Solved problem 4-3 for elastic and plastic sections.
This is a solved example 4.3 from Prof. Alan Williams’s reference manual. It is required to estimate three items Sx, Zx, and shape factor for the Sx, first, we need to evaluate the Y bar, Ymax, and Ix value, the next photos show the two methods for estimating the Ix value, these methods will lead to the same result.
The first method for Ix estimation.
The first method is to divide the T-section into a flange and stem, we will estimate the area and the Cg value for each area, and an outer datum is considered at the bottom of the stem. The first moment of area is used to estimate the distance of the elastic neutral axis from the datum line by setting the different values from the cg of both flanges and stem to the datum line.

The y bar distance is 14.883″, which is the distance from the lower datum. Now we will estimate Y1 top and y2 bottom from the elastic neutral axis to the upper and lower fibers. the total moment for inertia will be estimated for the two areas about the Elastic neutral axis, the steps can be reviewed in the next slide.

The total moment of inertia is equal to 2275.54 inch4. To get the elastic section modulus of elasticity we will use the next formula. we have to use y max value or the maximum value of y1 top or y2 bottom. Sx value or the elastic section modulus is 153 inch2.

The second method for Ix estimation.
This is the second method for estimating the Ix value for the T section. After finding the Ybar distance from the lower datum, we will estimate ix for the T section.
The T section can be represented by a big rectangle of width=15 inches and overall depth =20”, from which we can deduct two rectangles, of breadth=7.50” and height=17”. As shown in the next slide, we will estimate the moment of inertia for the T section about the neutral axis. Ix final value=2275.53 inch4 which is the same value as estimated earlier.
How to estimate Zx?
Now for the estimating of Zx, the following pictures, will show the procedure for Zx, or the Plastic neutral axis or P.N.A divides the area into two equal areas.
Assuming that we have y distance from the top fiber to the plastic section axis, we equate the two areas and get the value of y, since =plastic section modulus divides one area into two equal halves.
The assumption should fulfill that A1=A2=At/2, where At is the total area, the total area =.15*3+17*1.5=70.50 inch2. Half of this area=0.50*70.50=35.25 inch2. the plastic neutral axis cuts the T flange at a distance 2.35″ from the top, Please refer to the next slide image.
1-Once the location of the plastic neutral axis is finalized, we will determine y1 and y2.
2-Estimate the plastic section modulus from the relation Zx=At/2*(Y1+Y2), where (At) is the total area, y1 is the distance from the plastic neutral axis to the upper fiber, while y2 is the distance from the plastic neutral axis to the Lower fiber
Estimation of the Shape factor.
3- Estimate the shape factor by dividing Zx/Sx, Zx value=278 inch4, Sx=153 inch4, and the shape factor=278/153=1.817.
This is the pdf file used for the illustration of this post.
For a useful external link-Chapter 8 – Bending Members.
This is a link for the next post, 5-Solved problem 5-1 for Sx&Zx.