- Solved problem 10-1-how to get Bearing value-ASD-3-3.
- Using table 7-4 for the Nominal value of the bearing for inner bolts in ASD.
- Using table 7-5 for the Nominal value of the bearing for the external bolt in ASD.
- For the solved problem 10-1.The final selected value of the shear force that is to be applied.
Solved problem 10-1-how to get Bearing value-ASD-3-3.
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This is a review of the content of the solved problem 10.1, in which the shear force for the given connection might be applied to the bolts in the connection.
We have earlier introduced the LRFD design and the value of force and we are going to continue with the ASD calculation as required.
For the edge distance Lc =2″-0.50*dh, the hole diameter dh is (3/4+1/16)=13/16″. Lc=2″-0.50*13/16=1.59″>1.
While the internal clear distance =3″, the clear inner distance=3″-(13″/16)=2.1875″ repeated two times and followed by the edge distance.
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.
We need to calculate for the external edge bolt,1.20*lco*t*Fu. While for the inner distance bolts, 1.20*lci*t*Fu, for the external bolt, bearing value=1.2*1.59*(12/16)*(58)=83.0 kips.
While for the inner bolts=1.2*2.188*(12/16)*(58)=114.21 kips. Select the lowest value. We could calculate for the external edge distance, then for the internal, we can multiply by the ratio of(lci/lco). The min of (78.3,83.0,114.21) is 78.30 kips will be selected.
The sum of forces =4*78.30=313.12 kips, then for ASD Rn/Ω value Ω=2.0. Ω*Rn=(1/2)313=157 kips. the arrangement of forces per bolt is shown.
Using table 7-4 for the Nominal value of the bearing for inner bolts in ASD.
For our solved problem, our bolt diameter is 3/4″, 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 Rn. the value of Rn=313 kips, we will divide by Ω, which is=2, we get Rn/Ω=157 kips.
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, Rn/Ω=52.2 kips/inch, multiply by 3/4″ and multiply by 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 Nominal value of the bearing for the external bolt in ASD.
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 the 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- second part.
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, 15th ed, Chapter 4 – Bolted Connections.
The next post is A solved for bearing connections.
