Structural Concrete - Hassoun

54 Chapter 2 Properties of Reinforced Concrete
5.3
5.3
β(f cm28
)= √ = √ = 2.49
f cm28
∕1450 6556∕1450
β(t 0 )=
1
=
0.1 + t 0.2
0
1
= 0.488
0.2
0.1 + 28
φ 0 = φ RH β(f cm28
)β(t 0 )=(1.598)(2.49)(0.488) =1.94
β H = 37.5h e [1 +(0.012H) 18 ]+250 = 37.5(3)[1 +(0.012 × 75) 18 ]+250
= 379.4 ≤ 1500 days
( ) t − 0.3 t0
(
)
35 − 28 0.3
β c (t, t 0 )=
=
= 0.3
β H + t − t 0 379.4 + 35 − 28
φ 28 (t, t 0 )=φ 0 β c (t, t 0 )=1.94 × 0.3 = 0.582
J(t, t 0 )= 1 + φ(t, t 0 ) 1
=
E cmt0 E cm28
5,156,356 + 0.582
5,156,356 = 0.307 × 10−6 psi −1
Example 2.5 (in.−lb Units)
Use the CEB MC 90–99 model to calculate the shrinkage strain and creep compliance and coefficient
for the specimen given in Example 2.1.
Solution
Shrinkage Calculation
ε s (t, t c )=ε as (t)+ε ds (t, t c )
Calculation of ε as (t):
α as = 600 for rapidly hardening high − strength cements
(
) 2.5
fcm28 ∕1450
ε as0 (f cm28
)=−α as × 10 −6
6 + f cm28
∕1450
( ) 2.5 6556∕1450
=−600
× 10 −6 =−72.64 × 10 −6 in.∕in.
6 + 6556∕1450
β as (t) =1 − exp(−0.2(t) 0.5 )=1 − exp(−0.2(35) 0.5 )=0.694
ε as (t) =ε as0 (f cm28
)β as (t) =(−72.64 × 10 −6 )(0.694) =−50.41 × 10 −6 in.∕in.
Calculation of ε ds (t, t c ):
= 6 for rapidly hardening high − strength cements
α ds1
α ds2
= 0.12 for rapidly hardening high − strength cements
)
ε ds0 (f cm28
)=[
(220 + 110α ds1 exp(−α ds2 f cm28
∕1450)] × 10 −6
=[(220 + 110 × 6)exp(−0.12 × 6556∕1450)] × 10 −6 = 511.5 × 10 −6 in.∕in.