Answer:
provide 180 mm spacing
Explanation:
GIVEN DATA
rectangular beam: (b) = 300 mm, (d) = 575 mm
reinforced for flexure = 4Ф32 bars
WD = 30 kN /m, WL = 45 kN/m
Wu = 1.4 * 30 + 1.6 * 45 = 114 kN/m
i) concrete shear stress ( vc)
100 Ac / bd = (100 * u * [tex]\frac{\pi }{4}[/tex] * 32^2) / 300 * 575 = 1.865
from table 3.8
when: 100 Ac / bd = 1.865 then Vc = 0.778 N/mm^2
Ultimate shear force = (114 *5.5) / 2 = 313.5 kN
design shear stress = V / bd = (313.5 * 10^3) / (300 * 575) = 1.82 N/mm^2
v < 0.8[tex]\sqrt{22}[/tex] = 1.82 < 3.75
design link provided according to
Asv / sv = b(v-vc) / 0.87 fy = 300(1.82 - 0.778) / 0.87 (420)
ASv / Sv = 0.855
From table 3.13 :the value of Asv / sv can be calculated as
[tex]\frac{0.855 - 0.785}{0.897 - 0.785} = \frac{x - 200}{175 - 200}[/tex]
x = (-25) [ 0.625] + 200 = 184.375 mm
provide 180 mm spacing
