Control of cracking - Concrete Structures: Stresses and Deformations - page 398

Figure 11.7 Cross-section subjected to shrinkage and a bending moment, analysed
to determine crack width (Example 11.1).

0.3216 m 2 ; B = 0; I = 2.561 × 10 −3 m 4 . The decompression forces are M 1 =

0 and by Equation (7.43):

N 1 = − 0.3216(0.774 × 10 6 ) = − 249 kN.

Forces producing cracking, after deducting the decompression forces
are:

N = 0 − ( − 249) = 249 kN M = 40 kN-m.

Stress at the extreme fi bre, ignoring cracking is:

249
0.3216

40(0.15)
2.561 × 10 −3

σ c max = + = 3.12 MPa.

The steel ratio ρ = 1800/(1000 × 270) = 0.67 per cent; M /( Nd ) = 40/

(249 × 0.27) = 0.60. Entering the graph in Fig. 11.6 with the values of ρ

and M /( Nd ) gives: ( σ s2 / σ c max ) = 51; thus,

σ s2 = 51(3.12) = 160 MPa (23.2 ksi).

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