Concrete Beam Design to EC2

Civil Guide

Steps to Concrete Beam Design to EC2

  1. Determine Loading
  2. Determine Support Conditions 
  3. Check Required Reinforcement for bending
  4. Check Reinforcement for shear
  5. Check span-effective Depth Ratio

Design Example 1: Simply Supported Beam (Compression and Tension Steel)

A simply supported beam is to be designed, where the span is taken as 5000mm. The characteristic strengths of the concrete is fck = 32 N/mm2. The yield strength of steel reinforcement is 500 N/mm2 . The concrete beam support bearing is 150mm.

The beam section is 400mm x 400m.

Cover to reinforcement to be 30mm. Assume tension bar to be 20mm diameter and 8mm link.

Effective depth, d= 400 – 30 – 8 -20/2 = 352mm and breadth = 300mm.

Permanent load, g = 40 kN/m (including self weight)

Variable load, qk = 30 kN/m

The concrete beam design for this example is shown in the steps below:

Loading

Ultimate load, wu = (1.35gk + 1.5qk) kN/m

= (1.35 x 40 + 1.5 x 30) = 100 kN/m

therefore

maximum design moment M = \(\frac{w_uL^2}{8} = \frac{100\times 5.0^2}{8} = 313 kNm\)

Concrete Beam Design
Design Example 1

Reinforcement for Bending and Compression

\(K=\frac{M}{bd^2f_{ck}} = \frac{313 \times 10^6}{400×352^2 \times 32}= 0.198 >K_{bal} = 0.167\) . Therefore, compression reinforcement, A’s is required.

x = 0.45d = 0.45 x 352 = 158.4mm

d’/x = (30+8+10)/158.4 = 0.30 < 0.38, therefore fsc = 0.87 fyk.

Compression steel:

\(A’_s\frac{M-0.167f_{ck}bd^2}{f_{sc}(d-d’)}\)

\(=\frac{313 \times 10^6 -0.167 \times 32 \times 400 \times 352^2}{0.87 \times 500(352-48)} = 364 mm^2\)

Provide 4B20 bars, A’s = 1256 mm2.

Tension steel, \(A_s=\frac{0.167f_{ck}bd^2}{0.87f_{yk}z_{bal}} + A_s’\)

where, from the lever arm curve, la = 0.82.

Thus,

\(A_s = \frac{0.167\times 32 \times 400 \times 352^2}{0.87\times 500 \times (0.82 \times 352)} + 364\)

\(= 2110 + 364 =2474mm^2\).

Provide 4 B32 bars, area = 3217 mm2.

Note – Keep the same number of bars top and bottom, to allow ease of fixing of rebar/cages on site (i.e. don’t use 3 bars top and 4 bars at the bottom). Also, the top bars need be a minimum of 2 bar sizes lower than the bottom bars, which is why we used B20 bars for the top reinforcement instead of using a B12 or B16.

Span- Effective Depth Ratio

\(\rho = 100A_{s,req}/bd = (100\times 2474)/(400\times 352) = 1.75%\)

From span-ratio table, basic span/effective depth for a simply supported beam is 14.

Modification for ratio = 14 x (3217/2474) = 18.20

Span-effective depth ratio provided = 5000/352 = 14.2.

14.2<18.20 .. Therefore, it is okay.

Beam span-effective depth

Shear Link in Beam Design

Design for Shear

Maximum design shear V = wu x effective span/2 = 100 x 5.0/2 = 250 kN

Design shear at face of support VED = 250 – 100×0.15 = 235 kN

Maximum shear at face of support,

Crushing strength VRd,max of diagonal strut, assuming angle ϑ = 22°, cot ϑ = 2.5

VRd,max = 0.124 bw d (1-fck /250) fck

= 0.124 x 400 x 400 (1-32/250) x 32 x 10-3

= 554 kN (> 235 kN)

Therefore, angle ϑ = 22°, cot ϑ = 2.5 will be taken.

Shear Links

The required shear links in a beam is defined the the equation below:

Asw/s = Ved/0.78dfykcotθ  = 235 x 103 / (0.78 x 400 x 500 x 2.5) = 0.60

Referring to the shear reinforcement table (click on link). We can calculate the link spacing and size.

Provide 2 Legs 8mm shear links at 150mm spacing = 0.671 … OKAY

Minimum Shear Links

Asw,min/s = 0.08fck0.5bw/fyk  = 0.08 x 320.5 x 400 / (500) = 0.36

We have already provided 8mm links at 150mm spacing (0.671) which is adequate.

The shear resistance of the links can be calculated below:

Vmin = Asw/s x 0.78dfykcotθ = 0.671 x 0.78 x 550 x 500 x 2.5 x 10-3 = 360 kN

General Notes for shear link sizing

Asw = the cross-sectional area of two legs of the stirrup

S = the spacing of the legs

Z = the lever arm between the upper and lower chord members

Fywd = the design yield strength of the stirrup/link

Fyk = the characteristic strength of the stirrup reinforcement

VEd = the shear force due to the actions at the ultimate limit state

Vwd = the shear force in the stirrup

VRd,s = the shear resistance of the stirrups

VRd,max = the maximum design value of the shear which can be resisted by the concrete strut

 

Maximum longitudinal spacing between shear links = 0.75 d

Maximum transverse spacing between legs in a series of shear links = 0.75 d (≤600mm)

Main steel reinforcement bars should be within a 150mm distance to a shear leg (i.e, check the min distance between bottom/top bar to shear leg)

Design Example 2: Simply Supported Beam (Tension only)

A simply supported beam is to be designed, where the span is taken as 6500mm. The characteristic strengths of the concrete is fck = 32 N/mm2. The yield strength of steel reinforcement is 500 N/mm2 . The concrete beam support bearing is 150mm.

The beam section is 400mm x 600m.

Cover to reinforcement to be 30mm. Assume tension bar to be 25mm diameter and 8mm link.

Effective depth, d= 650 – 30 – 8 -25/2 = 600mm and breadth = 400mm.

Permanent load, g = 50 kN/m (including self weight)

Variable load, qk = 35 kN/m

The concrete beam design for this example is shown in the steps below:

Loading

Ultimate load, wu = (1.35gk + 1.5qk) kN/m

= (1.35 x 50 + 1.5 x 35) = 120 kN/m

therefore

maximum design moment M = \(\frac{w_uL^2}{8} = \frac{120\times 6.50^2}{8} = 635 kNm\)

Simply Supported Beam
Design Example 2

Reinforcement for Bending

\(K=\frac{M}{bd^2f_{ck}} = \frac{635 \times 10^6}{400x 600^2 \times 32}= 0.137 <K_{bal} = 0.167\) . Therefore, compression reinforcement, is not required.

Tension steel, \(A_s=\frac{M}{0.87f_{yk}z}\)

z = d[0.5 + \(sqrt\)(0.25 – K/1.134)

z = 600 x [0.5 + \(sqrt\)(0.25 – 0.137/1.134) = 515 mm 

Thus,

\(A_s = \frac{635 x 10^6}{0.87\times 500 \times (515)}\)

\(=2834mm^2\).

Provide 4 B32 bars, area = 3217 mm2.

Note – Although compression bars are not required, top bars are still needed to form the reinforcement cage

Span- Effective Depth Ratio

\(\rho = 100A_{s,req}/bd = (100\times 2834)/(400\times 600) = 1.18%\)

From span-ratio table, basic span/effective depth for a simply supported beam is 14.

Modification for ratio = 14 x (3217/2834) = 15.9

Span-effective depth ratio provided = 6500/600= 10.83.

10.83<15.9 .. Therefore, it is okay.

Beam span-effective depth

Shear Link in Beam Design

Design for Shear

Maximum design shear V = wu x effective span/2 = 120x 6.50/2 = 390 kN

Design shear at face of support VED = 390 – 120 x 0.15 = 372 kN

Maximum shear at face of support,

Crushing strength VRd,max of diagonal strut, assuming angle ϑ = 22°, cot ϑ = 2.5

VRd,max = 0.124 bw d (1-fck /250) fck

= 0.124 x 400 x 650 (1-32/250) x 32 x 10-3

= 900kN (> 372kN)

Therefore, angle ϑ = 22°, cot ϑ = 2.5 will be taken.

Shear Links

The required shear links in a beam is defined the the equation below:

Asw/s = Ved/0.78dfykcotθ  = 390 x 103 / (0.78 x 400 x 500 x 2.5) = 1.0

Referring to the shear reinforcement table (click on link). We can calculate the link spacing and size.

Provide 2 Legs 10mm shear links at 125mm spacing = 1.256 … OKAY

Minimum Shear Links

Asw,min/s = 0.08fck0.5bw/fyk  = 0.08 x 320.5 x 400 / (500) = 0.36

We have already provided 10mm links at 125mm spacing (1.256) which is adequate.

The shear resistance of the links can be calculated below:

Vmin = Asw/s x 0.78dfykcotθ = 1.256 x 0.78 x 650 x 500 x 2.5 x 10-3 = 796 kN

Design Example 3: Cantilever Beam (Tension only)

A cantilever beam is to be designed, where the span is taken as 1500mm. The characteristic strengths of the concrete is fck = 32 N/mm2. The yield strength of steel reinforcement is 500 N/mm2 . The concrete beam support bearing is 150mm.

The beam section is 400mm x 600m.

Cover to reinforcement to be 30mm. Assume tension bar to be 25mm diameter and 10mm link.

Effective depth, d= 650 – 30 – 10 -25/2 = 598 mm and breadth = 400mm.

Permanent load, g = 50 kN/m (including self weight)

Variable load, qk = 35 kN/m

The concrete beam design for this example is shown in the steps below:

Loading

Ultimate load, wu = (1.35gk + 1.5qk) kN/m

= (1.35 x 50 + 1.5 x 35) = 120 kN/m

therefore

maximum design moment M = \(\frac{w_uL^2}{2} = \frac{120\times 1.5^2}{2} = 135 kNm\)

Cantilever beam design example
Design Example 3

Reinforcement for Bending

\(K=\frac{M}{bd^2f_{ck}} = \frac{135\times 10^6}{400x 598^2 \times 32}= 0.03 <K_{bal} = 0.167\) . Therefore, compression reinforcement, is not required.

Tension steel, \(A_s=\frac{M}{0.87f_{yk}z}\)

z = d[0.5 + \(sqrt\)(0.25 – K/1.134)

z = 598x [0.5 + \(sqrt\)(0.25 – 0.03/1.134) = 582 mm 

Limit for z = 0.95d = 0.95 x 598 = 568mm (use 568mm for z)

Thus,

\(A_s = \frac{135x 10^6}{0.87\times 500 \times (568)}\)

\(=550mm^2\).

Provide 4 B16 bars, area = 804 mm2.

Note – Tension bar is on the top flange of the concrete beam (see bending moment diagram)

Span- Effective Depth Ratio

\(\rho = 100A_{s,req}/bd = (100\times 550)/(400\times 598) = 0.23%\)

From span-ratio table, basic span/effective depth for a cantilever beam is 6.

Modification for ratio = 6x (804/550) = 8.77

Span-effective depth ratio provided = 1500/598= 2.5.

2.5<8.77.. Therefore, it is okay.

Beam span-effective depth

Shear Link in Beam Design

Design for Shear

Maximum design shear V = wu x effective span = 120x 1.5 = 180 kN

Crushing strength VRd,max of diagonal strut, assuming angle ϑ = 22°, cot ϑ = 2.5

VRd,max = 0.124 bw d (1-fck /250) fck

= 0.124 x 400 x 650 (1-32/250) x 32 x 10-3

= 900kN (> 180kN)

Therefore, angle ϑ = 22°, cot ϑ = 2.5 will be taken.

Shear Links

The required shear links in a beam is defined the the equation below:

Asw/s = Ved/0.78dfykcotθ  = 180 x 103 / (0.78 x 400 x 500 x 2.5) = 0.5

Referring to the shear reinforcement table (click on link). We can calculate the link spacing and size.

Provide 2 Legs 10mm shear links at 250 mm spacing = 0.628… OKAY

Minimum Shear Links

Asw,min/s = 0.08fck0.5bw/fyk  = 0.08 x 320.5 x 400 / (500) = 0.36

We have already provided 10mm links at 250mm spacing (0.628) which is adequate.

The shear resistance of the links can be calculated below:

Vmin = Asw/s x 0.78dfykcotθ = 0.628 x 0.78 x 650 x 500 x 2.5 x 10-3 = 400 kN

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