Structural Concrete - Hassoun

50 Chapter 2 Properties of Reinforced Concrete
(Since the type of member is not defined)
[ ( )]
τ sh = 190.8(t c ) −0.08 (f cm28 ) −0.25 V 2
2k s
S
= 190.8(8) −0.08 (6556) −0.25 [2(1)(1.5)] 2
= 161.58 days
E cm607 =(1.167) 1 ∕ 2
E cm28
=(1.167) 1 ∕ 2
(4,615,240) =4,985,741 psi
(
) 1∕ 2 (
)
E cm(tc +τ sh ) = t c + τ
1∕ 2
sh
4 + 0.85 ( 8 + 161.58
) E cm28 = (4,615,240)
t c + τ sh
4 + 0.85(8 + 161.58
= 4,937,886 psi
ε shu =−ε su
E cm607
E cm(tc +τ sh )
= −(−798 × 10 −6 ) 4,985,741
4,937,886 = 806 × 10−6 in.∕in.
Determination of K h :
According to the Table 2.10, for H = 75%
( ) H 3 ( ) 75 3
K h = 1 − = 1 − = 0.578
100 100
Determination of S(t):
S(t) =tanh
√
t − tc
T sh
= tanh
√
35 − 8
161.58 = 0.387
ε s (t) =(ε shu )(K h )S(t) =(806 × 10 −6 )(0.578)(0.387) =180 × 10 −6 in.∕in.
Creep Calculation
Determination of q 1 :
Calculation of C 0 (t, t 0 ):
c =
w
w∕c = 345 = 750 lb∕yd3
0.46
J(t, t 0 )=q 1 + C 0 (t, t 0 )+C d (t, t 0 , t c )
q 1 = 0.6 0.6
=
E cm28
4,615,240 = 1.3 × 10−7 psi −1
q 2 = 86.8(c) 0.5 (f cm28
) −0.9 × 10 −6 = 86.8(750) 0.5 (6556) −0.9 × 10 −6
Q f (t 0 )=
= 0.873 × 10 −6
1
0.086(t 0 ) 2∕9 + 1.21(t 0 ) = 1
= 0.182
4∕9 0.086(28) 2∕9 + 1.21(28) 4∕9
Z(t, t 0 )= ln[1 +(t − t 0 )0.1 ]
√
t0
= ln[1 +(35 − 28)0.1 ]
√
28
= 0.150
r(t 0 )=1.7(t 0 ) 0.12 + 8 = 1.7(28) 0.12 + 8 = 10.54