 # 15A-Solved problem 3-8-1 for staggered Bolts

## Solved problem 3-8-1 for staggered Bolts.

In this post, we will start to discuss the concept of load transfer by fasteners to the estimation of the net area. We can consider this method included in the solved problem as a refinement for the selection of the final net area. We have solved problem 3-8-1 for the staggered Bolt that was introduced in the textbook steel structures fifth edition by Salmon.

### Introduction to solved problem 3-8-1 for staggered Bolts.

Calculate the governing net area for plate A of the single lap joint in Fig. 3.8.1 and show free body diagrams of portions of plate a with sections through each line of holes .assume that Plate b has an adequate net area and does not control the strength T.

The plate thickness is 5/8 inches and we add 1/8 inch to the diameter of each bolt to get the hole diameter, which will be equal to one inch.

In the next image, there is a detailed plan for the arrangements of fasteners for plate A which is under tension force P and its thickness is 5/8 inches. There is another plate B on the back side of plate A which is connected with plate A by using 10 bolts. Each bolt carries one-tenth of the load P.

It is required to estimate the minimum net area for given plate A, by considering the different routes and finding out what is the modified net area values under the effect of load transfer as we will say in more detail.

### The net area for the first vertical route 1-3-1 with no staggered bolts.

The first route 1-3-1 is a vertical line through three bolts taken at the right side of these bolts. if we have a look at the section, from the right side it is acted upon by force P and is balanced by the same value P.

If we have a look at the group of bolts for plate A in the same section, we will find that there are ten bolts each carrying P/10 so that the sum of these forces is equal to P and it is in compliance with the right section as shown for the solved problem 3-8-1 for staggered Bolts-fasteners lines.

We will deduct three holes of diameter one inch for each from the width of the plate then multiply by the plate thickness. the final value of the net area will not be modified due to load transfer since the section carries 100%p.

### The net area for the second route is 1-2-3-2-1 with staggered bolts.

The second route 1-2-3-2-1 is a staggered or zigzag line through five bolts taken at the right side of these bolts. If we have a look at the section, from the right side it is acted upon by force P and is balanced by the same value P.

If we have a look at the group of bolts for plate A in the same section, we will find that there are ten bolts each carrying P/10 so that the sum of these forces is equal to P and it is in compliance with the right section as shown for the solved problem 3-8-1 for staggered Bolts-fasteners lines.

For the value of the net area, since we have staggered lines we start to estimate the deduction length we have five holts for which the diameter is 1 inch, and the sum of deductions is 5 inches. On the other hand, we have 4 staggered lines for which we need to add four values of (S^2/4g) where s is the spacing between bolts in the force direction.
The g value is the value of spacing in the perpendicular direction to the force. the 4S^2/4g will give us a value of (+1.333) inches.

The net width will be (15+(-5+1.33)= 12.33 inches when multiplied by the plate thickness, which is 5/8 inches will give a net area of 7.08 inches2.
The section carries 100% of the force so no modification for the net area is to be done. The net area estimation for route 1-2-3-2-1 and the modified value.

This is a quote from The Author’s calculation and the detailed sections for solved problem 3-8-1 for staggered Bolts-fasteners lines and matches with our calculations.

### The net area for the third vertical route 2-2.

The third route 2-2 is a vertical line through two bolts taken at the right side of these bolts. if we have a look at the section, from the right side it is acted upon by force P and at the left side we have (7/10)*P, we have three bolts at 1,2,3 received 3/10*P of the load.

At the left side of the section, there are 7 bolts, each bolt carries P/10 of the load, please refer to the next image for more details. The net area for this section is 8.125 inch2, the section carries only a value of 7/10 P value.

### The net area for the fourth route 1-2-2-1 with staggered bolts.

The fourth route 1-2-2-1 is a staggered or zigzag line through four bolts taken at the right side of these bolts. If we have a look at the section, from the right side it is acted upon by force P and one bolt carries (1/10)*P the acting force on the section thus equals 9/10P.

For the left side of the section, we have Nine bolts each carrying P/10, and the section carrying 9/10*P which is the same value as the right-hand side section. the neta rea equals 7.29 inch2. A modification is made to the net area since the section carries (9P/10). The modified net area value will be equal to (10/9)*7.29 inch2.

### The net area for the fifth vertical route 4-5-6 with no staggered bolts.

The fifth route 4-5-6 is a vertical line through three bolts taken at the left side of these bolts. If we have a look at the section, from the right side it is acted upon by force P and there are eight bolts carrying 8*P/10 force, at the section the balance of the force and bolt forces are equal to (2/10)P.
Again from the left side, we have only two bolts having 2P/10 values.

The net area will be found to be equal to 7.50 inch2. The modified net area due to load transfer is 37.50 inch2.

### The net area for the sixth vertical route 7-8 with no staggered bolts.

The sixth route7-8 is a vertical line through two bolts taken at the right side of these bolts. If we have a look at the section, from the right side it is acted upon by force P and there are eight bolts carrying 8P/10 force, at the section the balance of the force and bolt forces are equal to (2/10)P.
Again from the left side, we have only two bolts having 2P/10 values.

The net area will be found to be equal to 8.125 inch2. The modified net area due to load transfer is 40.625 inch2.

### What is the final net area for the section of plate A?

The table shows the different values of the net area for each route, we are concerned with the minimum value of the net area, and the modified area is used for the sake of comparison to finalize the net area. As we can see the lowest value of net area equals 7.08 inch2, for the section that carries 100% P. The section is route 1-2-3-2-1. The full details are shown for each route ad its net area. 