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

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

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

We will check the solved problem 10-1, for the shear part from Prof. Alan Williams’s book. The first part will estimate the shear force applied to a connection based on the shear of bolts. We have a case of double shear. I have included a sketch to show the shear force distribution for the case of double shear quoted from the Engineering beginner site.

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 thick and Consists of four, grade A490,3/4″ diameter bolts. We have four, grade A490, Which are type B.3/4″ diameter bolts, with snug-tight, for which bolts are tightened by a spanner, and 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 the shear and bearing for bolts, we will start with god’s will to deal with the shear stress. the first thing to do is to find the area of one bolt dia 3/4 inch which is Grade A490-X type. the area is equal to 0.442 inch2.

### A solved problem 10-1- the first method to get Shear strength for a connection by using table J3.2.

We have two methods to evaluate the nominal shear strength, the first one is 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. This value is for single shear, we will multiply by two for the case of double shear.

Then multiply ( 2 Fnv) by the number of bolts by the area of each bolt. We get the value of Rnv which is equal to 297.02 kips. to get the design nominal shear strength multiplied by the phi value which is 0.75. The LRFd value of shear strength is 222.768 kips.

For the ASD value of shear strength multiply Rnv by (1/ omega). The omega value is 2. The ASD value is 148.152 kips. Please refer to the next slide image for more details.

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

Table 7-1 gives the value of shear Fnv value in both LRFD value and ASD value based on the diameter of bolts. For our solved problem 10-1 we have a 3/4 inch bolt group b Type-x.

The φ*Rnv for a double shear bolt is 55.70 kips per bolt, we will multiply by the number of bolts which are 4. The design shear strength is equal to 222.80 kips which is matching with the previous estimate based on Table J3-2.

The (1/Ω)*Rnv for a double shear bolt is 37.10 kips per bolt, we will multiply by the number of bolts which are 4. The ASD design shear strength is equal to 148.40 kips which is matching with the previous estimate based on Table J3-2.

A sketch shows the shear force distribution for both LRFD and ASD design values. The direction of shear force is opposite to the direction of applied force.

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

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