## Info

(7.51)

Ps 2a e [-(l + ß)^ + 5 + ß5 ' ]

When w _ wk, after some calculations we deduce where p=cs/(ktfctm)

setting wok

EsWk

Combining the above equations it results av

3.4•c•S2 +0.34ae •^•X[ö+ßö'-(l+ß)S] S 2S3

which numerically solved, gives the neutral axis position from which the reinforcement tension and its amount can be determined. If it is not the case, it is necessary to set in the A = 2.5 (1-5) and then reevaluating being the value A =0.5 practically impossible for bending problems. The procedure, aimed to the determination of the reinforcement amount and its tension corresponding to fixed crack width values and stress level, requires to assume the value of the bars diameter f

Alternatively, it is possible to set the tensional level as, for example equal to the limit one, and to evaluate the corresponding reinforcement amount ps and the maximal bar diameter. In this case, as the parameter p is defined, the neutral axis is calculated from (7.52), ps from (7.50) and the maximal diameter can be derived solving with respect to which gives imax

7.3.4.7.2 Approximated derivation

The procedure discussed above is quite laborious as it requires iteration. An alternative procedure, easier to be applied, is based on the assumption of a lever arm ho = 0.9d constant and independent from ^.Therefore, asAs0.9d=M and ps=0.185v/(pô)

The general formula for w = wk gives

For further simplification of the problem, assuming 5=0.9, therefore A = 0.243 and by definition v v v * = ■