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Part II: Use of Side Friction to Resist Overturning Moment
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- Subject: Part II: Use of Side Friction to Resist Overturning Moment
- From: "Robert Rogers" <robert.rogers(--nospam--at)woolpert.com>
- Date: Wed, 16 Dec 98 07:11:34 EST
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 ?
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