ISTANBUL TECHNICAL UNIVERSITY
Department of Civil Engineering
Section of Reinforced Concrete Structures
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REINFORCED CONCRETE - I
Double reinforced beam (By Using Tables)
Example 6:
Calculate the required reinforcement for the given section by using the table.
Material: C20/5220
Solution:
As = ks*Md/d - Nd K = bw*d2/Md = 0.25*0.462/24066 = 21.98*10-5<24.4*10-5
When we look at the table on page 3, this is less than the given underlined value for C20 concrete
and S220 steel. This underlined value corresponds to the maximum reinforcement ratio (ρmax=0.85
ρb) of the beam. Our section does not behave in a ductile way. To prevent this, instead of changing
the beam dimensions, we can use an alternative way : Adding compression reinforcement.
Let's find the maximum amount of tensile steel and the maximum amount of moment that can be
carried (supplying the required level of ductility) by that section:
M* = 0.25*0.4621/(24.4*10-5) = 216.8 kNm and the max reinforcement is
As* = ks*Md*/d = 7.08*216.8/0.46
As* = 3337 mm2
∆M = Md-M* = 240.66-216.8 = 23.86 kNm. This additional moment should be created by the
compression reinforcement.
Checking whether the compression reinforcement is yielding or not?
x = kx*d = 0.644*460 = 296.2 mm
εcu/x = εs'/(x-d') --- εs' = 3*(296.2-40)/296.2 = 2.59 %0
εyd = fyd/εs = 191/105 = 0.955 %0
εs’>εyd then σs' = fyd = 191 N/mm2
Moment equation: ∆M = Fs'*(d-d') = As'*fyd*(460-40)
As' = 23.86*106/(191*(460-40)) = 297.4 mm2
Horizontal equilibrium: Fs' = Fsadd 297.4 * 191 = Asadd Then: Asadd = 297.4 mm2
Compression Steel: As' = 297.4 mm2--- 3Φ12 (=339 mm2)
Tension Steel: As = As*+Asadd = 3337+297.4 = 3634.4 mm2 - 10Φ22 (=3801 mm2)