Civil Engineering Reference
In-Depth Information
Combining Equations (5.5) and (5.6) gives:
σ
r
−
σ
d
)
2
sin
2
α
r
cos
2
(
−
σ
+
ρ
f
+
σ
r
)(
−
σ
t
+
ρ
t
f
t
+
σ
r
)
=
(
α
r
(5.8)
Squaring Equation (5.7) gives
2
t
σ
r
−
σ
d
)
2
sin
2
α
r
cos
2
τ
=
α
r
(
(5.9)
Equating Equations (5.8) and (5.9) and taking square root results in:
(
τ
t
=±
−
σ
+
ρ
f
+
σ
r
)(
−
σ
t
+
ρ
t
f
t
+
σ
r
)
(5.10)
It can be seen that the applied shear stress is simply the square-root-of-the-product average
of the two stress quantities, (
−
σ
t
+
ρ
t
f
t
+
σ
r
).
Dividing Equation (5.5) by Equation (5.6) we have:
−
σ
+
ρ
f
+
σ
r
) and (
α
r
=
−
σ
+
ρ
f
+
σ
r
−
σ
t
+
ρ
t
f
t
+
σ
r
tan
2
(5.11)
Neglecting the tensile strength of concrete by taking
σ
r
=
0, Equations (5.5)-(5.7) are
simplified as follows:
−
σ
d
)sin
2
−
σ
+
ρ
f
=
(
α
r
(5.12)
−
σ
d
) cos
2
−
σ
t
+
ρ
t
f
t
=
(
α
r
(5.13)
τ
t
=
(
−
σ
d
)sin
α
r
cos
α
r
(5.14)
In the case of a reinforced concrete element subjected to pure shear
σ
=
σ
t
=
0. Then
−
σ
d
)sin
2
ρ
f
=
(
α
r
(5.15)
−
σ
d
) cos
2
ρ
t
f
t
=
(
α
r
(5.16)
τ
t
=
(
−
σ
d
)sin
α
r
cos
α
r
(5.17)
Equations (5.15)-(5.17) are identical to Equations (2.15)-(2.17) in Chapter 2. The difference
in the sign of
σ
d
is due to the difference in sign conventions. In Equations (5.15)-(5.17) the
sign convention in Section 4.1.1 requires the compressive stress in the concrete element to be
negative. In Equations (2.15)-(2.17), however, all stresses have been taken as absolute values
without sign.
When a RC 2-D element is subjected to pure shear (
σ
=
σ
t
=
0), and assuming zero tensile
stress (
σ
r
=
0), Figure 5.2(a) and (b) become the same as Figure 2.3(a) and (b), respectively.
5.1.3 Second Type of Equilibrium Equations
The three equilibrium equations, (5.1)-(5.3), can also be expressed in another form. Substi-
tuting (
σ
r
−
σ
d
) from Equation (5.7) into Equations (5.5) and (5.6) gives
(
−
σ
+
ρ
f
+
σ
r
)
=
τ
t
tan
α
r
(5.18)
(
−
σ
t
+
ρ
t
f
t
+
σ
r
)
=
τ
t
cot
α
r
(5.19)
