2.7 Variational Problems Depending on Higher Order Derivatives
151
(E I y
)
− q = 0
If EI is a constant, then there is
E I y
(4)
− q = 0
The general solution is
y =
q
24E I
x
4
+ c 1 x
3
+ c 2 x
2
+ c 3 x + c 4
Using the above boundary conditions, we obtain
c 2 = c 4 = 0, c 1 = −
q L
12E I
, c 3 =
q L
3
24E I
Thus, the deflection curve of the beam is
y =
qx
24E I
(x
3
− 2Lx
2
+ L
3
) (0 ≤ x ≤ L)
At this time, the total potential energy of the system is minimum, and its value is
J =
1
2
L
0
q
2
4E I
(x
2
− Lx)
2
−
q
2 x
12E I
(x
3
− 2Lx
2
+ L
3
)
dx = −
q
2 L
5
240E I
When the two ends of the beam is clamped, the boundary conditions are
y(0) = 0, y(L) = 0, y
(0) = 0, y
(L) = 0
At this time, the general solution remains unchanged, applying the above boundary
conditions to the general solution, we obtain
c 3 = c 4 = 0, c 1 = −
q L
12E I
, c 2 =
q L
2
24E I
Thus, the deflection curve of the beam is
y =
qx
2
24E I
(x − L)
2
(0 ≤ x ≤ L)
At this time, the total potential energy of the system is minimum, and its value is
J =
1
2
L
0
q
2
144E I
(6x
2
− 6Lx + L
2
)
2
−
q
2 x
2
12E I
(x − L)
2
dx = −
q
2 L
5
1440E I