Last Updated on December 21, 2024 by Maged kamel

Practice problem 5-2-2: find y bar, Zx, and Zy for the unsymmetric section.

Practice Problem-5-2-2 from the steel design handbook.

There is an unsymmetric flexural member consisting of three parts. The upper flange is 12×0.50 inches, the web part is 3/8”x 16”, and the lower flange is 7×1/2”. There are three requirements: a), b), and c).
For Part a), the y-bar or the distance from the top of the upper flange to the plastic neutral axis must be found.

First, we estimate the total area of the three parts: the area of the upper flange equals 12×1/2, the area of the web equals 16×3/8, and finally, the area of the lower flange equals 7×1/2. The total area is equal to 15.50 inch2.

First, we estimate the total area of the three parts: the area of the upper flange equals 12×1/2, the area of the web equals 16×3/8, and finally, the area of the lower flange equals 7×1/2. The total area is equal to 15.50 inch2.

The distance Y bar from the top of the shape to the P.N.axis.

In the next slide, since the area of the top flange is less than At/2, the plastic neutral axis cannot cross the upper flange so that it will be located below the upper flange by a distance; we call it y1.

After checking it, we can find the y1 value by equating the difference between At/2 minus the upper flange area by the product of (3/8 y1). y1 will be equal to 4.667 inches.

In the third slide, we will estimate the y-bar distance, which is the distance from the top of the upper flange to the P.N axis.

If we add ½ inches to y1, we will have a y-bar equal to 5.166 inches, approximated to 5.17 inches. We could check the lower portion’s area to ensure it equals At/2.

From the next slide, as required by item b), we need to find the Plastic moment in case A572 steel grade 50 is used. The yield stress Fy equals 50 ksi. We will estimate the compression force C, which equals At/2*Fy=7.75*50=387.50 kips.

The distance C1 from the compression force to the P.N axis.

We will estimate the C1 distance, which is the distance from the point of application of force c to the P.N.A. The value of c1 can be estimated by taking the first moment of area for the upper half of the section about the P.N.A.The moment of area equals 33.609 inch3.

When we divide the first moment of the area by At/2, we find that the value of C1 equals 4.34 inches.

The distance T1 from the Tension force to the P.N axis.

We move to the next slide to estimate the distance T1, which is the distance from the point of application of Force T to the P.N.A. Since we do not have symmetry, we expect a different value than c1.
Again, we will estimate the moment of area of the lower half of the section, which is the sum of the moment of area of the lower flange and the web area part below the P.N axis. The value of T1 equals 8.34 inches.

 In the next slide, we have the values of the C and T forces and c1 and T1 the two distances. Mp equals C*(C1+T1) or C*yct, which is the vertical distance between force C and force T.

The Plastic Moment about P.N axis-part b.

The Mp value equals (387.50×12.68); we divide by 12 to find the value of Mp in Ft-kips. The Mp value is equal to 409.46 Ft-kips, which is the requirement for item b).

The value of a section modulus about the minor axis y.

Due to symmetry about axis y, we will deal with the C-shape section, find the first moment of area about the y-axis, and divide by the sum of the three shapes. We will find that X—X-bar equals 1.592 inches.

Finally, we will estimate the yielding MomentMy (C*2x bar) and divide by Fy. The final value of the plastic section modulus is about Y, and Zy equals 24.676 inch3.

For bending members, please refer to Chapter 8-A Beginner’s Guide to the Steel Construction Manual, 15^th ed.

The next post-Post 6b- is Practice problem 5-2-3-verify Zx for W18x50.

The next post is “A Guide to Local Buckling Parameters for Steel Beams.