Civil Engineering Reference
In-Depth Information
Check equilibrium equations 20 and 21 :
(
ρ
f
+
ρ
t
f
t
)
1
=
0
.
0170(212
.
3)
+
0
.
0019(465
.
5)
=
3
.
609
+
0
.
884
=
4
.
493 MPa
(
ρ
f
−
ρ
t
f
t
)
1
=
0
.
0170(212
.
3)
−
0
.
0019(465
.
5)
=
3
.
609
−
0
.
884
=
2
.
725 MPa
−
σ
2
1
Equation 20
(
ρ
f
+
ρ
t
f
t
)
2
=
(
σ
+
σ
t
)
+
σ
=
(0
+
0)
−
(0
.
2285
−
4
.
725)
=
4
.
496 MPa
≈
4
.
493 MPa OK
−
σ
2
cos 2
1
12
sin 2
Equation 21
(
ρ
f
−
ρ
t
f
t
)
2
=
(
σ
−
σ
t
)
−
σ
α
1
+
2
τ
α
1
=
(0
−
0)
−
(0
.
2285
+
4
.
725) (0)
+
2(1
.
361)(1)
=
2
.
722 MPa
≈
2
.
725 MPa OK
1
2
)sin
12
(cos
2
sin
2
Equation 3
τ
t
=
(
σ
−
σ
α
1
cos
α
1
+
τ
α
1
−
α
1
)
=
(0
.
2285
+
4
.
725)(0
.
5)
+
1
.
361(0
.
5
−
0
.
5)
=
2
.
477 MPa
γ
t
2
=
α
1
+
γ
12
2
(cos
2
sin
2
ε
1
−
ε
2
)sin
α
1
cos
α
1
−
α
1
)
Equation 6
(
0
.
01672
2
=
(0
.
027416
+
0
.
00300)(0
.
5)
+
(0
.
5
−
0
.
5)
=
0
.
015208
.
γ
t
=
2(0
.
015208)
=
0
.
030416
.
6.1.12.3 Characteristics of Panels with Large
ρ
/ρ
t
Ratios
Figure 6.24 shows seven behavioral curves for panel M3, including: (a) applied shear stress
versus shear strain relationship; (b) concrete compressive stress-strain relationship; (c) con-
crete shear stress-strain relationship; (d) steel stress versus biaxial strain in the longitudinal
direction; (e) steel stress versus uniaxial strain in the longitudinal direction; (f) steel stress
versus biaxial strain in the transverse direction; and (g) steel stress versus uniaxial strain in
the transverse direction.
Examination of Figure 6.24 reveals five characteristics of panel M3 because of its large
ρ
/ρ
t
ratio of 9.0:
1. In a panel with transverse steel (
ρ
t
=
0
.
19%) much less than the longitudinal steel (
ρ
=
70%), the first yield point in the shear stress versus shear strain curve, designated by point
1, is due to the yielding of the transverse steel as shown in Figure 6.24(f) and (g). The steel
stresses in the longitudinal direction were much less than the yield strength as shown in
Figure 6.24(d) and (e).
2. In Table 6.1, the Hsu/Zhu ratio
1
.
ν
12
at first yielding (point 1) is 1.52, much less than
1.9. This is because the transverse steel ratio is very small (
ρ
t
=
0
.
0019), resulting in
ε
y
=
a large
B
value of 0.129 and a small smeared yield strain (
0
.
001554). Substituting
ν
12
value of 1.52.
3. In panel M3, the peak point 2 of the applied shear stress (Figure 6.24a) is accompanied by
the peaks of concrete properties in Figure 6.24 (b) and (c), indicating that the compressive
ε
sf
=
ε
t
=
0
.
001554 into Equation
7
a
gives a
