 # 15-Problem 3-7 for staggered bolts-tension members

## Problem 3-7 for staggered bolts-tension members.

### Brief description of problem 3-7 for staggered bolts.

In this post, we will review the solved problem 3-7 for staggered bolts-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

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An angle of 6x8x1/2 has a total of four lines of bolts. Two lines are on the short leg and the other two are located on the long leg.

The diameter of the bolt is 7/8 inches, and the steel used is A36 steel. it is required to find the LRFD design strength for the bolted angle and also the ASD design strength of the same angle. for more clarity please, open the image in a separate link.

The image shows the arrangement of the bolt lines when the angle is unfolded to become a plate.

We turn the unequal angle to a plate, and the height of the plate is equal to the sum of the legs from which we deduct the thickness of the angle, or it could be said this is the centerline length.

The height of the plate is equal to 6 plus 8 minus 1/2 which is the angle thickness. The total height will be equal to 13.5 inches as shown in the next slide image.

The detailed four lines of bolts are shown also in the image.

### Estimate the distance between the second and third bolt lines.

We will estimate the distance between the second bolt line and the third bolt line the distance is equal to 4 3/4 inches for problem 3-7 for staggered bolts.

The diameter of the bolt is 7/8 inches and we will add an additional 1/8 inch to account for damage and bunching the hole diameter will be equal to 1 inch.

The area of the unequal angle 6x8x1/2 is equal to 6.80 inches obtained from the table of angles and that area will be set equal to the gross area of the angle.

### Investigate the net area value of the first failure line for problem 3-7 for staggered bolts.

The first failure line to be checked is the line abdf which is a vertical line passing by two bolts. as shown in the next slide image.

The net area for the section is estimated by deducting the area of the two holes from the gross area of the angle.

The net area value will be found to be equal to 5.80 inch^2. The 100% of tension force P is resisted by the section, and no transfer of force to the other plate has been done.

The next image shows the arrangement of the bolts at the left of the first failure mode.

### Investigate the net area value of the second failure line-abcdf for problem 3-7 for staggered bolts.

As we can see the second failure line abcde has two parts as straight lines and other staggered lines bc and the Cg distance for the staggered lines, we will add two lengths by the values of S^2/4g1 plus S^2/4g2 s is common as equal to 1 1/2 inches as given.

We will estimate the deduction as equal to three hole lengths each hole equal to 1 inch so the distance to be deducted is 3 inches.

The next step is to estimate the length to be added by summing the S^2/4g1 plus s^2/4g2. The relevant values are shown in the next slide image. The net area can be found to be equal to 5.47 inch^2.

The sketch shows that we deduct the lengths of holes and add lengths of staggered holes and the sum is multiplied by the thickness of the angle.

### Investigate the net area value of the third failure line-abceg.

The Force P is transferred through bolts to the other connected plate, each bolt will receive a part of that load which equals the total load divided by the number of bolts. We have ten bolts so each bolt will transfer a value of P/10.

There is an important note that the failure line abceg excludes one bolt d, which already transferred p/10 to the connecting plate.

There will be an adjustment for the net area that will be estimated by multiplying by the ratio of (10/9). the values of S and g for the line of failure are given as 1 1/2″ and 21/2 inches.

The acting force in the section is P*(9/10) as shown in the next slide image.

We will deduct the three holes’ distance and then add the length due to staggered bolts for problem 3-7 for staggered bolts. the relative s and g are estimated then the net length is multiplied by the thickness of the angle.

### Investigate the net area value of the fourth failure line-abcdeg.

As we can see from the next image we have a fourth failure route which is (abcdeg). The route has three staggered lines, which are bc, cd, and de. For all these lines we will add a length equal to the sum of Si^2/4gi from I=1 to I =3. We will deduct the length of four holes.

The net area for the route is 5.065 inch^2, there will be no modification due to tension force P since no portion is transferred by bolts.

### What is the least net area for problem 3-7 for staggered bolts?

The lowest value of the net area is 5.065 inch^2 and it belongs to the final route. The effective area for the lowest value of the net area can be estimated by multiplying U by the Anet, U value is equal to one since the angle is connected for all the legs.

We can estimate the Pn due to yielding and rupture, these values are shown in the next slide. image.

The full data of the LRFd and Asd Strengths are shown in the next slide image.

This is the previous post solved 5-7 for block shear.

The next post is Problem 3-4-3 for staggered bolts-tension members-1/2.

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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