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Structural Concrete - Hassoun

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48 Chapter 2 Properties of Reinforced <strong>Concrete</strong><br />

The modulus of elasticity at 28 days should be calculated as follows:<br />

where<br />

{ √<br />

33000ω<br />

1.5<br />

c f<br />

′<br />

c<br />

E c =<br />

√<br />

0.043ω 1.5<br />

c f<br />

′<br />

c<br />

(in. − lb)<br />

(SI)<br />

ω c = concrete unit weight (kip/ft 3 )orkg/m 3<br />

f c ′ = specified concrete compressive strength at 28 days (ksi or MPa)<br />

(2.120)<br />

Example 2.1 (in.−lb Units)<br />

Calculate the shrinkage strain and creep compliance and coefficient for the concrete specimen given<br />

below. Use the ACI 209R-92 model.<br />

Given factors:<br />

Humidity = 75%<br />

h e = 2V/S = 2A c /u = 3in.<br />

f cm28<br />

= 6556psi<br />

w = 345 lb/yd 3<br />

w/c = 0.46<br />

a/c = 3.73<br />

t = 35 days<br />

t 0 = 28 days<br />

t c = 8days<br />

γ = 146 lb/ft 3<br />

Cement type III<br />

Moist-cured concrete<br />

Solution<br />

Shrinkage Calculation<br />

ε sh (t, t c )=<br />

t − t c<br />

f +(t − t c ) K ss K sh ε shu<br />

ε shu = 780 × 10 −6 in.∕in.<br />

According to Table 2.4, f = 35<br />

V<br />

S = 1.5in.<br />

( ) V<br />

K ss = 1.23 − 0.152 = 1.23 − 0.152(1.5) =1.002<br />

S<br />

For H = 75%,<br />

K sh = 1.40 − 0.01H = 1.40 − 0.01(75) =0.65<br />

t − t<br />

ε sh (t, t c )= c<br />

f +(t − t c ) K ss K sh ε shu<br />

35 − 8<br />

=<br />

35 +(35 − 8) (1.002)(0.65)(780 × 10−6 )= 222.3 × 10 −6 in.∕in.

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