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As the shear force
P
Q   0  
y
2
(Figure. 8.3), then

3
 d y  P
 3   .
 dx  x 0 2EJ z
(8.12)
Figure 8.3

The condition of
symmetry of loading and the reaction of a foundation implies that
the rotation angle of a section at the origin is zero
 dy 
   0     0 . (8.13)
 dx  x 0
Using the boundary conditions (8.10) - (8.13), we find
constant integrationC ,C ,C , C .
1 2 3 4
If (8.10) the first term of (8.9) becomes zero, and the
0
0
second term is zero only whenC  ,C  , therefore,
3 4
y   x  e   x C 1 cos x C  2 sin   x . (8.14)
Differentiated value (8.14), we obtain

dy   x   x
  C e cos x  sin   x  C  e cos x  sin   x .
dx 1 2
The condition (8.13) gives
C  C  0 , or C  C ,
1 2 1 2
so the equation of depressions takes the form

y   x  e   x C cos x  sin   x , (8.15)
1
and the equation of angle rotation

dy   x
   x    2e C  sin x . (8.16)
1
dx
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