Appendix: All Solutions to the Problems
919
y =
q
E I
a 0
2
+
∞
n=1
a n cos
nπx
l
(13)
where, the form of the coefficients a 0 and a n is
a 0 =
1
12l
l
0
(x
4
− 2lx
3
+ l
2 x
2
)dx =
l
4
360
(14)
a n =
2
l
l
0
f (x) cos
nπx
l
dx =
1
12l
l
0
(x
4
− 2lx
3
+ l
2 x
2
) cos
nπx
l
dx
=
1
12nπ
(x
4
− 2lx
3
+ l
2 x
2
) sin
nπx
l
l
0
−
1
12nπ
l
0
(4x
3
− 6lx
2
+ 2l
2 x) sin
nπx
l
dx
=
l
6n 2 π 2 (2x
3
− 3lx
2
+ l
2 x) cos
nπx
l
l
0
−
l
6n 2 π 2
l
0
(6x
2
− 6lx + l
2
) cos
nπx
l
dx
= −
l
2
6n 3 π 3 (6x
2
− 6lx + l
2
) sin
nπx
l
l
0
+
l
2
n 3 π 3
l
0
(2x − l) sin
nπx
l
dx
= −
l
3
n 4 π 4 (2x − l) cos
nπx
l
l
0
+
2l
3
n 4 π 4
l
0
cos
nπx
l
dx
= −
l
4
π 4 n 4 (cos nπ + 1) =
−2l
4
π 4 n 4 when n is the even number
0 when n is the odd number
(15)
Substituting Eqs. (14) and (15) into Eq. (12), and taking note that n only takes the
even function, we get
y =
q
E I
l
4
720
−
∞
n=1
l
4
8n 4 π 4 cos
2nπx
l
(16)
From the boundary condition y(0) = 0, we get
l
4
720
=
∞
n=1
l
4
8n 4 π 4 , namely
∞
n=1
1
n 4 =
π
4
90
, substituting it into Eq. (16), we get
y =
ql
4
8π 4 E I
∞
n=1
1
n 4
1 − cos
2nπx
l
(17)
It follows that Eq. (17) is the very analytical solutions of mechanics of materials.
Quod erat demonstrandum.
Using Eq. (25) of Example 9.6.12, the deflection of the beam at the middle is