9.6 The Hamilton Principles and Their Applications
573
E I y =
Pb
2
6l 3 [3alx
2
− (l + 2a)x
3
] +
P
6
(x − a)
3
+ c 3 x + c 4 (a ≤ x ≤ l) (16)
The four integral constants appearing in the above equation can be obtained by
the following conditions: y(0) = y(l) = 0, y
(0) = 0, at x = a, the deflections of
two parts of beam are equal, we get c 2 = 0, c 1 = c 3 = 0, c 4 = 0. Substituting these
values into Eqs. (15) and (16), the equations of the deflection curve can be obtained
E I y =
Pb
2
6l 3 [3alx
2
− (l + 2a)x
3
] (0 ≤ x ≤ a)
(17)
E I y =
Pb
2
6l 3 [3alx
2
− (l + 2a)x
3
] +
P
6
(x − a)
3
(a ≤ x ≤ l)
(18)
Expanding the equations of the deflection curve in the interval [0, l] into Fourier
cosine series, its form is
y =
P
E I
a 0
2
+
∞
n=1
a n cos
nπ x
l
(19)
where, the forms of the coefficients a 0 and a n are
a 0 =
2
l
a
0
b 2
6l 3 [3alx 2 − (l + 2a)x 3 ]dx +
2
l
l
a
b 2
6l 3
3alx 2 − (l + 2a)x 3 +
1
6
(x − a) 3
dx
=
2
l
l
0
b 2
6l 3 [3alx 2 − (l + 2a)x 3 ]dx +
2
l
l
a
1
6
(x − a) 3 ]dx
=
2
l
b 2
6l 3 [alx 3 −
1
4
(l + 2a)x 4 ]
l
0
+
b 4
12l
=
a 2 b 2
12l
(20)
a n =
2
l
l
0
b 2
6l 3 [3alx 2 − (l + 2a)x 3 ] cos
nπ x
l
dx +
2
l
l
a
1
6
(x − a) 3 cos
nπ x
l
dx
=
b 2
3l 3 nπ
[3alx 2 − (l + 2a)x 3 ] sin
nπ x
l
l
0
−
b 2
3l 3 nπ
l
0
[6alx − 3(l + 2a)x 2 ] sin
nπ x
l
dx
+
1
3nπ
(x − a) 3 sin
nπ x
l
l
a
−
1
nπ
l
a
(x − a) 2 sin
nπ x
l
dx
=
b 2
l 2 n 2 π 2 [2alx − (l + 2a)x 2 ] cos
nπ x
l
l
0
−
2b 2
l 2 n 2 π 2
l
0
[al − (l + 2a)x] cos
nπ x
l
dx
+
l
n 2 π 2 (x − a) 2 cos
nπ x
l
l
a
−
2l
n 2 π 2
l
a
(x − a) cos
nπ x
l
dx
= −
2b 2
n 3 π 3 l
[al − (l + 2a)x] sin
nπ x
l
l
0
−
2b 2 (l + 2a)
n 3 π 3 l
l
0
sin
nπ x
l
dx