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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+b=1000//consider 1 m width of slab
+D=100//depth of slab, in mm
+cover=20//in mm
+d=D-cover//effective depth, in mm
+W=7//uniformly distributed load, in kN/m^2
+dia=10//in mm
+s=100//spacing of 10 mm dia bars, in mm
+l=4//span, in m
+V=W*l/2//in kN
+Pt=1000*.785*dia^2/(s*b*d)*100//in %
+Tv=(V*10^3)/(b*d)//in MPa
+//for given Pt and M15 grade concrete
+Tc=0.37//in MPa
+//and for solid slabs
+k=1.3
+Tc=k*Tc//in MPa
+mprintf("Nominal shear stress in slab, Tv=%f MPa\nShear strength of slab, Tc=%f MPa. As Tc > Tv, no shear reinforcement is required", Tv, Tc)