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:
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
- 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
- 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/bwd = 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
- 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
- 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√fck (ρo/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 = 10-3√fck
ρo = 10-3√30
= 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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