Last Updated on September 17, 2026 by Maged kamel
Solved problem 10- 1: how to get bearing value-ASD-3-3.
This sketch compares the LRFD and ASD nominal bearing Load values, as well as the total values for each. The upper limit criterion is (2.4*db*tp*Fu); The thickness of the plate is termed tp=3/4″; the Bolt diameter db=3/4 inches. The ultimate stress of A36 plate is Fu=58 ksi, then the Product 2.4db*tp*Fu=2.4(3/4″)(3/4)(58)=78.30 kips.
The upper limit has controlled the design. As shown, the bearing Value for each Bolt, based on ASD, is 39.15 kips; this Value was obtained in the previous Post. The next step is to verify these values using both Tables 7-4 and 7-5 for the bearing-ASD values.
Use Table 7-4 to find the bearing-ASD values for the inner bolts.
For problem 10-1, determine the Bearing Value for the inner bolts e based on the ASD design. The Bolt diameter is 3/4″, so we use Table 7-4 for determining the bearing values for inner bolts. We use Group B for ASTM A-490. Then, by intersecting a Row and the relevant column, we obtain 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 solved problem 10-1, we use Table 3-5 for the external Bolt bearing Value. Estimate the bearing Value – ASD for the edge distance of 2″, Fu = 58 58 ksi, and the 3/bolt diameterrer.
We calculate the force at the intersection to be 52.2 kips/inch. We then multiply this Value by (3/4), the plate thickness, and by n=1, since we have one external Bolt—rn/ω= 39.15 kips.
The sum of all bearing values for all bolts is Rn/ω = 117.45 + 39.15 = 156.60 kips. Verify the ASD Value from the previous calculation using the same equation; the results match.
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, and ASD = 149 kips. Bearing strength for LRFD equals 235 kips, and for ASD equals 157 kips.
The lowest Value is 223 kips, and ASD = 149 kips, since, according to LRFD, failure will occur due to bearing when the Load reaches 223 kips.
This is the criterion for selection. We have reached the end of this solved problem 10-1 – The last part. Thank You.
The complete data, as estimated by the author, and the relevant Shear and bearing tables are shown on the next slide. The calculations are similar to our estimates of Shear and bearing values for the given connection.
You can view or download the PDF for this Post from the following Link.
You can buy the PDF for solved problem 10-1, which combines posts 5, 6, and 6A, via this Link for just three dollars.
This is a very useful source for designing various Steel elements: A Beginner’s Guide to the Steel Construction Manual, 15th ed., Chapter 4 – Bolted Connections.
This is a useful source for designing various Steel elements: A Beginner’s Guide to the Steel Construction Manual, 16th ed., Chapter 4 – Bolted Connections.
This is a Link to the first part of the same solved problem (Post 5).
This is a Link to the second part of the solved problem (Post 6).
The next Post is Post 7, a solved problem for bearing connections.