Lecture -3 Flexure Load Only

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Lecture-03Lecture 03Design of RC members for Flexure and Axial Loads

By: Prof Dr. Qaisar Ali

Civil Engineering DepartmentCivil Engineering Department

NWFP UET Peshawardrqaisarali@nwfpuet edu pk

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 1

[email protected]


Topics AddressedTopics AddressedBehavior of Homogenous Members Subjected toFlexure Load only

Reinforced Concrete Members subjected to FlexureReinforced Concrete Members subjected to FlexureLoad only

Si l i f d (t i l i f t) t lSingly reinforced (tension only reinforcement) rectangularmembers

BehaviorBehavior

Flexural capacity

Ductility Requirement

Prof. Dr. Qaisar Ali


2


Topics AddressedTopics AddressedDoubly reinforced (tension and compression reinforcement)rectangular members

Flexural capacity


T and Hollow members

Flexural capacity


Prof. Dr. Qaisar Ali 3


Behavior of Homogenous Members gSubjected to Flexure Load only

BehaviorElastic and inelastic stress

R (radius of curvature)Compression

f2

f

inelastic stress distribution in homogeneous members

N.A(zero stress line)

P

f1

fp

εmembers subjected to flexure load only

Tension ∆

fp

−εpε1 εp ε2

h

ε < εmax f < fmax

1fp

ε1

p

f = MyI Linear

Proportionality f = 7 5 f '

ε

only fp

b

p y

h

ε > εmax f > fmax

2f

p

ε2

p

f = 7.5 f

Non-linearProportionality

εp

max c


b

Proportionalitypf


RC members subjected to

B h i

RC members subjected to flexure load only

Behavior

Three stages before collapse

Un-cracked Concrete – Elastic Stage

Cracked Concrete (tension zone) – Elastic Stage

Cracked Concrete (tension zone) – Inelastic (UltimateStrength) Stage



RC members subjected to RC members subjected to flexure load only

Singly reinforced (tension only reinforcement) rectangular members

AP

Behavior

Un-cracked

A

f = MyI

f = 7.5 f 'max c

Concrete –

Elastic Stage

A

dh c

εc fc

Cl

c/3 fc

C

c/3

g dh

bnAs (n-1)As

εct ctf

la

Section A-A

TsTc

ctfT

T T + T C (1)f b


T = T + Tc s C = (12)f bcc


RC members subjected to jflexure load only

Singly reinforced (tension only reinforcement) rectangularmembers

Behavior

Un-cracked Concrete – Elastic Stageg

Stresses are computed as in case of homogeneous elastic concretebeam.

f = My/ I (with fmax = 7.5√fc′)




Singly reinforced (tension only reinforcement)Singly reinforced (tension only reinforcement)rectangular members

Flexural CapacityUn-cracked Concrete – Elastic Stage

Maximum moment capacity,

Mmax = fmaxc/I

In terms of moment couple (∑M = 0),

M = T × lMmax = T × laC = T (∑Fx = 0)

(½)fcbc = T


c = 2T/(fcb)

8



Singly reinforced (tension only reinforcement)Singly reinforced (tension only reinforcement) rectangular members A P

Behavior

Cracked A

Concrete –

(Tension Zone)

A

c

fc

Cl

c/3

dElastic Stage

dh

b nAs

T = A fs

l

S ti A A

d

s

a


Section A-A

T = A fs s C = (12)f bcc l = (d - c/3)a



Singly reinforced (tension only reinforcement)Singly reinforced (tension only reinforcement) rectangular members

Flexural Capacityp yCracked Concrete (tension zone) – Elastic Stage

In terms of moment couple (∑M = 0)

M = Tla = Asfs (d – c/3)

As = M/fs(d – c/3)

C T (∑F 0)C = T (∑Fx = 0)

(½)fcbc = Asfsc = 2Asfs/fcb {where fs = nfc and n =Es/Ec}


s s c { s c s c}

10



Singly reinforced (tension only reinforcement) g y ( y )rectangular members

Behaviorb

Cracked Concrete (tension zone) –

Inelastic (Ultimate Strength) Stage

cd hC

As

Tb εu

c a=β c1

c0.85f '

ca/2

C 0 85f 'abβc

Cd

A

c β1

T = A fs y

c

T = A fs s

C = 0.85f 'abcC

Prof. Dr. Qaisar Ali 11εs

AsT A fs yT A fs s




BehaviorCracked Concrete (tension zone) – Inelastic (UltimateStrength) Stage

D di th t f A th b f il iDepending on the amount of As, the member can fail inone of the following ways:

Under reinforced

Balanced

Over reinforced





B h iBehavior

Cracked Concrete (tension zone) – Inelastic (UltimateSt th) StStrength) Stage

Under reinforced: Steel yields before concrete crushes





BehaviorBehavior

Cracked Concrete (tension zone) – Inelastic (UltimateStrength) StageStrength) Stage

Balanced: Crushing of concrete and yielding of steeloccurs simultaneously.occurs simultaneously.





BehaviorBehavior

Cracked Concrete (tension zone) – Inelastic (UltimateStrength) StageStrength) Stage

Over reinforced: Concrete crushes before steel yields




Flexural Capacity of RC members subjected toflexure load only

The ACI Code

Design Assumptions of the ACI code for design ofDesign Assumptions of the ACI code for design ofmembers subject to flexure or axial loads or to combinedflexure and axial loads.




Flexural Capacity of RC members subjected toflexure load only

The ACI Code

The Flexural Strength: ACI R10 3 3 The nominalThe Flexural Strength: ACI R10.3.3 — The nominalflexural strength of a member is reached when the strainin the extreme compression fiber reaches the assumedstrain limit 0.003.




Singly reinforced (tension only reinforcement)g y ( y )rectangular members

Flexural Capacityp y

Mn = Asfy (d – a/2) [Nominal capacity]

ΦMn = ΦAsfy(d – a/2) [Ultimate capacity]n s y

a = Asfy/0.85fc′b

b εu c0.85f '

dc a=β c1c

a/2C = 0.85f 'abc

βcC


AsT = A fs yT = A fs s




D tilit R i tDuctility Requirements

In codes prior to 2002, maximum reinforcement requirement usedto be ensured by maximum reinforcement ratio ρ = 0 75ρbto be ensured by maximum reinforcement ratio ρmax 0.75ρb.

ρb = 0.85 β1(fc′/fy) εu/ (εu + εy) b εu

cd

A

c


As




D tilit R i tDuctility Requirements

However ACI 318-02 replaces εy by the net strain (εt) intension steel and ρ by ρ Transition zone TensionCompressiontension steel and ρb by ρmax.

ρmax = 0.85 β1(fc′/fy) εu/ (εu + εt)

{3√ (f ′)/f } ≥ 200/f

Transition zone Tensioncontrolled

Compression controlled

φ = 0.90

φ = 0.567 + 66.7ε t

Spiral

ρmin = {3√ (fc′)/fy} ≥ 200/fy

φ = 0.70

φ = 0.65

φ = 0.483 + 83.3ε t

Other


ε = 0.002t ε = 0.005t

Net tensile strain




M i Fl l C itMaximum Flexural Capacity

Maximum factored flexural capacity (ΦMn in in-kips) of singly reinforced RC rectangular beam for specified material strength and dimensions.

fc′ = 3 ksi b (in)

fy = 40 ksi 12 15 18

12 1063.45 (2.93) 1329.32 (3.66) 1595.18 (4.39)

d (in)

18 2392.77 (4.39) 2990.96 (5.49) 3589.16 (6.59)

20 2954.04 (4.88) 3692.55 (6.10) 4431.06 (7.32)

24 4253.82 (5.86) 5317.27 (7.32) 6380.73 (8.79)


30 6646.59 (7.32) 8308.25 (9.15) 9969.89 (10.98)

Note: The values in brackets represents Asmax in in2.



Doubly Reinforced (tension and compressionDoubly Reinforced (tension and compressionreinforcement) rectangular members

Flexural CapacityFlexural Capacity

Mu = ΦAsfy (d – a/2) + ΦAs′fs′ (d – d′)

Where, ΦAsfy (d – a/2) is equal to ΦMnmax (singly) for As = Asmax, s y ( ) q nmax (singly) s smax

{Mu – ΦMnmax (singly)} = ΦAs′fs′ (d – d′)

As′ = {Mu – ΦMnmax (singly)}/ {Φfs′ (d – d′)}

b εu

sε'd

d'

A's c a=β c1

0.85f 'c

C = A' f 's

d - d'

s A' f 's sa/2

C = 0.85f 'abccs


AstT = A fst

d d

A fys

(a) (b) (c) (e)(d)

y A' fys




Ductility RequirementsDuctility RequirementsCc + Cs = T [ ∑Fx = 0 ]

0.85fc′ab + As′fs′ = AstfyWhere,

Cc = Compression force due to concrete in compression region,

Cs = Compression force in steel in compression region needed to balance the t i f i dditi t th t i f id d b Atension force in addition to the tension force provided by Asmax (singly).

For amax = β1c = 0.85 × 0.375d, 0.85fc′ab is equal to Asmax (singly)fyAsmax (singly)fy + As′fs′ = Astfy


(Ast – As′fs′/fy) = Asmax (singly) (Ast – As′fs′/fy) ≤ Asmax (singly)

23




Ductility RequirementsDuctility Requirements

The ratios (d′/d) and minimum beam effective depths (d) for compressionreinforcement to yield

Minimum beam depths for compression reinforcement to yield

εt = 0.005t

fy, psi Maximum d'/d Minimum d for d' = 2.5" (in.)

40000 0.2 12.3

60000 0 12 21 5


60000 0.12 21.5

75000 0.05 48.8



T-BeamsT-Beams

Flexural Capacity

bεu 0.85f 'c0.85f '

/2 h /2f

(b-b )/2w

dc a=β c1

a/2

T A f

C = 0.85f 'abc w

hfh /2

f

C = 0.85f '(b - b )hc w f

f

1 2

xC

εs

AstT = A fst

(a)

y

bw

T = A fs y T = A fysf

(b) (c) (d) (e)

1 2




T-BeamsT-Beams

Flexural Capacity

ΦM ΦA f (d h /2)ΦMn1 = ΦAsffy (d – hf/2)

Asf, is the steel area which, when stressed to fy, is required to balancethe longitudinal compressive force in the overhanging portions of theflange that are stressed uniformly at 0.85fc′.

Asf =0.85fc′(b – bw)hf/fy

ΦM M ΦM ΦA f (d /2)ΦMn2 = Mu – ΦMn1 = ΦAsfy (d – a/2)

a = Asfy/ (0.85fc′bw)

Total capacity (ΦMn)= ΦMn1 + ΦMn2


Total capacity (ΦMn) ΦMn1 ΦMn2

Total steel area required (Ast) = Asf + As

26



T BeamsT-Beams

Flexural Capacity (Alternate Formulae)

ΦMn = Mu= ΦAstfy (d – x)

Ast = Mu/ {Φfy (d – x)}

a = {A f – 0 85f ′ (b – b )h }/0 85f ′ba = {Astfy – 0.85fc (b – bw)hf}/0.85fc bw

bεu 0.85f 'c0.85f '

h x

(b-b )/2w

d

A

c a=β c1

T = A fst y

hfxC


εs

AstT A fst

(a)

y

bw

(b) (c)



T BeamsT-Beams

Ductility RequirementsT = C1 + C2 [ ∑Fx = 0 ]

Astfy = 0.85fc′abw + 0.85fc′(b – bw)hf [ Ast = As (singly) + Asf ]

Astfy = 0.85fc′abw + AsffyAstfy 0.85fc abw Asffy

For εt = 0.005, a = β1c = 0.85 × 0.375d,

Astmax = 0.319 β1(fc′/fy)bwd + Asf ………………… (1)

So, for T-beam to behave in a ductile manner:

(As + Asf) ≤ Astmax


Therefore, the provided steel reinforcement (As + Asf) shall be less than Astmax given in equation (1).

28



Effective Flange width for T and L beam (ACI 8 10)Effective Flange width for T and L beam (ACI 8.10)




Effective Flange width for T and L beam (ACI 8 10)Effective Flange width for T and L beam (ACI 8.10)T-Beam: For T-beams, beff shall not exceed ¼ times the span length of beam, Further, the overhanging slab on either side of g g gbeam shall not exceed the following:

8hf

½ clear span to next beam½ clear span to next beam.

L-Beam: For L-beam, the effective overhanging slab width shall not exceed the following:

One-twelfth the span length of beam,

6hf

½ l t t b


½ clear span to next beam.

30



Hollow RC BeamHollow RC BeamFlexural Capacity

b εu

ca/2

C = 0.85f 'a(b -b )c o

h /2C = 0.85f 'b hc o f

fhf

21Cax

d

AstT = A fst y

bo T = A fs2 yT = A fys1

εs(a) (b) (c) (d) (e)




Hollow RC BeamHollow RC BeamFlexural Capacity

ΦMn1 = ΦAs1fy (d – hf/2)n1 s1 y ( f )Asf, is the steel area which, when stressed to fy, is required to balance thelongitudinal compressive force in the overhanging portions of the flangethat are stressed uniformly at 0.85fc′.

As1 =0.85fc′bohf/fyΦMn2 = Mu – ΦMn1 = ΦAs2fy (d – a/2)

a = As2fy/ {0.85fc′(b - bo)}

Total capacity (ΦMn)= ΦMn1 + ΦMn2

T t l t l i d (A ) A A


Total steel area required (Ast) = As1 + As2

32



Hollow RC BeamHollow RC Beam

Flexural Capacity (Alternate Formulae)

ΦM = M = ΦA f (d x)ΦMn = Mu = ΦAstfy (d – x)

Ast = Mu/ {Φfy (d – x)}

x = {bohf2/2 + (b – bo)a2/2}/ {(b –bo)a + bohf}{ o f ( o) } {( o) o f}

a = {Astfy – 0.85fc′bohf}/0.85fc′(b –bo)b εu

h

dc

bo

hf

Cax


AstT = A fst y

bo

(a) (b) (c)



Hollow RC Beam b εuHollow RC Beam

Ductility Requirements

F i f i l f

u

dc

T A fbo

hf

Cax

For summation of internal forces,

Astfy = 0.85fc′ba – 0.85fc′bo(a – hf)

For εt = 0 005 a = β1 × 0 375d we have A t = A t h ll therefore

εs

AstT = A fst y

(a) (b) (c)

For εt = 0.005, a = β1 × 0.375d, we have Ast = Astmax, hollow-, therefore,

Astmax, hollow = {0.85fc′bβ1 × 0.375d – 0.85fc′bo(0.375d – hf)}/ fy

Astmax, hollow = 0.319(fc′/fy)β1bd – 0.85(fc′/fy)bo(0.375d – hf), y y

Astmax, hollow = Asmax (singly) – 0.85(fc′/fy)bo(0.375d – hf)



1 Un-cracked Concrete Stage1. Un-cracked Concrete Stage

ft < frM < MCompression zone M < Mcrfc = ft << fc'

h d

b

Tension Zone

Strain Diagram

Stress Diagram

Compressive Stress

fc'e s o o e

ft = fr = 7.5 √fc'

fc

ft = fc


Tensile Stress

t r √ c

Stress-Strain Diagram for Concrete


C=0.5fc x (b x0.5h)

fc

c ( )

1/2 h

M

2/3 h

1/2 h

T=0.5ft x (b x0.5h)

bft

Stress diagram

C=T ; fc = ftM = 0.5fc x (b x 0.5h) x (2/3 h)

= 1/6 f x b x h2fc = ft = Mc/Igwhere c = 0 5hOR= 1/6 fc x b x h2

fc = ft = 6M/(bh2)

where c = 0.5hIg = bh3/12

OR

At ft = fr , where modulus of rupture, fr = 7.5 √fc’C ki M t C it M f I /(0 5h) (f b h2)/6


Cracking Moment Capacity, Mcr = fr x Ig/(0.5h) = (fr x b x h2)/6


2. Cracked Concrete (Tension Zone) -Elastic Stage

Compression zone

ft > frM > Mεc < 0.003

h d

M > Mcrfc = 0.45fc'fs =0.5 fy

εs = fs/Es

h

b Strain Diagram Stress Diagram

fs = 0.5 fy

f 0.45f '

Tension Zone Concrete Cracked Compressive Stress

fc'

ε

fy

0.5fy

0.45fc

Tensile Stress

ft = fr = 7.5 √fc'ε

Es

0.003


Tensile Stress

Stress- Strain Diagram for Concrete in Compression

Stress- Strain Diagram for Reinforcing steel in Tension


C=0.5fc × (bc)

fc

c ( )

1/2 h c

la = d – c/31/2 h

M

d

T= Asfs

b Stress diagram

In terms of moment couple (∑M = 0)M = Tla = Asfs (d – c/3)A M/f (d /3)As = M/fs(d – c/3)

C = T (∑Fx = 0)


(½)fcbc = Asfs

c = 2Asfs/fcb {where fs = nfc and n =Es/Ec}


3. Cracked Concrete (Tension Zone) -( )Ultimate Strength Stage

ft > >frM > >Mcr

Compression zone

ε = 0 003 crfs = fyfc = entire stress block until compression f ilh d

εc = 0.003

failureh

b

d

Strain Diagram Stress Diagram Compressive Stress

T = Asfyεs = fy/Es

bTension Zone Concrete Cracked

g Compressive Stress

fc'fy

E

εStress-Strain Diagram for Concrete in Compression

Stress-Strain Diagram for Reinforcing Steel in Tension

Es0.003 ε

Prof. Dr. Qaisar AliDr.

Qaisar Ali

p

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawarfc < fc′ 0.85fc′

C = 0.85fc′ab 1/2 h

M

ca = β1c

la = d – a/21/2 hd

T= As fy

bStress diagram

In terms of moment couple (∑M = 0)M = Tla = Asfy (d – a/2)A M/f (d /2)As = M/fy(d – a/2)

C = T (∑Fx = 0)


0.85fc ′ab = Asfy

a = Asfy/0.85fcb


Typical Stress Strain Curves for Concrete Typical Stress-Strain Curves for Concrete and Reinforcing Steel

Prof. Dr. Qaisar AliDr.

Qaisar Ali

Documents

Lecture -3 Flexure Load Only