3 Flexural Analysis/Design of Beam 3 Flexural Analysis/Design of Beam 3. Flexural Analysis/Design of Beam 3. Flexural Analysis/Design of Beam

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3 Flexural Analysis/Design of Beam 3 Flexural Analysis/Design of Beam 3. Flexural Analysis/Design of Beam 3. Flexural Analysis/Design of Beam

BENDING OF HOMOGENEOUS BEAMS

REINFORCED CONCRETE BEAM BEHAVIOR

DESIGN OF TENSION REINFORCED REC. BEAMSS G O S O O C C S DESIGN AIDS

PRACTICAL CONSIDERATIONS IN DESIGN PRACTICAL CONSIDERATIONS IN DESIGN REC. BEAMS WITH TEN. AND COMP. REBAR T BEAMS

T BEAMS

447.327

Theory of Reinforced Concrete and Lab. I

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3. Flexural Analysis/Design of Beam 3. Flexural Analysis/Design of Beam 3. Flexural Analysis/Design of Beam 3. Flexural Analysis/Design of Beam

Typical Structures Typical Structures

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3. Flexural Analysis/Design of Beam 3. Flexural Analysis/Design of Beam

BENDING OF HOMOGENEOUS BEAMS

BENDING OF HOMOGENEOUS BEAMS Concrete is homogeneous?

Concrete is homogeneous?

Reinforced Concrete is homogeneous?

The fundamental principles in the design and analysis of

reinforced concrete are the same as those of homogeneous structural material.

Two components Internal forces

at any cross section

Two components

normal to the section - flexure tangential to the section - shear at any cross section tangential to the section shear

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REINFORCED CONCRETE BEAM BEHAVIOR REINFORCED CONCRETE BEAM BEHAVIOR Basic Assumptions in Flexural Design

1. A cross section that was plane before loading remains plane under load

plane under load

F unit strain in a beam above and below the neutral axis are proportional to the distance from that axis;

axis are proportional to the distance from that axis;

strain distribution is linear G Bernoulli’s hypothesis ( not true for deep beams)

( not true for deep beams)

(8)

2. Concrete is assumed to fail in compression, when ε = ε (limit state) = 0 003

when ε_c = ε_cu(limit state) = 0.003

3. Stress-strain relationship of reinforcement is assumed to p be elastoplastic (elastic-perfectly plastic).

G Strain hardening effect is neglected

4. Tensile strength of concrete is neglected for calculation of flexural strength

of flexural strength.

(9)

f ) l f h l

Basic Assumptions in Flexural Design

cf.) Typical s-s curve of homogeneous material

proportional

not proportional not proportional

(10)

5. Compressive stress-strain relationship for concrete may be assumed to be any shape (rectangular

may be assumed to be any shape (rectangular, trapezoidal, parabolic, etc) that results in an acceptable prediction of strength.p p g

G Equivalent rectangular stress distribution

(11)

REINFORCED CONCRETE BEAM BEHAVIOR Behavior of RC beam under increasing load

(12)

REINFORCED CONCRETE BEAM BEHAVIOR Behavior of RC beam under increasing load

entMome

EI M y =

= ε

φ y EI

Curvature

(13)

REINFORCED CONCRETE BEAM BEHAVIOR Elastic uncracked Section

not homogeneous homogeneous

uncracked transformed section

A (n-1)A

section

A_s (n-1)A_s

In elastic range

s

c f

f F E^s f f

f

s s c c

f

f = =

= ε

ε ^F f_s = ^s f_c = nf_c

(14)

REINFORCED CONCRETE BEAM BEHAVIOR Example 3.1 (SI unit)

A rectangular beam 250

A_s = 1,520 mm²

650 600

A_s 1,520 mm

f_cu = 27 MPa (cylinder strength) f_r = 3.5 MPa (modulus of rupture) f_r 3.5 MPa (modulus of rupture) f_y= 400 MPa

(unit: mm) D25

Calculate the stresses caused by a

bending moment M = 60 kN·m (unit: mm)

bending moment M = 60 kN m

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REINFORCED CONCRETE BEAM BEHAVIOR Solution

250

2.0 105

7 84 8 Es

n = = × = ≈

650 600

3 7.84 8

8,500

c cu

n = E = f = ≈

650

(n −1)A_s = ×7 1,520 10,640= mm2

transformed area of rebars

D25

transformed section

(16)

Assuming the uncracked section

250 Assuming the uncracked section, neutral axis

2

( 1) bh n A d

dA + −

∫

650 600

( 1) 2

( 1) 342

x s

s

n A d Q ydA

y A dA bh n A

= = = + −

∫ + −

650 ∫

342 mm

=

3

2 bh h

I y dA ⎛ y bh⎞ + ⎜ ⎟

∫

D25 ²

12 2

( ) ( 1)

x

s

I y dA y bh

d y n A

= = +⎜⎝ − ⎟⎠

+ − −

∫

transformed section 6, 477 10⁶ ⁴

s

= × mm

(17)

Compressive stress of concrete

at the top fiber ²⁵⁰

c 3.17

x

f M y MPa

= I =

650 600600

Tension stress of concrete at the bottom fiber

D25

( ) 2.85

ct r

x

f M h y MPa f

= I − = <

Assumption of uncracked, transformed section is justified !! D25

(18)

Stress in the tensile steel ₂₅₀

( ) 19 12

f M d MP

600

( ) 19.12

s

x

f n M d y MPa

= I − =

650 600

Compare f_c and f_s with f_cu and f_y respectively!!!

D25 D25

(19)

REINFORCED CONCRETE BEAM BEHAVIOR Elastic Cracked Section

This situation is under service load state

• f > f • f < 11 f • f < f

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REINFORCED CONCRETE BEAM BEHAVIOR REINFORCED CONCRETE BEAM BEHAVIOR Elastic Cracked Section

To determine neutral axis

kd (1)₍₁₎

0 )

2 ( )

( kd −nA d − kd = kd

b _s

(21)

Tension & Comp. force

) (kd f b

C ^c ( ) T A f ⁽²⁾

2 b kd

C = f^c T = A_s f_s ⁽²⁾

(22)

Bending moment about C

) (

)

( jd A f jd T

M = T( jd) = A f ( jd) (3)

M _s _s (3)

) ( jd A

f_s = M ₍₄₎

F A_s( jd)

Bending moment about T

f ( )( ) ) 2

( f b kd jd jd

C

M = = ^c ⁽⁵⁾

2M

2

2 kjbd

f_c = M ⁽⁶⁾

F How to get k and j ?

(23)

bd A_s ρ =

Defining ^Then, A_s = ρbd ⁽⁷⁾

Substitute (7) into (1) and solve for k

k ( )² 2 ⁽⁸⁾

0 ) 2 (

)

( kd −nA d −kd = kd

b _s

n n

n

k = (ρ )² + 2ρ − ρ ⁽⁸⁾

/ 3 d d kd

f ) / 3

1

jd d kd j k

= −

cf.)

F 1 See Handout #3-3

j = − 3

F See Handout #3 3

Table A.6

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REINFORCED CONCRETE BEAM BEHAVIOR REINFORCED CONCRETE BEAM BEHAVIOR Example 3.2 (Quiz)

The beam of Example 3.1 is subjected to a bending

moment M=120 kN·m (rather than 60 kN·m as previously).

Calculate the relevant properties and stress right away!!

(25)

REINFORCED CONCRETE BEAM BEHAVIOR REINFORCED CONCRETE BEAM BEHAVIOR Flexural Strength

fav

α = ₍₉₎

fck

α

f_av = ave. compressive

(9)

av stress on the area bc

b f

C f bc ^{( )} C =α _ck ⁽¹⁰⁾

(26)

αα 0.72 f_ck ≤ 28 MPa

decrease by 0.04 for every 7 MPa 28 MPa ≤ f_ck ≤ 56 MPa β

0.56 f_ck > 56 MPack

0.425 f_ck ≤ 28 MPa

decrease by 0.025 for every 7 MPa 28 MPa ≤ f_ck ≤ 56 MPa 0.325 f_ck > 56 MPa

(27)

This values apply to compression zone with other cross sectional shapes (circular, triangular, etc)

However, the analysis of those shapes becomes complex.

Note that to compute the flexural strength of the section it Note that to compute the flexural strength of the section, it is not necessary to know exact shape of the compression stress block. Only need to know C and its location.

stress block. Only need to know C and its location.

These two quantities are expressed in α and β _.

(28)

G The higher compressive strength, the more brittle.

(29)

Tension failure (ε_u<0.003, f_s =f_y)

Equilibrium Equilibrium

T

C = αf_ckbc = A_s f_s (11) Bending moment

) (d c f

A Tz

M = = _s _s − β ⁽¹²⁾

or M = Cz = α f bc d_ck ( − βc) ⁽¹³⁾

(30)

REINFORCED CONCRETE BEAM BEHAVIOR REINFORCED CONCRETE BEAM BEHAVIOR

Tension failure (ε_u<0.003, f_s =f_y)

Neutral axis at steel yielding, f_s = f_y

d f f

A ρ

ck y ck

y s

f d f b

f f c A

α ρ

α ⁼

= ⁽¹⁴⁾

From Eq.(11)

Nominal bending moment

( ) ^{f d}^y

M A f d c bdf d ρ

β ρ ^⎛ β ^⎞

= ( − ) = ⎜ − ⎟

n s y y

ck

M A f d c bdf d

β ρ β f

= = ⎜ α ⎟

⎝ ⎠

2 1 f^y 2 1 0 59 f^y

f bd ρ f bd ρ

ρ ^⎛ β ^⎞ ρ ^⎛ ^⎞

= _y ⎜1 ⎟ = _y ⎜1 0.59 ⎟ ⁽¹⁵⁾

ck ck

f bd f bd

f f

ρ β ρ

= ⎜ − α ⎟ = ⎜ − ⎟

⎝ ⎠ ⎝ ⎠ ⁽¹⁵⁾

(31)

Compression failure (ε_u=0.003, f_s < f_y)

Hook’s law Hook s law

s s

s E

f = ε ⁽¹⁶⁾

from strain diagram

c E d

f = ε − ⁽¹⁷⁾

Equilibrium

E c f_s ε_u _s

d c−

(17)

f bc A f A E d c

α = = ε ⁻ ⁽¹⁸⁾

(32)

Compression failure (ε_u=0.003, f_s < f_y)

Solving the quadratic for c

2 + A E c − A E d = 0

bc

f _k ε ε

αf_ckbc + A_sε_uE_sc A_sε_uE_sd 0 ⁽¹⁹⁾ α

= c

(19) (20)

N i l b di t

s =

f ₍₂₁₎

Nominal bending moment

n =

M_n ⁽²²⁾

(33)

Balanced reinforcement ratio ρ_b

The amount of reinforcement necessary for beam fail to The amount of reinforcement necessary for beam fail to by crushing of concrete at the same load causing the steel to yield; (ε_u=0.003, f_s =f_y)

steel to yield; (ε_u 0.003, f_s f_y)

• ρ < ρ_b lightly reinforced, tension failure, ductile

• ρ = ρ_b balanced, tension/comp. failure

• ρρ > ρρ_b_b heavily reinforced, compression failure, brittle y p

(34)

Balanced reinforcement ratio ρ_b

Balanced condition

y

s f

f = _y f^y

ε = (23)

Substitute Eq. (23) into Eq.(17)

y

s f

f

s

y E ⁽²³⁾

c c E d

f_s =ε_u _s −

d c

y u

uε ε

ε

= + (24)

c

Substitute Eq. (24) into Eq.(11)

(25)

u

fck ε ρ α

s s

ckbc A f

f =

α

(25)

y u

u y

b ck

f ε ε

ρ = +

(35)

REINFORCED CONCRETE BEAM BEHAVIOR Example 3.3 (SI unit)

A rectangular beam 250

A_s = 1,520 mm²

650 600

A_s 1,520 mm

f_cu = 27 MPa (cylinder strength) f_r = 3.5 MPa (modulus of rupture) f_r 3.5 MPa (modulus of rupture) f_y= 400 MPa

(unit: mm) D25

Calculate the nominal moment M_n

at which the beam will fail (unit: mm)

at which the beam will fail.

(36)

Check whether this beam fail in tension or compression

1 520

A 1,520

0.0101 (250)(600)

As

ρ = bd = =

(0.72)(27) 0.003

0.0292 400 0.003 0.002

ck u

b

y u y

f f

α ε

ρ ρ

ε ε

= = = >

+ +

F The beam will fail in tension by yielding of the steel

(37)

Using Eq. (15) for tension failure

⎛ f ⎞

2 1 0.59

(0 0101)(400)

y

n y

ck

M f bd f

f

ρ ^⎛ ρ ^⎞

= ⎜ − ⎟

⎝ ⎠

⎛ ⎞

2 (0.0101)(400) (0.0101)(400)(250)(600) 1 0.59

332 kN m 27

⎛ ⎞

= ⎜⎝ − ⎟⎠

= 332 kN m⋅

(38)

DESIGN OF TENSION REINFORCED REC. BEAMS DESIGN OF TENSION REINFORCED REC. BEAMS

Korea’s design method is Ultimate Strength Design.

called as Limit States Design in the US and Europe

1. Proportioning for adequate strength 1. Proportioning for adequate strength 2. Checking the serviceability

:deflections/crack width compared against limiting values :deflections/crack width compared against limiting values

(39)

DESIGN OF TENSION REINFORCED REC. BEAMS Equivalent Rectangular Stress Distribution

DESIGN OF TENSION REINFORCED REC. BEAMS

is called as Whitney’s Block (Handout #3-1)

What if the actual stress block is replaced by an What if the actual stress block is replaced by an

equivalent rectangular stress block for compression zone.

Actual Equivalent

(40)

DESIGN OF TENSION REINFORCED REC. BEAMS DESIGN OF TENSION REINFORCED REC. BEAMS Equivalent Rectangular Stress Distribution

. equi ck

ck

actual f bc f ab C

C =α = γ =

αc Go to

P.25

γ = a

F

c a = β₁

α ^G

β1

γ = α

a

γdepends on α,β

a = βc

2 _G a = β₁c

β β₁ = 2

(41)

f_ck, MPa

< 28 35 42 49 56 ≤

α 0.72 0.68 0.64 0.60 0.56

β 0.425 0.400 0.375 0.350 0.325

β₁=2 β 0.85 0.801 0.752 0.703 0.654

γ= α/ β₁ 0.857 0.849 0.851 0.853 0.856

- γ is essentially independent of f_ck.

β 0 85 0 007 (f 28) d 0 65 ≤ β ≤ 0 85 - β₁ = 0.85 - 0.007×(f_ck -28) and 0.65 ≤ β₁ ≤ 0.85

(42)

- KCI 6.2.1(6) allows other shapes for the concrete stress block to be used in the calculations as long as they result g y in good agreement with test results.

- KCI6.2.1(5) makes a further simplification. TensileKCI6.2.1(5) makes a further simplification. Tensile stresses in the concrete may be neglected in the cals.

G Contribution of tensile stresses of the concrete below N.A. is very small.

“Concrete Stress Distribution in Ultimate Strength Design”Concrete Stress Distribution in Ultimate Strength Design

Handout #3-2 by Hognestad

(43)

DESIGN OF TENSION REINFORCED REC. BEAMS Balanced Strain Condition

steel strain is exactly equal to ε_y and

concrete simultaneously reaches y ε_u_u =0.003

Eq. (24)

d εu

d c

y u

uε ε +

=

Equilibrium C=T

0.85 0.85 1

s y b y ck ck

A f_{s y}f = ρρ_bbd ff_y = f abf_ck = ff_ck ββ₁cb

(44)

DESIGN OF TENSION REINFORCED REC. BEAMS Balanced Strain Condition

Balanced reinforcement ratio

y u

u y

ck y

b ck

f f d

f c f

ε ε

β ε β

ρ = 0.85 ₁ = 0.85 ₁ + ₍₂₆₎

Apply ε_u =0.003 and E_s=200,000 MPa

1 1

0.003 600

0.85 0.85

0 003 600

ck ck

b

y y y y

f f

f f f f

ρ = β = β

+ ⁽²⁷⁾

0.003 ^y

y y y

s

f f f f

+ E

(45)

DESIGN OF TENSION REINFORCED REC. BEAMS Underreinforced Beams

In actual practice, ρ should be below ρ_b for the reasons,

1. Exactly ρ= ρ_b, then concrete reaches the comp. strain limit and steel reaches its yield stress

reaches its yield stress.

2. Material properties are never known precisely.

3 Strain hardening can cause compressive failure although ρ may be 3. Strain hardening can cause compressive failure, although ρ may be

somewhat less than ρ_b.

4. The actual steel area provided, will always be equal to or larger than 4. The actual steel area provided, will always be equal to or larger than

required based on ρ.

5. The extra ductility provided by low ρ provides warning prior to failure.

(46)

DESIGN OF TENSION REINFORCED REC. BEAMS KCI Code Provisions for Underreinforced Beam

By the way, how to guarantee underreinforced beams?

F KCI Code provides,

(1) The minimum tensile reinforcement strain allowed at nominal strength in the design of beam.g g

(2) Strength reduction factors that may depend on the tensile strain at nominal strength. g

Note Both limitations are based on the net tensile strain ε_t of the rebar farthest from the compression face at the

of the rebar farthest from the compression face at the depth d_t. (d ≤ d_t)

(47)

(1) For nonprestressed flexural members and members with factored axial compressive load less than 0.1f_ckA_g, ε_tshall not be less than 0.004 (KCI 6.2.2(5))

Substitute d_t for d and ε_t for ε_y

ε _d _c

( ( ))

d c

y u

uε ε

ε

= +

f

F _t _u ^d^t ^c ε = ε c⁻

f ε

0.85 1 ^ck ^u b

y u y

f f

ρ β ε

ε ε

= + F ^0.85 ¹ ^ck ^u

y u t

f f

ρ β ε

ε ε

= +

The reinforcement ratio to

(48)

Maximum reinforcement ratio (KCI 2007) 85

0 85

0 β f^ck ε^u β f^ck ε^u

ρ ⁽²⁹⁾

004 .

85 0 . 0 85

.

0 ₁ ₁

max = +

= +

u

u y

ck t

u u y

ck

f f f

f

β ε ε

β ε

ρ ⁽²⁹⁾

Cf.) (prior to KCI 2007

ρb

ρ_max = 0.75 (28)

& ACI 2002) ρ_max ρ^b

ε_t = 0.00376 at ρ= 0.75ρ_bfor f_y=400 MPa.

(28)

F

0.00376 < 0.004 ,i.e., KCI 2007 is slightly conservative.

F

(49)

ε_u = 0.003 ε_u = 0.003 ε_u = 0.003

c

d_t

ε_t = 0.002 ε_t = 0.004 ε_t = 0.005

0 003

c c 0 003 c 0 003

Minimum

net strain Tension

controlled Compression

controlled

= =

+ 0.003

0.600 0.003 0.002

t

c

d = =

+ 0.003

0.429 0.003 0.004

t

c

d = =

+ 0.003

0.375 0.003 0.005

t

c d

net strain

for flexural controlled member controlled

member

(50)

(2) ^Transition

zone Tension

controlled Compression

controlled

(2)

Φ = 0.85

Φ = 0 70

Φ = 0.70 + (ε_t- 0.002) 0.85 Φ = 0.65 + (ε_t- 0.002) 200/3 Spiral

Other Φ = 0.70

Φ = 0.65

( _t ) /

Other

ε_t = 0.002 ε_t = 0.005

(51)

Nominal flexural strength

⎛ a ⎞

n s y 2

M = A f ⎛⎜d − a ⎞⎟

⎝ ⎠

A fs y

(30)

s yf

a = ₍₃₁₎

(52)

DESIGN OF TENSION REINFORCED REC. BEAMS Example 3.4 (the same as Ex.3.3)

A rectangular beam A_s = 1,520 mm²

250

A_s 1,520 mm

f_cu = 27 MPa (cylinder strength)

f_r = 3.5 MPa (modulus of rupture) ⁶⁵⁰ ⁶⁰⁰ f_r 3.5 MPa (modulus of rupture)

f_y= 400 MPa

Calculate the nominal strength M_n

using the equivalent stress block (unit: mm) ^D25

using the equivalent stress block. (unit: mm)

(53)

DESIGN OF TENSION REINFORCED REC. BEAMS Solution

Maximum reinforcement ratio f k ε

max 0.85 1

0.004

(27) 0.003

(0 85)(0 85) 0 0209 0 0101

ck u

y u

f f

A

ρ β ε

= ε

+

(27) 0.003

(0.85)(0.85) 0.0209 0.0101

(400) 0.003 0.004

As

= = > = bd

+

F This beam is underreinforced (tension controlled) and will fail yielding of the steel

(54)

DESIGN OF TENSION REINFORCED REC. BEAMS Solution

Depth of stress block

(1520)(400) A f (1520)(400) 0.85 (0.85)(27)(250) 106

s y ck

a A f mm

= f b = =

106

⎛ a ⎞ ⎛ ⎞

Nominal Strength

(1520)(400) 600 106 333

2 2

n s y

M = A f d⎛⎜⎝ − a ⎞⎟⎠ = ⎛⎜⎝ − ⎞⎟⎠ = kN m⋅

C thi ith E l 3 3 !!!

Compare this with Example 3.3 !!!

(55)

Nominal flexural strength (Alternative)

Eq(31) can be written w.r.t. ρ

ρ f d 0.85

y ck

a f d

f

= ρ ₍₃₂₎

Nominal flexural strength

f d f

⎛ ⎞ ⎛ ⎞

( ) ² ^{1 0.59}

1.7

y y

n y y

ck ck

f d f

M bd f d f bd

f f

ρ ρ

ρ ^⎛ ^⎞ ρ ^⎛ ^⎞

= ⎜ − ⎟ = ⎜ − ⎟

⎝ ⎠ ⎝ ⎠ ⁽³³⁾

A_s_s

a/2