### Solved problem 10-1-how to get Bearing value-ASD-3-3.

This is a review of the content of solved problem 10.1, in which the shear force for the given connection might be applied to the bolts in the connection. the content of this post is included in the video from time 9.3 till the end. We have earlier introduced the LRFD design and the value of force and we are going to continue with the ASD calculation as required.

This is a reminder of the bearing values for the edge bolt and the inner bolts as done in the previous post.

This is a sketch showing both the LRFD and ASD values of nominal bearing loads side by side and the total values for each. The upper limit criteria is 2.4*db*tp*Fu, tp=3/4″, Fu=58 ksi, then2.4db*tp*Fu=2.4*(3/4″)*(3/4)*(58)=78.30 kips.

The upper limit has controlled the design. As we can see that we have the value of the bearing based on the ASD for each bolt is equal to 39.15 kips. The next step is to verify these values by using both tables 7-4 and 7-5 for the bearing value-ASD values.

### Using Table 7-4 for the Bearing value-ASD for the inner bolts.

For our solved problem, for the Bearing value-ASD for the inner bolts. Our bolt diameter is 3/4″, so we use table 7-4, We have a group B for ASTM A-490, then by the intersection of a row and the relevant column, we get the value of Rnb.

From Table 7-4, for the ASD value, for the solved problem 10-1 we have inner spacing 3″ and Fu=58 ksi, 3/4″ plate thickness, the selected value is 52.2 kips/inch, since the value is per inch we will multiply by the value of the plate thickness which is 3/4″ and multiply by the number of inner bolts n=3. The final value for intermediate spacing of 3″, Rn/Ω=52.2 *3/4*n =52.2*3/4*3=117.45 kips.

### Using Table 7-5 for the Bearing value-ASD for the external bolts.

For the Solved problem 10-1, we proceed to table 3-5 for external bolt bearing value. Estimate the bearing value -ASD. For the edge distance=2″, Fu=58 ksi, and from the column of 3/4″ for bolt diameter.

We get at the intersection the force value of 52.2 kips/ inch, again multiply by (3/4), which is the plate thickness, multiply by n=1, since we have one external bolt, the value of Rn/Ω =39.15 kips.

The sum of all the bearing values of all the bolts can be finalized as Rn/Ω=117.45+39.15=157 kips. Check with the previous calculation for the ASD value by using the equation, the calculations are the same value.

### For the solved problem 10-1.The final selected value of the shear force that is to be applied.

The final step is to select the minimum value of the nominal shear strength and the bearing nominal strength. For the shear nominal value LRFD=223 kips, in ASD value=149 kips. While the bearing strength for LRFD=235 kips and for ASD=157 kips.

The lowest number is 223 kips and ASD=149 kips since when the load value reaches 223 kips according to LRFD a failure will occur due to bearing

This is the criteria for selection. We have come to the end of this solved problem 10-2- The last part. Thank You.

The full data as estimated by the author and the relevant tables for shear and bearing are included in the next slide image. the calculations are similar to our estimated calculations for shear and bearing values for the given connection.

This is a very useful source for the design of various Steel elements, **A Beginner’s Guide to the Steel Construction Manual, 15 ^{th} ed, Chapter 4 – Bolted Connections**.

This is a link to the first part of the same solved problem, how to get the shear value?

This is a link for the second part of the solved problem.

The next post is A solved for bearing connections.