Last Updated on September 17, 2026 by Maged kamel
Solved problem 12-1 for bearing connections.
The following slide image summarizes the content of this Post.
The following slide image reviews 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 is a 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 y = 36 ksi nd 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 Area. Assume that deformations at Bolt holes are a design consideration.
The overlap distance is 9″, with a CL distance of 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 failure modes. The first failure mode is 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 stacked and connected by 4 bolts. The bolts are 8″, with an Area of 0.60 in^2 for each Bolt, as given.
The upper plate is 1/2″, the lower plate thickness is 1/2″, and the plate width is given as 12″.
For tensile Yielding, Fy is multiplied by Area or the gross Area. In the seArea case, for tensile rupture, the net Area is multiplied by Fy.
We are aware that, because we place bolts during installation, we add 1/16″ + 1/16″ to the diameter of one Bolt, which is 7/8″; the final diameter considered 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.
For allowable design, use (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 designing for the net Area, we use Fu = si.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.
Bearing calculation for solved problem 12-1.
Before checking Bolt bearing capacity, we need to review the tables required for our estimation.
The diameter of bolts=7/8″, the Hole diameter is the Bolt diameter plus 1/16″, and the Hole 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 edge-distance requirement, refer to Table J3.4 for a 7/8″ Bolt; 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 are 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 (2db). We find that the outer clear distance for the exterior Bolt exceeds 2db, indicating that the nominal Value is governed by the upper-limit equation (2.4*d*t*fu). The details of the estimation are 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 is 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 S-bolt’s Shear strength. 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 comparing the Bolt’s nominal bearing loads with the corresponding bolts’ Shear strength. The bolts’ Shear strength is less than the nominal Load, so bearing and Shear strength govern the design.
In the next Post, we will finalize the calculations for the given connection. Thank you.
You can view or download the PDF for this Post from the following Link.
The next Post will be a solved problem 12-1-part 2: connection nominal Load.
This is a 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.