{(inches)}^2$	$d(inches)$	$I=\overline{I}+A\times d^2{(inches)}^4$
Web	$153.53$	$17\times \frac{3}{8}=6.375$	$0.00$	$153.53$
Top Flange	$0.078125$	$0.5\times 7.5=3.75$	$17.50$	$\begin{array}{l}I=0.078125+\left(3.75\times {17.50}^2\right)\\ I=1148.52\end{array}$
Bottom Flange	$0.078125$	$0.5\times 7.5=3.75$	$17.50$	$\begin{array}{l}I=0.078125+\left(3.75\times {17.50}^2\right)\\ I=1148.52\end{array}$
Sum				${\displaystyle \sum I=}2450.56$

Calculate the Elastic section modulus S with the following formula

$S=\frac{{\displaystyle \sum I}}{c}$

Where, C is the distance between the extreme fiber of the section and the neutral axis and is equal to

$c=\frac{h_w}{2}+b_f$

Here, $h_w$ is the height of the web and $b_f$ is the width of flange.

By substituting the values in the above equation, we have

$\begin{array}{l}c=\frac{h_w}{2}+b_f\\ c=\frac{17}{2}+0.5\\ c=8.5+0.5\\ c=9.00in.\end{array}$

Substitute the value of c in the following equation, $S=\frac{{\displaystyle \sum I}}{c}$

$\begin{array}{l}S=\frac{2450.56i{n}^4}{9.00in}\\ S=272.284i{n}^3.\end{array}$

Now, calculate the yield moment $M_y$ of the section with respect to the major principal axis as follows:

$M_y=F_y\times S$

Substitute the value of $F_y$ and $S$, we have

$\begin{array}{l}{M}_y=F_y\times S\\ M_y=50Ksi\times 272.284i{n}^3.\\ M_y=13614.22\text{in}\text{.}\text{kips}\text{.}\\ M_y=1134.52\text{ft}\text{.}\text{kips}\text{.}\end{array}$

Conclusion:

Therefore, the elastic section modulus, $S=272.284i{n}^3$ and the yield moment of the section with respect to the major principal axis is $M_y=1134.52\text{ft}\text{.}\text{kips}$.