Structural Concrete - Hassoun

2.13 Models for Predicting Shrinkage and Creep of Concrete 51
⎡
Q(t, t 0 )=Q f (t 0 ) ⎢
⎢
1 + Q ( ) −1∕r(t r(t0 )
f t0
⎤
0 )
[
] −1∕10.54
⎥⎥⎦ = 0.182 1 + 0.18210.54
= 0.148
Z(t, t
⎣
0 ) r(t 0 ) 0.150 10.54
( ) w 4
q 3 = 0.29q 2 = 0.29(0.873 × 10 −6 )(0.46) 4 = 0.011 × 10 −6
c
( ) a −0.7
q 4 = 0.14 × 10 −6 = 0.14(3.73) −0.7 × 10 −6 = 5.57 × 10 −8
c
( )
C 0 (t, t 0 )=q 2 Q(t, t 0 )+q 3 ln[1 +(t − t 0 ) 0.1 t
]+q 4 ln
t 0
=(0.873 × 10 −6 )(0.148)+(0.011 × 10 −6 )ln[1 +(35 − 28) 0.1 ]+(5.57 × 10 −8 )ln
= 0.15 × 10 −6 psi −1
Calculation of C d (t,t 0 ,t c ):
q 5 = 0.757|ε shu × 10 6 | −0.6
= 0.757|806 × 10−6 × 10 6 | −0.6
= 2.08 × 10 −6
f cm28
6556
S(t) =0.387
√
√
t
S(t 0 )=tanh 0 − t c 28 − 8
= tanh
T sh
161.58 = 0.338
[(
H(t) =1 − 1 − H ) ] [(
S(t) = 1 − 1 − 75
100
100
[(
H(t 0 )=1 − 1 − H
100
) ] [(
S(t 0 ) = 1 − 1 − 75
100
C d (t, t 0 , t c )=q 5
√
exp[−8H(t)] − exp[−8H(t0 )]
) ]
0.387 = 0.903
) ]
0.338 = 0.916
=(2.08 × 10 −6 ) √ exp[−8 × 0.903]−exp[−8 × 0.916] =0.0176 × 10 −6 psi −1
J(t, t 0 )=q 1 + C 0 (t, t 0 )+C d (t, t 0 , t c )
Creep coefficient
=(1.3 × 10 −7 )+(0.15 × 10 −6 )+(0.0176 × 10 −6 )=0.298 × 10 −6 psi −1
c c (t, t 0 )=J(t, t 0 )E cm28 − 1 =(0.298 × 10 −6 )(4615240)−1 = 0.375
( ) 35
28
Example 2.3 (in.−lb Units)
Using the GL 2000 model, calculate the shrinkage strain and creep compliance and coefficient for the
specimen given in Example 2.1.
Solution
Shrinkage Calculation
ε s (t) =ε shu β(h)β(t − t c )