- Practice problem-longitudinal weld of a C section.
- Estimate the minimum value of shear lag factor U based on CM-14.
- The estimation of the effective area-CM-14.
- Nominal strength for longitudinal weld of a C section.
- Shear lag factor U based on CM-15.
- Estimate the shear lag factor U based on CM-15.
- The estimation of the effective area-CM-15.
- Nominal strength for 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., this is case 2 for the shear lag factor table D3.1-CM-14. and also can be considered as case 4a based on CM-15 and related specification -2016.

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

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, here it is only for plates where the tension load is transmitted by longitudinal weld only. 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.

### Estimate the minimum value of shear lag factor U based on CM-14.

1-From the given Data for the ASTMA36, we can get the yield stress Fy=36 ksi and the ultimate stress Fult as 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, the area will be equal to 3.07 inch2. 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 minimum value of the shear lag factor U can be estimated as the ratio of the connected area by the longitudinal weld over the total area of the C section.

5- The connected area is equal to the height of the section by its thickness, which is the product of 6×0.314=1.884 inch2

6- The U value equals the connected area over the area of the section or=(1.844/3.07)=0.614.

### The estimation of the effective area-CM-14.

The net area of the -longitudinal weld of a C section is equal to the gross area of the C section which is equal to 3.07 inch2.

The U factor for the C channel can be estimated from Case # 2, U=(1- x̅ /L). The x̅ value from the table can be found as equal to 0.5 inches, and the length of connection is given as 5 inches.

We apply in the U equation, the shear lag factorU=(1-(0.50/5)=0.90 which is bigger than U min, and select the maximum value which is 0.90.

The effective area can be found by multiplying the U*net area and the value is equal to 2.763 inch2.

### Nominal strength for longitudinal weld of a C section.

We have two cases for the nominal strength, the first case is the case of tensile yielding for which we consider the area as the full area and the tensile strength is equal to yield stress multiplied by the area and can be found as (3.07*36)=110.52 kips.

While the second case for tensile strength is the case of rupture strength in which the value is found to be equal to the effective area multiplied by the ultimate stress. and can be found as (2.763*58)=160.254 kips.

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

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

We will select the lesser value as our final LRFD strength which equals 99.50 kips, this is an indication that the LRFD strength is governed by yielding.

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

Multiply the first value of the strength due to yield by the reduction value (1/omega) equal to (1/1.67), we can get the ASD strength due to yielding as equals 66.18 kips.

Multiply the second value of the strength due to rupture by the reduction value of (1/omega) equal to (1/2), and we can get the ASD strength due to rupture as equals 80.127 kips.

We will select the lesser value as our final ASD strength which equals 41 kips, this is an indication that the ASD strength is governed by yielding.

### 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 termed 4a and includes also other structural items that have also longitudinal weld unlink, the previous version of table D3.1 for Cm-14. The U value is estimated by the product of (3L^2/3l^2*w2) multiplied by (1-1- x̅/ L).

1-From the given Data for the ASTMA36, we can get the yield stress Fy=36 ksi and the ultimate stress Fult as 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, the area will be equal to 3.07 inch2. 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 minimum value of the shear lag factor U can be estimated as the ratio of the connected area by the longitudinal weld over the total area of the C section.

5- The connected area is equal to the height of the section by its thickness, which is the product of 6×0.314=1.884 inch2.6- The U value equals the connected area over the area of the section or U min=(1.844/3.07)=0.614.

### Estimate the 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 plate is termed 4a and includes also other structural items that have also longitudinal weld unlink, the previous version of table D3.1 for Cm-14. The U value is estimated by the product of (3L^2/3l^2*w2) multiplied by (1-1- x̅/ L). The second term is to be estimated.

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 which is close to the minimum u value which is 0.6137, choose U as equal to 0.61

### The estimation of the effective area-CM-15.

The net area of the -longitudinal weld of a plate section is equal to the gross area of the C section which is 3.07 inch2.

The effective area can be found by multiplying the U*net area and the value is equal to 1.685 inch2.

### Nominal strength for Practice problem-longitudinal weld of a C section

We have two cases for the nominal strength, the first case is the case of tensile yielding for which we consider the area as the full area and the tensile strength is equal to yield stress multiplied by the area and can be found as (1.875*36)=73.125 kips.

While the second case for tensile strength is the case of rupture strength in which the value is found to be equal to the effective area multiplied by the ultimate stress. and can be found as (1.685*58)=97.73 kips.

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

The LRFD strength value for yielding equals 99.468 kips, while the LRFD strength due to rupture as equal to 73.30 kips.

We will select the lesser value as our final LRFD strength which equals 73.30 kips, this is an indication that the LRFD strength is governed by rupture.

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

From the estimated calculation the selected ASD value is 48.87 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.