brief illustration -post 6a-bearing types connections

6a-Solved problem 10-1-how to get Bearing value-ASD?3-3.

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

Solved problem 10-1-bearing value-ASD.

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

The estimate for bearing values for bolts.

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 2.4dbtpFu, tp=3/4″, Fu=58 ksi, then2.4dbtpFu=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.

The sketch shows both LRFD and ASD nominal 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.

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

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.

ASD value for Bearing for exterior bolts from table 7-5.

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. Bearing strength for LRFD equals 235 kips and for ASD=157 kips.

Bearing values for the connection.

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.

Author selection for the bearing of the connection.

This is a very useful source for the design of various Steel elements, A Beginner’s Guide to the Steel Construction Manual, 15th 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.

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