574
9 Variational Principles in Mechanics and Their Applications
−
2l 2
n 3 π 3 (x − a) sin
nπ x
l
l
a
+
2l 2
n 3 π 3
l
a
sin
nπ x
l
dx
=
2b 2 (l + 2a)
n 4 π 4
cos
nπ x
l
l
0
−
2l 3
n 4 π 4 cos
nπ x
l
l
a
=
2b 2 (l + 2a)
n 4 π 4
(cos nπ − 1) −
2l 3
n 4 π 4
cos nπ − cos
nπ a
l
=
⎧
⎨
⎩
−
4b 2 (l+2a)
n 4 π 4
+ 2l 3
n 4 π 4
1 − cos nπ a
l
when n takes odd number
− 2l 3
n 4 π 4
1 − cos nπ a
l
when n takes even numder
=
0
w h e n n takes odd number with a = b = l/2
− 2l 3
n 4 π 4
1 − cos nπ a
l
when n takes even numder
(21)
Substituting Eqs. (20) and (21) into Eq. (19), we give
y =
P
E I
a
2 b
2
24l
−
∞
n=1
l
3
8n 4 π 4
1 − cos
2nπ a
l
cos
2nπ x
l
(22)
From y(0) = 0, we get
a
2 b
2
24l
=
∞
n=1
l
3
8n 4 π 4
1 − cos
2nπ a
l
(23)
Substituting Eq. (23) into Eq. (22), we get
y =
Pl
3
8π 4 E I
∞
n=1
1
n 4
1 − cos
2nπ a
l
1 − cos
2nπ x
l
(24)
Equation (24) is exactly Eq. (10). Quod erat demonstrandum.
When the concentrated force acts on the beam at the middle point, there is a =
b = l/2, Eq. (23) can be directly used to determine the deflection of the point. From
Eq. (23), we get
∞
n=1
1 − cos nπ
n 4
=
∞
n=1
2
(2n − 1) 4 =
π
4 a
2 b
2
3l 4 =
π
4
48
(25)
or
∞
n=1
1
(2n − 1) 4 =
∞
n=0
1
(2n + 1) 4 =
π
4
96
(26)