Design of a Singly Reinforced Beam to Eurocode 2 – Worked Example

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

The image below shows the Plan view of a reinforced concrete structure, use the data given below to design Beam 1

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 1, we will first distribute the slab load on the beam and then analyze the beam to compute its internal forces.

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

Analysis

Permanent action

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 1   = 0.5 x 1 x 2.5  = 1.258m²

Characteristic Permanent Load of Slab on Beam 1 = 1.258 x 6.38  =  8.02KN/m

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

Total Permanent Load on beam = 8.02 +2.4 = 10.42KN/m

Variable action

Variable load on slab = 5KN/m²

Area of Slab load Supported by Beam 1 = 0.5 x 1 x 2.5 = 1.258m²

Characteristic Variable Load of Slab on Beam 1 = 1.258 x 5 =  6.25KN/m

Total Design Load

Ultimate Load acting on Slab =1.35(10.42) + 1.5(6.25) =  23.44KN/m

Computation of Internal Forces:

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

M =  wL²/8  = 23.44 x 5² /8 = 73.25KNm

V   =   wL/2  = 23.44 x 5/2 = 58.6KN

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

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

$
K\,=\,\,\frac{73.25×10^6}{225×407^2×30}
$

= 0.066

Since K (0.066) < K’ (0.168); design as singly reinforced.

3) 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{0.066}{1.134}} \right)
$ = 381.9

Since 381.9 < 0.95d (386.7): use Z = 381.9

4. Calculate the area of steel

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

$
A_{st\,\,=\,\,\frac{73.25×10^6}{0.87x500x381.9}}
$

Ast = 440.9mm²

Provide 3Y16 (599.8mm²)

5) Check Whether area of tensile steel provided satisfies minimum area requirement

               A_smin = 0.26 (f_ctm/f_yk )b_t d

            = 0.26 (2.9/500) 225 x 407

          =   138.095mm²

         Since Ast > Asmin, minimum area requirement satisfied

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

      Asc  = 0.2×599.8

      =119mm²

        Provide 2T16

Shear Strength Design

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

V_Rdc = (0.12K(100_ρLf_ck)^1/3 + K1σ_cp) b_wd

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

ρL = Asl/bwd = 440.9/225 x 450 = 0.005

V_Rdc = (0.12 x 1.7 (100 x 0.005 x 30) ^1/3) 225 x 407

V_Rdc = 50.3KN

Since V_Rdc (50.3KN) is less than V_Ed (58.6KN) then shear reinforcement has to be designed for.

2. Calculate the shear resistance using ϴ  = 22

V_Rdmax(22) = 0.124b_wd(1 – f_ck/250)f_ck

= 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

$$
\frac{A_{sw}}{s}\,\,\,\,=\,\,\frac{V_{Ed}}{0.78xdxfyk\cot22}
$$

$$
\frac{A_{sw}}{s}\,\,\,\,=\,\,\frac{58.6×10^3}{0.78x407x500\cot 22}
$$

$\frac{A_{sw}}{s}\,\,\,\,$ = 0.15

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

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

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

= 0.19

Since A_smin/s is greater than As/s, shear reinforcement will be designed base on minimum area of reinforcement.

A_smin/s = 0.19

5) 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 = 0.19

s = 157/0.19

s = 796.9

6) Check whether maximum spacing requirement is satisfied

Smax = 0.75xd

= 0.75 x 407

= 305mm

provide Y10 @ 300mm spacing.

Deflection Check

Deflection Check

1. Calculate the actual span-effective depth ratio

Span/depth ratio = 5000/407   = 12.3

2. Calculate the limiting Span-effective depth ratio

l/d = K[11 + 1.5√f_ck ρ0/ρ + 3.2√f_ck (ρ^o/ρ – 1)3/2] if ρ ≤ ρ^o

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

ρ = Asprovided/b x d

ρ = 599.8/225 x 407

= 0.006

ρ^o = 10^-3√fck

ρ^o = 10^-3√30

= 0.005

K = 1 (for simply supported)

Since ρ >ρ^o  = then we will use

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

l/d = [11 + 1.5√30 0.005/0.006 + 3.2√30 √0.005/0.006]

= 21.8                   

Since actual span-effective depth ratio is less than the limiting span-effective depth ratio, the beam passes deflection check.

Click here to also study the design of doubly reinforced beam to Eurocode 2