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

Chapter 4 - Bolted Connections

2006, 2007, 2008, 2011 T. Bartlett Quimby


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


Report Errors or Make Suggestions

Purchase Hard Copy

Make Donation



Section 4.6

Shear Rupture

The limit state of shear rupture is introduced 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:

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 4.3.

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

Rn = FnvAb


  • Fnv 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 shear strength of a connection that is concentrically loaded is determined by:

Rn = FnvAbns


  • ns is the number of shear planes that transfer the complete load.

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

SCM Table J3.2 tabulates nominal shear stresses for the commonly used structural bolts.  It also provides provision for computing the nominal shear 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.  If threads exist in a shear plane then there is less area available to transfer the shear.  Instead of computing a reduced Ab, the SCM specification reduces Fnv.  Use the tabulated value for Fnv that applies to your design situation.

Combined Tension and Shear in Bearing-Type Connections

When this condition occurs, the tensile capacity is reduced as discussed in BGSCM 4.5.

Sample Spreadsheet Computation

This spreadsheet computes both the shear and tension strength limit states, including combined shear and tension.  The input consists of information about the bolts (type, size, tabulated stress, number) and loads (shear and tension) obtained previous to doing this calculation.

Bolt Strength Capacity (both Tension and Shear)      
Bolt: A325-N            
  Ab 0.4418 in          
  Fnv 54 ksi          
  Fnt 90 ksi          
  Ns 1 2 per bolt        
  Nb 8 8 bolts        
Total Shear Planes 24          
LRFD       ASD      

Tension, Tu

250 k/connection

Tension, Ta

200 k/connection
  Shear, Vu 200 k/connection Shear, Va 150 k/connection
fv 18.9 ksi   fv 14.1 ksi  
f 0.75     W 2    
  F'nt 75.1 ksi   F'nt 69.8 ksi  
  Rnt 33.2 k/bolt   Rnt 30.9 k/bolt  
  Rnt  530.8 k/connection Rnt  493.7 k/connection
  f Rnt = 398.1 k/connection Rnt / W = 246.9 k/connection
  Tu/f Rnt = 62.8% OK   Ta / (Rn / W ) = 81.0% OK  
  Rnv 23.9 k/shear plane Rnv 23.9 k/shear plane
  Rnv  572.6 k/connection Rnv  572.6 k/connection
  f Rnv = 429.4 k/connection Rnv / W = 286.3 k/connection
  Vu/f Rnv = 46.6% OK   Va / (Rn / W ) = 52.4% OK  

<<< Previous Section <<<        >>> Next Section >>>