# 14-Solved problem 5-7 for block shear.

Last Updated on April 29, 2024 by Maged kamel

## A Solved problem 5-7 for block shear.

### A Solved problem 5-7 for Block shear.

Topics included in our discussion are shown in the next slides. This is a list of the Block shear Equations by the code provision J4-5.

The provision is used to calculate the block shear strength for any plane.

Introduction: In the solved problems 5-7 for block shear of a tension member connected to a gusset plate through bolted connections, the types of modes of failure are shown from mode b to mode d.

Example 5-7, from Prof. Alan Williams’s book, states that the tension member connected to a guest plate is 1/2 inch thick, while the gusset plate is 1″ thick. it is required to find the block shear for the cionnection.
The steps are as follows:
1: The diameter for each bolt D will be 7/8+1/8=1″. We add 1/8″ to compensate for drilling and occurring damage.

2: The abbreviation g is the gauge distance, which is the perpendicular distance to the tension load, here is the horizontal distance=2″ as given, while, s, is the pitch distance in the direction of the load=3″.

### A Solved problem 5-7 for block shear, an inspection of mode b.

Mode b is the block shear first mode for the tension member. The gross area for shear Avg is divided into two parts. Each part has a height of 5 inches and a width of 0.50 inches. Agv=2*0.5*5=5.0 inch^2.

For the net area of shear or Anv, deduct 3 holes. The final value is 3.50 inch2. For the net area for tension, the width equals 2 inches, and the width equals 1/2 inch.

The gross area for tension is 1.0 inch^2; deduct the area of one hole to get the net area for tension, which will be equal to 0.50 inch2.

### How do we find the block shear Nominal load for mode b?

We list all the values of Agv, Anv, and Ant together with the yield stress value for the tension member, which is 36 ksi. The ultimate stress is 58 ksi.

In the case of shear yielding and tensile rupture, Multiply Agv by 0.6*Fy, and the product of UBs is Ant by Fy. The nominal load for this case is equal to 137.0 kips.

In the case of Shear and tensile rupture, Multiply Anv by 0.6*Fu and the product of UBs by Ant by Fy. The nominal load for this case is 150.80 kips. Select the minimum value, which is 137.0 kips.

### Estimate the value of Agv, Anv, and Ant for mode C.

Mode C is the block shear second mode for the tension member. The gross area for shear Avg is divided into two parts. Each part has a height of 5 inches and a width of 0.50 inches. Agv=2*0.5*5=5.0 inch^2.

Deduct 3 hole areas from the gross shear area to get the Anv, which is 3.50 inch2. The tension part of the block shear area dimension is 2 inches wide, and the width is 1/2 inch.

The gross area for tension is 2.0 inches^2; deduct the area of one hole to get the net area for tension, which is equal to 1.50 inches^2.

### How do we find the block shear Nominal load for mode C?

We list all the values of Agv, Anv, and Ant together with the yield stress value for the tension member, which is 36 ksi. The ultimate stress is 58 ksi.

Shear yielding and tensile rupture: Multiply Agv by 0.6*Fy and the product of UBs by Ant by Fy. The nominal load for this case is 195.0 kips.

In the case of Shear and tensile rupture, Multiply Anv by 0.6*Fu and the product of UBs by Ant by Fy. The nominal load for this case is 208 kips. Select the minimum value, which is 195 kips.

### Estimate the value of Agv, Anv, and Ant for mode d.

Mode d is the block shear third mode for the gusset plate. The gross area for shear Avg is divided into two parts. Each part has a height of 5 inches and a width of 1.0 inches. Agv=2*5*1=10.0 inch^2.

Deduct 1 1/2 hole areas from the gross shear area to get the Anv, which is 7.0 inch2. The tension part of the block shear area dimension is 2 inches wide, and the width is 1.00 inches.

The gross area for tension is 2.0 inches^2; deduct the area of one hole to get the net area for tension, which is equal to 1.00 inches^2.

### How do we find the block shear Nominal load for mode d?

We list all the values of Agv, Anv, and Ant together with the yield stress value for the tension member, which is 36 ksi. The ultimate stress is 58 ksi.

Shear yielding and tensile rupture: Multiply Agv by 0.6*Fy and the product of UBs by Ant by Fy. The nominal load for this case is 274 kips.

In the case of Shear and tensile rupture, Multiply Anv by 0.6*Fu and the product of UBs by Ant by Fy. The nominal load for this case is 301.60 kips. Select the minimum value, which is 274 kips.

The next slide image shows the p nominal for the three modes,b, c, and d. The smallest value of Pn is 137.0 kips which will govern the design

The LRFD value or the block shear design strength is 103 kips and the block shear allowable strength is 69 kips.

The next post is post number 15-Problem 3-7 for staggered bolts-tension members

There is a very useful external link: Block Shear Rupture, Chapter 3, Tension Members, Beginner’s Guide to the Steel Construction Manual, 14th ed.

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