Structural Concrete - Hassoun

11.18 Equation of Failure Surface 409
in Fig. 11.26. The axial load can be compressive or a tensile force. The equation is presented as
follows:
( ) ( )
Pn − P 1.5 ( ) 1.5
b Mnx Mny
+ + = 1.0 (11.43)
P 0 − P b M bx M by
where
P n = nominal axial strength (positive if compression and negative if tension) for given
eccentricity
P 0 = nominal axial load (positive if compression and negative if tension) at zero eccentricity
P b = nominal axial compressive load at balanced strain condition
M nx , M ny = nominal bending moments about x and y axes, respectively
M bx , M by = nominal balanced bending moments about x and y axes, respectively, at balanced strain
conditions
To use Eq. 11.4, all terms must have a positive sign. The value of P 0 was given earlier
(Eq. 10.1):
P 0 = 0.85f ′ c(A g − A st )+A st f y (11.44)
The nominal balanced load, P b , and the nominal balanced moment, M b = P b e b , were given in
Eqs. 11.6 and Eqs. 11.7, respectively, for sections with tension and compression reinforcement
only. For other sections, these values can be obtained by using the principles of statics.
Note that the equation of failure surface can also be used for uniaxial bending representing
the interaction diagram. In this case, the third term will be omitted when e x = 0, and the second
term will be omitted when e y = 0.
When e x = 0 (moment about the x-axis only),
( )
Pn − P b
+
P 0 − P b
(
Mnx
M bx
) 1.5
= 1.0 (11.45)
(This is Eq. 11.21, given earlier.) When e y = 0 (moment about the y axis only),
( ) ( ) 1.5 Pn − P Mny
b
+ = 1.0 (11.46)
P 0 − P b M by
Applying Eq. 11.4 to 11.2 and 11.4, P b = 453.4 K, M bx = 6810.8 K.in., e y = 10 in., and P 0 =
0.85(4)(14 × 22 − 8) + 8(60) = 1500 K.
( )
P n − 453.4
1.5
1500 − 453.4 + 10Pn
= 1.0
6810.8
Multiply by 1000 and solve for P n :
(0.9555P n − 433.2)+0.05626P 1.5
n = 1000
0.9555P n + 0.05626P 1.5
n = 1433.2
Let P n = 611 K, which is close to that obtained by analysis.