Control of cracking - Concrete Structures: Stresses and Deformations - page 403

598.3 in 2 270.4 in 3 118 500 in 4 . A = B = I =

The decompression forces are (Equations (7.40) and (7.41) ):

N 1 = 213.2 kip M 1 = 835.4 kip-in.

Forces producing cracking, after deducting the decompression
forces;

N 2 = − 213.2 kip; M 2 = 7000 − 835.4 = 6164.6 kip-in.

Stress at the extreme fi bre, ignoring cracking (Equations (2.19) and

(2.17) ), 0.868 ksi.

The tension is resisted by the prestressed and the bottom non-
prestressed steel; the total steel area resisting tension

σ c max =

= 1.74 + 2.33 =

4.07 in 2 and its centroid is at depth d = 43.6 in. The steel ratio ρ = 4.07/

(12 × 43.6) = 0.78 per cent. The value M /( Nd ) = 6164.6/( − 213.2 × 43.6)

= − 0.663. Entering the graph in Fig. 11.6 with the values of ρ and

M /( Nd ) gives: ( σ s2 / σ c max ) = 21.3. Thus,

σ s2 = 21.3(0.868) = 18.5 ksi.

The mean crack width (Equations (11.24) and (11.27) ) is:

16(0.9) 18.5 w m =

29 000

= 0.0092 in (0.23 mm).

A more accurate analysis, using the equation of Chapter 8 gives: c =

20.3 in; σ s2 in the bottom non-prestressed steel = 17.8 ksi; mean crack

width = 0.0088 in (0.22 mm).

Example 11.3 Overhanging slab: reinforcement to control
thermal cracking
Figure 11.10 represents top view and section in a reinforced concrete
slab extending as a cantilever from the

oor of a building. Transverse
cracks can occur in the cantilever due to temperature di

fl

erence
between the outside air and the interior heated building. It is required to
calculate the concrete stress which would occur, ignoring cracking, due

ff