8- Solved problem 5-part 1 for design shear strength for bolts

Last Updated on February 23, 2026 by Maged kamel

Solved problem 5-part 1 for design shear strength.

This post will introduce the solved problem 5-part 1 for design shear strength, quoted from the Unified Design of Steel Structures handbook, 3rd and 4th editions. To find the design shear strength for different types of bolts with different bolt diameters, develop a table showing the design shear strength or use the LRFD design for shear strength.

The following slide summarizes the post’s content and explains Practice Problems 1 and 2 in Unified Steel Design of Steel Structures.

The first type of bolt is type A 325-N, where N indicates that the threads are included in the shear plane of the bolt. Based on Table J3.2, the nominal shear strength is 54 ksi for a single shear plane. The shear limit state is 0.75.

The following slide image is a reminder of the Fnv, or nominal shear value, for Group A-325N and x bolts, and how we derived the expression.

Solved problem 5-part 1 for design shear strength for sizes 5/8″, 3/4″, and 7/8 inches for A325-N type.

We start with a 5/8″ Type A-325 N diameter bolt with a nominal shear strength of 54 ksi. We estimate the design shear strength per inch^2 by multiplying Φ*Fnv, which is equal to (0.75*54)=40.50 ksi

To get the shear strength for one bolt with a diameter of 5/8 inch, estimate the area of the bolt. The area is 0.307 in². We get the design shear value by multiplying (Φ*Fnv*A): (40.50*0.307)=12.40 kips.

For the second bolt with a diameter of 3/4″, Type A-325 N with a nominal shear strength of 54 ksi. We estimate the design shear strength per inch^2 by multiplying Φ*Fnv, which is equal to (0.75*54)=40.50 ksi

To get the shear strength for one bolt, estimate the area of the bolt with a diameter of 3/4 inch. The area is 0.442 in². We get the design shear value for a 3/4-diameter bolt type -325 N by multiplying (Φ*Fnv*A)=(40.50*0.442)=17.90 kips.

The third bolt has a 7/8″ Type A-325 N diameter with a nominal shear strength of 54 ki. We estimate the design shear strength per inch^2 by multiplying Φ*Fnv, which equals (0.75*54)=40.50 ksi.

To get the shear strength for one bolt, estimate the area of the bolt with a diameter of 7/8 inch. The area is 0.601 in². We get the design shear value for a 7/8-inch-diameter bolt type -325 N by multiplying (Φ*Fnv*A): 40.50*0.601 = 24.34 kips. Please refer to the next slide for more details.

Solved problem 5-part 1 for design shear strength for size 1″ for A325-N type.

The last bolt has a diameter of 1 inch, Type A-325 N with a nominal shear strength of 54 ksi. We estimate the design shear strength per inch^2 by multiplying Φ*Fnv, which equals (0.75*54)=40.50 ksi.

To get the shear strength for one bolt with a diameter of 1 inch, estimate the area of the bolt. The area is 0.785 in². We get the design shear value for a 1-inch-diameter bolt type -325 N by multiplying (Φ*Fnv*A): 40.50*0.785 = 31.80 kips.

Please refer to the next slide for more details. I have added the Nominal shear strength values for Problems 1 and 2 of the Unified Design of Steel Structures handbook, 3rd and 4th editions.

Solved problem 5-part 1 for design shear strength for sizes 5/8″, 3/4″, 7/8 and 1 inch for A325-N type using Table 7-1.

We can use Table 7-1 to verify our estimations of available bolt shear strength. We refer to Group A as N-type. We use the diameter of 5/8 inch with a single shear termed S. We can find that the design shear strength for the bolt is equal to 12.40 kips

We use the diameter of 5/8 inch with a single shear termed S. We can find that the design shear strength for the bolt is equal to 17.90 kips

We use a diameter of 7/8 inch with a single shear termed S. We can find that the design shear strength for the bolt is equal to 24.30 kips

for the last diameter of 1 inch with a single shear termed S. We can find that the design shear strength for the bolt is equal to 31.80 kips. The previous values match our calculations. Please refer to the next slide image.

Solved problem 5-part 1 for design shear strength for sizes 5/8″, 3/4″, and 7/8 inches for A325-X type.

The second type of bolt is type A 325-X, where X indicates that the threads are not in the bolt’s shear plane.

Based on Table J3.2, the nominal shear strength equals 68 ksi for a single shear plane. The shear limit state is 0.75.

We start with a bolt with a diameter of 5/8″ Type A-325X with a nominal shear strength of 68 ksi. We estimate the design shear strength per inch^2 by multiplying Φ*Fnv, which is equal to (0.75*68)=51 ksi

To get the shear strength for one bolt with a diameter of 5/8 inch, estimate the area of the bolt. The area is 0.307 in². We get the design shear value by multiplying (Φ*Fnv*A): (51*0.307)=15.70 kips.

For the second bolt with a diameter of 3/4″, Type A- to get the shear strength for one bolt, estimate the area of the bolt with a diameter of 3/4 inch; the area is equal to 0.442 inch2.

We get the design shear value for a 3/4-diameter bolt type -325 X by multiplying (Φ*Fnv*A): 51*0.442 = 22.50 kips.

To get the shear strength for one bolt, estimate the area of the bolt with a diameter of 7/8 inch. The area is 0.601 in². We get the design shear value for a 7/8-diameter bolt type -325 X by multiplying (Φ*Fnv*A)=(51*0.601)=30.65 kips.

Solved problem 5-part 1 for design shear strength for size 1″ for A325-x type.

The last bolt has a diameter of 1 inch, Type A-325 X, with a nominal shear strength of 68 ksi. We estimate the design shear strength per inch^2 by multiplying Φ*Fnv, which equals (0.75*68=51.0 ksi.

To get the shear strength for one bolt with a diameter of 1 inch, estimate the area of the bolt. The area is 0.785 in². We get the design shear value for a 1-inch-diameter bolt type -325 X by multiplying (Φ*Fnv*A): 51*0.785 = 40 kips. Please refer to the next slide for more details.

Please refer to the next slide for more details. I have added the Nominal shear strength values for Problem 2 of the Unified Design of Steel Structures handbook, 3rd and 4th editions.

Solved problem 5-part 1 for design shear strength for sizes 5/8″, 3/4″, 7/8 and 1 inch for A325-X type using table 7-1.

To verify our estimations, we can use Table 7-1 to determine the available shear strength of bolts. We refer to group A with X-type and use a 5/8-inch diameter with single shear, termed S. We find that the bolt’s design shear strength is 15.70 kips.

We repeat the same steps for the other diameters: 3/4″,7/8″, and 1″. All the values match our previous calculations.

Thank you very much, and I look forward to seeing you in the next post.

The PDF for this post can be viewed or downloaded from the following link.

This is the link to the following post, part two. The second part will include the design shear strength for A490-N and A490x bolts.

This link is for the previous post, ‘A Solved Problem 12-1 Part 2 for Bearing Connections.

This is a very useful source for designing various Steel elements: A Beginner’s Guide to the Steel Construction Manual, 15th ed, Chapter 4—Bolted Connections.

This is a very useful source for the design of various Steel elements, A Beginner’s Guide to the Steel Construction Manual, 16^th ed, Chapter 4 – Bolted Connections.