10.2 Continuous Beam
397
λ
4
= ρ
Ap
2
E I
Solution of Eqs. (10.6), (10.7) and (10.8) is
Y 1 = C 1 sin λx 1 + C 2 cos λx 1 + C 3 sinh λx 1 + C 4 cosh λx 1
(10.9)
and
Y 2 = D 1 sin λx 2 + D 2 cos λx 2 + D 3 sinh λx 2 + D 4 cosh λx 2
(10.10)
Eight constants associated with Eqs. (10.9) and (10.10) are to be evaluated, and
they can be done on the basis of boundary conditions given by Eq. (10.5).
Applying the end conditions at two extreme supports, that is, at x 1 = x 2 = 0, we
obtain
C 2 = C 4 = D 2 = D 4 = 0
(10.11)
Applying the condition at x 1 = L 1 , y 1 = 0 and at x 2 = L 2 , y 2 = 0, we obtain
C 1 sin λL 1 = −C 3 sinh λL 1 = C
(10.12)
and
D 1 sin λL 2 = −D 3 sinh λL 2 = D
(10.13)
Based on the constants evaluated as given by Eqs. (10.11)–(10.13), Eqs. (10.9)
and (10.10) may be written as
Y 1 = C
sin λx 1
sin λL 1
−
sinh λx 1
sinh λL 1
(10.14)
and
Y 2 = D
sin λx 2
sin λL 2
−
sinh λx 2
sinh λL 2
(10.15)
Using the condition given by Eq. (10.2a), we get
C
sin λL 1
sin λL 1
+
sinh λL 1
sinh λL 1
= D
sin λL 2
sin λL 2
+
sinh λL 2
sinh λL 2
or
C = D
(10.16)