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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+sigma_cbc=5//in MPa
+sigma_st=230//in MPa
+MF=1.4//modification factor
+//let a be span to depth ratio
+l=1//span, in m
+a=MF*7
+D=l*1000/a//in mm
+D=105//assume, in mm
+//to calculate loading
+self_weight=25*(D/10^3)*1.5//in kN/m
+finish=0.5*1.5//in kN/m
+live_load=0.75*1.5//in kN/m
+W=self_weight+finish+live_load//in kN/m
+lef=l+0.23/2//effective span, in m
+M=W*lef/2//in kN-m
+//check for depth
+d=(M*10^6/(0.65*1500))^0.5//in mm
+dia=12//assume 12 mm dia bars
+D=d+12/2+15//<105, hence OK
+D=100//assume, in mm
+d=D-dia/2-15//in mm
+//main steel at mid-span
+Ast=M*10^6/(sigma_st*0.9*d)//in sq mm
+s1=1500*0.785*12^2/Ast//>3d = 237 mm
+s1=235//assume, in mm
+Ads=0.12/100*1000*D//distribution steel, in sq mm
+//assume 6 mm dia bars
+s2=1000*0.785*6^2/Ads//in mm
+s2=235//assume, in mm
+Tbd=0.84//in MPa
+Ld=dia*sigma_st/4/Tbd// in mm
+Ld=821//round-off, in mm
+Tv=W*10^3/1500/d//in MPa
+As=1500*0.785*12^2/235//in sq mm
+pt=As/1500/d*100//in %
+Tc=0.316//in MPa
+//as Tc>Tv, no shear reinforcement required
+mprintf("Summary of design\nThickness of slab = %d mm\nCover = 15mm\nMain steel = 12 mm dia @ %d mm c/c\nProvide development length of %d mm in the beam from face of beam\nDistribution steel = 6 mm dia @ %d mm c/c",D,s1,Ld,s2)