Structural Concrete - Hassoun

2.13 Models for Predicting Shrinkage and Creep of Concrete 53
Example 2.4 (in.−lb Units)
Using the CEB 90 model, 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 cs0 )β s (t − t c )
Calculation of ε cs0 :
ε cs0
= ε s (f cm28 )β RH
β sc = 8
ε s (f cm28
)=
=
[
160 + 10 ( ) (
β sc 9 − f cm 28
1450
[
(
160 + 10 (8)
9 − 6556
1450
For H = 75%,
β RH =−1.55β arh
( ) H 3 ( ) 75 3
β arh = 1 − = 1 − = 0.578
100 100
β RH =−1.55β arh =−1.55 × 0.578 =−0.896
ε cs0
Calculation of β s (t − t c ):
)]
× 10 −6
)]
× 10 −6 = 518.3 × 10 −6 in.∕in.
= ε s (f cm28
)(β RH )=(518.3 × 10 −6 )(−0.896) =−464.4 × 10 −6 in.∕in.
h e = 2A c
= 2 × 1.5 = 3.0in.
u
√
√
t − t √√√√
β s (t − t c )=
c
350(h e ∕4) 2 +(t − t c ) = 35 − 8
( ) 2
= 0.347
350 3
4 +(35 − 8)
ε s (t, t c )=(ε cs0 )β s (t − t c )=(−464.2 × 10 −6 )(0.347) =−161 × 10 −6 in.∕in.
Creep Calculation
Calculation of E cmt0 and E cm28
:
J(t, t 0 )= 1
E cmt0
+ φ 28 (t, t 0 )
E cm28
E cm28
Calculation of φ(t, t 0 ):
t 0 = 28 days ⇒ E cmt0 = E cm28
√
√
f
= 3,118,310 3 cm28
6556
1450 = 3,118,310 3 = 5,156,356 psi
1450
φ RH = 1 + 1 − H∕100
0.46 3 √
he ∕4 = 1 + 1 − 75∕100
0.46 3 √
3∕4
= 1.598