Last Updated on September 17, 2026 by Maged kamel
- Nominal Shear strength and bearing tear-out for bearing connections.
- The nominal Shear strength for a Bolt is Fnv.
- What is the Tensile stress and Shear strength of the A307 Bolt?
- What are the Tensile stress and Shear strength of the A325 Bolt?
- What is the Tensile stress and SGroup strength A490e Fut = 150.0
- Table J-2-Nominal Shear strength of Fasteners.
- The values of nominal Shear strength and nominal tensile strength for bolts in the RCSC.
- Bearing and tear -out of a simple connection- A Tale of tear-outs.
Nominal Shear strength and bearing tear-out for bearing connections.
The Post content has two items. The first is the relationship between ultimate stress and nominal Shear strength, while the second describes bearing and tear-out. The next slide shows these items.

The nominal Shear strength for a Bolt is Fnv.
The Nominal Shear strength, denoted as Fnv, is based on Ultimate strength. The relationship is Fnv = Fult * 0.625As, as previously discussed while discussing Shear in beams, and we consider the Shear stress factor to be approximately 0.60.
However, for bolts, the Code considers the Nominal strength = Fu* 0.625, again multiplied by two factors. The first factor is Ca, which depends on the connection length. If the connection length is 38 inches, Ca equals 0.90; otherwise, it equals 0.75.
Meanwhile, Cb is for the types of threads, whether included or excluded, for high-strength tensile bolts, whether X or N.

For ta, it is e.75 if the connection is >8″; it’s e. if the connection is Orrb”.Ortb, it is e if the threaded part is excluded or the Shear line does not pass through the Bolt threads t equals 0.80 if the threads are included, meaning the Shear plane cuts the threaded portion of the Bolt.
What is the Tensile stress and Shear strength of the A307 Bolt?
Bolt A307 is A carbon steel Bolt ranging from 1/4 inch to 4 inches in diameter.

The next slide image shows a detailed estimate of the tensile stress and Shear strength of A307 bolts.

What are the Tensile stress and Shear strength of the A325 Bolt?
Let us refer back to the equation: the Shear if Fult = 120 ksi hile the Shear factor 0.625*ca=0.90 if the connection is >38″, cb=1 for threaded portion is excluded, the multiplication will give =120*0.625*0.9*1=67.50 KSI.

However, if we consider the threaded inclusion, we multiply by 0.80 again, resulting in 67.5 × 0.80 = 54 ksi.
What is the Tensile stress and SGroup strength A490e Fut = 150.0
For Group A, ASTM A 490, Fut = 150.00 ks; the tensile stress Fut and Shear stress Fnv are shown in the next slide image.

Table J-2-Nominal Shear strength of Fasteners.
Table J3.2 is divided into two parts: nominal tensile strength Fut= 0.75*Fult, whether threading is included or excluded.
For Group C, per ASTM F3043, the tensile stress is Fnt = 0.75*200 = 150 ksi. Please refer to the next slide image of Table J3.2.

The values of nominal Shear strength and nominal tensile strength for bolts in the RCSC.
Refer to the Bolt Council table in the Research Council on Structural Connections, or ASTM A325or Group A; the static tension is 90 ksi. Tension in Group B is 113 ksi. For fatigue due to repeated loads, such as in bridges, refer to Section 5.5.
Here is the Shear condition for connection = 38″: Fnv = 68 ksi and 54 Ksi; refer back to Table J3.2. The exact figures are there. If the connection is >38″, then Fnv will be 45 ksi and 56 ksi for the threaded, excluding the Shear plane.
Here, for the threaded, excluded case, the first case for a connection <=38″ has Fnv = 1 and 84 ksi. According to J3.2, we wind two figures: Fnv = 68 ksi and 84 ksi. That is for the included thread with a connection>38″.
Again, for Group B, Fult = 150 ksi multiplied by 0.625*0.90*1 = 84 ksi gives Fnv for the x condition, where threads are not included; the N condition, where threads are included, Fnv = 150*0.563*0.80 = 67.50 = 68 ksi.

For the bearing and tear-out, LRFD and ASD values are φ=0.75, while Ω=2.00 for the nominal strength of the connected material.

The different equations presented in the specifications are provided below. For case (i) of bearing, when elongation of 0.25″ is a design criterion, then Rn=2.40*d*t*Fult.
For case ii of bearing, when elongation of 0.25″ is not a design crit: Rn = 3*d*t*FultValuee for Rn is given, Rn=3*d*t*Fult.
For the tear-out case, if deformation at the Bolt Hole is a design criterion, R = 1.0; Rn = 1.2LctFult. If deformation is not a design criterion, Rn = 1.5Let*Fult.

For long slotted connections with the slot perpendicular to the direction of the force, the bearing Value is Rn = 2.30dtfult, while for tear-out, Rn = 1.0Lc*Fult. Here is a condition c for connections made using bolts that pass entirely through an unstiffened box member.

Table J3.3 gives the Hole dimensions for various Bolt sizes, whether oversized, short-slot, or long-slot. HeTheeole diameter is bigger than the diameter of the Bolt by 1/16″ for the standard, as shown in the table.

Table 3.1 contains nominal bolt-hole dimensions for standard-diameter, oversized, shot-slotted, or long-slotted bolts. The bearing Value is based on the slot condition.

Table J3.4, introduced earlier, determines the minimum edge distance from the Bolt’s centerline.

Bearing and tear -out of a simple connection- A Tale of tear-outs.
The following images are quoted from the specs and include crucial information about tear-out.
Then, a discussion of the tear-out is shown. The tear-out for LH external: One part will split like a wedge. The inclination angle is based on the external clear distance.

Bearing depends on the Hole elongatio It differsnt from the slip connection, which we will discuss; acriticalal slip elongation, it is prevented I quote when deformation at the Hole elongation will not exceed 1/4″ When high tensile stress occurs in the net section, other relations can be estimated if they are not design criteria.

If not a design criterion, then Rn=1.5*Lc*t* Fult. Again, this is a tear-out slide for the Rn Value.
The change is that the Rn values changed between the 2010 and 2016 specifications: Rn upper Value = 2.4*dt*Fult and Rn = 1.2*lct*Fult. Whether the deformation meets the design criteria, or whether the deformation is 0.25″ or not, the equation is used to check the upper limit, which is 2.4*d *t*Fult.


This Link lets you download A Tale of Tearouts from Specwise. includes an interesting solved problem on connections and how to determine Shear bearing values.
This is the PDF file for this Post, which you can view or download using the button below.
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 very useful source for designing various Steel elements: A Beginner’s Guide to the Steel Construction Manual, 16th ed., Chapter 4 – Bolted Connections.
The following Post, 6, presents the solution to problem 10-1 for the Shear force Value.