Structural Concrete - Hassoun

11.14 Biaxial Bending 397
5 no. 10
5 no. 10
Figure 11.22 Example 11.16.
4. Check the chosen section by statics similar to Example 11.3.
a. Determine the value of a using the general equation Aa 2 + Ba + C = 0 with e ′ = e + d −
h/2 = 38.5 in., A = 0.425 f c ′ b = 27.2, B = 2A(e′ − d) =1033.6, C = A ′ s (f y − 0.85f c ′)(e′ − d +
d ′ )−A s f y e ′ =−6941.2. Solve to get a = 5.82 in. and c = a/0.85 = 6.85.
b. Check f s ′:
( )
f s ′ c − d = 87 ′ ( ) 6.85 − 2.5
= 87
= 55.26 ksi
c
6.85
Let f s ′ = 57 ksi.
c. Recalculate a:
C = A ′ s (f ′
s − 0.85f ′ c )(e′ − d + d ′ )−A s f y e ′ =−7351
Solve now for a to get a = 6.13 and c = 7.21 in.
d. Check f s ′:
( )
f s ′ c − 2.5
= 87 = 56.83 ksi
c
Calculate
C c = 0.85(4)(6.13)(16) =333.5K C s = A ′ s (f s ′ − 0.85f c ′ )=6.35(57 − 0.85 × 4)
= 340.4 K T = A s f y = 6.35(60) =381 K
e. Let P n = C c + C s − T = 292.9 K.
5. Determine φ: ε t = [(d t − c)/c] 0.003 = 0.00511. Because ε t = 0.00511 > 0.005, φ = 0.9.
6. Because φP n = 0.9(292.9) = 263.6 K > 257 K, the section is adequate.
11.14 BIAXIAL BENDING
The analysis and design of columns under eccentric loading was discussed earlier in this chapter,
considering a uniaxial case. This means that the load P n was acting along the y-axis (Fig. 11.23),
causing a combination of axial load P n and a moment about the x-axis equal to M nx = P n e y or