Structural Concrete - Hassoun

Summary 183
There are two cases:
Case 1 When d and A s (or ρ) arenot given:
1. Let a ≤ t (as singly reinforced rectangular section). If a = t is assumed, then
(
M u =(totalflange) =φ(0.85f c)bt
′ d − t ) (
= φA
2 s f y d − t )
2
Solve for d and then for A s :
d =
2. If a is assumed to be less than t, then
M u
φ(0.85f ′ c)bt + t 2 A s =
M u
φf y (d − t∕2)
d =
M u
φ(0.85f ′ c)ba + a 2
and A s =
M u
φf y (d − a∕2)
Case 2 When d is given and A s is required (one unknown):
1. Check if ‘a’ is greater or less than t by considering the moment capacity of the flange (bt).
(
(flange)φM n = φ(0.85f c)bt
′ d − t )
2
If φ M n > M u (external), then a < t (rectangular section).
If φ M n < M u (external), then a > t (T-section).
2. If a < t, calculate R n = M u / φ bd 2 and then calculate ρ (or determine ρ from tables or Eq. 4.2):
( √ )
ρ = 0.85f c
′ 1 − 1 − 2R n
f y 0.85f c
′
Then calculate A s = ρbd.
3. If a > t:
a. Calculate C f and A sf :
A sf = 0.85f ′ ct b − b w
f y
= C f
f y
(flange)
Then calculate M uf (flange) = φC f (d − t/2).
b. Calculate M uw (web) = M u − M uf . Calculate R n ;(web)= M uw /(φb w d 2 ); then find ρ w (use
the equation or tables). Calculate A sw (web) = ρ w b w d.
c. Total A s = A sf (flange) + A sw (web). Total A s must be less than or equal to A s,max and greater
than or equal to A s,min .
d.
( 0.8f
′
) ( √ )
ρ w = c
1 − 1 − 2R n
f y
0.85f c
′
e. Check that ρ w = A s /b w d ≥ ρ min (ρ w = steel ratio in web) or A s > A s,min ,whereA s,min = ρ min
(b w d). Check that A s ≤ max A s , or check ε t = (d t –c)/c ≥0.005.