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

Solved problem 5-part 1 for design shear strength.

In this post, we will introduce a Solved problem 5-part 1 for design shear strength quoted from the Unified Design of Steel Structures handbook. It is required to develop a table showing the design shear strength for different types of bolts with different bolt diameters and it is required to find the design shear strength or use the LRFD design for shear strength.

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 equals 54 ksi for a single shear plane. The limit state of the shear design is 0.75.

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 bolt with a diameter of 5/8″ 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 dia 5/8 inch, the area is equal to 0.307 inch2. 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 dia 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 N, by multiplying (Φ*Fnv*A)=(40.50*0.442)=17.90 kips.

The third bolt with a diameter of 7/8″ Type A-325 N with a nominal shear strength of 54 ki. 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 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 -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 with 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 is equal to (0.75*54)=40.50 ksi.

To get the shear strength for one bolt estimate the area of the bolt with dia 1 inch, the area is equal to 0.785 inch2. 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.

Solved problem 5-part 1 for design shear strength for sizes 5/8″, 3/4″, and 7/8 and 1 inches for A325-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 A 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 12.40 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 17.90 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 24.30 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 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 shear plane of the bolt.

Based on Table J3.2 the nominal shear strength is equal to 68 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-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 estimate the area of the bolt with dia 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- to get the shear strength for one bolt estimate the area of the bolt with dia 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 dia 7/8 inch, the area is equal to 0.601 inch2. 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 with 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 is equal to (0.75*68=51.0 ksi.

To get the shear strength for one bolt estimate the area of the bolt with dia 1 inch, the area is equal to 0.785 inch2. 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.

Solved problem 5-part 1 for design shear strength for sizes 5/8″, 3/4″, and 7/8 and 1 inches for A325-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 A with X-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 repeat the same steps for the other diameters 3/4″,7/8 inch, and 1″. All the values match our previous calculations.

Thanks a lot and see you in the next post.

This is the next post-part 2 link.

This is a link for the previous post-A solved problem 12-1 part 2 for bearing connections.

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.

This is a very useful link: A Beginner’s Guide to the Steel Construction Manual, 15^th ed, Chapter 4 – Bolted Connections