## Solved Problem 5-2 For the plastic moment value for W10x60.

It is required to estimate the Plastic moment Mp for the given section, W10x60 of A992.

### Using the first option by considering the W section as a group of plates.

I have considered, as a first option, the W section as consisting of three plates, for the sake of comparison, after evaluating the Zx and Mp. for the first option as an assembly of plates for the solved problem 5-2. The upper and lower plate dimension is (10.10″)*0.68″, while the intermediate plate dimension is 8.84″*0.42″.

The following pictures show in detail the sequence for that calculation.

Due to the symmetry of the section, the P.N.A will be in the middle. for A1+A2 areas, where A1 is the upper plate area, its value=6.868 inch2, while A2 is the area of half of the web, adding them together will give AT/2, where AT is the total area. the value of the AT/2=8.7244 inch2. The next step is to estimate the Cg distance For A1+A2, from the first moment of areas about the P.N.A. The next step is to estimate the Cg distance For A1+A2, from the first moment of areas about the P.N.A.

We consider the formula AT/2*y bar=A1*y1+A2*y2, we have=Ybar=(6.868*4.76+1.8564*2.21)/8.7244, which will be Y bar =4.22″, as we can see from the next slide.

Zx can be estimated as Zx=AT/2*(Y1bar+Y2 bar), the AT/2=8.7244 inch2, while Y1 bar=y2 bar=4.22″, then Zx=8.7244*(2*4.22)=73.63 inch3.

For the formula Mp=Zx*Fy, then the Plastic moment can be calculated as Mp=73.63*50=3681.50 inch-kips, to convert that value to Ft-kips, just divide by 12, at the end the plastic moment value MP=307.0 Ft- kips for the solved problem 5-2 as an option- a).

### Can we use Table 1.1 to get Zx and Mp?

Using table 1-1 to get the plastic moment will give us only the Zx value and the dimension of the sections as the W10x60 section from AISC table 1-1.

The flange width=10.10″ the tf=0.68″ and the overall depth=10.20″ and the web thickness=0.42″. The Zx value is 74.60 inch3, we need y plastic to estimate the Mp value, and we cannot find the Y plastic in that table.

The only solution is to consider W10x60 as composed of two Wt sections. Each Wt section is Wt 5×30. We need to proceed to another table.

The plastic section modulus or Z value of w10x60 can be found in Table 1-1. The Z value =74.60 inch3.

### Using the second option by considering the W section as two Wt sections.

The data for the needed Wt section of WT5x30 can be found in Table 1-8. We can get the Y bar from the section property of WT5x30, the value Y bar=0.884″, which is the distance from the plastic neutral axis to the upper top flange, as we can see from the next slide.

The total area of the Wt 5×30 =8.84 inch2.

To get the plastic moment we join the two Wt 5×30 to create W10x60, the compression force C is acting in the plastic Neutral axis of Wt which is=0.884″ as we get from the previous data of WT5x30. The acting Tensile force T due to symmetry acts at the plastic neutral axis but at a distance =0.844″ from the bottom.

The distance between C and T which we call it Yct is the whole depth(-) 2*yp. the overall depth we can get from table1-1 fr W10x60, which is=10.20″. The compression force C=T=At/2*Fy=(17.70/2)*50==442.50 kips, yct=10.20″-2*0.884=8.432″.

The Plastic moment=442.50*8.432/12=310.93 Ft.kips and can be approximated to 311.0 Ft.kips.

There is another way by considering Mp=Zx*Fy=At/2*yct, which gives us the same result as Mp=311.0 ft. kips.

The following calculation, accompanied by a sketch from the next slide shows the value of Zx and Mp. For the sake of comparison between options a and b, the values of Mp are written side by side.

This is a link to download the PDF file used for the illustration of this post.

For the buckling concept, please refer to this link from Prof T. Bart Quimby, P.E., Ph.D., F.ASCE site.

For more solved problems: please refer to post 6A-Practice problem 5-2-2, Find y bar, Zx, and Zy.

This is another post: 6b-Practice problem 5-2-3-verify Zx for W18x50.

This is the link for the next post, 7-Local buckling parameters.