Retaining Walls Based Entirely On the Theory of Friction
Retaining Walls Based Entirely On the Theory of Friction
Pedro J Dozal
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16), and that value gives us the maximum required. To s^lve the above equation let co = y -f- K. Then substituting this value in (1) and replacing values on B, G and H we have that: v + -5- (4 cos 2 /^ -}- sin (2, _'_j cot s) Calculating K we have that K = sin /\- And substituting value of x in A cos /\ cos 2 ^ cos ( \ -J- ?) sin ci tan a = x A sin /\ angle a can be calculated from tables. Second Condition. Maximum Horizontal Component. 39 The horizontal component of the unbalanced force is L c...os a: Now; A Y cos L cos a = 2 cos A x 2 (cos (A ?) 4- sm ZA sm ?) x3 sm ? A 2 x cos (A, cp) sin / Where: A 8 2 A x sin / + x x A sin cos y. = V A* : -2 A x sin /, -f 58 calling; cos (A - ?) 4- s i n . '. _. Sin ? = B, sin ? = C. Cos ( A - a) sin A = F. sfn _ = H, x == o A we have that: L COS a -- 1 2 w Differentiating and simplifying; d (L cos a) __ A~ y cos d co 2 COS a ~ M * C + 4 a)3 G H + " )J (F 2BH 3C)4-2coB F (1 2a>H-f o)* and equating to zero for a maximum we have that; - to 4 C+ 4 co ;i C H -f- ur (F 2 B H 3 G) + 2 co B F = This equation can be solved by trial (see paragraph 11, pag.
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