Last Updated on August 25, 2026 by Maged kamel
Solved problem 12-1 for bearing connections.
The following slide image shows a summary of the content of this post.
The following slide image shows a review of the different modes of failure for bearing connections, quoted from The Applied Strength of Materials Book by Robert Mott, chapter 13.
A solved problem for bearing connection 12-1-Part 1.
Our new topic will be the discussion of problem 12-1 from Prof. McCormack’s book, which has been solved.
Determine the design strength φ*Pn and the allowable strength Pn/Ω for the bearing type connection shown in Fig.12-5. The steel is A36, where (Fy = 36 ksi) and Fu=58 ksi; the bolts are 7/8″ in A325; the bolts are standard sizes; and the threads are excluded or X from the shear plane. Assume that deformations at bolt holes are a design consideration.
The overlap distance is length =9″, where the CL distance=3″ between bolts. The length is in the direction of the load, while the transverse CLL distance =6″. The edge distance is 3″ and 3″ from each side.
Tensile Failure nominal load by yielding for solved problem 12-1.
For this connection, we have two modes of failure. The first mode of failure is due to yielding, as shown in sec 1-1, and the second is rupture, as shown in sec II-II. Please refer to the next image for section 1-1.
We have a single-shear plane. Two plates are placed on top of each other and connected by 4 bolts. The bolts are 8″, where the Area for each bolt Area= 0.60 in^2, as given.
The upper plate is 1/2″, the lower plate thickness is 1/2″, and the plate width is given as 12″.
For the tensile yielding Fy to be multiplied by Area or the gross Area. In the seArea case, for tensile rupture, the net Area is multiplAreaby Fult.
We are aware that, due to placing bolts during installation, we add 1/16″ + 1/16″ to the diameter of one bolt, which is 7/8″; the final value considered for the diameter is 7/8 + 2*(1/16) = 1″.
For the first case, the gross Area = 12 × 1/2 × Area in².
Then, the tensile yielding = Ag* Fy; we have Fy = 36 ksi and Fult = 58 ksi, Ag = 6 in^2, then Rn = Ag*Fy = 36*6 = 216 kips. For LRFD, φ=0.90, φ*Rn=0.90*216=194.4 kips.
While for allowable design (1/Ω)*Rn, where Ω=1.67.
Tensile Failure nominal load by rupture for solved problem 12-1.
Area case b) where Area gross = 2*diameter*thickness of the plate. Anet =12*1/2- 2*Area*1/2.
The net Area, Anet, is 5 inches². Since we are deaArea with the net Area, we use FuAreaf 58 ksi.Rn=5*58=290 kips.
Remember from the tension area lecture that Aeff = U*Anet; U = 1 for plates, so Aeff = 1*5 = 5 in^2. But now φ=0.75, and Ω=2 .φ*Rn=0.75*290=217.50 kips. While Rn/Ω=290/2=145 kips.
The calculation of bearing for the solved problem 12-1.
Before checking the bolt bearing capacity, we need to review the list of tables required for our estimation.
The diameter of bolts=7/8″, the hole diameter is the bolt diameter plus 1/16″, and the diameter hole equals 15/16″—the inner spacing is 2 2/3 db.
To check the requirement of the edge distance, there is Table J3.4 for a 7/8″ bolt; the required edge distance is 1/18″
To check the requirement of the edge distance, refer to Table J3.4 for a 7/8″ bolt, where the required edge distance is 1/18″. The minimum spacing between bolts is 2.32 inches for db = 7/8 inches.
We have a distance from CL to the outer edge of 3″, which satisfies the case since this distance is greater than 1 1/8″, as given in the table.
For inner bolts, the spacing is 3 inches, which is greater than the minimum of 2.33 inches.
Nominal bearing value for the external bolts.
The next step is to estimate the nominal bearing value for the external bolts. We estimate the clear distance to be 2.531 inches.
We compare this value with (2db). We find that the outer clear distance for the exterior bolt is greater than 2db, indicating that the nominal value is governed by the upper-limit equation (2.4*d*t*Fu). The details of the estimation can be found on the next slide.
Nominal bearing value for the inner bolts.
We will estimate the nominal bearing value for the inner bolts. We estimate the clear distance to be 2.06 inches.
We compare this value with the value of (2db). We find that the outer clear distance for the exterior bolt exceeds2 dbb, indicating that the nominal value is governed by the upper-limit equation (2.4*d*t*fu). The details of the estimation can be found on the next slide.
The upper limit is 2.4d*b*t*Fult, db=7/8″ *t=1/2 for a plate, which is common between the left and right sides of the equation.
The upper limit value will be equal to 2.4*7/8*1/2*58=60.90 kips. The nominal load, based on equation (1.2*lci*t*fu), is shown; as we can see, it exceeds the upper limit and is not considered in the design.
Nominal strength value for bolts.
The next step is to estimate the shear strength of the bolts. There are two ways to estimate. The first way is to use Table J3.2 to determine Fnv for the bolt type. In our problem, the bolt is 7/8-inch X type A. The Fnv value is 68 ksi, a single-shear value for one bolt.
I have included two sketches for comparison of the bolt’s nominal bearing loads with the corresponding bolts, but with shear strength. It can be seen that the bolts’ shear strength is less than the nominal load, as bearing and shear strength govern the design.
In the next post, we will finalize the calculations for the given connection. Thank you.
The PDF for this post can be viewed or downloaded from the following link.
The next post will be A solved problem 12-1-part 2-connection nominal load.
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 very useful source for the design of various Steel elements: A Beginner’s Guide to the Steel Construction Manual, 16th ed, Chapter 4 – Bolted Connections.