A Beginner's Guide to the Steel Construction Manual, 13th ed. (old)

Chapter 4 - Bolted Connections

© 2006, 2007, 2008 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 4.9

Example Problem 4.1

The example problems presented in this section refer to a spreadsheet solution.  Click on the link below to get the file:

Chapter 4: Excel Spreadsheet Solutions

Given: A lap splice is connected with (11) A325-X bolts as shown in Figure 4.9.1.1.  Assume that the applied load consists of 40% dead load and 60% live load.  The steel plates are ASTM A36 stel, the bolts are fully pretensioned 3/4" diameter ASTM A325-X bolts in standard holes.  The faying surface is class A.

Figure 4.9.1.1
Lap Splice
Click on hotlinks for larger views

Wanted:  Determine the capacity of the connection based on bolt strength as specified below.  Consider both LRFD and ASD.  Express your results in terms of service load levels.

  1. Determine the slip capacity of the connection.
  2. Determine the bearing capacity of the connection.

Solution:

Since the load combination only includes live and dead load, ASCE 7 load combination 2 (for both LRFD and ASD) will be the controlling case.  The relationship between factored and unfactored load is:

Pu = 1.2D +1.6L = 1.2(0.40Ps,eq) + 1.6(0.60Ps,eq) = 1.44 Ps,eq

The ASD case is simply the D + L:

Pa = D +L = (0.40Ps,eq) + (0.60Ps,eq) = 1.0 Ps,eq

a)  Slip Capacity - (See specification section J3.8)

rn = mDuhscTbNs = 11.1 kips/per bolt

where:  m = 0.35, Du = 1.13, hsc = 1.00, Tb = 28 k (from Table J3.1), and Ns = 1 (each bolt has only 1 shear plane).

The connection has 11 bolts, so, for the connection:

Rn = (11.1 k/bolt)(11 bolts) = 122 kips/connection

For LRFD, the capacity is determined by:

Pu = 1.44 Ps,eq < fRn = 1.0(122 kips/connection) = 122 kips/connection

Ps,eq < 84.6 k/connection (LRFD)

For ASD, the capacity is determined by:

Pa = 1.0 Ps,eq < Rn/W = (122 kips/connection)/1.5

Ps,eq < 81.2 k/connection (ASD)

Observations:

  • In this case, LRFD allows more load than ASD. 
  • These numbers are NOT the ACTUAL theoretical slip forces.  These are the values that the specification allows to guarantee that slip will not occur.

b)  Bearing Capacity - (See specification J3.6)

The nominal capacity of a single shear plane is computed by:

rn = FnAb = 26.5 kips/shear plane

Where Fn = 60 ksi (from Table J3.2) and Ab = .4418 in2 (area of a 3/4" diameter circle).

The connection has 11 bolts, all in single shear, so, for the connection:

Rn = (26.5 k/shear plane)(11 shear planes) = 292 kips/connection

For LRFD, the capacity is determined by:

Pu = 1.44 Ps,eq < fRn = 0.75 (292 kips/connection) = 219 kips/connection

Ps,eq < 152 k/connection (LRFD)

For ASD, the capacity is determined by:

Pa = 1.0 Ps,eq < Rn/W = (292 kips/connection)/2.0

Ps,eq  < 146 k/connection (ASD)

Observations:

  • In this case, LRFD allows more load than ASD. 
  • The resulting values are loads that can be safely carried without causing shear failure in the bolts.

Solution Summary

The following values are the capacities of the CONNECTION based on two bolt limit states. The total capacity of the connection is also a function of the capacity of members in the connection.

  LRFD   ASD  
Limit State fRn Ps,eq Rn/W Ps,eq
    (k) (k) (k) (k)
Slip   122 84.6 81.2 81.2
Shear Rupture 219 152 146 146
    Answer   Answer

Observations:

  • Notice that the bearing values are larger the slip values.  This is to be expected.  Slip will occur before the bolts fail in shear.  The load level that you want the connection to carry will depend on whether or not you want to allow slip to occur in your connection.

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