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# Retaining wall sliding stability

• To: <seaint(--nospam--at)seaint.org>
• Subject: Retaining wall sliding stability
• From: <William.Sherman(--nospam--at)CH2M.com>
• Date: Tue, 17 Jul 2007 17:14:53 -0600

Section 1806.1 of the 2006 IBC requires retaining walls to be designed for a safety factor of 1.5 against lateral sliding.  This raises several questions:

1.  Section 1801.2 indicates that the allowable stress design load combinations in 1605.3 should be used with design formulas in this chapter.  If section 1605.3.1 is used, must 0.6D be used in combination with the 1.5 safety factor?  (That would appear to be overkill!)

2.  A literal reading would also imply that this safety factor applies to load combinations that include seismic forces.  (However, there is a "Code Interpretation" on the ICC website that says that the 1.5 safety factor does not apply to seismic loads!)

3.  The code does not define how the safety factor is to be calculated.  Many have traditionally used the ratio of (resisting forces) / (driving forces) to represent the safety factor.  But EM 1110-2-2502, Retaining and Flood Walls, by the USACE, defines the safety factor against sliding as the ratio of (friction + cohesion) / (net sliding force at base).  These are different values when passive pressure is included.  For example:

If H = driving horizontal force = 100; F = friction = 100; and P = passive = 35:

Traditional SF = (F + P)/H = 135/100 = 1.35

USACE SF = F/(H-P) = 100/(100-35) = 1.54

Does the wall meet the code or not?

The USACE method seems more rational where a retaining wall extends some distance below grade but has a more limited differential soil height across it.  Take an extreme case of a narrow wall placed in a trench below grade with only a few inches of differential soil height.  The soil would essentially be at rest with nearly equal pressures on both sides - but it might be defined as non-code compliant per the first method, since a resisting force of 1.5 times the full driving force would be required; however, the second method would only require 1.5 times the small differential force, a much more rational design.