9.6 The Hamilton Principles and Their Applications
577
E I y =
qx
5
120
+ c 1 x
3
+ c 2 x
2
+ c 3 x + c 4
(11)
For the simple beam with linearly varying load q(x), the boundary conditions are
y(0) = y(l) = 0, y
(0) = y
(l) = 0, from which the integral constants can be
determined c 4 = c 2 = 0, c 1 = −
ql
2
36
, c 3 =
7l
4
360
. Substituting these integral constants
into Eq. (11), the deflection curve expression of the beam can be obtained
y =
q
360l E I
(3x
5
− 10l
2 x
3
+ 7l
4 x)
(12)
Expanding the deflection curve Eq. (12) in the interval [0, l] into Fourier sine
series, its form is
f (x) =
q
E I
∞
n=1
b n sin
nπ x
l
(13)
where, the form of the coefficient b n
b n =
2
l
l
0
1
360l
(3x 5 − 10l 2 x 3 + 7l 4 x) sin
nπ x
l
dx
= −
1
180nπ l
(3x 5 − 10l 2 x 3 + 7l 4 x) cos
nπ x
l
l
0
+
l
0
1
180nπ l
(15x 4 − 30l 2 x 2 + 7l 4 ) cos
nπ x
l
dx
=
1
180n 2 π 2 (15x 4 − 30l 2 x 2 + 7l 4 ) sin
nπ x
l
l
0
−
l
0
1
3n 2 π 2 (x 3 − l 2 x) sin
nπ x
l
dx
=
l
3n 3 π 3 (x 3 − l 2 x) cos
nπ x
l
l
0
−
l
0
l
3n 3 π 3 (3x 2 − l 2 ) cos
nπ x
l
dx
= −
l 2
3n 4 π 4 (3x 2 − l 2 ) sin
nπ x
l
l
0
+
l
0
2l 2
n 4 π 4 x sin
nπ x
l
dx
= −
2l 3
n 5 π 5 x cos
nπ x
l
l
0
+
l
0
2l 3
n 5 π 5 cos
nπ x
l
dx
= −
2l 4
n 5 π 5 cos nπ +
2l 4
n 6 π 6 sin
nπ x
l
l
0
=
(−1) n+1 2l 4
n 5 π 5
(14)
Substituting Eq. (14) into Eq. (13) obtains
y =
q
E I
∞
n=1
(−1)
n+1 2l
4
n 5 π 5
sin
nπ x
l
=
2ql
4
π 5 E I
∞
n=1
(−1)
n+1
n 5
sin
nπ x
l
(15)
Equation (15) is exactly Eq. (9). Quod erat demonstrandum.