- Case 2 for block Shear-Coped Beam Problem.
- What are the gross sections for the given beam?
- The value of shear yielding.
- Gross and net area for tension-UBs=1/2 and tension rupture.
- Estimate the shear rupture and tension rupture.
- Estimate the LRFD value of the block shear force Case 2 for the block Shear-Coped Beam Problem.
- Estimate the ASD value of the block shear force Case 2 for block Shear-Coped Beam Problem.

### Case 2 for block Shear-Coped Beam Problem.

In this post, I will include a solved problem for Case 2 for block Shear-Coped Beam problem where the UBS=1/2. The solved problem number #10.8 from the Unified Design of Steel Handbook.

It is required to determine the resistance to block shear of the coped beam W16x40 grade A992, which is the same coped beam in the first solved problem in the previous post But with a different steel grade which is ASTM A992.

The horizontal distance between the beam edge and the center line of the first line of bolts is 1.25 inches, and the vertical distance between the edge of the beam to the first line of bolts is 2 inches. While the vertical spacing between bolts is 3 inches.

There are two vertical lines of bolts, the horizontal edge distance to the first line of bolts is 1 1/4″. the horizontal spacing between the lines of bolts is 3 inches. Please refer to the next slide image for more details. The bolt diameter is 5/8 inches. because we have more than one line of bolts this case is case 2, where the UBS value is equal to 1/2.

#### What are the gross sections for the given beam?

For W16x40, the web width is 0.305 inches, the breadth of the flange is 7 inches, and the flange width is 0.505 inches. The yield stress for A992 is 50 ksi, while the ultimate stress is 65 ksi.

For the hole diameter, we add 1/8 of inches to the bolt diameter and the hole diameter will be equal to 6/8 inches.

2-There is one section that is subjected to shear, and that is the vertical section thru bolts. There is also one section that is subjected to tension, and that is the lower section.

#### The value of shear yielding.

3-We estimate the gross area for the vertical section subjected to shear which can be found equal to (11×0.305)=3.355 inch2. We can estimate the shear yielding value which is the product of the gross area by (0.6*Fy), where Fy=50 ksi. The shear-yielding value is equal to 100.65 kips.

#### Gross and net area for tension-UBs=1/2 and tension rupture.

4-We estimate the gross area for the horizontal section subjected to tension which can be found equal to (4.25×0.305)=1.296 inch2.

To get the net area for tension or Ant we deduct the area for one and a half of a bolt hole, the net area Ant can be found to equal 0.9531 inch2. This is case 2 for the block Shear-Coped Beam Problem. For the tension rupture force multiply (ubs*Ant*Fu) which is (1/2*0.9351*65)=30.976 kips.

#### Estimate the shear rupture and tension rupture.

We can get the value of shear yielding and tension rupture. To estimate the shear rupture, we need to find the net area for shear which is the gross area for shear minus (3.5 holes area). The hole diameter of the bolts is 6/8 inches, the web thickness is 0.305 inches and the height is 11 inches. The net area for shear is equal to 2.554 inch2. Multiply by (0.6*Fy) to get the shear rupture force value which is 99.60 kips. for the tension rupture, we estimated earlier as equal to 30.976 kips.

#### Estimate the LRFD value of the block shear force Case 2 for the block Shear-Coped Beam Problem.

We list all the data which we have obtained we group into two groups. The first group is the case of shear yielding and tension rupture. The second group will be the Shear rupture and tension rupture. We can find that the failure is controlled by shear rupture and tension rupture since their sum is less than the shear yielding and tension rupture.

We select the minimum 130.576 kips, we multiply by phi which is equal to 0.75 we get the LRFD value which is 98 kips, please refer to the next slide image for more details.

#### Estimate the ASD value of the block shear force Case 2 for block Shear-Coped Beam Problem.

We list all the data which we have obtained we group into two groups. The first group is the case of shear yielding and tension rupture. The second group will be the Shear rupture and tension rupture. We can find that the failure is controlled by shear rupture and tension rupture since their sum is less than the shear yielding and tension rupture.

We select the minimum 130.576 kips, we divide by omega which is equal to 2 we get the ASD value which is 65.30 kips, please refer to the next slide image for more details.

I have come to an end for Case 2 for block Shear-Coped Beam Problem. Thanks a lot.

For more details regarding block shear, please refer to the post -Quickstart to the introduction to block shear resistance.

For an introduction to coped beams, please refer to the previous posts, post 13-The relation between Block shear and coped beams.

Post-19-Solutions for Block Shear-Coped Beam Problem.

Post-20-Aisc Tables 9-3a,b, and C for Block shear-coped beam.

For a more detailed illustration of block shear, there is a very useful external link- **Tensile yielding and tensile rupture**. **A Beginner’s Guide to the Steel Construction Manual, 14 ^{th} ed.**