# 6A-Practice problem 5-2-2 Find y bar, Zx, and Zy.

## 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”. The lower flange is 7×1/2”. There are three requirements a), b), and c).
For Part a) It is required to find the y-bar or the distance from the top of the upper flange to the plastic neutral axis.

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 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, we add ½ inches to y1 we will have y- bar equal to 5.166 inches which is approximated to 5.17 inches. We could check the area of the lower portion to make sure that 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. We divide the first moment of area y the At/2, we can find the value of C1 as equal to 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. We expect to have a different value compared to c1 since we do not have symmetry.
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 C, T  forces values and c1, T1 the two distances, Mp is equal to 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, this is the requirement by 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 shapes section and 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 My which is (C*2x bar), and then divide by Fy. The final value of plastic section modulus about Y, Zy is equal to 24.676 inch3.

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

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

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

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