RAM Connection v23.0.1 now calculates bolt capacities, for AISC 360-16 design code, utilizing the criteria outlined in the User Note of Section J3.6. This note specifies that the effective strength of an individual bolt should be considered as the lesser of its shear strength (per Section J3.6) and its bearing strength (which includes tearout, as per Section J3.10). The overall strength of a bolt group is determined by summing the effective strengths of each individual bolt. According to the AISC, the "User Note model" is designed to produce results consistent with fundamental engineering principles, which may not be the case with other models.
As part of this enhancement, RAM Connection now distinguishes between bearing and tearout strengths as separate limit states for evaluating bolted connections.
One of the key advantages of incorporating the bolt tearout limit state for the 2016 AISC code is the program's ability to calculate the true clear distance "lc" in the direction of the applied force for each individual bolt. While the AISC Examples Manual provides a conservative calculation, it is also mentioned that with access to a computer program, this distance can be precisely determined based on the plate's geometry and the actual load direction. In relevant cases, clear distances can be calculated as a result of the analysis using the instantaneous center of rotation (IC) method.
Upon modeling within the software, the following examples illustrate the benefits of these more accurate calculations. Additionally, this document provides a comprehensive explanation of the geometries obtained through bolt group analysis, focusing exclusively on the bolted aspects of the connections.
EXAMPLE II.A-1B (AISC Examples Manual v.15)
Given: Verify the available strength of an all-bolted double-angle connection for an ASTM A992 W18´50 beam, as shown in Figure II.A-1B-1, to support the following beam end reactions:
Use ASTM A36 double angles that will be shop-bolted to the beam.
Strength of the bolted connection at the beam side
Bolts shear. The program calculates the following strength per bolt:
φRn = 2 * (φ*Fnv*Ab) Eq. J3-1
= 2 * (0.75*54000[lb/in2]*0.601[in2])
= 48.681[kip]
For a double angle connection there is no eccentricity in the beam side. The effective bolt group factor is equal to the total number of bolts. And the result is:
φRn = C*φRn
= 5*48.681[kip]
= 243.405[kip]
Bolt bearing on angles. The program calculates the following strength per bolt:
φrn = 2 * (φ*2.4*d*tp*Fu) Eq. J3-6a
= 2 * (0.75*2.4*0.875[in]*0.625[in]*58000[lb/in2])
= 114.188[kip]
For a double angle connection there is no eccentricity in the beam side. The effective bolt group factor is equal to the total number of bolts. And the result is:
φRn = C*φRn
= 5*114.188[kip]
= 570.938[kip]
Bolt tearout on angles. For a double angle connection there is no eccentricity in the beam side, then it is assumed that the connection is concentrically loaded. For this case, the program calculates the strength for the bolt group summating the individual strength of each bolt.
For this purpose, from the analysis of the concentrically loaded bolt group, the following figure shows the clear distances “lc” for all the bolts in the connection at the beam web.
As can be seen, the “lc” distances for the angles are evaluated as the free distance over the line of action of the force between a bolt hole and the nearer edge of the angle. Since the automatic detection of these distances can become a time-consuming process for the software, the shape of the bolt holes has been set to octagons (8 sides) that give very approximate and acceptable values of “lc”.
With previous results, the program calculates the following strength for the group:
φrn = 2 * (φ*1.2*Lc-end*tp*Fu) Eq. J3-6c
= 2 * (0.75*1.2*1.55[in]*0.625[in]*58000[lb/in2])
= 101.144[kip]
φrn = 2 * (φ*1.2*Lc-end*tp*Fu) Eq. J3-6c
= 2 * (0.75*1.2*1.55[in]*0.625[in]*58000[lb/in2])
= 101.144[kip]
φrn = 2 * (φ*1.2*Lc-end*tp*Fu) Eq. J3-6c
= 2 * (0.75*1.2*1.55[in]*0.625[in]*58000[lb/in2])
= 101.144[kip]
φrn = 2 * (φ*1.2*Lc-end*tp*Fu) Eq. J3-6c
= 2 * (0.75*1.2*1.55[in]*0.625[in]*58000[lb/in2])
= 101.144[kip]
φrn = 2 * (φ*1.2*Lc-end*tp*Fu) Eq. J3-6c
= 2 * (0.75*1.2*1.15[in]*0.625[in]*58000[lb/in2])
= 75.031[kip]
φRn = ∑(rni,eff)
= 479.605[kip]
Bolt bearing on beam web. The program calculates the following strength per bolt:
φrn = φ*2.4*d*tp*Fu Eq. J3-6a
= 0.75*2.4*0.875[in]*0.355[in]*65000[lb/in2]
= 36.343[kip]
φRn = C*φRn
= 5*36.343[kip]
= 181.716[kip]
Bolt tearout on beam web. With no eccentricity in the beam side, the program calculates the “lc” distances for the beam web as the free distance over the line of action of the force between a bolt hole and the nearer edge of the beam web.
With previous results, the program calculates the following strength for the group:
φrn = φ*1.2*Lc-end*tp*Fu Eq. J3-6c
= 0.75*1.2*2.35[in]*0.355[in]*65000[lb/in2]
= 48.814[kip]
φrn = φ*1.2*Lc-end*tp*Fu Eq. J3-6c
= 0.75*1.2*2.35[in]*0.355[in]*65000[lb/in2]
= 48.814[kip]
φrn = φ*1.2*Lc-end*tp*Fu Eq. J3-6c
= 0.75*1.2*2.35[in]*0.355[in]*65000[lb/in2]
= 48.814[kip]
φrn = φ*1.2*Lc-end*tp*Fu Eq. J3-6c
= 0.75*1.2*2.35[in]*0.355[in]*65000[lb/in2]
= 48.814[kip]
φrn = φ*1.2*Lc-end*tp*Fu Eq. J3-6c
= 0.75*1.2*2.35[in]*0.355[in]*65000[lb/in2]
= 48.814[kip]
φRn = ∑(rni,eff)
= 244.068[kip]
Bolt group effective strength. Since there is no eccentricity for the beam side, the program determines the effective strength of the group by summating the critical strength of five limit states (bolt shear, bolt bearing on angles, bolt tearout on angles, bolt bearing on web and bolt tearout on web) of each individual bolt in the group. For this case, it turns out that the critical strength of the mentioned limit states is the bearing on the beam web for each bolt.
Bolt 1 : Bearing (Beam web)
φrn = φ*2.4*d*tp*Fu Eq. J3-6a
= 0.75*2.4*0.875[in]*0.355[in]*65000[lb/in2]
= 36.343[kip]
Bolt 2 : Bearing (Beam web)
φrn = φ*2.4*d*tp*Fu Eq. J3-6a
= 0.75*2.4*0.875[in]*0.355[in]*65000[lb/in2]
= 36.343[kip]
Bolt 3 : Bearing (Beam web)
φrn = φ*2.4*d*tp*Fu Eq. J3-6a
= 0.75*2.4*0.875[in]*0.355[in]*65000[lb/in2]
= 36.343[kip]
Bolt 4 : Bearing (Beam web)
φrn = φ*2.4*d*tp*Fu Eq. J3-6a
= 0.75*2.4*0.875[in]*0.355[in]*65000[lb/in2]
= 36.343[kip]
Bolt 5 : Bearing (Beam web)
φrn = φ*2.4*d*tp*Fu Eq. J3-6a
= 0.75*2.4*0.875[in]*0.355[in]*65000[lb/in2]
= 36.343[kip]
Strength of the group
φRn = ∑(rni,eff)
= 181.716[kip]
Strength of the bolted connection at the column side
Bolts shear. The program calculates the following strength per bolt:
φRn = 2 * (φ*Fnv*Ab) Eq. J3-1
= 2 * (0.75*54000[lb/in2]*0.601[in2])
= 48.681[kip]
For a double angle connection there is no eccentricity in the support side. The effective bolt group factor is equal to the total number of bolts. And the result is:
φRn = C*φRn
= 5*48.681[kip]
= 243.405[kip]
Bolt bearing on angles. The program calculates the following strength per bolt:
φrn = 2 * (φ*2.4*d*tp*Fu) Eq. J3-6a
= 2 * (0.75*2.4*0.875[in]*0.625[in]*58000[lb/in2])
= 114.188[kip]
For a double angle connection there is no eccentricity in the beam side. The effective bolt group factor is equal to the total number of bolts. And the result is:
φRn = C*φRn
= 5*114.188[kip]
= 570.938[kip]
Bolt tearout on angles. For a double angle connection there is no eccentricity in the support side, then it is assumed the connection is concentrically loaded. For this case, the program calculates the strength for the bolt group summating the individual strength of each bolt:
For this purpose, from the analysis of the concentrically loaded bolt group, the following figure shows the clear distances “lc” for all the bolts in the connection at the column side. Note that the demand for the bolt group analysis is the vertical load.
As can be seen, the “lc” distances for the angles are evaluated as the free distance over the line of action of the force between a bolt hole and the nearer edge of the angle or the free distance between adjacent bolt holes. Since the automatic detection of these distances can become a time-consuming process for the software, the shape of the bolt holes has been set to octagons (8 sides) that give very approximate and acceptable values of “lc”.
With previous results, the program calculates the following strength for the group:
φrn = 2 * (φ*1.2*Lc-end*tp*Fu) Eq. J3-6c
= 2 * (0.75*1.2*0.781[in]*0.625[in]*58000[lb/in2])
= 50.977[kip]
φrn = 2 * (φ*1.2*Lc-end*tp*Fu) Eq. J3-6c
= 2 * (0.75*1.2*2.063[in]*0.625[in]*58000[lb/in2])
= 134.578[kip]
φrn = 2 * (v*1.2*Lc-end*tp*Fu) Eq. J3-6c
= 2 * (0.75*1.2*2.063[in]*0.625[in]*58000[lb/in2])
= 134.578[kip]
φrn = 2 * (φ*1.2*Lc-end*tp*Fu) Eq. J3-6c
= 2 * (0.75*1.2*2.063[in]*0.625[in]*58000[lb/in2])
= 134.578[kip]
φrn = 2 * (φ*1.2*Lc-end*tp*Fu) Eq. J3-6c
= 2 * (0.75*1.2*2.063[in]*0.625[in]*58000[lb/in2])
= 134.578[kip]
φRn = ∑(rni,eff)
= 589.289[kip]
Bolt bearing on support. The example does not specify data for the column, but the joint can be modeled using a W14x90 section with A992 Gr50 steel. The program calculates the following strength per bolt:
φrn = 2 * (φ*2.4*d*tp*Fu) Eq. J3-6a
= 2 * (0.75*2.4*0.875[in]*0.71[in]*65000[lb/in2])
= 145.373[kip]
φRn = C*φRn
= 5*145.373[kip]
= 726.863[kip]
Bolt tearout on support. Not specified data for the support in the example. However, it is not mentioned that the connection is near the column end or a free edge, condition that allow to neglect this check.
Bolt group effective strength. Since there is no eccentricity for the support side, the program determines the effective strength of the group by summating the critical strength of five limit states (bolt shear, bolt bearing on angles, bolt tearout on angles, bolt bearing on support and bolt tearout on support) of each individual bolt in the group. For this case, it turns out that the critical strength of the mentioned limit states is the bolt shear for each bolt.
Bolt 1 : Shear
φRn = 2 * (φ*Fnv*Ab) Eq. J3-1
= 2 * (0.75*54000[lb/in2]*0.601[in2])
= 48.681[kip]
Bolt 2 : Shear
φRn = 2 * (φ*Fnv*Ab) Eq. J3-1
= 2 * (0.75*54000[lb/in2]*0.601[in2])
= 48.681[kip]
Bolt 3 : Shear
φRn = 2 * (φ*Fnv*Ab) Eq. J3-1
= 2 * (0.75*54000[lb/in2]*0.601[in2])
= 48.681[kip]
Bolt 4 : Shear
φRn = 2 * (φ*Fnv*Ab) Eq. J3-1
= 2 * (0.75*54000[lb/in2]*0.601[in2])
= 48.681[kip]
Bolt 5 : Shear
φRn = 2 * (φ*Fnv*Ab) Eq. J3-1
= 2 * (0.75*54000[lb/in2]*0.601[in2])
= 48.681[kip]
Strength of the group
φRn = ∑(rni,eff)
= 243.405[kip]