Structural Concrete - Hassoun

11.19 SI Example 411
3. Compute the nominal balanced load for biaxial bending, P bb :
tanα = M ny
M nx
= P n e x
P n e y
= e x
e y
= 8 12
α = 33.7 ∘
P bx − P by
90 ∘ = ΔP b
90 ∘ − α ∘ or
ΔP b = 25.8 K
676.1 − 634.8
90
P bb = P by + ΔP b = 634.8 + 25.8 = 660.6K
=
ΔP b
90 − 33.7
4. Compute P n from the equation of failure surface:
( )
P n − 660.6
2198.4 − 660.6 + Pn × 12 1.5 ( )
Pn × 8 1.5
+
= 1.0
8973 5557.3
Multiply by 1000 and solve for P n :
(0.65P n − 429.85)+0.0489P 1.5
n + 0.0546P 1.5
n = 1000
0.65P n + 0.1035P 1.5
n = 1429.85
By trial, P n = 487 K. Because P n < P bb , it is a tension failure case for biaxial bending, and
thus P 0 =−2198.4 K (to keep the first term positive).
1000
(
Pn − 660.9
−2198.4 − 660.9
)
+ 0.0489P 1.5
n + 0.0546P 1.5
n = 1000
0.35P n + 0.1035P 1.5
n = 769.1
P n = 429 K and P u = 0.65P n = 278.8 K
Note: The strength capacity, φP n , of the same rectangular section was calculated using the Bresler
reciprocal equation (Example 11.19), Parme method (Example 11.20), and Hsu method (Example 11.21)
to get φP n = 421.5, 455, and 429 K, respectively. The Parme method gave the highest value for this
example.
11.19 SI EXAMPLE
Example 11.22
Determine the balanced compressive forces P b , e b ,andM b for the section shown in Fig. 11.31. Use
f c ′ = 30 MPa and f y = 400 MPa (b = 350 mm, d = 490 mm).
Solution
1. For a balanced condition, the strain in the concrete is 0.003 and the strain in the tension steel is
ε y = f y /E s = 400/200,000 = 0.002, where E s = 200,000 MPa.
A s = A ′ s = 4(700) =2800 mm2
2. Locate the neutral axis depth, c b :
( )
600
c b =
d
600 + f t (where f y is in MPa)
y
(
=
600
600 + 420
)
(490) =288 mm
a b = 0.85c b = 0.85 × 288 = 245 mm