The length of all longitudinal bars is computed along their geometric axis.

Top bars

- marginal bars over support: L = t0/3 + lbd ; where lbd is the development length of about 40

- interior bars over support: L = t0/3 + bpb + t0/3

- interior bars over middle zone: L = t0/3 + 2 x (overlapping distance with the adjacent bars (left and right)). The overlapping distance may be 300..400 mm for constructive bars or about 40 in case of tension bars.

t0/3 t0/3 t0/3 t0/3 t0/3 t0/3 t0/3 t0/3 t0/3

bpb t0 bpb t0 bpb t0 bpb

LENGTH OF BARS

interior face of the marginal beam

lbd

1

Bottom bars

- marginal bars must provide a similar development length inside the marginal supporting beam: L = lbd + t0 + bpb ; where lbd is the development length of about 40 , similar as for top marginal bars. Over the interior supporting beam the bars extend to the opposite face of the support. - interior bars extend from one face of the supporting beam to the other face of the supporting beam: L = bpb + t0 + bpb

Links or stirrups

The length of transvers bar (link or stirrup) is computed at the interior face of the bar. Consequently the total length of the stirrup is: L = 2 x (bsb 2 x cnom) + 2 x (hsb 2 x cnom) + length of hooks. The length for one hook is about 10 link, so for two hooks the length is 2 x 10 link. Remark: links or stirrups are represented on the elevation of the beam on each zone of t0/3. They may be represented as their real number of pieces or alternatively, only 3 or 4 pieces over each zone.

cnom

cnom

hsb

bsb

2

For the middle zone of the span (t0/3), where the shear force diminishes as value and even become zero, it is common not to perform a complete shear design but to provide links or stirrups with the same diameter as in the adjacent left and right zones. The spacing over the middle zone of the span is chosen greater with 50 . 100 mm than the spacing of the adjacent zones, but not exceeding the maximum acceptable distance smax between links or stirrups.

3