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

Last Updated on September 17, 2026 by Maged kamel

Solved problem 5-part 1 for design Shear strength.

This Post introduces the solution to Problem 5, Part 1: design Shear strength, quoted from the Unified Design of Steel Structures handbook, 3rd and 4th editions. To find the design Shear strength for different Bolt types and 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 Bolt type is A325-N-N, where N indicates that the threads are included in the Bolt’s shear plane. 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 reminds us of Fnv, or the 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 5/8″, 3/4″, and 7/8 “bolts for A325-N.

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 find the Shear strength for one Bolt with a diameter of 5/8 inch, estimate the Bolt Area. 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 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, 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, is Type A-325 N, and has 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 find the Shear strength for a 1-inch-diameter Bolt, estimate the Bolt area. 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 55, Part 1, Pat1Part 1deign Shear strength for sizes 5/8″, 3/4″, 7/8 “, and 1” for A325-N type using Table 7-1.

We can use Table 7-1 to verify our estimates 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, the design Shear strength for the Bolt is 31.80 kips. The previous values match our calculations. Please refer to the image on the next slide.

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

The second Bolt type is A325-X, where X indicates 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 find the Shear strength for one Bolt with a diameter of 5/8 inch, estimate the Bolt area. 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 A325-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 find the Shear strength for a 1-inch-diameter Bolt, estimate the Bolt area. 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 added the Nominal Shear strength values for Problem 2 from 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” 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 as X-type and use a 5/8-inch-diameter single-shear Bolt, 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, and I look forward to seeing you in the next Post.

You can view or download the PDF for this Post 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 useful source for designing various Steel elements: A Beginner’s Guide to the Steel Construction Manual, 16^th ed. Chapter 4 – Bolted Connections.