Concrete Design to Eurocode 2

Concrete Design

to Eurocode 2

Jenny Burridge

MA CEng MICE MIStructE

•

Introduction to the Eurocodes

•

Eurocode

•

Eurocode 1

•

Eurocode 2

· Materials · Cover · Flexure · Shear · Deflection

•

Further Information

• BS EN 1990 (EC0) : Basis of structural design

• BS EN 1991 (EC1) : Actions on Structures

• BS EN 1992 (EC2) : Design of concrete structures

• BS EN 1993 (EC3) : Design of steel structures

• BS EN 1994 (EC4) : Design of composite steel and concrete structures

• BS EN 1995 (EC5) : Design of timber structures

• BS EN 1996 (EC6) : Design of masonry structures

• BS EN 1997 (EC7) : Geotechnical design

• BS EN 1998 (EC8) : Design of structures for earthquake resistance

• BS EN 1999 (EC9) : Design of aluminium structures

•

BS EN 1990 (EC0): Basis of structural design

•

BS EN 1991 (EC1): Actions on Structures

•

BS EN 1992 (EC2): Design of concrete structures

• BS EN 1993 (EC3): Design of steel structures

• BS EN 1994 (EC4): Design of composite steel and concrete structures • BS EN 1995 (EC5): Design of timber structures

• BS EN 1996 (EC6): Design of masonry structures

•

BS EN 1997 (EC7): Geotechnical design

• BS EN 1998 (EC8): Design of structures for earthquake resistance • BS EN 1999 (EC9): Design of aluminium structures

•

The Eurocodes contain Principles (P) which comprise:

◦

General statements and definitions for which there is

no alternative, as well as:

◦

Requirements and analytical models for which no

alternative is permitted

•

They also contain Application Rules, which are generally rules

which comply with the Principles

•

The Eurocodes also use a comma (,) as the decimal marker

•

Each Eurocode part has a National Annex which modifies the

main text of the Eurocode

National Annex

The National Annex provides:

•

Values of Nationally Determined Parameters (NDPs)

(NDPs have been allowed for reasons of safety, economy and durability)

• Example: Min diameter for longitudinal steel in columns

φφφφ_min = 8 mm in text φφφφ_min = 12 mm in N.A.

•

The decision where main text allows alternatives

• Example: Load arrangements in Cl. 5.1.3 (1) P

•

The choice to adopt informative annexes

• Example: Annexes E and J are not used in the UK

•

Non-contradictory complementary information (NCCI)

•

Introduction to the Eurocodes

•

Eurocode

•

Eurocode 1

•

Eurocode 2

· Materials · Cover · Flexure · Shear · Deflection

•

Further Information

Published 27 July 2002

Structures are to be designed, executed and maintained so that,

with appropriate forms of reliability, they will:

•

Perform adequately under all expected actions

•

Withstand all actions and other influences likely to occur during

construction and use

•

Have adequate durability in relation to the cost

•

Not be damaged disproportionately by exceptional hazards

The code sets out the following:

•

Basis for calculating design resistance of materials

•

Combinations of actions for ultimate limit state

•

Persistent

•

Transient

•

Accidental

•

Seismic

•

Combinations of actions for serviceability limit state

For one variable action:

1.25 G

_k

+ 1.5 Q

_k Provided:

1. Permanent actions < 4.5 x variable actions 2. Excludes storage loads

Eurocode

Design values of actions, ultimate limit state – persistent and transient design situations (Table A1.2(B) Eurocode)

Comb’tion expression reference

Permanent actions Leading variable action

Accompanying variable actions

Unfavourable Favourable Main(if any) Others

Eqn (6.10) γ_G,j,sup G_k,j,sup γ_G,j,inf G_k,j,inf γ_Q,1 Q_k,1 γ_Q,i Ψ_0,i Q_k,i Eqn (6.10a) γ_G,j,sup G_k,j,sup γ_G,j,inf G_k,j,inf γ_Q,1Ψ_0,1Q_k,1 γ_Q,i Ψ_0,i Q_k,i Eqn (6.10b) ξ γ_G,j,supG_k,j,sup γ_G,j,inf G_k,j,inf γ_Q,1 Q_k,1 γ_Q,i Ψ_0,i Q_k,i Eqn (6.10) 1.35 G_k 1.0 G_k 1.5 Q_k,1 1.5 Ψ_0,i Q_k,i Eqn (6.10a) 1.35 G_k 1.0 G_k 1.5 Ψ_0,1 Q_k 1.5 Ψ_0,iQ_k,i Eqn (6.10b) 0.925x1.35G_k 1.0 G_k 1.5 Q_k,1 1.5 Ψ_0,i Q_k,i

•

Introduction to the Eurocodes

•

Eurocode

•

Eurocode 1

•

Eurocode 2

· Materials · Cover · Flexure · Shear · Deflection · Axial

•

Further Information

Eurocode 1 has ten parts:

•

1991-1-1

Densities, self-weight and imposed loads

•

1991-1-2

Actions on structures exposed to fire

•

1991-1-3

Snow loads

•

1991-1-4

Wind actions

•

1991-1-5

Thermal actions

•

1991-1-6

Actions during execution

•

1991-1-7

Accidental actions due to impact and explosions

•

1991-2

Traffic loads on bridges

•

1991-3

Actions induced by cranes and machinery

•

1991-4

Actions in silos and tanks

Eurocode 1 Part 1-1: Densities, self-weight and imposed loads

•

Bulk density of reinforced concrete is 25 kN/m

3

•

The UK NA uses the same loads as BS 6399

•

Plant loading not given

•

Introduction to the Eurocodes

•

Eurocode

•

Eurocode 1

•

Eurocode 2

· Materials · Cover · Flexure · Shear · Deflection

•

Further Information

BS EN 1990 BASIS OF STRUCTURAL DESIGN BS EN 1991 ACTIONS ON STRUCTURES BS EN 1992 DESIGN OF CONCRETE STRUCTURES

Part 1-1: General Rules for Structures

Part 1-2: Structural Fire Design

BS EN 1992 Part 2: Bridges BS EN 1992 Part 3: Liquid Ret. Structures BS EN 1994 Design of Comp. Struct. BS EN 13369 Pre-cast Concrete BS EN 1997 GEOTECHNICAL DESIGN BS EN 1998 SEISMIC DESIGN BS EN 13670 Execution of Structures BS 8500 Specifying Concrete BS 4449 Reinforcing Steels BS EN 10080 Reinforcing Steels

Eurocode 2 Relationships

•

Code deals with phenomena, rather than element types

•

Design is based on characteristic cylinder strength

•

Does not contain derived formulae (e.g. only the details of the

stress block is given, not the flexural design formulae)

•

Unit of stress in MPa

•

One thousandth is represented by %

o

•

Plain or mild steel not covered

•

Notional horizontal loads considered in addition to lateral loads

•

High strength, up to C90/105 covered

Concrete properties (Table 3.1)

•

BS 8500 includes C28/35 & C32/40

•

For shear design, max shear strength as for C50/60

Strength classes for concrete

f

_ck

(MPa)

12 16 20 25 30 35 40 45 50 55 60 70 80

90

f

_ck,cube

(MPa) 15 20 25 30 37 45 50 55 60 67 75 85 95 105

f

_cm

(MPa)

20 24 28 33 38 43 48 53 58 63 68 78 88

98

f

_ctm

(MPa)

1.6 1.9 2.2 2.6 2.9 3.2 3.5 3.8 4.1 4.2 4.4 4.6 4.8

5.0

E

_cm

(GPa)

27 29 30 31 33 34 35 36 37 38 39 41 42

44

f

_ck

= Concrete cylinder strength

f

_ck,cube

= Concrete cube strength

f

_cm

= Mean concrete strength

f

_ctm

= Mean concrete tensile strength

E

_cm

= Mean value of elastic modulus

Product form Bars and de-coiled rods Wire Fabrics

Class

A B C A B C Characteristic yield strength fyk or f0,2k (MPa) 400 to 600 k = (ft/fy)k ≥1,05 ≥1,08 ≥1,15 <1,35 ≥1,05 ≥1,08 ≥1,15 <1,35 Characteristic strain at maximum force, εεεεuk (%) ≥2,5 ≥5,0 ≥7,5 ≥2,5 ≥5,0 ≥7,5 Fatigue stress range

(N = 2 x 106) (MPa) with an upper limit of 0.6fyk

150 100

•

In UK NA max. char yield strength, f

_yk

, = 600 MPa

•

BS 4449 and 4483 have adopted 500 MPa

BS EN 1992-1-1 & Cover

Nominal cover, c

_nom

Minimum cover, c

_min

c_min = max {c_min,b; c_min,dur ; 10 mm}

Axis distance, a

Fire protection

BS EN 1992-1-1 & Cover

Minimum cover, c

_min

a

Axis Distance Reinforcement cover Axis distance, a, to centre of bar a = c + φφφφ_m/2 + φφφφ_l

BS EN 1992-1-2 Structural fire design

Scope

Part 1-2 Structural fire design gives several methods for fire engineering

µµµµ_fi = N_Ed,fi/ N_Rd or conservatively 0.7

For grades of concrete up to C50/60, ε_{cu= 0.0035} ηηηη = 1 λλλλ = 0.8 fcd = αααα_cc fck/ γγγγ_c = 0.85 fck/1.5 = 0.57 fck fyd = fyk/1.15 = 435 MPa

Design flowchart

The following flowchart outlines a design procedure for rectangular beams with concrete classes up to C50/60 and class 500 reinforcement

Determine K and K’ from:

Note: δδδδ= 0.8 means 20% moment redistribution. ck 2 f d b M K ==== & '==== 0.6δδδδ −−−−0.18δδδδ 2 −−−−0.21 K

Carry out analysis to determine design moments (M)

δδδδ K’ 1.00 0.208 0.95 0.195 0.90 0.182 0.85 0.168 0.80 0.153 0.75 0.137 0.70 0.120

It is often recommended in the UK that K’ is limited to 0.168 to ensure ductile failure

Beam is overreinforced -compression steel needed Is K ≤ K’ ?

Beam is under-reinforced - no compression steel needed

Flow chart for under-reinforced beam

Calculate lever arm Z from:

_z

d

[[[[

₁

₁

₃

_.

₅₃

_K

]]]]

₀

_.

₉₅

_d

2

++++

−−−−

≤≤≤≤

====

Check minimum reinforcement requirements:

d b f d b f A _t yk t ctm min , s 0.013 26 . 0 ≥≥≥≥ ≥≥≥≥

Check max reinforcement provided A_s,max ≤≤≤≤ 0.04A_c (Cl. 9.2.1.1) Check min spacing between bars > φφφφ_bar > 20 > A_gg + 5

Check max spacing between bars

Calculate tension steel required from:

z f M A yd s ====

Flow chart for over-reinforced beam

Calculate lever arm Z from:

[[[[

₁

₁

₃

_.

₅₃

_'

]]]]

2

K

d

z

====

++++

−−−−

Calculate excess moment from: 2

((((

_'

))))

2

bd

f

K

K

M

====

_ck

−−−−

Calculate compression steel required from:

((((

₂

))))

yd 2 2 s d d f M A −−−− ====

Calculate tension steel required from:

yd sc 2 s yd s ' f f A z f M M A ==== −−−− ++++

Check max reinforcement provided A_s,max ≤≤≤≤ 0.04A_c (Cl. 9.2.1.1) Check min spacing between bars > φφφφ_bar > 20 > A_gg + 5

Strut inclination method

θθθθ

cot

sw s Rd,

z

f

ywd

s

A

V

====

θθθθ

θθθθ

νννν

αααα

tan

cot

1 max Rd,

====

cw w

₊₊₊₊

cd

f

z

b

V

21.8

°°°° < θθθθ < 45°°°°

We can manipulate the Expression for the concrete strut: When cot θ θ θ θ = 2.5 (θ θ θ θ = 21.8°) V_Rd,max = 0.138 b_w z f_ck (1 - f_ck/250) Or in terms of stress: v_Rd = 0.138 f_ck (1 - f_ck/250) where v_Rd = V_Rd/(b z) = V_Rd/(0.9 bd) When v_Rd > v_Ed cot θ θ θ θ = 2.5 (θθθθ = 21.8°)

When v_Rd < v_Ed we can rearrange the concrete strut expression:

θθθθ = 0,5 sin-1_[v

Rd/(0.20 fck(1 - fck/250))]

We can also manipulate the reinforcement expression to give: A_sw/s = v_Ed b_w/(f_ywd cot θθθθ) f_ck v_Rd (when cot θθθθ = 2.5) 20 2.54 25 3.10 28 3.43 30 3.64 32 3.84 35 4.15 40 4.63 45 5.08 50 5.51

Shear

Design flow chart for shear

Yes (cot θθθθ = 2.5)

Determine the concrete strut capacity v_Rd when cot θθθθ = 2.5

v_Rd = 0.138f_ck(1-f_ck/250)

Calculate area of shear reinforcement:

A_sw/s = v_Ed b_w/(f_ywd cot θθθθ)

Determine θθθθ from:

θθθθ = 0.5 sin-1_[(v

Ed/(0.20fck(1-fck/250))]

Is v_RD > v_Ed? No

Check maximum spacing of shear reinforcement :

s_{,max} = 0.75 d

For vertical shear reinforcement Determine v_Ed where:

Deflection

The deflection limits are:

•

Span/250 under

quasi-permanent

loads to avoid

impairment of appearance and general utility

•

Span/500 after construction under the

quasi-permanent

loads to avoid damage to adjacent parts of the structure.

Deflection requirements can be satisfied by the following

methods:

•

Direct calculation (Eurocode 2 methods considered to be

an improvement on BS 8110) .

EC2 Span/effective depth ratios

l/d is the span/depth ratio

K is the factor to take into account the different structural systems

ρρρρ₀ is the reference reinforcement ratio = √√√√f_ck 10-3

ρρρρ is the required tension reinforcement ratio at mid-span to resist the moment due to the design loads (at support for cantilevers)

ρρρρ’ is the required compression reinforcement ratio at mid-span to resist the moment due to design loads (at support for cantilevers)





























−

+

+

=

2 3 0 ck 0 ck

3

,

2

1

5

,

1

11

ρ

ρ

ρ

ρ

f

f

K

d

l

_{if ρ ≤ ρ} 0 (7.16.a)













+

−

+

=

0 ck 0 ck

'

12

1

'

5

,

1

11

ρ

ρ

ρ

ρ

ρ

f

f

K

d

l

_{if ρ > ρ} 0 (7.16.b)

EC2 Span/effective depth ratios

Structural system

K

Simply supported beam, one- or two-way simply

supported slab

1.0

End span of continuous beam or one-way

spanning slab continuous slab or two-way slab

over continuous over one long side

1.3

Interior span of beam or one-way or two-way

spanning slab

1.5

Slab supported without beams (flat slab) (based

on longer span)

1.2

EC2 Span/effective depth ratios

18.5

Percentage of tension reinforcement (A_s,req’d/bd)

S p a n t o d e p th r a ti o ( l/ d )

Flow Chart

Is basic l/d x F1 x F2 x F3 >Actual l/d? Yes

No

Factor F3 accounts for stress in the reinforcement F3 = 310/σσσσ_s

where σσσσ_s is tensile stress under quasi-permanent load Note: A_{s,prov ≤ 1.5 As,req’d} (UK NA)

Check complete Determine basic l/d

Factor F2 for spans supporting brittle partitions > 7m F₂ = 7/l_eff

Factor F1 for ribbed and waffle slabs only F₁ = 1 – 0.1 ((b_f/b_{w) – 1) ≥ 0.8}

Increase

A_s,prov

or f_ck

Column design process

Determine the actions on the column Determine the effective length, l₀ Determine the first order moments

Determine slenderness, λλλλ Determine slenderness limit, λλλλ_lim

Is λλλλ ≥≥≥≥ λλλλ_lim? Yes No

Column is not slender, M_Ed = M₀₂

Column is

slender

Calculate A_s(eg using column chart)

Effective length

Actions

Effective length, l₀

First order moments Slenderness, λλλλ Slenderness limit, λλλλ_lim

Is λλλλ ≥≥≥≥ λλλλ_lim? Yes No Not slender, M_Ed = M₀₂ Slen-der Calculate A_s Detailing l₀= l l₀= 2l l₀ = 0,7l l₀ = l / 2 l₀ = l l /2 <l₀< l l₀> 2l       + + ⋅       + + 2 2 1 1 45 , 0 1 45 , 0 1 k k k k

l

₀

= 0,5l⋅

Braced members: Unbraced members:               + + ⋅       + + + ⋅ ⋅ + k k k k k k k k ₂ 2 1 1 2 1 2 1 1 1 1 1 ; 10 1 max

l

₀

= l⋅

θ M θ

Effective length (2)

1

.

0

2

≥≥≥≥

∑

∑

∑

∑

====

b

l

E

l

E

k

b c c

I

I

(From PD 6687: Background paper to UK NA) Where:

Ib,Ic are the beam and column uncracked

second moments of area

l_b,l_c are the beam and column lengths

From Eurocode 2:

k = (θθθθ

/ M)⋅⋅⋅⋅

(E

ΙΙΙΙ

/ l)

Alternatively...

Actions

Effective length, l₀

First order moments Slenderness, λλλλ Slenderness limit, λλλλ_lim

Is λλλλ ≥≥≥≥ λλλλ_lim? Yes No Not slender, M_Ed = M₀₂ Slen-der Calculate A_s Detailing

Effective length (3)

l_o = Fl

How to…Columns has a look up table Actions

Effective length, l₀

First order moments Slenderness, λλλλ Slenderness limit, λλλλ_lim

Is λλλλ ≥≥≥≥ λλλλ_lim? Yes No Not slender, M_Ed = M₀₂ Slen-der Calculate A_s Detailing

Design moment

The design moment M

_Ed

is as

follows:

M₀₁ = Min {|M_top|,|M_bottom|} + e_i N_ed M₀₂ = Max {|M_top|,|M_bottom|} + e_i N_ed e_i = Max {I_o/400, h/30, 20}

M₂ = N_ed e₂

For stocky columns:

M_Ed = M₀₂

There are alternative, methods for calculating eccentricity, e₂, for slender columns

Actions

Effective length, l₀

First order moments Slenderness, λλλλ Slenderness limit, λλλλ_lim

Is λλλλ ≥≥≥≥ λλλλ_lim? Yes No Not slender, M_Ed = M₀₂ Slen-der Calculate A_s Detailing

Slenderness

Second order effects may be ignored if they are less than 10% of the corresponding first order effects

Second order effects may be ignored if the slenderness, λλλλ < λλλλ_lim

Slenderness λλλλ = l₀/i where i = √√√√(IIII/A) hence

for a rectangular section λλλλ = 3.46 l₀ / h for a circular section λλλλ = 4 l₀ / h

With biaxial bending the slenderness should be checked separately for each direction and only need be considered in the directions where λλλλ_lim is exceeded

Actions

Effective length, l₀

First order moments Slenderness, λλλλ Slenderness limit, λλλλ_lim

Is λλλλ ≥≥≥≥ λλλλ_lim?Yes No Not slender, M_Ed = M₀₂ Slen-der Calculate A_s Detailing

λλλλ

_lim

= 20

⋅⋅⋅⋅A⋅⋅⋅⋅B⋅⋅⋅⋅C/√√√√n

Slenderness Limit

where:

A = 1 / (1+0,2ϕϕϕϕ_ef)

ϕϕϕϕ_ef is the effective creep ratio;

(if ϕϕϕϕ_ef is not known, A = 0,7 may be used)

B = √√√√(1 + 2ωωω)ω ωω = Aωω _sf_yd / (A_cf_cd)

(if ωωωω is not known, B = 1,1 may be used)

C = 1.7 - r_m

r_m = M₀₁/M₀₂

M₀₁, M₀₂ are first order end moments,

 

M₀₂ ≥≥≥≥ M ₀₁

(if r_m is not known, C = 0.7 may be used)

n = N_Ed / (A_cf_cd)

Actions

Effective length, l₀

First order moments Slenderness, λλλλ Slenderness limit, λλλλ_lim

Is λλλλ ≥≥≥≥ λλλλ_lim? Yes No Not slender, M_Ed = M₀₂ Slen-der Calculate A_s Detailing

Slenderness limit – factor C

Actions

Effective length, l₀

First order moments Slenderness, λλλλ Slenderness limit, λλλλ_lim

Is λλλλ ≥≥≥≥ λλλλ_lim? Yes No Not slender, M_Ed = M₀₂ Slen-der Calculate A_s Detailing 105 kNM 105 kNM 105 kNM -105 kNM _{105 kNM} r_m = M₀₁/ M₀₂ = 0 / 105 = 0 C = 1.7 – 0 = 1.7 r_m = M₀₁/ M₀₂ = 105 / -105 = -1 C = 1.7 + 1 = 2.7 r_m = M₀₁/ M₀₂ = 105 / 105 = 1 C = 1.7 – 1 = 0.7

•

Introduction to the Eurocodes

•

Eurocode

•

Eurocode 1

•

Eurocode 2

· Materials · Cover · Flexure · Shear · Deflection

•

Further Information

Design aids from the UK concrete sector

Concise Eurocode 2

RC Spreadsheets ‘How to’ compendium

www.eurocode2.info ECFE – scheme sizing Worked Examples Properties of concrete

TCC Courses

•

Eurocode 2 half-day course for building

designers

•

Background to Eurocode 2 for building designers

(one day)

•

Eurocode 2 with design workshops for building

designers (one day)

•

Background to Eurocode 2, including liquid

retaining structures (one day)

•

Design of Concrete Bridges to Eurocodes (one

day)

Other Resources

Updated Detailing Manual

Updated

Recent Concrete Industry

Design Guidance is

written for Eurocode 2

Design Guidance

TR 64 Flat Slab TR43 Post-tensioned Slabs TR58 Deflections

•

Introduction to the Eurocodes

•

Eurocode

•

Eurocode 1

•

Eurocode 2

· Materials · Flexure · Shear · Deflection · Axial

•

Further Information

•

Worked Example

Worked Example

Cover = 40mm to each face

f_ck = 30

Check the beam for flexure, shear and deflection

G_k = 75 kN/m, Q_k = 50 kN/m

10 m

600

Determine K and K’ from:

Beam is under-reinforced - no compression steel needed

Is K ≤ K’ ? Yes ck 2 f d b M K ==== 21 . 0 18 . 0 6 . 0 ' & ==== δδδδ −−−− δδδδ 2 −−−− K

Carry out analysis to determine design moments (M)

Solution - Flexure

ULS = (75 x 1.25 + 50 x 1.5) = 168.75 kN/m M_ult = 168.75 x 102_/8 = 2109 kNm d = 1000 - 40 - 10 - 16 = 934 δδδδ K’ 1.00 0.205 0.95 0.193 0.90 0.180 0.85 0.166 0.80 0.151 0.75 0.135 134 . 0 30 934 600 10 2109 2 6 ==== ×××× ×××× ×××× ==== K

Calculate lever arm Z

[[[[

K

]]]]

d

d

z

1

1

3

.

53

0

.

95

2

++++

−−−−

≤≤≤≤

====

Check max reinforcement provided Check min reinforcement provided

Check min spacing between bars

Check max spacing between bars Calculate tension steel

z f M A yd s ====

Solution - Flexure

[[[[

]]]]

d

z

95

.

0

806

134

.

0

x

53

.

3

1

1

2

934

≤≤≤≤

====

−−−−

++++

====

2 6 s 6015 mm 806 x 435 10 x 2109 ==== ==== A Provide 8 H32 (6430 mm2₎

Space between bars = 35mm > φ φ φ φ ⇒⇒ OK⇒⇒

Design flow chart for shear

Determine the concrete strut capacity v_Rd Determine v_Ed where: v_Ed = V_Ed/(b_wd) Shear force: V_Ed= 168.75 x (10/2 - 0.934) = 686.1 kN Shear stress:v_Ed= V_Ed/(b_wd) = 686.1 x 103_{/(1000 x 600)} = 1.14 MPa

Solution - Shear

f_ck v_Rd (when cot θθθθ = 2.5) 20 2.28 25 2.79 28 3.08 30 3.27 32 3.46 35 3.73 40 4.17 45 4.58 50 4.96

Design flow chart for shear

Yes (cot θθθθ = 2.5) Determine the concrete strut capacity v_Rd

Area of shear reinforcement:

A_sw/s = v_Ed b_w/(0.9 f_ywd cot θθθθ) Determine v_Ed where:

v_Ed = V_Ed/(b_wd)

Is v_RD > v_Ed?

Check maximum spacing of shear reinforcement : s_l,max = 0.75 d Shear force: V_Ed= 168.75 x (10/2 - 0.934) = 686.1 kN Shear stress:v_Ed= V_Ed/(b_wd) = 686.1 x 103/(1000 x 600) = 1.14 MPa v_Rd = 3.27 MPa v_Rd > v_Ed ∴∴ cot ∴∴ θθθθ = 2.5 A_sw/s = 1.14 x 600 /(0.9 x 435 x 2.5) A_sw/s = 0.70 mm

Try H10 links with 2 legs.

A_sw = 157 mm2

s < 157 /0.70 = 224 mm

⇒

⇒⇒

Determine basic l/d

Solution - Deflection

Reinforcement ratio: ρρρρ = A_s/bd = 6430 x 100/(600 x 934) = 1.15%

Basic span-to-depth ratios

(for simply supported condition)

12 14 16 18 20 22 24 26 28 30 32 34 36 0.30% 0.80% 1.30% 1.80%

Percentage of tension reinforcement (As/bd)

S p a n t o d e p th r a ti o ( l/ d ) fck = 20 fck = 25 fck = 28 fck = 30 fck = 32 fck = 35 fck = 40 fck = 45 fck = 50 14.9

Is actual l/d < (l/d).j₁.j₂? j₁ = 1.0 Yes Determine basic l/d No Is b_f > 3b_w

Beam > 7m & support brittle partitions? No j₂ = 1.0 Check complete

Solution - Deflection

Reinforcement ratio: ρρρρ = A_s/bd = 6430 x 100/(600 x 934) = 1.15% Req’d l/d = 14.9 x 1.0 = 14.9 Actual l/d = 10000/934 = 10.7 Basic l/d > Actual l/d