# 9c- Practice problem-longitudinal weld of a c section

## Practice problem-longitudinal weld of a C section.

This is a Practice problem for the longitudinal weld of a C section. It is case 2 for the shear lag factor table D3.1-CM-14 and can also be considered case 4 based on CM-15 and related specifications 2016.

In this post, we will solve two solutions: the first is based on CM-14, and the second is based on CM-15.

Refer to Table 1-5 part-1 to get the flange thickness and the depth for the given C6x10.50 section.

From Part 2 of Table 1-5, we get the x-bar distance.

### Shear lag factor U based on CM-14.

As we can see from the next picture, for Table D3.1, shear lag factors for connections to tension members, for longitudinal weld, it is case 4, but it is only for plates where the tension load is transmitted by longitudinal weld only. We will use case 2 instead.

The shear lag factor depends on the width of the joint w and establishes a relation between the length of connection L and width W.

1-From the given Data for the ASTMA36, we can get the yield stress Fy=36 ksi and the ultimate stress Fult equal to 58 ksi.

2—From the given section of C6x10.50, we can find the following data. The first item is the area of the given section, which will be equal to 3.07 inches2.

The breadth of the flange is bf=2.03 inches, the flange thickness is 0.343 inches, and the width of the web equals 0.314 inches. As for the overall depth, it will be equal to 6 inches.

3- The length of connection is given by the given section as equal to 5 inches.

4- The U value equals (1-xbar/L)=(1-0.5/5)=0.90.

### LRFD strength for Practice problem-longitudinal weld of a plate section-CM-14.

The gross area is the area of the C channel, which equals 3.07 inches2, while the effective area equals 0.9* 3.07 = 2.763 inches2.

The LRFD strength value for yielding equals 99.47 kips, while the LRFD strength due to rupture equals 120.19 kips.

We will select the lesser value as our final LRFD strength, which is 99.50 kips.

#### ASD strength for Practice problem-longitudinal weld of a plate section-CM-14

The gross area is the area of the C channel, which equals 3.07 inches2, while the effective area equals 0.9* 3.07 = 2.763 inches2.

The ASD strength value for yielding equals 66.18 kips, while the ASD strength due to rupture equals 80.13 kips.

We will select the lesser value as our final ASD strength, which is 66.18 kips.

### Shear lag factor U based on CM-15.

Referring to Table D3.1 for shear lag factor based on CM-15. The item number for shear lag factor U for the longitudinal weld to a C channel is case 4a. The U value is estimated by the product of (3L^2/3l^2*w2) multiplied by (1-1- x̅/ L).

The connection length is given by the given section as equal to 5 inches.

The first term (3L^2/3l^2*w^2 equals (3*5^2/(3*5^2+5^2) to be multiplied by (1-(0.50/5), the final value of U=0.608.

#### LRFD strength for Practice problem-longitudinal weld of a C section-CM-15.

The LRFD strength value for yielding equals 99.468 kips, while the LRFD strength due to rupture is equal to 81.24 kips.

We will select the lesser value as our final LRFD strength, which equals 81.24 kips. This indicates that the LRFD strength is governed by rupture.

#### ASD strength for the Practice problem-longitudinal weld of a C section-CM-15.

From the estimated calculation the selected ASD strength for the Practice problem-longitudinal weld of a C section-CM-15 is 54.14 kips, the full data is shown in the next slide image.

The next post is post #9d, about how to get an Effective area for a staggered bolted angle.

Chapter 3 – Tension Members– A Beginner’s Guide to Structural Engineering is a great external resource.

Scroll to Top