Civil Engineering Reference
In-Depth Information
5.4.7.1 Equal Strain Condition
Suppose that the longitudinal steel and transverse steel have the same stress
f
=
f
t
throughout
the loading history, then the strains in both directions must be equal, i.e.
ε
=
ε
t
(5.137)
Inserting the compatibility equations, (5.41) and (5.42) , into Equation (5.137) we have
ε
r
cos
2
α
r
+
ε
d
sin
2
α
r
=
ε
r
sin
2
α
r
+
ε
d
cos
2
α
r
(5.138)
Regrouping the terms results in
ε
r
−
ε
d
) (cos
2
sin
2
(
α
r
−
α
r
)
=
0
(5.139)
45
◦
, except when
Equation (5.139) is satisfied only if
α
r
=±
ε
r
=
ε
d
, the special case of
uniform tension in biaxial stresses.
Let us now limit our study to non-prestressed elements (
ρ
p
=
ρ
tp
=
0). The three equilibrium
45
◦
:
equations, (5.124)-(5.126), can be further simplified by taking
α
r
=±
σ
d
2
+
ρ
f
m
σ
1
=
(5.140)
σ
d
2
+
ρ
t
f
t
m
t
σ
1
=
(5.141)
m
t
σ
1
=−
σ
d
2
(5.142)
Multiply Equation (5.140) by
m
t
and Equation (5.142) by
m
. Then, we can subtract the
latter from the former to eliminate the variable
σ
1
and obtain
ρ
=
−
σ
d
2
f
m
+
m
t
m
t
(5.143)
Similarly, multiplying Equation (5.141) by
m
t
and Equation (5.142) by
m
t
, and then
subtracting the latter from the former give:
ρ
t
=
−
σ
d
2
f
t
m
t
+
m
t
m
t
(5.144)
Dividing Equation (5.143) by Equation (5.144) and taking
f
=
f
t
result in the equal strain
condition:
ρ
ρ
t
=
m
+
m
t
(5.145)
m
t
+
m
t
The equal strain condition in Eq. (5-145) was first derived by Fialkow (1985). This equa-
tion expresses the ratio of the percentage of longitudinal steel to the percentage of transverse
steel in terms of the three
m
-coefficients for proportional loadings. If a 2-D element is de-
signed according to this equal strain condition, the yielding of the steel is expected to occur
simultaneously in the longitudinal and the transverse steel.
