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Q   x   4y  3 K EJ  4  2 K EJ  4M  K 
z
2
0
0
0
4
3
z
3
d y *   x
Q K  .
0 1 3
dx
A partial solution (8.27) at different external loads
assumes different shapes.
While acting the centered moment M (Figure 8.6) at the
distance a from the origin
M
a

y *   x  K   x a  , when x  .
3 

 2 EJ z
While acting the centered moment P at the distance
b from the origin
P
y *   x  K    x b  , when x  .
b

4 
 3 EJ z
While acting the uniformly distributed load of intensityq ,
which begins in the section x 
c
q
y *   x   1 K  1    x c    , when x  .
c

4 4 EJ
z
On the left end of the beam the two primary options of
four are always known. The others are determined from the
boundary conditions at the other end of the beam.
For example, if the left end of the beam is rigidly fixed,
then y  ;   .
0
0
0 0
If the left end of the beam is hingedly dropped up
0
0
then y  ; M  .
0 0
If the left end is not fixed and is free from concentrated
loads, then M  ; Q  .
0
0
0 0
If on the left end of the beam the concentrated loads M
л
0
and P are applied, then M  ;Q  .
0
л 0 0

Questions for self-assessment

108