Appendix: All Solutions to the Problems
917
+
2
n 2 π 2
l
a
[bx − l(x − a)] sin
nπx
l
dx
= −
2bl
n 3 π 3 x cos
nπx
l
a
0
+
2bl
n 3 π 3
a
0
cos
nπx
l
dx
−
2l
n 3 π 3 [bx + l(x − a)] cos
nπx
l
l
a
+
2(b − l)l
n 3 π 3
l
a
cos
nπx
l
dx
=
2bl
2
n 4 π 4 sin
nπx
l
a
0
+
2(b − l)l
2
n 4 π 4 sin
nπx
l
l
a
=
2l
3
n 4 π 4 sin
nπa
l
(14)
Substituting Eq. (14) into Eq. (13), we obtain
y =
2Pl
3
π 4 E I
∞
n=1
1
n 4 sin
nπc
l
sin
nπx
l
(15)
Equation (15) is just Eq. (10). Quod erat demonstrandum.
9.7 Find the deflection curve of the clamped-end beam carrying the uniform load q,
as shown in problem 9.7 figure, the length of the beam is l, the flexural rigidity of
the beam is E I .
Solution Let the deflection curve equation of the beam have the form of trigonometric series
y =
∞
n=1
a n
1 − cos
2nπx
l
(1)
The potential energy produced by load q is
V = −
l
0
qydx = −
l
0
q
∞
n=1
a n
1 − cos
2nπx
l
dx
(2)
The bending strain energy of the beam is
U =
1
2
E I
l
0
y 2 (x)dx =
1
2
E I
l
0
∞
n=1
4n 2 π 2
l 2 a n cos
2nπ
l
x
2
dx =
4π 4 E I
l 3
∞
n=1
n 4 a 2
n
(3)
Equation (4) uses the formula
l
0
cos
2mπ
l
x cos
2nπ
l
xdx =
l
2
δ mn
(4)