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Part II: Use of Side Friction to Resist Overturning Moment

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 I appreciate the opinions with regard to this subject matter.  I apologize 
 for the length of this follow-up, but perhaps I may be enlightened by a 
 helpful respondent.  Some things which make me suspicious of these 
 calculations are:
 
 (1) The geotech report says absolutely nothing about this issue.  The      
     calculation assumes a side friction value of 0.50 ksf.  The report only 
     states to design the spread footings for an allowable soil bearing     
     pressure of 2,500 psf (typical for the medium to stiff clay soil found at 
     the site in question) and that it may be cast against undisturbed earth 
     or engineered fill.
 
 (2) The overturning resistance by friction was calculated as follows:
 
 
                                 +++++  <----- 60 Kips
         Footing is              +   +
         7.5 ft. wide         3' +   +
                                 +   +          F
                        /|\ ++++++   +++++++    |
                         |  + 2'           +    |
                         |  ++++++++++++++++   \|/
                            <---- 7.5'----->
                         F                       
 
    Friction Resisting Force = F = (2 ft.)*(7.5 ft.)*(0.5 KSF)= 7.5 Kips 
    Resisting Moment Due to Friction = (7.5 Kips)*(3.75 ft.)*2 = 56.25'K
 
    And then for the sides of the footings a methodology similar to an      
    eccentric load on a weld pattern is utilized:
 
           friction = (Resisting Moment Due to Side Friction)*(c)
                         ------------------------------------
                                        (Ip)
 
    where (Ip) is the polar moment of inertia and (c) is the distance to 
    the extreme fiber:
 
 
           0.500 ksf =            (M) * (7.5' / 2)
                       ---------------------------------------
                           {(1/12)* 7.5' * 2' (7.5^2 + 2^2)}      
 
    Thus the Resisting Moment to friction is: 10 'K
 
    Thus the total resisting moment due to friction is:
 
                           (56.25'K)+(2)(10'K)  = 76.2'K
 
 
 The traditional treatment of friction is a normal force multiplied by a 
 friction coefficient to calculate a force resisting movement.  If I 
 understand this correctly (which I may not), to achieve a 7.5 Kips 
 resisting force on the side of the footing the normal force which must be 
 applied would be:
 
                        7.5 Kips
                --------------------------   =  Normal force
                u = Coefficient of Friction
 
 Assuming u=0.5 the normal force would have to be 15.0 Kips (applied 
 uniformly to the side of the footing to ensure 0.5 ksf).  Now once the 
 footing rotates, allowable soil pressure is exceeded, and the footing 
 starts to pull up on one side and push down on the other, then we get into 
 the discussion about the shear strength of the soil, cohesion, angle of 
 internal friction, failure planes, etc. .
 
 Frankly I don't understand how you obtain this much lateral force on the 
 sides of the footing (which is only 5 ft. deep to the bottom of the 
 footing) to generate this type of resisting moment.  Am I on line or can 
 someone show me the light ?