 # 5- A solved problem 10-1-for bearing connection-1-3.

## A solved problem 10-1-for bearing connection-

### The content of video.

We are going to check the solved problem 10-1 for bearing connection from Prof. Alan Williams’s book, for which it is required to estimate the shear force for a simple bearing connection, the solved problem will be dealt with in two parts, the first part is to estimate the Nominal shear value of the bolts of the connection based on bolts shear. That was a part of the video that has a subtitle and a closed caption in English.

You can click on any picture to enlarge then press the small arrow at the right to review all the other images as a slide show.

### A solved problem 10-1-for bearing connection-How to get the value of shear for a bearing connection?

We are going to check the solved problem 10-1, for the shear part from Prof.Alan Williams’s book. in the first part will estimate the shear force applied to a connection based on the shear of bolts.

We will start, check the solved problem 10-1. The title is Bolts in Shear and Bearing with deformation a design consideration. So the item which we are dealing with is the 0.25″ deformation.

The connection is shown in figure 10.7, which is a double angle connected with a guest plate. The two angles are Back-to-back and a guest plate, each angle is 4x4x with 7/16″ thickness.

The guest plate is 3/4″ inch thickness and Consists of four, grade A490,3/4″ diameter bolts. We have four, grade A490, Which is type B.3/4″ diameter bolts, with snug-tight, for which bolts are tightened by a spanner, no pre-tension is done.

The threads are excluded, then the connection is Type B and Type X, from the shear planes. Deformation around the bolt is around is a design criterion or design consideration and the bolt spacing is as indicated, the inner spacing of bolts is 3″ and the edge distance is 2″.

Assuming that the angles and gusset plate are satisfactory, As I understand, that no need to check the angle, which is under tension force against bearing calculations regarding spacing and diameter of bolts.

The only requirement is to determine shear and bearing for bolts, we will start with god’s will to deal with the shear stress.

### A solved problem 10-1- the first method to get Shear strength for a connection by using table J 3.2.

We have table J.3.2, for the A solved problem 10-1-for-bearing connection which we have included in the last video. Let us review our given data, we have bolts that are classified as in group B with ASTM-A-490 and Excluded.

Our case is the fourth item as shown in the table. We have the highest value of the Nominal shear value is 84 ksi, which is estimated from multiplying 150*0.625*0.90 =84.375 ksi. While for the Nominal shear value for the included bolt multiplied by 0.80.

We have two methods to evaluate the nominal shear strength, the first one is by using table J-3.2.

The second method is by using table 7-1 as shown. In the first method, we use table J 3.2, we get the nominal shear values and compare them with table 7-1, which is for the available shear strength of the bolts.

For our solved problem, our bolt diameter is 3/4″, we use table 7-1, We have a group b for ASTMA-490, then by the intersection of a row and the relevant column, we get the value of Rn.

In the case of LRFD, the φ*Rn=55.7 kips, while the next value to the left is the (1/Ω)*Rn =37.10 kips for the ASD design.

We have 3/4″ double shear D. The estimated value are φ*Rn=55.7 kips, and (1/Ω)*Rn =37.10 kips for only one bolt.

We have 4 bolts, so we will multiply these values by 4. In order to check these values, we can use table j3.2. The advantage of table 7-1, that it gives the area of bolts.

While table J-3.2 is giving the nominal shear strength in KSi, to convert to a force value, we need necessary items which are the number of bolts and the area. We have a strength=84 ksi

Type B, multiply by area which is=0.442 inch2 and then multiply by 4, which are the number of bolts, but again we have a double shear, we will multiply by 2. finally we have=84*0.442*4*2=297.023 kips,

For the LRFD value, our φ=0.75, then (φ*Rn)=0.75*297.02=222.77 kips, which is this is LRFD value, which is very close to the value obtained by table 7-1.

### A solved problem 10-1-for-bearing connection- the first method to get Shear strength for a connection by using table 7-1.

While for the case of ASD. We have (1/Ω)*Rn=297.02/2=148.5 kips. Compared to the value of 148.4 obtained by using table 7-1, very close for the A solved problem 10-1-for bearing connection.

A sketch shows the shear force distribution for both LRFd and Asd design coefficients.

The shown figures are the values for the shear for the connection for solved problem 10-1, quoted from the Author’s book.

The next post is Solved problem 10-1-how to get Bearing value-LRFD-2-3

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