 # 19-Solutions for Block Shear-Coped Beam Problems

## Solutions for Block Shear-Coped Beam Problems.

In this post, we will introduce solutions for Block Shear-Coped Beam Problems. We have two solved problems for block shear-coped beams. The first solved problem is quoted from solved problem#5.13 from Prof. Alan Williams‘s handbook.
The solved problem shows how to estimate the block shear-coped beam value when the factor for block shear  UBS equals 1.00, which is the case where we have a uniform stress distribution for tension force.

### First solved problem for Block Shear-Coped Beam-UBS=1.00

It is required to determine the resistance to block shear of the coped beam W16x40 grade A36.
The horizontal distance between the beam edge and the center line of the first line of bolts is 1.50 inches, and the vertical distance between the edge of the beam to the first line of bolts is 1.5 inches. While the vertical spacing between bolts is 3 inches as shown in the next slide image. The bolt diameter is 3/4 inches.

#### Gross and net area for shear-UBs=1.

1. 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 A36 is 36 ksi, while the ultimate stress is 58 ksi.

For the hole diameter, we add 1/8 of inches to the bolt diameter and the hole diameter will be equal to 7/8 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.

3-We estimate the gross area for the vertical section subjected to shear which can be found equal to =(7.50×0.305)=2.29 inch2. To get the net area for shear or Anv we deduct the area for two holes and a half, the net area Anv can be found equal to 1.625 inch2.

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

4-We estimate the gross area for the horizontal section subjected to tension which can be found equal to (1.50×0.305) =0.4575 inch2. To get the net area for tension or Ant we deduct the area for half of a bolt hole, the net area Ant can be found to equal to 0.3241 inch2. This is the case of Block Shear-Coped Beam-UBS=1.00.

#### Estimate the shear yielding and tension rupture-LRFD.

After estimating the gross area for shear Agv and the net area for shear Anv together with the net area for tension Ant, we can get the value of Phi. *Pn for the case of shear yielding and tension rupture.

The block shear-coped beam Phi value equals 0.75, we will multiply (0.60*Fy*Agv) plus (Ubs*Ant*Ful) by the Phi value.

The result can be found to be equal to 51.20 kips.

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

For the case of the shear rupture and tension rupture-LRFD., the block shear-coped beam Phi value equals 0.75, we will multiply (0.60*Fult*Anv) plus (Ubs*Ant*Ful) by the Phi value. The result can be found to be equal to 56.46 kips as shown in the next slide image.

We will select the min value of (51.20,56.42) which can be found equal to 51.20 kips, which is the LRFD value for block shear-coped beam for UBS=1.

For the ASD value, we can find that for the case of shear yielding and tension rupture it is equal to 34.13 kips, while for the case of shear rupture and tension rupture it is equal to 37.64 kips, we will select the minimum value which is kips for UBS=1.

### The Second solved problem for Block Shear-Coped Beam-UBS=1/2.

In the Second solved problem for Block Shear-Coped Beam-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, but with 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 as shown in the next slide image. The bolt diameter is 5/8 inches.

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

1-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.

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. To get the net area for shear or Anv we deduct the area for three holes and a half, the net area Anv can be found equal to 2.554 inch2.

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

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 the case of Block Shear-Coped Beam-UBS=1/2.

We refer to the block shear equation, listing all the estimated values for the gross area for shear, the net area for shear, and the net area for tension.

#### Estimate the shear yielding and tension rupture-LRFD.

After estimating the gross area for shear Agv and the net area for shear Anv together with the net area for tension Ant, we can get the value of Phi. *Pn for the case of shear yielding and tension rupture.

The block shear-coped beam Phi value equals 0.75, we will multiply (0.60*Fy*Agv) plus (Ubs*Ant*Ful) by the Phi value. note that UBs value is equal to 0.50. The result can be found to be equal to 131.625 kips.

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

For the case of the shear rupture and tension rupture-LRFD., the block shear-coped beam Phi value equals 0.75, we will multiply (0.60*Fult*Anv) plus (Ubs*Ant*Ful) by the Phi value. The result can be found to be equal to 97.94 kips as shown in the next slide image.

We will select the min value of (131.625,97.94) which can be found equal to 97.94 kips, which is the LRFD value for block shear-coped beam for UBS=1/2.

For the ASD value, we can find that for the case of shear yielding and tension rupture it is equal to 65.81 kips, while for the case of shear rupture and tension rupture it is equal to 65.29 kips, we will select the minimum value which is 65.29 kips for UBS=1/2.