Structural Concrete - Hassoun

584 Chapter 16 Continuous Beams and Frames
2. If a rotation θ occurs at the plastic hinge at the fixed end, A, the rotation at the sagging hinge is
C = 2θ. The deflection of C under the load is (L/2)θ (Fig. 16.25).
W e = external work = ∑ ( ) Lθ
P u Δ = P u
2
W i = internal work = ∑ M u θ = M u1
(θ)+M u2
(2θ)
If the two sections at A and C have the same dimensions and reinforcement, then M u1
= M u2
= M u ,
and W i = 3M u θ. Equating W e and W i ,
L
M u1
+ 2M u2
= P u
2 = 3M u and M u = P u L
6
Example 16.4
Calculate the collapse moments at the critical sections for the beam shown in Fig. 16.26 due to a uniform
load w u .
Solution
1. The number of plastic hinges is two.
2. For a deflection at C = 1.0, the rotation at A, θ A ,is1/a; θ B = 1/b; and
θ c = θ A + θ B = 1 a + 1 b = a + b
ab
= L ab
3. External work is
Internal work is
Equating W e and W i ,
If both moments are equal, then
W e = ∑ ( ) 1 × L
w u Δ = w u = w u L
2 2
W i = ∑ M u θ = M u1
θ A + M u2
θ c
( ) ( 1 1
= M u1
+ M
a
u2
a + 1 )
b
w u = 2 L
w u = 2M u
L
(
Mu1
a + M u 2
a + M )
u 2
L − a
[ 2
a + 1 ]
= 2M u
L − a L
[ ] (2L − a)
a(L − a)
(16.9)
(16.10)
4. To determine the position of the plastic hinge at C that produces the minimum value of the collapse
load w u , differentiate Eq. 16.9 with respect to a and equate to 0:
If M u1
= M u2
= M u ,then
δw u
δa = 0
− (
Mu1
a 2 + M u 2
a 2 − M u 2
(L − a) 2 )
= 0
2
a 2 = 1
(L − a) 2 or a = L(2 −
√
2) =0.586L