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Eccentrically loaded foundation (Meyerhoff theory)

  • Thread starter dss975599
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  • #1
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Homework Statement


For this question , I tried to use L' = L -2e and B '= B , i ended up getting different final answer when i use L ' = L , B'= B -2e
In the notes , we can see that if the eccentricity were in the direction of length of the foundation , L '= L-2e , and B'= B ...
But , in this question , I am not sure whether the eccentricity were in the direction of length of the foundation or the width of foundation

Homework Equations




The Attempt at a Solution


Why cant I use L' = L -2e = L-2(0.25) = and B '= B to determine the answer. Since this question is a square foundation , I'm confused whether to use L' = L -2e = L-2(0.25) = and B '= B or L ' = L , B'= B -2e ... Can anyone explain please ?
 

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Answers and Replies

  • #2
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I think this goes back to the definitions of B and B'.

B and B' are always the shorter lengths of the plan dimensions of the foundation.

So, although you compute L'= L - 2e, L' will now be smaller than B'=B=L. So automatically, we reassign B' = L-2e and L'=B.
 
  • #3
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Let's say my original L = 3.2m , B = 3.0m , e= 0.2m, , the eccentricity is in the direction of L ... So , L ' = L-2e = 3.2-2*0.2 = 2.8m , B ' = B = m ......So, we need to reassign old L' = B' = 2.8 ? whereas the old B' = 3 = L' ?
 
  • #4
227
58
Let's say my original L = 3.2m , B = 3.0m , e= 0.2m, , the eccentricity is in the direction of L ... So , L ' = L-2e = 3.2-2*0.2 = 2.8m , B ' = B = m ......So, we need to reassign old L' = B' = 2.8 ? whereas the old B' = 3 = L' ?
In this case: L' = 2.8 m and B'= 3.0 m , but remember that B' must be smaller of the two. So, L'= 3.0m and B' = 2.8 m
 
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