Structural Concrete - Hassoun

772 Chapter 19 Introduction to Prestressed Concrete
3. The alternative approach presented by the ACI Code is that V c may be taken as the smaller value
of V ci and V cw .
a. Based on the flexural-shear cracking strength,
V ci =
(
0.6λ √ f ′ c
)
b w d p +
Calculate each item separately:
(0.6λ √ √
5000
f c ′ )b w d p = 0.6 × 1 × × 6 × 20.7 = 5.3K
1000
(
V d + V i M )
cre
M max
V d = unfactored dead load shear = 1.288(24 − 1.67) =28.8K
(Eq. 19.51)
M max = maximum factored moment at section (except for weight of beam)
Factored load = 1.2 × 0.9 + 1.6 × 1.3 = 3.13 K∕ft
]
M max = 3.13
[24 × 1.67 − (1.67)2 = 121 K∕ft = 1453 K ⋅ in.
2
Therefore,
V i = 3.13(24 − 1.67) =69.9K
M cre = I y t
(6λ √ f ′ c + f pe − f d )
I = 66, 862 in. 4 y t = 19.2in.
f pe = compressive stress due to prestressing force
= F A + Fey b
I
= 306.2
372 + 306.2(1.5)(20.8) = 0.966 ksi
66, 862
f d = dead load stress = M D y b
I
]
M D =(1.288)
[24 × 1.67 − (1.67)2 = 49.8K⋅ ft = 598 K ⋅ in.
2
f d =
M cre =
598 × 20.8
= 0.186 ksi
66, 862
66, 862 √
19.2 [6(1.0) 5000 + 966 − 186] =4193 K ⋅ in.
V ci = 5.3 + 28.8 + 69.9
( ) 4193
= 235.8K
1453
and V ci must not be less than (1.7λ √ f ′ c )b w d p =(1.7 × 1 × √ 5000)×6 × 20.7 = 14.9K
b. Shear strength based on web-shear cracking is
V cw =(3.5λ √ f ′ c + 0.3 f pc )b w d p + V p
f pc = 306.2
372
= 0.823 ksi
d p = 20.7in. or 0.8h = 0.8 × 40 = 32 in. (Eq. 19.53)