REINFORCED CONCRETE - ILecture :

SIMPLE BENDING

Res.Ass. Idris Bedirhanoglu

room : Reinforced Concrete Group Room:526

and Structure and Earthquake Engineering Lab.

url : www.ins.itu.edu.tr/ibedirhanoglu

e-mail : ibedirhanoglu@.ins.itu.edu.tr

topics:Beam under bending moment-calculate with

Equivalent rectangular stress blockRC tables

Balanced beam

Beam under bending moment and small axialloads-calculate with

Equivalent rectangular stress blockRC tables

Double reinforced beam under bending moment-calculate with

Equivalent rectangular stress blockRC tables

Example 1:

• As=? ( = = 10 %) total area of the required bars.• =? max. compression strain

• Obtain the reinforcement ratio =?• Compare the obtained reinforcement ratio with the minimum

reinforcement ratio given in the TS500 Code for beams min.

materialsConcrete: C20Steel : S420 k1=0.85dimensionsbw=300 mm h=500 mmd’=50 mmεs εsu

εcρ

ρ

q=11 kN/mg=22.5 kN/m

L=4000 mm

h

bw

As

εc

εs

Design load: p = 1.4*g + 1.6*q = 49.1 kN/mMaximum Bending Moment: Md = p*l2 / 8 = 49.1 * 42 / 8 =98.2 kNm

• Moment Equilibrium : • Mr = Md (equality of internal and external moment)

• Fc * (d-a/2) = Md (with reference to the point Fs acts on)

• 0.85*fcd*a*bw*(d-a/2) = Md• 0.85*13*a*300*(450-a/2) = 98.2*106

• a = 72 mm x = a/k1 = 72/0.85 = 84.7 mm• Using the strain diagram : x = kx * d

εc σc

Fc

Fs

h

bw

As εs

N/Ax a Fc=0.85*fcd*a*bw

Fs=AS*fyd

0.85fcd

Solution:

σc

εcεco=0.002

0.85fcd

εcu=0.003

σs

εsεsu=0.01

fyd

εyd

Es

concrete steel

Meterial models

N/Afs1fs2fs3fs4

= fs

fcσc

first: we should calculate x

ε

V=0.85fcd*bw*a

N/Afs1fs2fs3fs4

= fs

fcσc

0.85fcd

a

bw

a=k1*x

(Fc=V)

N/Afs1fs2fs3fs4

= fs

fcσc 0.85fcd

x

bw V=integration (Fc=V)

k1=0.85

2.32%ε450*)ε(ε

εx c

sc

c=⇒

+=

now: from strain diagram we can calculate εc εc

εs

d-x

xSimilar trianglesεc/x=εs/d-x

Calculate Fc ?

Compression force Fc is obtained as,Fc = 0.85 * 13 * 72 * 300 = 238680 N

Writing the horizontal equilibrium we calculate the required amount of steel reinforcement :Fc – Fs = 0 → 238680 = As * 365 → As = 653.9 mm2

Selection of reinforcement : 6 Φ12 (= 679 mm2) or 5 Φ14 (= 770 mm2) Note: check whether wehave enough place to settle the bars! (RC table page 4)

Checking the minimum reinforcement ratio given by TS500 :ρmin = 0.8 * fctd / fyd = 0.8 * 1 / 365 = 0.00219 (TS500 page 23)

ρ = As / (bw * d) = 679 / (300*450) = 0.00503 ρ > ρmin

ρmin : In reinforced concrete beams a certain minimum amount of steelmust be used in order to develop a resisting moment greater than theinitial cracking moment. Before tension cracking occurs on the beam, most of the tensile force is carried by concrete and as soon as theconcrete cracks this tensile force is transferred to steel.

Example 2:

Calculate the required reinforcement for the question above by using RC tables.

Solution:

Md= 98.2 kNm C20/S420

As=ks * Md / d – Nd / σs K = bw*d2/Md = 0.3 * 0.452 / 98.2 = 61.86 * 10-5

For C20 concrete and S420 steel K=64.2 or K=61.7 (take the closest one) , ks=2.99, kx=0.213, j=0.916, εc=2.7%, εs=10% .x = kx * d = 95.85 mm

As we have no axial forces, N=0. Then the required reinforcement area is,As = 2.99 * 98.2 / 0.45 = 652.5 mm2 ⇒ Choose 5 Φ14 (= 770 mm2) or 6 Φ12 (= 679 mm2)

5 Φ14

500 mm

300 mm

N/A

x

BALANCED BEAMExample 3:Since the R-C cross-section given in the figure belongs to balanced beam, calculate the values :h=?, M=?, ρb=?, ρmax=?, Asmax=?

Materials: C16 concrete and S220 steelk1= 0.85 , d' = 30 mm, bw=250 mm, Asb=3400 mm2

Solution: Strain and stress diagrams are given below.

Writing the horizontal equilibrium: Fc - Fs = 00.85*11*0.645*d*250=3400*191---d=430 mmBeam height: h=d+d' = 430+30 = 460 mma= 0.645*d=0.645*430=277.4 mm

εyd= 191/ (2*105) = 0.000955x = kx*dFrom strain diagram :kx=εc / (εc+εs)=3/(3+0.955) =0.759x = kx*d = O.759*d

Rectangular stress block depth:a=k1*x=0.85*0.759*d = 0.645*d

• Writing the moment equilibrium:

• Md=0.85*a*bw*(d-a/2)=Fs*(d-a/2)• 3400 * 191 * (430-277.4/2) = Md• Md = 189.17 kNm

• Balanced reinforcement ratio : ρb=Asb/(bw*d) = 3400/(250*430) = 0.0316 • Maximum reinforcement raito: ρmax= 0.85*ρb= 0.85 * 0.0316 = 0.0269• Asmax=ρmax*bw*d=2890 mm2

Beam subjected to bending moment and axial force

Example 4:

A beam under aggressive environmental conditions is subjected to bending moment and axial force. Obtain the required reinforcement at the ultimate state cansidering εcu=0.003 and εsu= 0.010 .Materials: C20 concrete and S220 steel , k1=O.85.

Solution:

Writing the moment equilibrium:

Md=Fc*(d-a/2)-Nd*(h/2-d)=O.85*fcd*a*bw*(d-a/2)-Nd*(h/2-d)0.85*13*a*400*(540-a/2)=150*106 + 300*103*(300-60)a=102.8 mm ---- x=102.8/0.85=120.9 mm

Let's check the strain ratios :Under the assumption that εc=εcu=0.003x = kx*d ----- 120.9=540*εc/(εc+εs) =540 1< 31 (3+εs) ----εs=10.4 %o > εsu=10 %oThis is not possible, we need to make another assumption:

Assume that: εs = εsu = 10 %o

Then, 120.9=540*εc/(εc+εs)=540*εc/(εc+10) ----εc=2.88 %0 < εcu=3 %o This one is OK.

From the equilibrium of horizontal forces: Fc - Fs = Nd0.85*13*102.8*400 - As*191=300000454376 - 300000=As*191 --- As=808.3 mm2

Choosing the reinforcement from table: 6Φ14 (= 924 mm2)

Double reinforced beam

Example 5:A beam cross-section subjected to bending moment and axial force is given below. Calculate the needed compression and tension steel areas.Materials: C25 concrete and S220 steel.

Solution:

Evaluating the x value: kx=εc/(εc+εs)=3/(3+4)=0.4285x=kx*d=0.4285*450=192.8 mm

Checking whether the compression reinforcement is yielding or not?

εc/x=εs'/(x-d') --- εs'=3*(192.8-50)/192.8=2.22 %0εyd=fyd/Es=191/105=0.955 %0 εs>εyd then σs=fyd=191 N/mm2

a=k1*x=0.85*192.8=163.88 mm

Fs'=As'*191Fc=0.85*17*163.88*250=592016.5 N Fs=As*191

Writing the moment equilibrium:

Md+Nd*(h/2-d')=Fc*(d-a/2)+Fs'*(d-d')202.5*106+105*103* (250-50)=592016.5*(450-163.88/2)+As'*191*(450-50) As'=73.3 mm2 --- (minimum reinforcement 24Φ12=226 mm2)

Writing the horizontal forces equilibrium:

Fc+Fs'-Fs = Nd592016.5+73.3*191-As*191=105000 As=2623.1 mm2 --- 7Φ22 (=2661 mm2)

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^-5When we look at the table on page 1, 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*0.462/(24.4*10-5)=216.8 kNm and the max reinforcement isAs*=ks*Md*/d=7.08*216.8/0.46As* = 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 mmεcu/x=εs'/(x-d') --- εs'=3*(296.2-40)/296.2=2.59 %0εyd=fyd/εs=191/105=0.955 %0εs’>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)