9.6 The Hamilton Principles and Their Applications
575
q(x)
O
x
y
l
q(x)
Fig. 9.14 The deflection of the simple beam subjected to the linear load
Substituting Eq. (26) into Eq. (24), the deflection of the beam in the middle point
can be obtained
y| x=
l
2
=
Pl
3
8π 4 E I
∞
n=1
4
(2n − 1) 4 =
Pl
3
192E I
(27)
When the load acts at the middle point of the beam, from Eq. (17) the equation
for the deflection curve of the beam can be obtained
y =
Px
2
48E I
(3l − 4x) (0 ≤ x ≤ l/2)
(28)
When taking x = l/2, Eq. (27) can be obtained. This shows that the results of the
two kinds of solutions are the same.
Example 9.6.13 Find the deflection curve of the simple beam subjected to the linear
load q(x) =
qx
l
, As shown in Fig. 9.14, the length of the beam is l, the flexural
rigidity of the beam is EI, q is a load constant.
Solution Let the deflection curve of the beam be in the form of sine trigonometric
series
y =
∞
n=1
a n sin
nπ x
l
(1)
The potential energy produced by the load q(x) is
V = −
l
0
q(x)ydx = −
l
0
q
l
x ydx = −
l
0
q
l
∞
n=1
a n x sin
nπ x
l
dx
(2)
The bending strain energy of the beam is