chapter_1

UEMX2343REINFORCED CONCRETE DESIGN IIUEMX2343REINFORCED CONCRETE DESIGN II

Topic 1:Re-visit Basic Design & FundamentalsTopic 1:Re-visit Basic Design & Fundamentals

1.1 Limit States Design Philosophy Ultimate Limit State (ULS)

- require structure to withstand against collapse- ensure safety of the building occupants/structure itself- must consider possibility of buckling or overturning and

accidental damage. Serviceability Limit State (SLS)

- concern functioning of the structure or members, comfortof the people & appearance of the structure.

- considered parameters are: deflection, cracking, durability,excessive vibration, fatigue, fire resistance & any otherspecial circumstances.

Ultimate Limit State (ULS)- require structure to withstand against collapse- ensure safety of the building occupants/structure itself- must consider possibility of buckling or overturning and




Ultimate Limit State (ULS)- require structure to withstand against collapse- ensure safety of the building occupants/structure itself- must consider possibility of buckling or overturning and




1.2 Characteristic Materials Strengths & Loads Characteristic material strengths: Characteristic material strengths:

5% of test specimenstrengths fall belowthe characteristicstrength, fk

sff mk 64.1strength,sticCharacteri

5% of test specimenstrengths fall belowthe characteristicstrength, fk

1.2 Characteristic Materials Strengths & Loads Characteristic loads or actions: Characteristic loads or actions:

5% of probability ofbeing exceededonce in a year

Characteristic loads or actions:

sF F mk 64.1action,sticCharacteri BS 648: Weights of BuildingMaterials EN1991-1-1: Actions onStructures Reynoldss ReinforcedConcrete Designers Handbook

1.2 Characteristic Materials Strengths & Loads Partial factor of safety for materials :

EN 1992-1-1: cl. 2.4.2 states several types of partial factor ofsafety.

m m

kfsafetyoffactorpartial

strengthsticcharacteristrengthDesign

Partial factor of safety for materials :


m



What is the partial safety factor forconcrete & steel bars under ULS?

Partial factor of safety for materials :


What is the partial safety factor forconcrete & steel bars under ULS?

1.2 Characteristic Materials Strengths & Loads According to EN 1992-1-1: cl. 2.4.2.4: According to EN 1992-1-1: cl. 2.4.2.4: According to EN 1992-1-1: cl. 2.4.2.4:

1.2 Characteristic Materials Strengths & Loads Partial factor of safety for actions :

Refer to EN1990 (NA): Table NA.A1.2(B)

ff actionsticcharacteriueaction valDesign

Partial factor of safety for actions :

Refer to EN1990 (NA): Table NA.A1.2(B) m



f actionsticcharacteriueaction valDesign

1.2 Characteristic Materials Strengths & Loads What is Notional Load?

- loads exist due to inaccuracies and imperfections(dimension & alignment) of the constructed structure.

According to BS 8110-1: 1997: cl. 3.1.4.2:- Ultimate design notional load = 1.5% of characteristic

dead load of the structure between mid-height of thestorey below and either mid-height of the storey above orthe roof surface.

- The design ultimate wind load should not be taken as lessthan this value when considering load combinations.

What is Notional Load?- loads exist due to inaccuracies and imperfections

(dimension & alignment) of the constructed structure. According to BS 8110-1: 1997: cl. 3.1.4.2:

- Ultimate design notional load = 1.5% of characteristicdead load of the structure between mid-height of thestorey below and either mid-height of the storey above orthe roof surface.


What is Notional Load?- loads exist due to inaccuracies and imperfections

(dimension & alignment) of the constructed structure. According to BS 8110-1: 1997: cl. 3.1.4.2:

- Ultimate design notional load = 1.5% of characteristicdead load of the structure between mid-height of thestorey below and either mid-height of the storey above orthe roof surface.


1.3 Load Combinations Ultimate limit state (ULS):

- For persistent or transient design situations (fundamentalcombinations)

- For accidental design situations

- For seismic design situations

Ultimate limit state (ULS):- For persistent or transient design situations (fundamental

combinations)



EN 1990:Eq. 6.10

EN 1990:Eq. 6.11b

Ultimate limit state (ULS):- For persistent or transient design situations (fundamental

combinations)



EN 1990:Eq. 6.11b

EN 1990:Eq. 6.12b

1.3 Load Combinations Value of factor for different types of buildings (EC 1990

(NA): Table NA.A1.1)

1.3 Load Combinations Serviceability limit state (SLS):

- Characteristic combination

- Frequent combination

- Quasi-permanent combination

EN 1990:Eq. 6.14b

Serviceability limit state (SLS):- Characteristic combination


- Quasi-permanent combination

EN 1990:Eq. 6.14b

EN 1990:Eq. 6.15b

Serviceability limit state (SLS):- Characteristic combination


- Quasi-permanent combinationEN 1990:Eq. 6.16b

1.4 Frame Structure Analysis Load analysis for braced framed structure:

max. = max. loading as per ULS load combination;min. = 1.35Gk

Load analysis for braced framed structure:


Load analysis for braced framed structure:


1.4 Frame Structure Analysis The analysis involves:

- Sketching of deflection profile of the structure- Defining the rotational stiffness for each member- Determining the distribution factor of the members at

each joint- Analyzing the frame structure using moment distribution

method- Sketching the BMD & SFD

The analysis involves:- Sketching of deflection profile of the structure- Defining the rotational stiffness for each member- Determining the distribution factor of the members at

each joint- Analyzing the frame structure using moment distribution

method- Sketching the BMD & SFD

Gosh! Wats hetalking about?

Yawn! Rotationalstiffness? Moment

distribution? I formattedmy brain after exam.

1.4 Frame Structure Analysis Load analysis for unbraced framed structure:

Max. = max. load as pereach ULS load combinationMin. = 1.35Gk(each load combinationmust consider those 3 loadarrangement)

Max. = max. load as pereach ULS load combinationMin. = 1.35Gk(each load combinationmust consider those 3 loadarrangement)

1.5 Section Analysis Rectangular Section Stress-strain diagram for a rectangular reinforced concrete

member:

For fck < 50 Mpa: = 1 (defining the effective strength);cc = 0.85 (ranging 0.8 ~ 1.0: account for long term effect)

1.5 Section Analysis Rectangular SectionHorizontal Force Equilibrium:

qbdfbdf

qqqkqbdfAf

bdfAfbdfAfbdfdkbf.bdf

AfAfdkbf.

ckck

ssu

scksllyk

cksllykcksllykckuck

ck

sllyksllykuck

)steelinforcenett()concreteinforceecompressiv(Therefore,

1Eq.......'87.087.0567.0/considerThen,

/'87.0/87.0/5670 withsidesbothDivide

'87.087.05670

,

,,

,,

qbdfbdf

qqqkqbdfAf


AfAfdkbf.

ckck

ssu

scksllyk

cksllykcksllykckuck

ck

sllyksllykuck


1Eq.......'87.087.0567.0/considerThen,


'87.087.05670

,

,,

,,

qbdfbdf

qqqkqbdfAf


AfAfdkbf.

ckck

ssu

scksllyk

cksllykcksllykckuck

ck

sllyksllykuck


1Eq.......'87.087.0567.0/considerThen,


'87.087.05670

,

,,

,,

1.5 Section Analysis Rectangular Sectionq is a combined reinforcing index. Its an unifying parameterbetween singly & doubly reinforced sections.where,q is a combined reinforcing index. Its an unifying parameterbetween singly & doubly reinforced sections.where,

ck

sllyks

ck

sllyks bdf

Afq

bdfAf

q'

',,

Moment Equilibrium:

2Eq.......8705670501

,

5670alsoor,5670 thatknow we1,Eq.From

)'('87.0)2/(5670

2

,

d')(dA'f.).

q.q(fbdM.

qkqk.

ddAfdkddkbf.M

slyk,lckEd

uu

sllykuuckEd

2Eq.......8705670501

,

5670alsoor,5670 thatknow we1,Eq.From

)'('87.0)2/(5670

2

,

d')(dA'f.).

q.q(fbdM.

qkqk.

ddAfdkddkbf.M

slyk,lckEd

uu

sllykuuckEd

1.5 Section Analysis Rectangular SectionDrop the compression bar term and rewrite Eq. 2:

3Eq.......5670501

5670501

2

2

).

q.q(fbdM

).

q.q(fbdM

ck

Ed

ckEd

uss

c

Ed kqqkfbd

Mq 567.0)'(87.0567.0

211567.0 2

3Eq.......5670501

5670501

2

2

).

q.q(fbdM

).

q.q(fbdM

ck

Ed

ckEd

Rearrange Eq. 3 to quadratic equation of q.

Besides, we also known that:

uss

c

Ed kqqkfbd

Mq 567.0)'(87.0567.0

211567.0 2

From Eq. 1

).

q.q(fbdMK

ck

Ed

56705012 0.167, singly reinf.> 0.167, doubly reinf.

Rearrange Eq. 3 to quadratic equation of q.

Besides, we also known that:

1.5 Section Analysis Rectangular SectionFor ku,max = 0.45; = 0.8; qmax = 0.204; Kmax = 0.167For ku,min = 0.125; = 0.8; qmin = 0.057; Kmax = 0.054 A reinforced concrete member undergoes large plastic

deformation prior to failure is said to be ductile. Good ductility is achieved by limiting quantity of tensioned

steel to smaller value below = 0.013. BS8110 limits ku = 0.50 (79% of balance failure)AS3600 limits ku = 0.40 (67% of balance failure)MS EC2 limits ku = 0.45 (71% of balance failure) However, max. lever arm, z is limited to 0.95d. Hence,

A reinforced concrete member undergoes large plasticdeformation prior to failure is said to be ductile.

Good ductility is achieved by limiting quantity of tensionedsteel to smaller value below = 0.013.

BS8110 limits ku = 0.50 (79% of balance failure)AS3600 limits ku = 0.40 (67% of balance failure)MS EC2 limits ku = 0.45 (71% of balance failure) However, max. lever arm, z is limited to 0.95d. Hence,

A reinforced concrete member undergoes large plasticdeformation prior to failure is said to be ductile.

Good ductility is achieved by limiting quantity of tensionedsteel to smaller value below = 0.013.

BS8110 limits ku = 0.50 (79% of balance failure)AS3600 limits ku = 0.40 (67% of balance failure)MS EC2 limits ku = 0.45 (71% of balance failure) However, max. lever arm, z is limited to 0.95d. Hence,

125.095.02/8.0

95.0

u

u

kddkd

dz

value for ku,min

1.5 Section Analysis Rectangular Section

0.054~0.167Kifsteelncompressionofor5670501

054.0,057.0;8.0;125.0For167.0,204.0;8.0;45.0For

0.057~0.204with567.0)'(87.0567.0

211567.0

2

minmaxmin,

maxmaxmax,

2

).

q.q(fbdMK

KqkKqk

qkqqkfbd

Mq

ck

Ed

u

u

uss

c

Ed

0.054~0.167Kifsteelncompressionofor5670501

054.0,057.0;8.0;125.0For167.0,204.0;8.0;45.0For

0.057~0.204with567.0)'(87.0567.0

211567.0

2

minmaxmin,

maxmaxmax,

2

).

q.q(fbdMK

KqkKqk

qkqqkfbd

Mq

ck

Ed

u

u

uss

c

Ed

ck

sllyks

ck

sllyks bdf

Afq

bdfAf

q'

'; ,,

1.5 Section Analysis Rectangular Section

uscss

lyk

excesssc

RdEdexcess

sckRd

ckEd

kddAAA

ddfMA

-MMMqqfbdM

.fbdMK

43.0'/steel,tension

yieldedsteelncompressiowith)'(87.0

beamcapacityexcessbytaken87.0204.0;167.0Hence,requiredbarncompressio;1670/If

1

,

1

max

21

2

uscsslyk

excesssc

RdEdexcess

sckRd

ckEd

kddAAA

ddfMA

-MMMqqfbdM

.fbdMK

43.0'/steel,tension

yieldedsteelncompressiowith)'(87.0

beamcapacityexcessbytaken87.0204.0;167.0Hence,requiredbarncompressio;1670/If

1

,

1

max

21

2

1.5 Section Analysis Flanged Section Effects of hogging and sagging moment on flanged beam:

1.5 Section Analysis Flanged Section Effective width, beff according EC2: cl. 5.3.2.1

1.5 Section Analysis Flanged Section If depth of stress block in flange (kud hf): design thesection as per singly reinf. section.

armlever87.0 fyMA Edsl

1.5 Section Analysis Flanged Section If depth of stress block out of flange (kud > hf): design thesection as per following:

analyze the moment equilibrium of concrete compressionarea separately decomposed flanged beam into 2 parts.

If depth of stress block out of flange (kud > hf): design thesection as per following:


If depth of stress block out of flange (kud > hf): design thesection as per following:


1.5 Section Analysis Flanged Section

yield As,flange result yield As,web result Flange:

yield As,flange result yield As,web result

armlever87.0

armlever87.0)2/(5670

,

,

fyM

A

Afyhdhbbf.Mflange

flanges

flangesffwckflange


yield As,flange result yield As,web result Web:

yield As,flange result yield As,web result

567.0211567.0 Kq

MMM

web

flangeEdweb

websflangessl AAA ,,

1.5 Section Analysis Flanged Section Shear reinforcement, deflection and durability parts refer

to my RC I additional lecture note. For spacing requirement:

- Min. spacing EC2: cl. 8.2- Max. spacing EC2: cl. 7.3.3 & Table 7.3N

Shear reinforcement, deflection and durability parts referto my RC I additional lecture note.

For spacing requirement:- Min. spacing EC2: cl. 8.2- Max. spacing EC2: cl. 7.3.3 & Table 7.3N

1.6 Slabs 1-way spanning slab:

- Ly/Lx > 2.0, and/or- The slab only supported by two beams at oppositelocation of each other .

1-way spanning slab:- Ly/Lx > 2.0, and/or- The slab only supported by two beams at opposite

location of each other .

AdministratorLine

AdministratorLine

1.6 Slabs 2-ways spanning slab:

- Any slab surrounded by four side beams andLy/Lx 2.0 .

2-ways spanning slab:- Any slab surrounded by four side beams and

Ly/Lx 2.0 .

1.6 Slabs 1-way spanning slab Ultimate bending moment and shear forces in one-way

spanning slabs:

(Source: BS 8110: 1997: Part 1: Table 3.12)

1.6 Slabs 1- way spanning slab Condition to use that table:

a) In a one way slab, the area of each bay is not less than 30m2.

b) Live load qkmust not exceed 1.25 times dead load.c) Live load qk must not exceed 5 kN/m2.d) Spans must be approximately equal.e) Loads must be uniformly distributed.

Condition to use that table:a) In a one way slab, the area of each bay is not less than 30

m2.b) Live load qkmust not exceed 1.25 times dead load.c) Live load qk must not exceed 5 kN/m2.d) Spans must be approximately equal.e) Loads must be uniformly distributed.

Definition of panels and bay

1.6 Slabs 2 ways spanning slab (simply) Simply supported slab spanning in two directions with four

sides support:- maximum moments per unit width:

- Bending moment coefficients () :

Simply supported slab spanning in two directions with foursides support:- maximum moments per unit width:


2xsxsx nlm 2

xsysy nlm

Simply supported slab spanning in two directions with foursides support:- maximum moments per unit width:


(Source: BS 8110: 1997: Part 1: Table 3.13)

1.6 Slabs - 2 ways spanning slab (restrained) Restrained slab spanning in two directions:

2xsxsx nlm 2

xsysy nlm

n = 1.35Gk + 1.5Qksx & sy = coefficientfrom BS8110-1: 1997:Table 3.14

2xsxsx nlm

2-ways restrained slab:Moment:

2xsysy nlm

n = 1.35Gk + 1.5Qksx & sy = coefficient fromBS8110-1: 1997: Table 3.14n = 1.35Gk + 1.5Qksx & sy = coefficient fromBS8110-1: 1997: Table 3.14

1.6 Slabs - 2 ways spanning slab (restrained)Case 1: Four edges continuousCase 2: One short edgediscontinuousCase 3: One long edgediscontinuousCase 4: Two adjacent edgesdiscontinuousCase 5: Two short edgesdiscontinuousCase 6: Two long edgesdiscontinuousCase 7: Three edges discontinuous(one long edge discontinuous)Case 8: Three edges discontinuous(one short edge discontinuous)Case 9: Four edges discontinuous

Case 1: Four edges continuousCase 2: One short edgediscontinuousCase 3: One long edgediscontinuousCase 4: Two adjacent edgesdiscontinuousCase 5: Two short edgesdiscontinuousCase 6: Two long edgesdiscontinuousCase 7: Three edges discontinuous(one long edge discontinuous)Case 8: Three edges discontinuous(one short edge discontinuous)Case 9: Four edges discontinuous

Case 1: Four edges continuousCase 2: One short edgediscontinuousCase 3: One long edgediscontinuousCase 4: Two adjacent edgesdiscontinuousCase 5: Two short edgesdiscontinuousCase 6: Two long edgesdiscontinuousCase 7: Three edges discontinuous(one long edge discontinuous)Case 8: Three edges discontinuous(one short edge discontinuous)Case 9: Four edges discontinuous

Fig: Different condition of the continuous slabs

2-ways restrained slab:

xvxsx nlV Shear at supports:

n = 1.35Gk + 1.5Qksx & sy = coefficient fromBS8110-1: 1997: Table 3.14

xvxsx nlV xvysy nlV

n = 1.35Gk + 1.5Qksx & sy = coefficient fromBS8110-1: 1997: Table 3.14

1.6 Slabs - 2 ways spanning slab (restrained) Arrangement of the reinf. at the middle and edge strips are

shown in figure below.

In the edge strips, nominal reinf. is necessary

Arrangement of the reinf. at the middle and edge strips areshown in figure below.


Arrangement of the reinf. at the middle and edge strips areshown in figure below.


1.6 Slabs - 2 ways spanning slab (restrained) Torsion reinf. must be provided at discontinuous corners:

- consists of top & bottom mats in both directions of thespan.

- other rules as illustrated in the figure below.

Torsion reinf. must be provided at discontinuous corners:- consists of top & bottom mats in both directions of the

span.- other rules as illustrated in the figure below.

provide 75% ofmid spans Aslprovide 75% ofmid spans Asl provide half of

the 75% of midspans Asl

1.6 Slabs Design the longitudinal reinf., perform shear reinf.

requirement checking, deflection and spacing checking as perrectangular section with considering b = 1m.

Spacing requirements, as per in EC2:

Design the longitudinal reinf., perform shear reinf.requirement checking, deflection and spacing checking as perrectangular section with considering b = 1m.

Spacing requirements, as per in EC2:

1.6 Slabs Simplified rules for curtailment of bars in solid slab spanning

in one direction:

Guided with EC2: cl. 9.3(1) & 9.3.1

Simplified rules for curtailment of bars in solid slab spanningin one direction:


Simplified rules for curtailment of bars in solid slab spanningin one direction:


chapter_1

Documents

year characteristic

excessive vibration

people appearance

otherspecial circumstances

building occupantsstructure

test specimenstrengths

fksff mk

sticcharacteri bs