 # 16-Problem 3-4-3 for staggered bolts-tension members-1/2.

## Problem 3-4-3 for staggered bolts-tension members-1/2.

In this post, we will review the solved problem 3-4-3 for staggered bolt-tension members. This problem is quoted from Prof. Segui’s handbook. the main aim is to investigate the different failure roots of failures and the corresponding net areas.

Prof. Segui has taken into consideration the actual force acting on each net area and made an adjustment to the estimated net area as we will see in the next calculations.

An MC9x 23.9 is connected by 3/4 inches bolts as shown in figure P-3.4-3.ASTMA572 steel is used. a) determine the design strength b) Determine the allowable strength.

The next image shows the arrangement of the bolts and both spacing and the vertical distances between bolts. Please click on the image to expand it for more clarity.

We can find a brief illustration of the properties of the MC section. The data for the area, breadth of the flange, the thickness of the web, and flange of the section are shown in the next image. The yield stress and the ultimate stress values are also shown.

The next image shows the number of blot lines and the spacing between each bolt. We add 1/8 inch to the diameter of the bolt to get the diameter of the hole and we use its value for the deduction. We can see that the hole diameter is 7/8 inches.

### Problem 3-4-3 for staggered bolts-First failure line-ABCD.

We are checking the various possibilities for the route of the failure line, we start the investigation of the first failure line which is line abcd , please refer to the previous image for more details.

The net area can be estimated as equal to the gross area minus the area of holes. The area for holes area can be estimated as the product of (n*db*tweb), n=2 since we have two bolts, dh=7/8″, and the thickness of the web=0.4″. The final net area can be estimated as equal to 6.32 inches2. The section abcd is acted upon by 100%P.

### Problem 3-4-3 for staggered bolts-2nd failure line-ABFCD.

We are checking the various possibilities for the route of the failure line, we check the second failure line which is line ABFCD, please refer to the next image for more details.

We have two zigzag lines BF and FC. These two lines have the same S value which is the horizontal spacing between bolts in the direction of the applied force. While the vertical distance are the same for these two lines which are 2 1/2″.

The net area can be estimated as equal to the gross area minus the sum of deductions and additions multiplied by the thickness of the web to avoid any errors. The deduction in inches for hole area can be estimated as the product of (n*db)=3*7/8=21/8″. We add the sum of the value of (S^2/4g) for the two lines. We have equal s and g for these lines. The addition is equal to +1.25.

The sum value=-1.375″. we will multiply this value by the web thickness and then add it to the gross area to get the final net area, net area can be estimated as equal to 6.47 inch2. the section abcd is acted upon by 100%P. Problem 3-4-3 for the staggered bolts-net area of the 2nd failure line.

### Problem 3-4-3 for staggered bolts-3rd failure line-AEFGD’.

We check the third failure line which is line AEFGD’, please refer to the next image for more details. Each bolt transfers the applied force by a ratio of force value by the bolt’s number which is equal to P/7. Section AEFGD’ excludes two bolts that transfer 2P/7 to the Gusset plate. The section is fit for only 5/7P which is the remaining value of force due to the sum of forces.

The net area can be estimated as equal to the gross area minus the area of holes. The area of holes can be estimated as the product of (n*db*tweb), n=3, since we have three bolts, dh=7/8″, and the thickness of the web=0.4″. The final net area can be estimated as equal to 5.97 inch2.

The section is acted upon by (5/7)*P. A modification can be done to the net area by multiplying by the ratio of 7/5. the value of net area equals 8.358 inch2.

We come to the end of the first part, we will continue in the second part. Thank you.

The next post is problem 3-4-3 for the MC section –LRFD and ASD value-2/2.

For a more detailed illustration of block shear, there is a very useful external link-Block Shear Rupture. A Beginner’s Guide to the Steel Construction Manual, 14th ed. Scroll to Top
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