Design of RC Flange beam to BS 8110 – Worked Example

This article presents a work example on the design of flanged beam to BS 8110-1-1997. The Beam 2 (shown in the plan below) that shall be designed here as a flanged section had already been designed as a rectangular section here. This worked example shall make bare the importance of designing beams as flanged rather than as a rectangular section when appropriate.  

Design Data:

Imposed load on slab =   5KN/m²

Finishes =   1.5KN/m²

Unit weight of concrete   =   24KN/m³

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

Characteristic Strength of main reinforcement (fy)  =  460N/mm²

Characteristic Strength of Shear reinforcement (fyv) = 460N/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 = 1.4(13.5) + 1.6(8.59) = 32.3KN/m

Computation of Internal Forces:

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

M = w x L²/8 =   32.3 x 8²/8 = 258.5KNm

V   = W x L/2 =    32.3 x 8/2   = 129.24KN

Effective flange width:

The section of the beam is L, hence flange width equals:

web width + (whichever is smaller between) Lz/10 or actual flange width

225 + 8000/10 = 1025mm

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

To calculate this, let’s assume the stress block to resist compression is within the flange

$ K\,=\,\,\frac{M}{bfd^2f_{cu}} $

$ K\,=\,\,\frac{258.5×10^6}{1025×407^2×30} $

= 0.05

Since K (0.05) < K’ (0.156); design as singly reinforced.

In the previous worked example here where the beam is designed as a rectangular section, the beam has to be doubly reinforced. However, in this worked example where the flange is considered, the capacity of the compression fiber of the section is augmented and no further reinforcement are needed in the compression zone, hence the beam is to be designed as singly reinforced.

 

3) Calculate the lever arm (Z)

$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{0.9}} \right) $

 

$ Z\,\,=\,\,407\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.05}{0.9}} \right) $

Since 382 < 0.95d (386.7): use Z = 382

 

4. Calculate the area of steel

$ A_{s\,\,=\,\,\frac{M}{0.87f_{y}Z}} $

 

$ A_{s\,\,=\,\,\frac{258.5×10^6}{0.87x460x382}} $

As = 1546mm²

Provide 8Y16 (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.

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

               A_smin = 0.0013bd

            = 0.0013 x 225 x 407

          =   131.625mm²

         Since Ast > Asmin, minimum area requirement satisfied

 

6)  For practical purpose of forming a reinforcement cage. Provide minimum compression reinforcement as hanger bars

                                                          Asc   = 0.013Ac

                                                              Asc = 0.0013 x 225 x 450

                                                             Asc    =131.6mm²

                                                               Provide 2T16(399.9mm²)

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

Shear Strength Design

1. Calculate the Shear Stress
   • v = V/bd
     • $\frac{129.24×10^3}{225×407}\,\,\,\,$
       • = 1.4N/mm²
2. Check whether the concrete section can resist the shear force without shear reinforcement.

 

$ v_{c\,\,=}\,\,\text{0.79}\left( \frac{100As}{b_vd} \right) ^{\text{1/}3}\frac{\left( \frac{400}{407} \right) ^{\text{1/}4}}{\gamma _m} $

 

$ v_{c\,\,=}\,\,\text{0.79}\left( \frac{100 x 1546}{225 x 407} \right) ^{\text{1/}3}\frac{\left( \frac{400}{407} \right) ^{\text{1/}4}}{1.5} $

v_c = 0.8N/mm²

 

Since V_c (0.8N/mm²) is less than V (1.4N/mm²) then shear reinforcement has to be designed for.

 

3. Check the form of shear links to be provided according to table 3.7.

0.5v_c =   0.5 x 0.8    = 0.4N/mm²

v_c + 0.4 =  0.8 + 0.4  =1.2 N/mm²

Since  v > v_c+0.4; shear links must be designed and the shear stress (v) must be checked not to be more than 0.8√fcu or 5N/mm²

0.8√fcu = 0.8 x√30 = 4.32N/mm²       

Since v (1.4) is less than 0.8√fcu then the beam section is adequate and shear reinforcement only needs to be designed.

4. Calculate the required shear links.

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

$
Asv_{}=\,\,\frac{b_v\,\,x\,\,s_v\left( v-v_c \right)}{0.95fy_v}\,\,
$

$
Asv_{}=\,\,\frac{225\,\,x\,\,s_v\left( 1.4-0.8 \right)}{0.95 x 460}\,\,
$

Substituting 157mm² for Asv and making sv the subject of the formular

s_v = 497mm

 

5) Check whether maximum spacing limit is satisfied

Smax = 0.75xd

= 0.75 x 407

= 305mm

Since 497mm is greater than maximum spacing limit, maximum spacing limit governs the design.

Hence: provide Y10 @ 300mm spacing.

Click here to also study a worked example on the design of flange beam to Eurocode 2

Deflection Check

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

From table 3.9 of BS 8110, the basic Span/depth ratio for a simply supported beam is 20.

Tension reinforcement modification factor

$
\text{0.55}+\frac{477-\,\,f_s}{120\left( \text{0.9\,\,}+\,\,\frac{M}{bd^2} \right)}
$

where:

$
f_s\,\,=\,\,\frac{2f_yA_{sreq}}{3A_{sprov}}
$

$
f_s\,\,=\,\,\frac{2 x 460 x 1546}{3 x 1599}
$

fs = 296.4

Substitute fs and values of other parameters into the tension modification formular

$
\text{0.55}+\frac{477-\,\,296.4}{120\left( \text{0.9\,\,}+\,\,\frac{258.5×10^6}{225 x 407^2} \right)}
$

= 0.74

Compression reinforcement modification factor

$
\frac{1+\frac{100Asprov}{bd}}{3+\frac{100Asprov}{bd}}
$

$
\frac{1+\frac{100 x 399.8}{225 x 407}}{3+\frac{100 x 399.8}{225 x 407}}
$

= 1.1

Modified basic span-effective depth ratio = basic span-effective depth ratio x M.F tension rebars x M.f compression rebars

Modified basic span-effective depth ratio = 20 x 0.74 x 1.1

= 16.28

Since Actual L/d (19.6) > modified Basic L/d (16.28); the beam fails deflection check.

What would you do to make the beam pass deflection check?