Design of a Singly Reinforced Beam to BS 8110 – Worked Example

This article presents a worked example on the design of singly reinforced concrete beam to BS 8110:1:1997.

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

Design Data:

Imposed load on slab =   5KN/m²

Finishes =   1.5KN/m²

Unit weight of concrete   =   24KN/m³

Compressive strength of concrete (fcu)   =   25N/mm²

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

Characteristic Strength of Shear reinforcement (fyv) = 460N/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

Dead 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 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 dead Load on beam = 8.02 +2.4 = 10.42KN/m

Live Load

live load on slab = 5KN/m²

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

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

Total Design Load

Ultimate Load acting on Slab =1.4(10.42) + 1.6(6.25) =  24.43KN/m

 

Computation of Internal Forces:

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

M =  wL²/8  = 24.43 x 5² /8 = 76.9KNm

V   =   wL/2  = 23.44 x 5/2 = 61KN

 

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_{cu}}
$

$
K\,=\,\,\frac{76.9×10^6}{225×407^2×25}
$

 

= 0.08

Since K (0.08) < K’ (0.156); design 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{0.08}{0.9}} \right)
$

Since 365 < 0.95d (386.7): use Z = 365

 

4. Calculate the area of steel

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

 

$
A_{s\,\,=\,\,\frac{76.9×10^6}{0.87x460x365}}
$

 

As = 525.8mm²

Provide 3Y16 (599.8mm²)

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 compression reinforcement as secondary reinforcement which will serve as hanger bars

      Asc  = 0.002×599.8

      =202.5mm²

        Provide 2T12 (224.9mm²)

 

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

 

Shear Strength Design

1. Calculate the Shear Stress
   • v = V/bd
     • $\frac{61×10^3}{225×407}\,\,\,\,$
       • = 0.67N/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 525.8}{225 x 407} \right) ^{\text{1/}3}\frac{\left( \frac{400}{407} \right) ^{\text{1/}4}}{1.5}
$

v_c = 0.52N/mm²

 

Since V_c (0.52N/mm²) is less than V (0.67N/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 0.52    = 0.26N/mm²

v_c + 0.4 =  0.52 + 0.4  =0.92 N/mm²

Since 0.5v_c <  v < v_c+0.4; minimum shear links is required according to table 3.7

 

 

4. Calculate the required minimum 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

$
Using: Asv_{\min}=\,\,\frac{\text{0.4}x\,\,b_v\,\,x\,\,s_v}{0.9fy_v}\,\,
$

$
Asv_{\min}=\,\,\frac{\text{0.4}x\,\,225\,\,x\,\,s_v}{0.9 x 460}\,\,
$

Since A_sv = 157mm² ; make s_v the subject of the formular

s_v = 763mm

 

 

5) Check whether maximum spacing limit is satisfied

Smax = 0.75xd

= 0.75 x 407

= 305mm

Since 763mm is greater than maximum spacing limit, maximum spacing limit governs the design.

Hence: 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

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

Since basic span-effective depth ratio (20) is greater than the actual span-effective depth ratio, the beam passes deflection check.

 

 

Reference(s)

BS 8110:1:1997 – Code of practice for design and construction