flange beam design · flange beam flanged beams occur when beams are cast integrally with and...

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Flange Beam Design

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Semester 1 2016/2017

Department of Structures and Material Engineering

Faculty of Civil and Environmental Engineering

University Tun Hussein Onn Malaysia

Introduction

▪ Main reinforcement

▪ Shear reinforcement

▪ Traverse reinforcement

Key

Different

Design Procedures

Step Task Standard

1 Determine design life EN 1990:2002 Table 2.1

2 Determine beam size EN 1992-1-1: Table 7.4N

EN 1992-1-2: Table 5.5 & 5.6

3 Determine the effective width EN 1992-1-1: Sec.5.3.2.1

4 Determine design actions on beam EN 1991-1-1

5 Durability and characteristic strengths EN 1992-1-1: Sec. 3 & 4

6 Determine nominal cover EN 1991-1-1: Sec.4.4.1

7 Calculate moment and shear force EN 1992-1-1: Sec.5

8 Design of flexural reinforcement EN 1992-1-1: Sec.6.1, 9.2.1.1

9 Design of shear reinforcement EN 1992-1-1: Sec.6.2

10 Design for traverse

11 Check deflection EN 1992-1-1: Sec.7.4

12 Check cracking EN 1992-1-1: Sec.7.3

13 Detailing EN 1992-1-1: Sec.8 & 9.2

Moment

▪ Negative moment - it should be noted that when T-beam is

subjected to negative moment, the slab at top of web will be in

tension while the bottom web is in compression. This usually

occurs at interior support of continuous beam.

Moment

Flange Beam

▪ Flanged beams occur when beams are cast integrally 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-beam.

▪ The effective width of flange, beff is given in Sec. 5.3.2.1 of

EC2 and should be based on the distance lo.

where;

beff = effective flange

width

bw = breadth of the web

of the beam.

hf = thickness of the

flange.

beff beff

bw bw

hf

h

T-Beam L-Beam

Flange Beam

▪ The design procedure of flange beam depends on where the

neutral axis lies.

▪ The neutral axis may lie in the flange or in the web.

▪ There are three cases that should be considered:

- Neutral axis lies in flange (M < Mf)

- Neutral axis lies in web (M > Mf but < Mbal)

- Neutral axis lies in web (M > Mbal)

beff

bw

hf

d

x

beff

bw

hf

d

x

Flange Beam

▪ The effective width of flange, beff is given in Sec. 5.3.2.1 of

EC2.

▪ beff should be based on the distance lo between points of zero

moment as shown in the figure below.

Flange Beam

▪ The effective flange width, beff for T-beam or L-beam may be

derived as:

where:

,eff eff i wb b b b= +

, 0.2 0.1 0.2eff i i o ob b l l= +

,eff i ib b

Example 1

▪ Based on figure below, determine the effective flange width,

beff of beam B/1-3.

3000 4500

25

00

40

00

1 2 3

A

B

C

200 x 500

200 x 500 200 x 500

200 x 500 200 x 500

20

0 x

50

02

00

x 5

00

20

0 x

50

02

00

x 5

00

FS1 (150 thk.)

FS2 (150 thk.) FS3 (150 thk.)

20

0 x

50

0

Example 1

▪ lo (distance between points of zero moment)

▪ Effective flange width, beff

3000 mm 4500 mm

lo = 0.85 x 3000 = 2550 mm

lo = 0.85 x 4500 = 3825 mm

lo = 0.15 x (3000 + 4500) = 1125 mm

1 2 3

,eff eff i wb b b b= +

Example 1

▪ Span 1-2

beff1 = 0.2(1250) + 0.1(2550)

= 505 mm < 0.2lo = 510 mm < b1 = 1250 mm

beff2 = 0.2(2000) + 0.1(2550)

= 655 mm > 0.2lo = 510 mm < b2 = 2000 mm

beff = (505 + 510) + 200 = 1215 mm < 3250 mm

2500 4000

b1 = 2500/2 = 1250 b2 = 4000/2 = 2000

beff,1 beff,2

bw = 200 mm

beff

b = 1250 + 2000 = 3250 mm

A B C

Example 1

▪ Span 2-3:

beff1 = 0.2(1250) + 0.1(3825)

= 632.5 mm < 0.2lo = 765 mm < b1 = 1250 mm

beff2 = 0.2(2000) + 0.1(3825)

= 782.5 mm > 0.2lo = 765 mm < b2 = 2000 mm

beff = (632.5 + 765) + 200 = 1597.5 mm < 3250 mm

2500 4000

b1 = 2500/2 = 1250 b2 = 4000/2 = 2000

beff,1 beff,2

bw = 200 mm

beff

b = 1250 + 2000 = 3250 mm

A B C

Example 1

▪ Dimension of flange beam:

Span 1-2 Span 2-3

beff = 1215mm

bw = 200 mm

hf = 150 mm

beff = 1598 mm

bw = 200 mm

hf = 150 mm

Design For Flexural

▪ Design procedure for flange beam:

1) Calculate Mf,

2) If M ≤ Mf, neutral axis in the flange and hence provide

tensile reinforcement only

( )0.567 0.5f ck f fM f bh d h= −

2

Ed

ck

MK

bd f=

0.5 0.25 0.951.134

Kz d d

= + −

,0.87

Ed

s req

yk

MA

f z=

Design For Flexural

3) If M > Mf, neutral axis in the web

, compression reinforcement is not required

, compression reinforcement is required

2

bal f ckM f bd=

( )

( )

0.1 0.36

0.87 0.5ck w f

s

yk f

M f b d d hA

f d h

+ −=

−

0.167 0.567 1 12

w f w f

f

b h b h

b d b d

= + − −

( )

( )'

'0.87f

s

yk

M MA

f d d

−=

−

( )'

0.167 0.567

0.87ck w ck f w

s s

yk

f b d f h b bA A

f

+ −= +

balM M

balM M

Design For Shear

▪ Design procedure:

1) Determine design shear force, VEd

2) Determine the concrete strut capacity for cot θ =1.0 and

cot θ =2.5 (θ = 22o and θ = 45o respectively)

If VEd > VRd,max cot θ = 1.0 (θ = 45o), redesign the section11

If VEd < VRd,max cot θ = 2.5, use cot θ = 2.5 (θ = 22o), and

calculate the shear reinforcement as follows,

( ),max

0.36 1 / 250

cot tanw ck ck

Rd

b df fV

−=

+

0.78 cotsw Ed

yk

A V

s f d = ( ); cot 2.5 =

Cl.6.2.3 (3)

Design For Shear

If VRd,max cot θ = 2.5 < VEd < VRd,max cot θ = 1.0

11

3) Maximum spacing

4) Calculate the minimum links,

( )10.5sin

0.18 1 / 250Ed

w ck ck

V

b df f − = −

0.78 cotsw Ed

yk

A V

s f d =

0.08sw w ck

yk

A b f

s f= Cl.9.2.2(5)

0.75s dMaximum

spacing

Design for Traverse

▪ Design procedure:

1) Determine distance Δx=0.5(L/2) which is the greatest

longitudinal shear stresses.

2) Change of moment over distance Δx

3) Change in longitudinal force

4) Calculate longitudinal shear stress

23

32

wLM =

0.5 2w

d

f

M b bF

d h b

− =

−

0.27d

ed ctk

f

Fv f

h x

=

Traverse shear reinforcement is required

Design for Traverse

5) Check concrete strut capacity in flange

6) Traverse shear reinforcement

7) Calculate minimum traverse reinforcement area

8) Check additional longitudinal reinforcement

( ),max

0.4 1 / 250

cot tanck ck

ed

f fv

−=

+

0.78 cotsf ed f

f yk

A v h

s f d =

,min 0.26 ctm

s f

yk

fA bh

f

=

,

0.5 cot

0.87Ed

s td

yk

VA

f

=

Deflection

▪ To control deflection to a maximum of span/250.

▪ Procedure:

1) Calculate ρo = √fck 10-3

2) Calculate ρ = As,req / bd

3) Calculate ρ’ = As’,req / bd

4) Determine K and calculate l/d

5) Calculate modification factor MFflange, MFspan and MFarea

6) For adequate deflection control, (l/d)actual < (l/d)allow

3/2

11 1.5 3.2 1 ; o o

ck ck o

lK f f

d

= + + −

'

'

111 1.5 ;

12o

ck ck o

lK f f

d

= + + −

Cl.9.2.2(5)

Deflection

▪ MFflange

- In Cl. 7.4.2(2), bf/bw > 3, Mfflange = 0.8

- Or bw/bf ≤ 0.3, Mfflange = 0.8, and for any, it should be

determined using the following graph.

Deflection

▪ Factor for structural system can be determined from Table

7.4N:

Cracking

▪ Crack control for beam design can be directly referred to

Cl.7.3.

▪ For a convenient, crack control without direct calculation is

preferable, Cl. 7.3.3, Table 7.2N and Table 7.3N.

▪ Procedure:

1) Calculate steel stress for limiting crack width, wk = 0.3mm

2) Determine maximum bar size or bar spacing

3) For adequate crack control,

Sprov .< Smax

+ = +

,

,

0.3435

1.35 1.5s reqk k

s

k k s prov

AG Qf

G Q ATable 7.2N

Table 7.3N

Cracking

▪ Maximum bar spacing for crack control (Table 7.3N)