916
Appendix: All Solutions to the Problems
Solve for a n
a n =
2Pl
3
π 4 n 4 E I
sin
nπc
l
(9)
Thus, the deflection equation of the beam can be obtained
y =
∞
n=1
2Pl
3
π 4 n 4 E I
sin
nπc
l
sin
nπx
l
(10)
Now to prove that Eq. (10) is the analytical solution of mechanics of materials.
Proof It follows from the mechanics of materials that for the above problem, the
deflection curve equation of the beam is
y =
Pbx
6l E I
(l
2
− b
2
− x
2
) (0 ≤ x ≤ a)
(11)
y =
Pbx
6l E I
(l
2
− b
2
− x
2
) +
P(x − a)
3
6E I
(a ≤ x ≤ l)
(12)
Expanding them into the Fourier sine series, the form is
y =
P
E I
∞
n=1
b n sin
nπx
l
(13)
where, the form of the coefficients b n is
b n =
2
l
a
0
bx
6l
(l
2
− b
2
− x
2
) sin
nπx
l
dx
+
2
l
l
a
bx
6l
(l
2
− b
2
− x
2
) +
(x − a)
3
6
sin
nπx
l
dx
= −
b
3nπl
(l
2 x − b
2 x − x
3
) cos
nπx
l
a
0
+
b
3nπl
a
0
(l
2
− b
2
− 3x
2
) cos
nπx
l
dx
−
1
3nπ
b
l
(l
2 x − b
2 x − x
3
) + (x − a)
3
cos
nπx
l
l
a
+
1
3nπ
a
0
b
l
(l
2
− b
2
− 3x
2
) + 3(x − a)
2
cos
nπx
l
dx
=
b
3n 2 π 2 (l
2
− b
2
− 3x
2
) sin
nπx
l
a
0
+
2b
n 2 π 2
a
0
x sin
nπx
l
dx
+
1
3n 2 π 2 [b(l
2
− b
2
− 3x
2
) + 3l(x − a)
2
] sin
nπx
l
l
a