Structural Concrete - Hassoun

568 Chapter 16 Continuous Beams and Frames
Figure 16.13
Structures with ties.
to compressive stresses that must not exceed one-third the yield strength of the steel bars f y under
service loads or 0.55f y under factored loads. The low stress is assumed because any rotation at the
hinge tends to bend the bars and induces secondary flexural stresses. It is generally satisfactory to
keep the compression stresses low rather than to compute secondary stresses. The areas of bars A
and B are calculated as follows:
Area of bars A∶ A s1
= R 1
0.55f y
(16.1)
Area of bars B∶ A s2
= R 2
(16.2)
0.55f y
where R 1 and R 2 are the components of the resultant R in the direction of the inclined bars A and B
using factored loads. The components R 1 and R 2 are usually obtained by statics as follows:
H + R 2 sin θ = R 1 sin θ and R 2 = R 1 − H
(16.3a)
sin θ
Also, (R 1 + R 2 )cos θ = P u ,so
R 1 =
P u
cos θ − R 2 =
P [
u
cos θ − R 1 −
H ]
sin θ
R 1 = 1 [
Pu
2 cos θ + H
]
(16.3b)
sin θ
The inclined hinge bars transmit their force through the bond along the embedded lengths
in the frame columns and footings. Consequently, the bars exert a bursting force, which must be
resisted by ties. The ties should extend a distance a = 8D (the larger bar diameter of bars A and B)
in both columns and footings. The bursting force F can be estimated as
F = P u Ha
tan θ + (16.4)
2 0.85d