Civil Engineering Reference
In-Depth Information
simultaneously by an increasing load. This second type of problem will be treated in Sections
5.4.5-5.4.7.
5.4.3.1 Characteristics of Equations
An efficient algorithm was discovered from a careful observation of the three characteristics
of the twelve equations:
1. Equations
3
and
6
are independent, because each contains one variable,
τ
t
and
γ
t
,
respectively, which are not involved in any other equations.
2. Equations 7 and 8 for the concrete material law involve only three unknown variables in
the
d-r
coordinate (
σ
d
,
ε
d
, and
ε
r
). If the strains
ε
d
and
ε
r
are given, then the stresses
σ
d
can be calculated from these two equations.
3. The longitudinal steel stresses
f
and
f
p
in equilibrium Equation 1 are coupled to the
longitudinal steel strain
ε
in compatibility Equation 4 through the longitudinal steel
stress-strain relationship of Equations 9 and 11 . Similarly, the transverse steel stresses
f
t
and
f
tp
in equilibrium Equation 2 are coupled to the transverse steel strain
ε
t
in com-
patibility Equation 5 through the transverse steel stress-strain relationships of Equation
10
and
12
.
5.4.3.2
ε
as a Function of
f
and
f
p
In view of characteristic 3, the longitudinal steel strain
ε
can be expressed directly as a function
of the longitudinal steel stresses
f
and
f
p
by eliminating the angle
α
r
from Equations 1 and
ε
r
cos
2
α
r
=
ε
r
−
ε
r
sin
2
4 . Inserting
α
r
into Equation 4 gives
ε
r
−
ε
ε
r
−
ε
d
sin
2
α
r
=
(5.109)
Substituting Equation (5.109) into Equation 1 results in
σ
−
ρ
f
−
ρ
p
f
p
ε
=
ε
r
+
ε
r
−
ε
d
−
σ
d
(5.110) or
13
ε
,
f
, and
f
p
in Equation
13
can be solved simultaneously with the two stress-strain
relationships of Equations
9
and
11
for longitudinal steel.
5.4.3.3
ε
t
as a Function of
f
t
and
f
tp
Similarly, the transverse steel strain
ε
t
can be expressed directly as a function of the transverse
steel stresses
f
t
and
f
tp
by eliminating the angle
α
r
from Equations
2
and
5 . Inserting
ε
r
sin
2
α
r
=
ε
r
−
ε
r
cos
2
α
r
into Equation 5 gives
ε
r
−
ε
t
ε
r
−
ε
d
cos
2
α
r
=
(5.111)
