Chapter 8, Problem 8.1PFS

Find the values of S and Z and the shape factor about the horizontal x—x axes, for the sections shown in the accompanying figures.

8-1.

To determine

The values of S and Z and the shape factor about the horizontal x-x axes for the given section.

Answer to Problem 8.1PFS

  $194.587in{.}^3$ , $350.93in{.}^3$ and 1.80

Explanation of Solution

The cross-section is shown below:

  

The depth of neutral axis from bottom is given by

  $\begin{array}{l}\overline{y}=\frac{A_1{y}_1+A_2{y}_2}{A_1+A_2}\cdots \cdots \cdots \cdots \left(1\right)\\ where,\overline{y}isthedepthofneutralaxisfrombottom\\ A_{1}isareaof\sec tion1\\ A_{2}isareaof\sec tion2\\ y_{1}isdepthofcenterofgravityof\sec tion1\\ y_{2}isdepthofcenterofgravityof\sec tion2\end{array}$

The area of both section 1 and section 2 are

  $\begin{array}{l}{A}_1=14\times 3\\ A_1=42in{.}^2\\ A_2=17\times 2\\ A_2=34in{.}^2\end{array}$

The depth of centre of gravity both section 1 and section 2 are

  $\begin{array}{l}{y}_1=17+\frac{3}{2}\\ y_1=18.5in.\\ y_2=\frac{17}{2}\\ y_2=8.5in.\end{array}$

Now, substituting the values in equation (1),

  $\begin{array}{l}\overline{y}=\frac{A_1{y}_1+A_2{y}_2}{A_1+A_2}\\ \overline{y}=\frac{42\times 18.5+34\times 8.5}{42+34}\\ \overline{y}=14.026in.\end{array}$

The distance of centre of gravity from section 1 from the Neutral Axis (N.A.) is

  $\begin{array}{l}{h}_1=y_1-\overline{y}\\ h_1=18.5-14.026\\ h_1=4.474in.\end{array}$

The distance of centre of gravity from section 2 from the Neutral Axis (N.A.) is

  $\begin{array}{l}{h}_2=\overline{y}-y_2\\ h_2=14.026-8.5\\ h_2=5.526in.\end{array}$

The moment of inertia is

  $\begin{array}{l}{I}_x=\left[\left(\frac{B\times t_f^3}{12}\right)+\left(A_1\times h_1^2\right)\right]+\left[\left(\frac{t_w\times d_w^3}{12}\right)+\left(A_2\times h_2^2\right)\right]\\ I_x=\left[\left(\frac{14\times 3^3}{12}\right)+\left(42\times {4.474}^2\right)\right]+\left[\left(\frac{2\times {17}^3}{12}\right)+\left(34\times {5.526}^2\right)\right]\\ I_x=872.2+1857.08\\ I_x=2729.28in{.}^4\end{array}$

The elastic section modulus is

  $\begin{array}{l}{S}_x=\frac{I_x}{c}\\ S_x=\frac{2729.28}{14.026}\\ S_x=194.587in{.}^3\end{array}$

The location of plastic neutral axis is

  

The area of flange is greater than the half of the sectional area. Hence the plastic neutral axis is located within the flange as shown in the above figure.

The depth of plastic neutral axis from top of the section x is

  $\begin{array}{l}14\left(x\right)=\left(17\times 2\right)+14\left(3-x\right)\\ 14\left(x\right)=34+42-14x\\ 28x=76\\ x=2.71in.\end{array}$

The area of both section 1 and section 2 are

  $\begin{array}{l}{A}_1=14\times 2.71\\ A_1=38in{.}^2\\ A_2=14\times \left(3-2.71\right)\\ A_2=4.06in{.}^2\\ A_3=2\times 17\\ A_3=34in{.}^2\end{array}$

The depths of centre of gravity both section 1 and section 2 from plastic neutral axis are

  $\begin{array}{l}{y}_1=\frac{x}{2}\\ y_1=\frac{2.71}{2}\\ y_1=1.355in.\\ y_2=\frac{3-x}{2}\\ y_2=\frac{3-2.71}{2}\\ y_2=0.145in.\\ y_3=\left(3-x\right)+\frac{17}{2}\\ y_3=\left(3-2.71\right)+\frac{17}{2}\\ y_3=8.79in.\end{array}$

The plastic section modulus is

  $\begin{array}{l}{Z}_x=A_1{y}_1+A_2{y}_2+A_3{y}_3\\ Z_x=\left(38\times 1.355\right)+\left(4.06\times 0.145\right)+\left(34\times 8.79\right)\\ Z_x=350.93in{.}^3\end{array}$

The shape factor is

  $\begin{array}{l}Shapefactor=\frac{Z_x}{S_x}\\ Shapefactor=\frac{350.93}{194.587}\\ Shapefactor=1.80\end{array}$

Conclusion:

The values of S and Z and the shape factor about the horizontal x-x axes are $194.587in{.}^3$ , $350.93in{.}^3$ and 1.80 respectively.