Structural Concrete - Hassoun

368 Chapter 11 Members in Compression and Bending
Note that the value of f s ′ − 0.85f c ′ must be a positive value. If this value is negative, then let
f s ′ − 0.85f c ′ = 0.
6. Substitute a in the equation of step 2 to obtain P n . The moment M n can be calculated:
M n = P n e
7. Check if compression steel yields as assumed. If ε ′ s ≥ ε y , then compression steel yields; otherwise,
f s ′ = E s ε ′ s. Repeat steps 2 through 5. Note that ε ′ s =[(c − d ′ ∕c]0.003,ε y = f y ∕E s ,and
a = β 1 c.
8. Check that tension controls. Tension controls when e > e b or P n < P b . Example 11.3 illustrates
this procedure.
9. The net tensile strain, ε t , in this section, is normally greater than the limit strain of 0.002 for a
compression-controlled section (Fig. 11.4). Consequently, the value of the strength reduction
factor, φ, will vary between 0.65 (or 0.75) and 0.90. Equation 11.1 or 11.2 can be used to
calculate φ.
Example 11.3
Determine the nominal compressive strength, P n , for the section given in Example 11.2 if e = 20 in. (See
Fig. 11.8.)
ε
ε
ε
ε
Solution
Figure 11.8
Example 11.3: Tension failure.
1. Because e = 20 in. is greater than d = 19.5 in., assume that tension failure condition controls (to be
checked later). The strain in the tension steel, ε s , will be greater than ε y and its stress is f y . Assume
that compression steel yields f s ′ = f y , which should be checked later.
2. From the equation of equilibrium (Eq. 11.10),
P n = C c + C s − T
where
C c = 0.85f c ′ ab = 0.85 × 4 × 14a = 47.6a
C s = A ′ s (f y − 0.85f c ′ )=4(60 − 0.85 × 4) =226.4K
T = A s f y = 4 × 60 = 240 K
P n = 47.6a + 226.4 − 240 = 47.6a − 13.6
(I)