## Shear

Design shear force, VEd

At d from face of end support

At d from face of 1st interior support

Shear resistance, VRd,c

Vrj.c = (0.18 / yc)k (100pfCk)0333 bwd > 0.0035k1 5fCk° 5bwd where k = 1 + (200 / d)05 < 2.0 as d < 200 mm k = 2.0 P = A6l / bd

Assuming 50% curtailment (at end support) = 50% x 754 / (144 x 1000) = 0.26% VRd.c = (0.18 / 1.5) x 2.0 x (100 x 0.26 / 100 x 30)033 x 1000 x 144 = 0.12 x 2 x 1.97 x 1000 x 144 = 0.47 x 1000 x 144 = 68.1 kN / m

where

But VR,cmin = 0.035k 5fck05 b^

k = 1 + (200 / d)05 < 2.0; as before k = 2.0 VRd,cmin = 0.035 x 215 x 30°5 x 1000 x 144 = 0.54 x 1000 x 144 = 77.6 kN / m VRd.c = 77.6 kN / m

OK; no shear reinforcement required at end or 1st internal supports

■•■H12 @ 150 B1 & H12 @ 175 T1 OK By inspection, shear at other internal supports OK.

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