8A- Solved problem 5-part 2 for design shear strength.

Last Updated on June 26, 2025 by Maged kamel

Solved problem 5-part 2 for design shear strength.

In this post, we will introduce solved problem 5-part 2 for design shear strength quoted from the Unified Design of Steel Structures handbook. The problem requires developing a table showing the design shear strength for different types of bolts with varying bolt diameters, determining the design shear strength, or applying the LRFD design method for shear strength.

We have included the design shear strength for A325 N and X bolts in part 1.

The first type of bolt is type A 490-N, where N indicates that the threads are included in the bolt’s shear plane. Based on Table J3.2, the nominal shear strength equals 68 ksi for a single shear plane.

The limit state of the shear design is 0.75.

The following slide provides a summary of the post’s content and explains the content of Practice Problems 3 and 4 in the Unified Steel Design of Steel Structures.

The following slide image is a reminder of the Fnv or the nominal shear value for Group A-490-N and x bolts, and how we have derived the expression.

Solved problem 5-part 2 for design shear strength for sizes 5/8″, 3/4″, and 7/8 inches for A490-N type.

We start with a bolt with a diameter of 5/8″ Type A-490 N 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, estimate the area of the bolt with a diameter of 5/8 inch; the area is equal to 0.307 inch2. 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-490 N 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, 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 -490 N, by multiplying (Φ*Fnv*A)=(51*0.442)=22.50 kips.

The third bolt with a diameter of 7/8″ Type A-490 N 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, estimate the area of the bolt with dia7/8 inch; the area is equal to 0.601 inch2. We get the design shear value for a 7/8 inch diameter bolt type -490 N, by multiplying (Φ*Fnv*A)=51*0.601=30.65 kips. Please refer to the next slide for more details.

Solved problem 5-part 2 for design shear strength for size 1″ for A490-N type.

The last bolt is a diameter of 1 inch, Type A-490 N, 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, estimate the area of the bolt with a diameter of 1 inch; the area is equal to 0.785 inch2. We get the design shear value for a 1-inch diameter bolt type -490-N, 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 Problems 3 and 4 of the Unified Design of Steel Structures handbook, 3rd and 4th editions.

Solved problem 5-part 2 for design shear strength for sizes 5/8″, 3/4″, 7/8 and 1 ” for A490-N type using table 7-1.

To verify our estimations, we can use Table 7-1 for the available shear strength for bolts. We refer to Group B type with n-type. We use the diameter of 5/8 inch with single shear termed S. We can find that the design shear strength for the bolt is equal to 15.70 kips.

We use the diameter of 5/8 inch with single shear termed S. We can find that the design shear strength for the bolt is equal to 22.50 kips.

We use a diameter of 7/8 inch with single shear termed S. We can find that the design shear strength for the bolt is equal to 30.70 kips.

For the last diameter of 1 inch with single shear termed S, we can find that the design shear strength for the bolt is equal to 40 kips. The previous values match our calculations. Please refer to the next slide image.

Solved problem 5-part 2 for design shear strength for sizes 5/8″, 3/4″, and 7/8 inches for A490-X type.

The second type of bolt is type A 490-X, where X indicates that the threads are not in the shear plane of the bolt.

Based on Table J3.2, the nominal shear strength is equal to 84 ksi for a single shear plane. The limit state of the shear design is 0.75.

We start with a bolt with a diameter of 5/8″ Type A-490 X with a nominal shear strength of 84 ksi. We estimate the design shear strength per inch^2 by multiplying Φ*Fnv, which is equal to (0.75*84)=63 ksi

To get the shear strength for one bolt, estimate the area of the bolt with a diameter of 5/8 inch; the area is equal to 0.307 inch2. We calculate the design shear value by multiplying (Φ*Fnv*A) = (63*0.307) = 19.34 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 -490 X, by multiplying (Φ*Fnv*A)=63×0.442=27.80 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 equal to 0.601 inch2. We get the design shear value for a 7/8 diameter bolt type -490 X, by multiplying (Φ*Fnv*A)=(63*0.601)=37.87 kips.

Solved problem 5-part 2 for design shear strength for size 1″ for A490-x type.

The last bolt with a diameter of 1 inch, Type A-490 X, with a nominal shear strength of 84 ksi. We estimate the design shear strength per inch^2 by multiplying Φ*Fnv, which is equal to (0.75*84=63 ksi.

To get the shear strength for one bolt, estimate the area of the bolt with a diameter of 1 inch; the area is equal to 0.785 inch2. We get the design shear value for a 1-inch diameter bolt type A-490 X by multiplying (Φ*Fnv*A)=63*0.785=49.50 kips. Please refer to the next slide for more details.

Solved problem 5-part 2 for design shear strength for sizes 5/8″, 3/4″, 7/8 and 1 ” for A490-X type using Table 7-1.

To verify our estimations, we can use Table 7-1 for the available shear strength for bolts. We refer to Group B as X-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 15.70 kips. We repeat the same steps for the other diameters: 3/4″, 7/8 inch, and 1″. All the values match our previous calculations.

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

This is a link to the previous post, solved problem 5-part-1.

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