Appendix: All Solutions to the Problems
915
9.6 As shown in problem 9.6 figure, let the beam of length l be simply supported on
both ends, the flexural rigidity of the beam is E I , a concentrated load P acts on the
beam at a certain point C. Find the deflection of the beam.
Solution Let the deflection of the beam be
y =
∞
n=1
a n sin
nπx
l
(1)
The deflection of point C is
y =
∞
n=1
a n sin
nπc
l
(2)
The potential energy produced by the load P is
V = −Py(c) = −P
∞
n=1
a n sin
nπc
l
(3)
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
n
2
π
2
l 2 a n sin
nπ
l
x
2
dx =
π
4 E I
4l 3
∞
n=1
n
4 a
2
n
(4)
Equation (4) uses the formula
l
0
sin
mπ
l
x sin
nπ
l
xdx =
l
2
δ mn
(5)
The total potential energy of the beam is
π = V + U = −P
∞
n=1
a n sin
nπc
l
+
π
4 E I
4l 3
∞
n=1
n
4 a
2
n
(6)
When the beam is in a state of stable equilibrium, Π gets the minima, there is
∂Π
∂a n
= 0 (n = 1, 2, . . .)
(7)
or
−P sin
nπc
l
+
π
4 E I
2l 3 n
4 a n = 0
( 8 )
915
9.6 As shown in problem 9.6 figure, let the beam of length l be simply supported on
both ends, the flexural rigidity of the beam is E I , a concentrated load P acts on the
beam at a certain point C. Find the deflection of the beam.
Solution Let the deflection of the beam be
y =
∞
n=1
a n sin
nπx
l
(1)
The deflection of point C is
y =
∞
n=1
a n sin
nπc
l
(2)
The potential energy produced by the load P is
V = −Py(c) = −P
∞
n=1
a n sin
nπc
l
(3)
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
n
2
π
2
l 2 a n sin
nπ
l
x
2
dx =
π
4 E I
4l 3
∞
n=1
n
4 a
2
n
(4)
Equation (4) uses the formula
l
0
sin
mπ
l
x sin
nπ
l
xdx =
l
2
δ mn
(5)
The total potential energy of the beam is
π = V + U = −P
∞
n=1
a n sin
nπc
l
+
π
4 E I
4l 3
∞
n=1
n
4 a
2
n
(6)
When the beam is in a state of stable equilibrium, Π gets the minima, there is
∂Π
∂a n
= 0 (n = 1, 2, . . .)
(7)
or
−P sin
nπc
l
+
π
4 E I
2l 3 n
4 a n = 0
( 8 )
