A Beginner's Guide to the Steel Construction Manual, 14th ed.

Chapter 4 - Bolted Connections

2006, 2007, 2008, 2011 T. Bartlett Quimby

Overview

Mechanics of Load Transfer

Finding Forces on Bolts

Hole Size and Bolt Spacing

Tensile Rupture

Shear Rupture

Slip Capacity

Chapter Summary

Example Problems

Homework Problems

References


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Section 4.5

Tensile Rupture

The limit state of tensile rupture is introduced in the specification in SCM J3.6 (SCM pg 16.1-125).

The Limit State:

The basic limit state follows the standard form.  The statement of the limit states and the associated reduction factor and factor of safety are given here:

LRFD ASD
Pu < ftRn Pa < Rn/Wt
Req'd Rn = Pu / ft < Rn Req'd Rn = Pa Wt < Rn
Pu / (ftRn< 1.00 Pa / (Rn/Wt) < 1.00
ft = 0.75 Wt = 2.00

The values of Pu and Pa are the LRFD and ASD factored loads, respectively, applied to the bolt. These forces are computed using the mechanics principles discussed in BGSCM 4.2.

In this case Rn is the nominal tensile strength of a bolt is computed using SCM equation J3-1:

Rn = FntAb

Where:

  • Fnt is obtained from SCM Table J3.2 (SCM pg 16.1-120)
  • Ab is the nominal cross sectional area of the bolt (pdb2/4), ignoring threads.

The nominal tensile strength of a connection that is concentrically loaded equals

Rn = FntAbnb

Where:

  • nb is the number of the bolts in the connection.

For connections that are not concentrically loaded the tensile force in the worst case bolt is determined by structural mechanics and compared against the capacity of a single bolt.

SCM Table J3.2 tabulates nominal tensile stresses for the commonly used structural bolts.  It also provides provision for computing the nominal tensile stress for other types of threaded fasteners.

You will notice that SCM Table J3.2 has rows for conditions including or excluding threads from the shear planes.  This is not an issue for bolt tension so the values of Fnt are the same for each case for a given bolt material.

Combined Tension and Shear in Bearing-Type Connections

As mentioned in the overview, when a bolt is subject to both imposed tension and shear, the interaction of these two forces must be accounted for when determining the capacity of the fastener.  SCM J3.7 does this.

In bearing type connections, the simplified approach for dealing with the interaction is to reduce Fnt.  The bolt then must satisfy the tensile rupture limit state (as modified by SCM J3.7) and the unmodified shear rupture limit state.

When shear is present, Fnt is replaced with F'nt in the limit state equations above.  Rn becomes:

Rn = F'ntAb

There are different equations for LRFD and ASD.  Let's look at these equations.

SCM equations J3-3a and J3-3b are essentially the same equation with appropriate application of the reduction factor and factor of safety. 

The term frv represents that required shear stress in the bolts.  This is computed as the shearing force (from LRFD combinations or from ASD combinations) on a shear plane divided by the cross sectional area of the shear plane.  For a concentrically loaded connection this becomes:

LRFD:  frv = Vu / (nsp*Ab)        ASD:  frv = Va / (nsp*Ab)

Where:  nsp = total number of shear planes resisting the V.

For connections that are not concentrically loaded, then the shear stress, fv, in each bolt can be determined by methods of structural mechanics (i.e. Elastic Vector Method or IC Method).

The inequalities in the SCM equations denote that the upper limit of the F'nt is Fnt Do NOT try to solve this inequality for frv!  That would only tell you the value of frv that would make F'nt equal to Fnt.  Not a very useful exercise.   Another way to write these equations is:

LRFD:  F'nt = min[(1.3 Fnt - frv Fnt / (f Fnv)), Fnt]

ASD:  F'nt = min[(1.3 Fnt - frv W Fnt / Fnv), Fnt]

One thing to note is that you can always use Rn = F'ntAb when computing tensile capacity of a fastener since F'nt defaults to Fnt when frv gets small or goes to zero (i.e. when there is no applied shear force present).

Sample Spreadsheet Computation

Since the combined tension and shear problem requires both tension and shear information, the most convenient approach is to write a spreadsheet that does both shear and tension as well as shear and tension combined.  As a result the sample spreadsheet computation is presented in the next section after the shear strength computation is presented.

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