Last Updated on June 18, 2024 by Maged kamel

## Nominal shear strength for S41x121-use CM#14.

### Compute the Nominal shear strength for S41x121.

We have a given S41x121 steel section of A572 grade 65; we must compute the nominal shear strength. In this example, we will apply CM#14 and the related specification 2010.

First, we will use Table 2-4 to find the yield stress Fy value for A572 steel grade 65. This type of steel has a yield stress Fy=65 ksi and an ultimate stress Fu=80 ksi.

The next question concerns the table to get the full data for S41x121; in part 1, we will find that Tables 1-3 are used to get the information for sections.

The data for S21x121 can be obtained from Table 1-3 from part 1. The flange width bf is 8.05 inches, and the thickness is 1.09 inches.

The depth is d=24.50 inches, and the web thickness is 0.80 inches. The most critical parameter is h/tw, which we will find in part 2 of the same table.

### Find h/tw from Table 1-3, part 2.

The ratio for h over the thickness of the web is equal to 25.90. The distance h is between the web’s two filleted portions. The next slide image shows part 2 of Table 1-3.

### Find the limiting h/w based on Fy.

The limiting h/tw and the yield stress are related, as shown in the next slide image. Based on the Fy of the given S41x121 for grade 65, the limiting h/w equals 47.30.

The following slide image shows the relation of h/tw and the nominal shear value. There are four zones; the first zone is from zero to 47.30, marked as point 1 on the x-axis, which is obtained from the relation of 2.24*sqrt(E/Fy); in this zone, Cv equals 1, and the φv equals 1. The nominal shear value is given by the equation for nominal shear Vn=Cv*0.60Fy*d*tw and can be written as 1*0.60Fy*Aw.

The second Zone is from h/tw equals 47.30 to 53.69, marked as point 2 on the x-axis, and is obtained from the relation of 1.1*sqrt(kv *E/Fy); in this zone, Cv equals 1, and the φv equals 0.90. the Kv factor equals 5.34 in specification 2016. The nominal shear value is given by the equation Vn=Cv*0.60Fy*d*tw and can be written as 1*0.60Fy*Aw.

The third Zone is from h/tw equals 53.69 and up , Cv1 is less than 1, and the φv equals 0.90. The nominal shear value is given by the equation Vn=Cv1*0.60Fy*d*tw; the Cv1 equals 1.1*sqrt(kv *E/Fy)/h/tw or 53.69/(h/tw).

For section M10x7.50, h/tw is 71; in zone 3, the Cv1 value will equal 53.69/71=0.756. Please refer to the next slide image.

### The nominal shear strength for M10x7.5.

The area of the web equals d*tw or 9.99*0.13 or 1.2987; the cv1 value equals 0.7562, and the nominal shear equals 0.7562*(0.60*Fy)*(Aw)=0.7562*0.6*65*11.2987=38.30 Kips.

### Excel plot for h/tw versus Vn.

An Excel plot for the relation between h/tw and the nominal shear Vn shows the four zones for shear. Zone 3 starts from Vn equals 50.41 kips to Vn equals 13.60 for h/tw=200; our section is shown in the graph with H/tw=71.0 and vn=38.30 kips. Thanks a lot.

As an external resource for the shear stress limit state.

The previously solved problem was checked based on CM#14.

The next post is a solved problem-Nominal shear strength for M10x7.50.