Structural Concrete - Hassoun

718 Chapter 18 Stairs
b. M u = φA s f y (d − a/2); assume a = 0.8 in.
Check:
a =
A s =
11.71 × 12
0.9 × 60(7 − 0.4) = 0.4in.2
A sf y
0.85f c ′ b = 0.4 × 60
= 0.78in. c = 0.92in.
0.85 × 3 × 12
Minimum A s = 0.0033 × 12 × 8 = 0.32in. 2 < 0.4in. 2
Use no. 4 bars spaced at 6 in. (A s = 0.4 in. 2 ). For 5-ft-wide stairs, use 10 no. 4 bars.
d t = 7in. c = 0.92in.
Net tensile strain,
ε t = d t − c = 0.0198in.
c
ε t > 0.005 φ = 0.9
c. Transverse reinforcement must be provided to account for shrinkage.
A s = 0.0018 × 12 × 8 = 0.18in. 2 ∕ft
Use no. 4 bars spaced at 12 in. (A s = 0.2 in. 2 ).
d. If the slab will be cast monolithically with its supporting beams, additional reinforcement must
be provided at the top of the upper and lower landings. Details of stair reinforcement are shown
in Fig. 18.19.
4. Minimum slab thickness for deflection is
L 17.2 × 12
= = 8.26in.
25 25
(for a simply supported slab). In the case presented here, where the slab ends are cast with the
supporting beams and additional negative reinforcement is provided, minimum thickness can be
assumedtobe
L
= 7.4in.