Design of Beams to BS 8110

BEAM DESIGN PROCEDURE Step Task

Standard

1

Determine design life

2

Assess actions on the beam

BS 6399-1,2,3

3

Determine which combinations of actions apply

BS 8110-1-1997-Cl 2.4.3.1

4

Determine loading arrangements

BS 8110-1-1997-Cl 2.4.3

5

Assess durability requirements and determine concrete strength

BS 5328-1-1997-Cl 3.1.5.

6

Check cover requirements for appropriate fire resistance period

BS 8110-1-1997-Table 3.4

7

Calculate min. cover for durability, fire and bond requirements

BS 8110-1-1997-Table 3.3

8

Analyze structure to obtain critical moments & shear forces

BS 8110-2-1997-Cl 3.4.4.1.

9

Design flexural reinforcement

BS 8110-1-1997-Cl 3.4.4.4

10

Check shear capacity

BS 8110-1-1997-Cl 3.4.5

11

Check deflection

BS 8110-1-1997-Cl 3.4.6

12

Check spacing of bars

BS 8110-1-1997-Cl 3.12.11

• Concrete is strong in compression but weak and unreliable in tension. • Reinforcement is required to resist tension due to moment. • A beam when loads applied, Apply load

compression

tension

• Concrete at the top resists compression and the steel resists tension at bottom. • Design is based on the strength of the section calculated from the stress distribution at collapse.(at ultimate condition, not in serviceability conditions) • Therefore beam section design for the ultimate state. • An elastic section analysis is later carried out for checking the serviceability limit states.

Assumptions and stress-strain diagrams 1. The strains in the concrete and reinforcement are derived assuming that plane sections remain plane;

2. The stresses in the concrete in compression are derived using either (a) the design stress-strain curve with ϒm=1.5 or (b) the simplified stress block where the depth of the stress block is 0.9 of the depth to the neutral axis Note that in both cases the strain in the concrete at failure is 0.0035;

3. The tensile strength of the concrete is ignored; 4. The stresses in the reinforcement are derived using ϒm=1.05. 5. Where the section is designed to resist flexure only, the lever arm should not be assumed to be greater than 0.95 of the effective depth.

b

0.0035

0.45 fcu

0.45 fcu k2 x C=k1 bx

ϵ0

Neutral Axis T

d

(a)

(b)

(c)

0.95 fy (d)

(a) Section; (b) Strain; (c) rectangular parabolic strain diagram; (d) simplified stress diagram

Stress-strain curve for concrete

Stress-strain curve for reinforcement

Moment of Resistance – Simplified stress block According to the beam section and the strain and stress diagrams, The concrete stress is, 0.67 fcu / ϒm = 0.67 fcu / 1.5 = 0.45 fcu The concrete strain is 0.0035. The steel stress is fy / 1.05 = 0.95fy According to the simplified stress diagram the internal forces are,

C = force in the concrete in compression = 0.45 fcu x 0.9b x 0.5d = 0.201 fcu bd For the internal forced to be in equilibrium C = T ; MR = Moment of Resistance = Cz = 0.201 fcu bd x 0.775d = 0.156 fcu bd2 Where the constant K=0.156, MR = Kfcu bd2

T = force in the steel in tension = 0.95 fy As z = lever arm = d – 0.5 x 0.9 x 0.5d = 0.775d

START

Procedure for determining flexural reinforcement

Carry out analysis of beam to determine design moments(M) Determine k from K = M/(bd2fcu ) Is K< K’ ?

NO

Compression reinforcement required

YES

No compression reinforcement required Use following equations to calculate r/f area

 K  z  d 0.5  0.25    0.95d 0.9  

Use following equations to calculate r/f area  K'  z  d 0.5  0.25   0.9  

x  (d  z ) 0.45

x  (d  z ) 0.45

As '  ( K  K ' ) f cubd 2 0.95 f y (d  d ' )

As  M 0.95 f y z

As  K ' f cubd 2 0.95 f y z  As '

Check for maximum and minimum reinforcement requirements for tension and compression reinforcement

Design of flanged beams Flanged beams occur where beams are cast integral with and support a continuous floor slab. Part of the slab adjacent to the beam is counted as acting in compression to form T and L shape beams

The effective breadth b of flanged beams is given by 1. T-beams – web width bw+lz/10 or the actual flanged width if less 2. L-beams – web width bw+lz/5 or the actual flanged width if less lz is the distance between points of zero moment (which for a continuous beam, may be taken as o.7 times the effective span) The design procedure depends on where the neutral axis lies. The neutral axis may lie in the flange or in the web. If it is in web it needs to check whether the section needs compression reinforcement.

Neutral Axis is in flange To satisfy the criteria the actual neutral axis depth (0.9X) should not exceed flange depth hf. The moment of resistance of the section for the case when 0.9X = hf ,

MR = 0.45 fcu b hf (d- hf/2) If the applied moment M is lesser than the moment of resistance of the flange MR neutral axis lies within the flange.

Neutral Axis is in web Equation in the code is derived using the simplified stress block with X=o.5d; depth of stress block = 0.9X = 0.45d

As 

M  0.1 f cubw d (0.45d  h f ) 0.87 f y (d  0.5h f )

This applies only when X is less than 0.5d. If otherwise the section should design for the compression reinforcement also.

START

Designing of beams with flanged sections

Carry out analysis of beam to determine design moments(M) Find the Moment of Resistance of the flange section(MRF) Is M > MRF ?

NO

Neutral axis is in flange

YES Design is same as for a rectangular beam

Neutral axis is in web Find the Moment of Resistance of the section when neutral axis depth is d/2 -(MR) Is M > MR ? YES Compression reinforcement required

NO

No compression reinforcement required Use following equations to calculate r/f area M  0.1 f cubw d (0.45d  h f ) As  0.87 f y (d  0.5h f )

Shear reinforcement in beams Action of shear reinforcement Concrete is weak in tension, and so shear failure is caused by a failure in diagonal tension with cracks running at 45o to the beam axis. Shear reinforcement is provided by bars which cross the cracks, and theoretically either vertical links of inclined bars will serve this purpose.

(b)

(a) (a) Inclined bars and links ;

Design shear stress in any cross section;  

(b) vertical links

V bv d

The design concrete shear stress is given in Table 3.8. After compare the values it can find the form and area of shear reinforcement using Table 3.7

Design shear resistance of beams START Find the design shear stress -

Find the design concrete shear stress -

Minimum links should be provided in all beams of structural importance

Minimum links for whole length of beam

Provide links or links combined with bent-up bars, not more than 50% of the shear resistance provided by the steel may be in the form of bent-up bars

Check for deflection of beams

START Basic span/eff. depth ratio(s/d) - TABLE 3.9 If span >10m ; (Table 3.9)x(10/span) Modification factor for tension reinforcement(MFT) - TABLE 3.10

Actual span/effective depth ratio (Act. s/d)

Modification factor for compression reinforcement(MFC) - TABLE 3.11 Allowable span/effective depth ratio (All. s/d = (s/d) x MFT x MFC )

Act. s/d < All. s/d NO Design is NOT OK

YES Design is OK

Basic span/effective depth ratio Table 3.9 – Basic span/effective depth ratio for rectangular or flanged beams Support conditions

Rectangular section

Flanged beam with (bw/b < 0.3)

Cantilever

7

5.6

Simply supported

20

16.0

Continuous 26 20.8 For values greater than 0.3, linear interpolation between the values given in Table 3.9 for rectangular sections and for flanged beams with bw/b of 0.3 may be used

Modification factors for tension reinforcement Modification factors for tension reinforcement is given in Table 3.10 of the code. These values were derived from the equation; (477  f s ) Modification _ Factor  0.55   2.0 M   120 0.9  2  bd  Where  M is the design ultimate moment at the center of the span or, at the support .

The design service stress(fs) in the tension r/f in a member is found by following 2 f y As ,req 1 equation,

fs 

3 As , prov



b

Modification factor for compression reinforcement Modification factor for compression reinforcement is given in Table 3.11 These values were derived from the equation;

1  100

A' s , prov bd

A' s , prov   3  bd 

   1.5 

100 A' s, prov bd

Factor

0.00

1.00

0.15

1.05

0.25

1.08

0.35

1.10

0.50

1.14

0.75

1.20

1.0

1.25

1.5

1.33

2.0

1.40

2.5

1.45

>3.0

1.50

Check for the deflection

•Allowable span/eff. depth can be calculated by multiplying basic span/eff. depth from Table 3.9 by the modification factors for tension and compression reinforcement. •Then it is compared with the actual span-to-effective depth ratio. •If allowable s/d is greater than actual s/d, beam is satisfies the deflection criteria. •If otherwise not