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

Chapter 5 - Welded Connections

© 2006, 2007, 2008, 2009 T. Bartlett Quimby

Introduction to Welding

Finding Forces in Welded Connections

Effective Areas and Size Limitations of Welds

Effective Areas of Base Metal

Strength Limit State

Designing Welds

Chapter Summary

Example Problems

Homework Problems

References


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Section 5.8

Example Problem 5.5

Last Revised: 11/04/2014

The example problems presented in this section have a spreadsheet solution.  You will need this file to follow along with the presented solutions.  You can click on the following link to get the file:

Chapter 5:  Excel Spreadsheet Solutions

In addition to the spreadsheet solution this problem also has a downloadable hand solution:  Example5_5.pdf

Example Problem 5.5:  This example illustrates the application of the two methods, or cases, that AISC provides for computing forces in welds when the applied force in out of the plane of the welds. 

Given: A connection plate is welded to the face of a W section column flange and is bolted to the web of a W section beam as shown in Figure 5.8.5.1.  The connection plate is ASTM A36 steel that is 1/4" thick.  The W sections are ASTM A992 steel.  The weld electrode is F7.  A = 2 inches and B is to 9 inches.  The weld size, a1, is 1/4 inch.  The beam reaction transferred by the connection plate is 3 parts dead load and 5 parts live load.

Figure 5.8.5.1
Out-of-Plane Eccentrically Loaded Connection
Click on image for larger view

Wanted: 

a)  Determine the maximum beam reaction based on weld strength of the fillet welds that connect the plate to the W section.  Use LRFD and express your result in terms of comparable service level loads.

b) Repeat the problem using the coefficients on SCM pg 8-66

Solution:

In this problem, the two welds (one on either side of the connection plate) support the beam reaction which is transferred to the plate via the bolted connection to the beam.  So, the beam reaction is located 2 inches from the face of the column.

Before we get too deep into the problem, the composite load factors (CLF) are computed because the problem implies that the results of LRFD and ASD are to be compared.  The CLFs allow us to see the loads at a common basis for comparison.

In part (a) of the problem, we use basic mechanics to compute the stresses in the welds.

The entire weld has a uniform stress in the plane of the column flange that equals the beam reaction divided by the area of the weld.  The direction of this reaction force is upward.

The moment on the weld group causes stresses that are normal to the column flange surface.  The maximum tension and compression stresses occur at the top and bottom of the welds, respectively.

The worst case stress in the welds is the vector sum of the stress due to the translational effect (Reaction divided by the area of the weld) and the stress due to the moment on the weld group (Mc/I).

Since we don't know the magnitude of the reaction force, we find out what maximum stress that a unit load causes.  We then scale the unit force up by the ratio of the maximum allowed stress to the maximum stress caused by a unit load.  This is probably best seen in the hand solution.

The resulting Rn is then used in the LRFD and ASD contexts to find the maximum allowed reaction.

One thing to note about the solution in part (a) is that it does not take into account the angle of the stress relative to the axis of the weld and is, as a result, conservative.

In this case, the angle of stress to axis of weld is 53.1 degrees.  Applying SCM Equation J2-5 to the allowable stress, this would change the compute Rn from 80.17 kips to 108.6 kips, which is more in line with what you get in part (b) of the problem.

For part (b) of the problem we used the coefficient found in the SCM on page 8-66.  This method considers the welds to be eccentric in plane and accounts for the angle of stress to axis of weld.  The result is a higher connection capacity.

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