Reinforced Concrete flange beams

Design of Reinforced Concrete Flange Beam to Eurocode 2 – Worked Example

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This article presents a worked example on the design of reinforced concrete flange beam to EC2.

We shall use the same example as that used in the design of doubly reinforced concrete to EC2. We shall design the beam as flange beam and observe the outcome of the design.

The image below shows the Plan view of a reinforced concrete structure, use the data given below to design Beam 2

Design Data:

Variable load on slab =   5KN/m2

Finishes  =   1.5KN/m2

Unit weight of concrete   =   24KN/m3

Compressive strength of concrete (fck)   =   30N/mm2

Characteristic Strength of main reinforcement (fyk)  =  500N/mm2

Characteristic Strength of Shear reinforcement (fyv) =  500N/mm2

To design Beam 2, we will first distribute the slab loads on the beam and then analyze the beam to compute its internal forces. As shown in fig (2) below the tributary area of slab loads on beam 2 is a trapezium, hence we compute and distribute the slap loads on it as follows:

Yield line pattern showing tributary area distributing Slab loads to supporting beams

However, to understand the process of distributing slab load to beams in detail, read, “Distribution of Slab loads to beams.”

Analysis

Permanent Load

Characteristic Self-weight of slab = 0.2 x 24 = 4.8KN/m2

Partition Load on Slab = 1.5KN/m2

Characteristic Permanent Load on Slab   = 4.8 + 1.5 = 6.38KN/m2

Area of Slab load Supported by Beam 2   = (0.5 x (8 + 3) x 2.5)/8 = 1.73m2

Permanent Load of Slab on Beam 2 = 1.72 x 6.38 = 10.97KN/m

Self-weight of beam = 0.23 x 0.45 x 24    = 2.48KN/m

Total Permanent Load on beam = 10.97 +2.48 = 13.5 KN/m

Variable Load    

Variable load on slab = 5KN/m2

Area of Slab load Supported by Beam 2   = (0.5 x (8 + 3) x 2.5)/8   =   1.73m2

Characteristic Variable Load of Slab on Beam 2    = 1.258 x 5   = 8.59KN/m

Total Design Load

Ultimate Load acting on beam 2 = 1.35(13.5) + 1.5(8.59) = 31.11KN/m

Computation of Internal Forces:

The beam is assumed to be simply supported for ease of analysis.

M = w x L2/8 =   31 x 82/8 = 248KNm

V   = W x L/2 =    31 x 8/2   = 124KNm

Design

flexural strength design

  1. Calculate the effective depth

Assumptions

Cover  =   25mm

Main reinforcement diameter = 16mm

Diameter of links =  10mm

Effective depth = h-c-ᴓ-ᴓ/2

         =450-25-10-16/2

        = 407mm

2. Effective flange width:

beff = beffj + bw

But beffj = 0.2bi + 0.1Lo ≤ (0.2lo, bi)

   = 0.2 x (5000 – 225)/2   + 0.1 x 0.85 x 8000

= 0.2 x 2387.5 + 0.1 x 6800

         = 1157.5 ≤ (0.2 x 6800, 2387.5)

          = 1157.5 ≤ (1360, 2387.5)

beffj    = 1157.5

Substitute the value of beffj into the main equation to calculate beff

beff =  1157.5 + 225 ≤ b

= 1382.5 ≤ (1157.5 x2 + 225)

=  1382.5 ≤ (2540)

beff    =   1382.5

3. Check whether section is to be designed as singly or doubly reinforced beam

$
K\,=\,\,\frac{M}{bd^2f_{ck}}
$

$
K\,=\,\,\frac{248×10^6}{1382.5×450^2×30}
$

= 0.22

Since K (0.22) < K’ (0.168); design as singly reinforced.

NB: The same beam which was initially designed as doubly reinforced here is now declared to be designed as singly reinforced. This is the effect of the flange width which has augmented the compressive strength of the beam so that no reinforcement is required to resist compressive stress.

4) Calculate the lever arm (Z)

$
Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right)
$

$
Z\,\,=\,\,407\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.22}{1.134}} \right)
$

Since 393.6 < 0.95d (386.7): use Z = 386.7

5) Calculate the area of steel required

$
A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}}
$

$
A_{st\,\,=\,\,\frac{248×10^6}{0.87x500x386.7}}
$

Ast = 1474.9mm2

Provide 8T16 (1599mm2)

NB: You will observe that the area of tension reinforcement is large, this can be attributed to the 8m long span of the beam, and more overriding, the assumption that the beam is simply supported which then result in large span moment.

6) Check whether the area of tensile steel provided satisfies minimum area requirement

               Asmin = 0.26 (fctm/fyk )bt d

            = 0.26 (2.9/500) 225 x 407

          =   138.095mm2

         Since Ast (1474.9mm2 )> Asmin (138.095mm2), hence minimum area requirement satisfied.

6) For practical purpose of forming a reinforcement cage, provide compression reinforcement as secondary reinforcement which will serve as hanger bars

      Asc  = 0.2×1474

      =294.9mm2

        Provide 2T12

Vertical Shear Design

  1. Check whether the concrete section can resist the shear force without shear reinforcement

VRdc = (0.12K(100ρLfck)1/3 + K1σcp) bwd

K     =  (1 +√200/d) =  1 + (200/407)0.5 = 1.7

ρL = Asl/bw= 1474/225 x 407 = 0.016

VRdc = (0.12 x 1.7 (100 x 0.016 x 30) 1/3) 225 x 407

VRdc = 75.3KN

Since VRdc (75.3KN) is less than VEd (124KN) then shear reinforcement has to be designed for.

2. Calculate the shear resistance using ϴ  = 22

VRdmax(22) = 0.124bwd(1 – fck/250)fck

= 0.124 x 225 x 407 (1 – 30/250)30

= 299.8KN

Shear resistance of links at ϴ  = 22 is adequate

3. Calculate Asw/s at ϴ = 22

$
\frac{A_{sw}}{s}\,\,\,\,=\,\,\frac{V_{Ed}}{0.78xf_{yk}d\cot 22}
$

$
\frac{A_{sw}}{s}\,\,\,\,=\,\,\frac{124×10^3}{0.78x407x500\cot 22}
$

$\frac{A_{sw}}{s}\,\,\,\,$ = 0.3

4. Check whether $\frac{A_{sw}}{s}\,\,\,\,$ satisfy the minimum requirement specified by the code.

$$
\frac{A_{sw\min}}{s}\,\,\,\,=\,\,\frac{0.08x\sqrt{f_{ck}}xb_w}{f_{yk}}
$$

$$
\frac{A_{sw\min}}{s}\,\,\,\,=\,\,\frac{0.08x\sqrt{30}x225}{500} $$

= 0.2

Since Asmin/s (0.2) is less than As/s (0.3), shear reinforcement is greater than the required minimum area of reinforcement.

Calculate shear link reinforcement spacing requirement

Assume two-legged shear reinforcement of 10mm is to be used.

Area of 10mm shear reinforcement = 78.58mm2

Area of two-legged 10mm links = 2x 78.58   = 157mm2

157/s = 0.3

s= 157/0.3

s = 497.9mm

6) Check whether maximum spacing requirement is satisfied

Smax = 0.75xd

= 0.75 x 407

= 305mm

Since the maximum spacing (305mm) is less than the calculated spacing of links (497.9mm), the maximum spacing limit governs the design.

provide Y10 @ 300mm spacing.

Longitudinal Shear Design

  1. Calculate the shear stress in the flange

$
v_{ED\,\,=\,\,}\frac{\varDelta F_d}{h_f\,\,X\,\,\varDelta x}
$

$
\varDelta F_d\,\,=\,\,\frac{\varDelta M}{d\,\,-\,\,h_f/2}\,\,X\,\,\frac{b_{eff}\,\,-\,\,\frac{b_w}{2}}{b_{eff}}
$

∆X = L/4 = 8000/4 = 20000

∆M = 3WL2/32

 = 3 x 31 x 82/32

=186KNM

$\varDelta F_d\,\,=\,\,\frac{186 X 10^8}{407\,\,-\,\,200/2}\,\,X\,\,\frac{1157.5}{1382.5}
$

∆Fd= 507.3KN

$
v_{ED\,\,=\,\,}\frac{507.3 X 10^3}{200\,\,X\,\,2000}
$

2.Check the strength of the compressive structs.

The compressive struct is checked for the minimum strength assuming ϴf = 26.5

Using $
v_{ED}\,\,=\,\,vf_{cd}\sin \theta _f\cos \theta _f
$

v = 0.6(1 – fck/250)

Substitute v into the main equation

$
v_{ED}\,\,=\,\,0.6(1 – 30/250) X 30/1.5 Sin26.5 Cos26.5
$

4.2N/mm2

Since 4.2N/mm2 is greater than 1.27N/mm2, compressive structs is adequate.

3. Check whether transverse reinforcement is required.

if vEd ≤ 0.4 fctd; no transverse shear reinforcement required.

0.4 fctd = 0.27fctk

For C25/30, fctk(0.05) = 2.03

0.4 fctd = 0.27 x 2.03

=0.55

Since vEd > 0.4 fctd; transverse shear reinforcement is required.

4. Design for transverse reinforcement

$
\frac{A_{sf}\,\,X\,\,f_{yd}}{s_f}\,\,=\,\,\frac{V_{Ed}X\,\,h_f\,\,}{\cot \theta _f}
$

$
\frac{A_{sf}\,\,X\,\,0.86 X 500}{s_f}\,\,=\,\,\frac{1.27\,\,200\,\,}{\cot 26.5}
$

Asf/sf = 0.29

Assume the diameter of rebar to be 12mm

Area of rebar = 113

sf = 113/0.29

Sf = 389

Provide Y12@300mm spacing

Deflection Check

  1. Calculate the actual span-effective depth ratio

Span/depth ratio = 8000/407   = 19.7

2. Calculate the limiting Span-effective depth ratio

l/d = K[11 + 1.5√fck ρ0/ρ + 3.2√fcko/ρ – 1)3/2] if ρ ≤ ρo

l/d = K[11 + 1.5√fck ρo/ρ + 3.2√fck √ρo/ρ ] if ρ > ρo

ρ = Asprovided/b x d

ρ = 1599/225 x 407

= 0.017

ρo = 10-3fck

ρo = 10-330

= 0.005

K = 1 (for simply supported)

Since ρ > ρo  = then we will use

l/d = K[11 + 1.5√fck ρo/ρ + 3.2√fck √ρo/ρ ]

l/d = [11 + 1.5√30 0.005/0.019 + 3.2√30 √0.005/0.019]

= 14.3mm

Since actual span-effective depth ratio (19.7) is greater than the limiting span-effective depth ratio (14.3), the beam fails deflection check.

NB: Since the beam fails deflection check, it shall be resized by increasing its depth so as to reduce the span-depth ratio. Another effective solution is to design the beam as encastre beam fixed at both ends; this will reduce its overall deflection as well as its sagging moment which will result to the decrease in the area of tensile steel provided for flexural strength of the member.

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