Structural Concrete - Hassoun

586 Chapter 16 Continuous Beams and Frames
Example 16.5
For the beam shown in Fig. 16.25, calculate the collapse moments using the equilibrium method.
Solution
Two plastic hinges will develop at A and C. Referring to Fig. 16.25e, the reaction at A is (P u ∕2)+(M u1
L)
and the reaction at B is (P u ∕2)−(M u1
∕L).
Considering the equilibrium of beam BC and taking moments about C,
(
Pu
2 − M )
u (L )
1
= M
L 2
u2
L
M u1
+ 2M u2
= P u
2
which is the same equation obtained in Example 16.3. When M u1
= M u2
= M u ,then
3M u = P u
L
2
or
M u = P u
L
6
Example 16.6
Calculate the collapse moments for the beam shown in Fig. 16.26 by the equilibrium method.
Solution
1. Two plastic hinges will develop in this beam at A and C. Referring to Fig. 16.26d, the reaction
at A = w u (L∕2)+(M u1
∕L) and the reaction at B = w u (L∕2)−(M u1
∕L). The load on BC is w u b
acting at b/2 from B, andb = (L − a). Considering the equilibrium of segment BC and taking
moments about C, (
L
w u
2 − M )
u 1
b −(w
L
u b) b 2 = M u 2
If M u1
= M u2
= M u ,then
(
b
w u
2 (L − b) =M u 1 + b )
= M u
(2L − a)
L L
w u = 2M u
L
× 2L − a
a(L − a)
which is similar to the results obtained in Example 16.4.
M u = w u L a(L − a)
×
2 2L − a
2. The position of a can be determined as before, where a = 0.586L, M u = 0.0858 w u L 2 ,and
w u = 11.66(M u /L 2 ).
16.12 ROTATION OF PLASTIC HINGES
16.12.1 Plastic Hinge Length
The assumption that the inelastic rotation of concrete occurs at the point of maximum moment
while other portions of the member act elastically is a theoretical one; in fact, the plastic rotation