Beam Design to BS 8110

Basic Structural Element Design to BS 8110-1:1997

2.0

Beam Design

Bending Moment Diagramm 1000 800 ) m N 600 k ( 400 t n 200 e m 0 o M -200 g n -400 i d n -600 e B -800 -1000 0

2

4

6

8

10

12

Length (m)

Bending Moment Diagramm

Figure 2.1: Bending Moment Diagram (Span 4-6)

Shear Force Diagramm 500 ) 400 m 300 N k ( 200 t n 100 e m 0 o M -100 g n i -200 d n e -300 B -400 -500

443kN

438kN

0

2

4

6

8

10

12

Length (m)

Shear Force Diagramm

Figure 2.2: Shear Force Diagram (Span 4-6)

Edifice Consultants Pvt.Ltd

Page 13


Reference

Calculations

Out Put

Design of Hogging Moment at LHS Support

M=804Nm Beam Size 350mmx750mm and Span of the Beam in 12.9m Cover to reinforcement is 25mm. Assume T25 bars and T10 Links are to be used. Hence Effective Depth (d) d = 750 − 30 − 10 − Cl.3.4.4.4 BS 8110:1-1997

k=

M f bd2 cu

=

25 2

= 697.5mm

804 × 106 30 × 350 × 697.52

= 0.157 > 0.156

Hence compression reinforcement is required. d′ = 30 + 10 +

25 2

= 52.5mm

Compression reinforcement requirement can be found as follows. A′s

=

=

2 (K - K′)f cu × b × d 0.95f y × (d − d′ ) 2 (0.157 − 0.156) × 30 × 350 × 697.5

0.95 × 460 × (697.5 − 52.5) = 18.12mm2

Area of tension reinforcement can be found as follows.

 

  0.9 



0.156 

z = d 0.5 + 0.25 −

z = d 0.5 + 0.25 −



As

=

0.95f y z

=


2 K′f cubd

K′

0.9

 = 0.78d 

+ A′

s

2 0.156 × 30 × 350 × 697.5 0.95 × 460 × 0.78 × 697.5

+ 18.12 = 3358mm2

Page 14


Reference

Calculations

Out Put

Provide 4T25+4T25 at support. (See Figure 7.3 ) 2 A s,provided = 3928mm

Table 3.25 BS 8110:1-1997

4T25+4T25 At support 4

Checks for Minimum area of reinforcement

100A

s = 100 × 3928 = 1.49 > 0.13 bh 350 × 750

Hence minimum steel requirement is satisfied. Clause 3.12.6

Checks for Maximum area of reinforcement

BS 8110:1-1997

Neither the area of tension reinforcement nor compression

Minimum r/f Ok.

Maximum r/f Ok

reinforcement should exceed 4% of cross sectional area of the concrete.

Figure 7.3: Beam Section at support 4/6


Figure 6.4: Beam Section at mid span of 4/6

Page 15


Reference

Calculations

Out Put

Design of Hogging Moment at RHS Support

M=784.8kNm Beam Size 350mmx750mm Cover to reinforcement is 25mm. Assume T25 bars and T10 Links are to be used. Hence Effective Depth (d) d = 750 − 30 − 10 − Cl.3.4.4.4 BS 8110

k=

M f bd2 cu

=

25 2

= 697.5mm

784.8 × 106 30 × 350 × 697.52

= 0.154 < 0.156

Hence the beam can be designed as a singly reinforced beam.

 

  0.9 



0.154 

z = d0.5 + 0.25 −

z = d0.5 + 0.25 −



As =

M 0.95f y z

=

k

0.9

 = 0.79d < 0.95d 

784.8 × 10

6

0.95 × 460 × 0.79 × 697.5

= 3259mm2

Provide 4T25+4T25 at support. (See Figure 7.3 ) 2 A s,provided = 3928mm

Table 3.25 BS 8110:1-1997

4T25+4T25 At support 6


100A

s = 100 × 3928 = 1.49 > 0.13 bh 350 × 750

Hence minimum steel requirement is satisfied. Cl. 3.12.6


BS 8110:1-1997


Minimum r/f Ok.

Maximum r/f Ok



Page 16


Reference

Calculations

Out Put

Design for Sagging Bending Moment between supports

M=859.5kNm(as per analyzed results with SAP 2000) At Mid Span Top of the beam in compression and can be designed as a flange beam. (“T” Beam) Hence effective width of the flange beam (“T” beam) Cl.3.4.1.5 BS 8110:1-1997

lz = 0.7 × 12900 = 9030mm b w = 350 +

9030 5

= 2156mm

Moment of resistance of the Flange “T” beam when the Neutral Axis is at the slab bottom level,

 h  = 0.45f cubwhf ×  d − f 

M



2 

 

= 0.45 × 30 × 2156 × 125 ×  697.5 −

125  2

 

= 2310.29kNm > 859.5kNm Hence Neutral axis is within the bottom flange. Cl.3.4.4.4 BS 8110

k=

M f bd2 cu

=

859.5 × 106 30 × 2156 × 697.52

= 0.026 < 0.156

Hence the beam can be designed as a singly reinforced beam.

 

  0.9 



0.026 

z = d0.5 + 0.25 −

z = d0.5 + 0.25 −



k

0.9

 = 0.97d > 0.95d 

Hence z =0.95d As =

M 0.95f y z

=

859.5 × 10

6

0.95 × 460 × 0.95 × 697.5

= 2947mm2

Provide 4T25+4T25 at mid span. (See Figure 7.4 )

4T25+4T25

2 A s,provided = 3928mm

At mid span


Page 17


Reference

Table 3.25 BS 8110:1-1997

Calculations

Out Put


100A

s = 100 × 3928 = 1.49 > 0.13 bh 350 × 750

Hence minimum steel requirement is satisfied.

Clause 3.12.6


BS 8110:1-1997


Minimum r/f Ok.


Maximum r/f Ok.

Checks for shear

Maximum shear at LHS Support= 443kN Cl.3.4.5.2 BS 8110:1-1997

Cl.3.4.5.2 BS 8110:1-1997

vmax =

443 × 10

3

(350 × 697.5)

= 1.81N/mm2

Maximum permissible shear stress is lesser of 0.8 f cu or 5N/mm2

0.8 30 = 4.38N/mm > 1.80N/mm 2

2

Hence maximum shear is O.K

Maximum Shear Ok

Shear force at distance “d” from the face of the support V = 430kN d

Hence , shear stress at distance “d” from the face of the support V = 430kN d

v=

430 × 10

3

(350 × 697.5)


= 1.75N/mm2

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Reference

Cl. 3.4.5.4 BS 8110:1-1997

Calculations

Out Put

Concrete shear Stress

The area of tension reinforcement at a distance “d” from the

2 face of the support is 3574mm Table 3.8

100As

BS 8110:1-1997

bd

 400   d 

=

100 × 3928 (350 × 697.5)

1 4

= 1.59 < 3

1

 400  4 = = 0.87 < 1  697.5 

Hence design concrete shear stress is,

vc

1 100As   3

= 0.79 × 

  bd 

1 400   4

×

  d 

×

1 δm

1

 f cu 

×

  25 

3

1 1 1 1 30  3  = 0.79 × (1.59) 3 × (1) 4 × ×  1.25  25 

= 0.77N/mm2

Table 3.7 BS 8110:1-1997

Spacing of the shear links

2 v c + 0.4 = 0.77 + 0.4 = 1.17N/mm v c + 0.4 < v < 0.8 f cu d

Assume T 10 Links at supports. Number of shear legs is 2.

 π × 102  2 A sv = 2 ×   = 157mm  4  Hence maximum possible shear link spacing at the support is,

Sv ≤

A sv × 0.95 × f yv (v − v c )bv

=

157 × 0.95 × 460 (1.75 − 0.77) × 350

= 200mm

Hence Provide T10 links at 150mm C/C. Cl. 3.4.5.5 BS 8110:1-1997

Maximum possible spacing of links is 0.75d = 523.125mm

T10@ 150mm Links

Hence maximum spacing is satisfied.


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Reference

Calculations

Cl. 3.4.1.6

Checks for Slenderness limit for lateral stability

BS 8110:1-1997

Clear distance between restraints is 12150mm.

Out Put

60bc = 60 × 350 = 21000mm > 12150mm 250b2 250 × 3502 c = = 43907mm > 12150mm d 697.5

Hence slenderness limit for lateral stability is satisfied.

Cl.3.4.6

Checks for Deflection

BS 8110:1-1997

Mid span of the section is designed as a flanged section.

b b Table 3.9 BS 8110:1-1997

350

=

2156

w

= 0.16 < 0.3

For simply supported beam,



  = 16  Effective Depth  Span

Basic 

For a Continuous beam,



  = 20.8  Effective Depth  Span

Basic 

Cl.3.4.6 .4

Span of the beam is 12.9m.

BS 8110:1-1997

 10  Hence modification factor for span is    12.9  In this beam end condition at support 6 can be regarded as simply supported and end condition at support 4 can be regarded as continuous.



 16 + 20.8 = 18.4 = Effective Depth 2   Span

Basic 

Actual

  12900 Span  Effective Depth  = 697.5 = 18.49   

  = Basic Effective Depth  

Allowable 


Span

  Span   × F1 × F2 Effective Depth  

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Reference

Calculations

Out Put

F1 -Modification Factor for Tension reinforcement

F2 - Modification Factor for Compression reinforcement Table 3.10 BS 8110:1-1997

f s =

2 3

× f y ×

A s,req A s,pro

×

1 βb

βb = 1

f s =

2 3

× 460 ×

2947 3928

1

× = 230.08N/mm2 1

6 M 859.5 × 10 = = 0.81 2 2 × bd 2156 697.5 F1 = 0.55 +

(477 − f s ) (477 − 230.08) = 0.55 + = 1.75 < 2 M 120(0.9 + 0.81) 120(0.9 + 2 ) bd

No Compression reinforcement is provided. Hence F2 = 1



  = Basic  Effective Depth 

Allowable 

Allowable

Span

  Span   × F1 × F2  Effective Depth 

  Span 10 = × × × = 24.96 18.49 1.75 1  Effective Depth  12.9   

   Span > Actual     Effective Depth   Effective Depth 

Allowable 

Span

Hence deflection check is satisfied The deflection of the beam is checked in SAP 2000 under Serviceability limit state and found to be 13.6mm which is less than 20mm , the allowable deformation in general to avoid cracking of brittle partition walls Deflection Ok


Page 21


Reference

Calculations

Out Put

Detailing of the reinforcement

Curtailment of Bars Curtailment of bars is done according to the guidelines specified in BS 8110-1:1997 and standard method of detailing manual. Procedure of curtailment of bars at right hand side of support 4 is described here. At mid span provided reinforcement for maximum bending is 4T25+4T25. It is desired to curtail 4T25 bars. As per Figure 3.24 of the BS 8100-1:1997 at least 30% of reinforcement for maximum sagging moment should be continued throughout the beam. In this case curtailment percentage is 50%. Hence area of continuing reinforcement is ,

3928 × 0.5 = 1964mm2 For the sectional equilibrium of simplified stress block of a rectangular section at ultimate state (See Figure 2.3) Compression of concrete =Tension of bottom steel

0.67f cu δm

× b × 0.9x = 0.95 × f y × A s

0.67 × 30 1.5

× 350 × 0.9x = 0.95 × 460 × 1964

Hence X =203.3mm x

=

d

203.3 697.5

Z = d−

= 0.291 < 0.5

0.9x 2

= 697.5 −

0.9 × 203.3 2

= 606.02kNm

Hence moment of resistance (MOR) is,

M = 0.95f y AsZ = 0.95 × 460 × 1964 × 606.02 = 524.42kNm


Page 22


Figure 2.3: Simplified stress block of a rectangular section at ultimate state

Curtailment of Reinforcement 1000 800 600 ) m 400 N k ( 200 t n e m 0 o M g -200 n i d n e -400 B

262.2kNm

-600

524.4kNm

-800 -1000 0

2

4

6

8

10

12

Length (m)

Bending Moment Diagramm

MOR of 4T25

1/2 MOR of 4T25

Figure 7.4: Bending moment diagram and MOR


Page 23


Reference

Calculations

Out Put

Bending moment diagram of the beam segment and Moment or Resistance corresponds to 4T25 is drawn in Figure 6.6 The point where MOR line cuts the bending moment diagram is referred to as the Theoretical Cut Off Point. (TCP). Cl.3.12.9.1

In a flexural member , every bar should be extended beyond

BS 8110:1-1997

the TCP for a distance equal to greater of 1.0

Effective depth of the member =702.5mm

2.0

Twelve times the bar size

=12x25 = 300mm

The point of physically cutting off the bars is referred as (PCP) In addition the least distance of, 3.0 Full anchorage bond length =40x25 = 1000mm 4.0 At PCP actual shear < Half shear capacity 5.0 At PCP moment< Half moment at TCP Condition 4 is not critical and hence not checked. Distance to TCP from centerline of LHS is 3.6m. Distance to PCP as per condition “5” from TCP is 1m. Hence Distance to PCP from centre line of Support is 3.61=2.6m Hence 4T 25 is curtailed 2m from the centre line of the support.


Page 24


Reference

Calculations

Out Put

Anchorage

Anchorage length required beyond the face of the column is given by Table 3.27

L = 40 × φ = 40 × 25 = 1000mm

BS 8110:1-1997

Stress in reinforcement is,

= 0.95 × 460 ×

3315 3928

= 368.80N/mm2

Assume bend starts at the centre line of column. Stress at the start of the bend is,

= 368.80 ×

(1000 − 350) 1000

= 239.72N/mm2

3.12.8.25.2

Bending stress inside the bend

BS 8110:1-1997

2f Fbt cu ≤ rφ 1 + 2( φ

a b

)

a = 30 + 25 = 55mm b 239.72 × 491 r × 25

≤

2 × 30 1 + 2(25 ) 55

r = 149.80mm Hence use a radius of 150mm. The bend is started as close as possible to the edge of the column. Hence length from the face of the column to start of the bend is

= 700 − 25 − 150 = 525mm


Page 25


Reference

Cl. 3.12.8.23 BS 8110:1-1997

Calculations

Out Put

Effective anchorage length of 90 Bend 

= 4 × 150 = 600mm = 12 × 25 = 300mm Hence Vertical length required is,

= 1000 − 175 − 300 = 525mm

C.l.3.12.11

Spacing of the reinforcement

BS 8110:1-1997 C.l.3.12.11

Minimum distance between bars

BS 8110:1-1997

Clear distance between bars is

=

350 − 2(25 + 10) − 4 × 25 3

= 60mm

Minimum distance between bars should not be less than

= 5+h

agg

= 5 + 20 = 25mm

C.l.3.12.11.2.3

Maximum Clear Horizontal distance between bars in tension

BS 8110:1-1997

with 0% of moment redistribution is 155mm. Hence maximum spacing of bars in tension is satisfied.

C.l.3.12.11.2.6

Bars near side faces of beams exceeding 750mm overall depth

BS 8110:1-1997

Spacing in not exceeding 250mm near the face of the beam. The distribution should be a distance of two third of the beam’s overall depth measured from its tension face.

Cl.3.12.5.4

Minimum size of bars in side faces of beams to control cracking

=

s b b = f y

160 × 350 460

= 11.03mm

Hence Provide T12 bars.


Page 26

Beam Design to BS 8110

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