Assignment 6: Design of Concrete BeamsQuestion 1:Question Two: SolutionWhere Asv = area of steel in shear, fsv = yield strength of steel in shear do = depth to lowest reinforcing bars, θv = shear crack angle (assume 45ᵒ) s = spacing between shear reinforcementThe total resisted shear is then Vu = Vuc + VuAsv= 22/7*3*3*3=84.86mm2
Fsv=400MPaDo=25mmΘv=45ᵒS=200mmVus=84.86*4oo*1000000*25*cot45/200Vus=4.234*1000Question 3 (no marks): Solution: Use the formula for determining the length of reinforcement as given belowWHEREBY,• Ab = area of the bar • db = bar thickness• 2a = Two times the distance to the cover.
• Assume k1 = 1.0 and k2 = 2.4. The standards provides more details for selecting these values,THUS, integrating values into the formula;Ab=22/7*6*6*2=226.195mm2Fsy=500MPaK1=1K2=2.42a=2*25=50Fc=40MPadb=12mmLsy.t=1*2.4*500*226.195/{(2*25+12)(401/2)=692.2>25*1*12=300Hence 692.2>300Take length BC to be 700mm.Question 4 (6 marks):
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