Plain & Reinforced Concrete-1 · Plain & Reinforced Concrete-1 Behavior Doubly Reinforced Beams (contd…) Case-I Both Tension & Compression steel are yielding at ultimate condition

Plain & Reinforced Concrete-1

CE-314

Lecture # 13

Flexural Analysis and Design of Beams

(Ultimate Strength Design of Beams)

Sixth TermCivil Engineering

Plain & Reinforced Concrete-1Doubly Reinforced Beams“Beams having both tension and compression reinforcement to allow the depth of beam to be lesser than minimum depth for singly reinforced beam but still having tension-controlled behavior”

By using lesser depth the lever arm reduces and to develop the same force more area of tension steel (besides extra compression steel) is required, so solution is costly.Ductility will be increased by providing compression steel.Hanger bars can also be used as compression steel reducing the cost up to certain cost. For high rise buildings the extra cost of the shallow deep beams is offset by saving due to less story height.

Plain & Reinforced Concrete-1Doubly Reinforced Beams (contd…)

Compression steel may reduce creep and shrinkage of concrete and thus reducing long term deflection.

Doubly Reinforced Beam

Plain & Reinforced Concrete-1Behavior of Doubly Reinforced BeamsTension steel always yields in D.R.B.There are two possible cases:

1. Case-I Compression steel is yielding at ultimate condition.

2. Case-II Compression steel is NOT yielding at ultimate condition.

Plain & Reinforced Concrete-1Behavior Doubly Reinforced Beams

Cc

T = Asfs

N.A.

εcu= 0.003

Strain Diagram Internal Force Diagram

εs

h

cd

b 0.85fca

Whitney’s Stress Diagram

(d-d’)fs

d

εs’ fs’Cs

d – a/2

T = Asfs

Cs=As’fs’Cc=0.85fc’ba

fs=Esεs

fs’=Esεs’

Plain & Reinforced Concrete-1Behavior Doubly Reinforced Beams (contd…)Case-I Both Tension & Compression steel are yielding at ultimate condition

fs = fy and fs’=fy

Location of N.A.Consider equilibrium of forces in longitudinal direction

sc CCT +=

yscys f'Aba'f85.0fA +=

( )b'f85.0f'AA

ac

yss −=1β

ac =and


cd'c

0.003'εs −=

εcu= 0.003

Strain Diagramεs

c εs’d’

BD E

C

A

∆ ABC & ADE

−

=c

d'c0.003'εs

1

1s β

βc

d'c0.003'ε ×

−

=

−

=a

d'βa0.003'ε 1s

If εs’ ≥ εy compression steel is yielding. If εs’ < εy compression steel is NOT yielding.

(1)


Cc

T = Asfy

N.A.

Internal Force Diagram

(d-d’)

Cs

d – a/2

T = total tensile force in the steel

21 TTT +=T1 is balanced by Cs

T2 is balanced by Cc

s1 CT =

c2 CT =


Cc

T = Asfy

N.A.

Internal Force Diagram

(d-d’)

Cs

d – a/2

Moment Capacity by Compression Steel

( ) ( )'''1

ddfAddCM yssn −=−=

( )'ddT1 −=Moment Capacity by Concrete

−=

−=

2adT

2adCM 2cn2

( )

−−=

21adTT

( )

−−=

2' adfAA yss


Total Moment Capacity

21 nnn MMM +=

( ) ( )

−−+−=

2''' adfAAddfAM yssysn

( )b'f85.0f'AA

ac

yss −=Where

Plain & Reinforced Concrete-1Case-II Compression steel is not yielding at ultimate condition. Similarly, even the tension steel may not be yielding

fs = fy or not and fs’< fy

'εE'f ss ×=

bffAfAa

c

ssss

'85.0''−

=1β

ac =and

ad'βa600'f 1

s−

=

Location of N.A.

fs = aad −1600

β

Mn = As′ fs′ (d − d′ ) + 0.85 fc′ b a (d – a/2)or As′ fs′ (d − d′ ) + (As – As′ ) fs (d – a/2)


BALANCED STEEL RATIO FOR DOUBLY REINFORCED SECTIONS

The following symbols may be used in the development of the expression for the balanced steel ratio in case of doubly reinforced sections.

= total balanced tension steel ratio in case of doubly reinforced beams, limiting steel ratio dividing the under-reinforced and the over-reinforced behaviorρb = balanced tension steel ratio in case of corresponding singly reinforced beamsρ′ = compression steel ratio =

bρ

dbAs′


Force in tension steel to balance

steel compression

+Force in tension steel to balance

concrete compression

=Total force in tension steel


T = T1 + T2or T = Cc + CsAs fy = 0.85 fc′ b a + As′ fy

In the above equation, the compression steel is assumed to be yielding. If it is not actually yielding, adjustment is made later on.

Divide the above equation throughout by bdfy.

dbA

da

ff

dbA s

y

cs ′+

′= 85.0

ρρ ′+′

=da

ff

y

cb 85.0

From the strain diagram for the balanced failure,

sy

s

EfE

da

003.0003.0

1 += β


The above formula is only valid if both the steels are yielding. Tension steel is always yielding at the balanced condition by the definition. However, the compression steel may or may not be yielding. If the compression steel has not yielded, the balanced steel ratio may be determined using Fig. 4.5 as follows:

ρβρ ′++

′=

60060085.0 1yy

cb ff

f

ρρρ ′+= bb

εcu = 0.003

εy

c

d − c

BA

C

ED

F Gd′

c−d′εs′

H J

From the similarity of triangles HBG and AJE, we get,

HGHB

= AEAJ or d

HB′ =

dyε+003.0

HB = dd

y ′×+ ε003.0

Now, εs′ = 0.003 − HB = 0.003 − )003.0( ydd ε+×′

fs′ = Es [ 0.003 − )003.0( ydd ε+×′

]

= 600 − )600( yfdd

+×′

≤ fy

The effectiveness of the compression steel of magnitude As′ is reduced in the ratio fs′ / fy when the compression steel is not yielding. The balanced steel ratio, for the cases where compression steel is not yielding, is written as follows:

bρ = ρb + ρ′ ×y

s

ff ′

From the strain diagram for the tension-controlled failure, the following may be written:

Putting this value of a/d in equation for replaced with , the following is obtained:

dda

83

1 ×= β

bρ

maxρ

ρβρ ′+×= d8385.0 1max

maxρ = ρmax for singly reinforced sections with any extreme strain + ρ′

If the compression steel is not yielding, the formula may be written as follows:

y

smam f

f ′×′+= ρρρ max

MINIMUM TENSION STEEL RATIO FOR COMPRESSION STEEL YIELDING

ydcc

ε003.0

=′−

From the strain diagram,

c εy = 0.003 c − 0.003 d′c (0.003 − εy ) = 0.003 d′

dcy

′−

=ε003.0

003.0 day

′−

=ε

β003.0

003.01

Fig.4.7. Ultimate Strain Diagram To Determine Minimum Steel Ratio For Compression Steel Yielding.

εcu = 0.003

εs′ = εy c

d − c

BA

C

ED

F Gd′c−d′

Summing the forces in the horizontal direction:

As fy = 0.85 fc′ b a + As′ fyDivide the above equation throughout by bdfy.

bdA

da

ff

bdA s

y

cs ′+

′= 85.0

ρε

βρ ′+

−′′

=yy

ccy d

dff

003.0003.085.0 1

ρβ ′+

−′′

=yy

c

fdd

ff

60060085.0 1

ANOTHER CHECK FOR YIELDING OF COMPRESSION STEEL

The condition for yielding of the compression steel when maximum tensile steel ratio is to be used may be evaluated for different grades of steel as follows:

For fy = 280 MPa, d′ / d ≤ 0.2For fy = 420 MPa, d′ / d ≤ 0.1125

ANALYSIS OF DOUBLY REINFORCED SECTIONS

Data: i) Dimensions like b, d, d′ and Lii) fc′, fy and Es

iii) Areas of tension and compression steels, As and As′

Required: Moment capacity, φb Mn

1. Check the beam as a singly reinforced section to see whether the provided compression steel will be used for strength or not. For this purpose, ρ may be checked and it must be more than ρmax for singly reinforced sections.

for singly reinforced sections 2.Assume both tension and compression steels

to be yielding and calculate the depth of N.A. (c) and depth of equivalent rectangular stress block (a).

and c = a / β1

3.Check for yielding of both the steels.εy = fy / Es

y

c

ff ′

=8385.0 1max βρ

bffAA

ac

yss

′

′−=

85.0)(

If εs ≥ εy ⇒ fs = fyIf εs′ ≥ εy ⇒ fs′ = fy

Alternatively calculate and make sure that ρ ≥ for yielding of the compression steel. Similarly d′ / d ratio may be evaluated for checking of the compression steel, as explained earlier.

4.If the tension and/or the compression steels are not yielding, write he expressions for fsand/or fs′ in terms of unknown a and solve to get its new value.

ccd

s−

= 003.0εcdc

s′−

=′ 003.0ε

cyρcyρ

T = Cs + Cc ⇒ tension failure

As = As′ + 0.85 fc′ b a

a2 + (As + As′) a − β1 (Asd + As′d′) = 0

Calculate a from this equation and then calculate fs′ and fs from the earlier two equations.

ada

f s′−

=′ 1600β

aad

f s−

= 1600β

aad −1600 β

ada ′− 1600

β

′

60085.0 bfc

5.Calculate the strength reduction factor (φb) depending on the value of the depth of N.A. and the extreme tensile strain.

6.Calculate the flexural strength depending upon the whether the steels are yielding or not.

Both Steels Are Yieldingfs = fs′ = fyMn = (As − As′) fy (d − a / 2) + As′ fy (d − d′)Mu = φb Mn

Any Steel Is Not Yielding

fs′ =

fs =

Mn = (Asfs − As′ fs′) (d − a / 2) + As′ fs′ (d − d′)

Mu = φb Mn

ada ′− 1600 β

aad −1600 β

Example 4.1A doubly reinforced rectangular section has the following section properties:As′ = 568 mm2 [2 − #19(US)]As = 3060 mm2 [6 − #25(US)]b = 300 mm d′ = 60 mmfy = 300 MPaCalculate the design flexural strength for the following two conditions:i) C− 20 concrete and d = 525 mm.ii) C− 35 concrete and d = 225 mm.

Solution:Case (i)

As′ = 570 mm2

As = 3060 mm2

b = 300 mm d′ = 60 mmd = 525 mmfy = 300 MPafc′ = 20 MPa Es = 200,000 MPaφb Mn = ?

ρ = = 0.0194

ρ′ = = 0.0036

ρmax = singly reinforced sections

= = 0.0181

= ρmax + ρ′assuming the compression steel to be yielding

= 0.0181 + 0.0036 = 0.0217

)525)(300()3060(

)525)(300()568(

y

c

ff ′

8385.0 1β

30020

8385.0 2 ××

maxρ

= 0.0146ρ > ρmax for singly reinforced section, analyze as doubly reinforced section.ρ > , compression steel is expected to be yielding.ρ ≤ , tension steel is expected to be yielding and φb = 0.90.

ρβρ ′+

−′′

=yy

ccy fd

dff

60060085.0 1

0036.0300600

60052560

3002085.0 2 +

−×××=

cyρ

maxρ

a = = = 146.6 mm

c = a / β1 = 146.6 / 0.85 = 172 mmεy = fy / Es = 300 / 200,000 = 0.0015

εs = = = 0.00616

> εy and 0.005, tension steel is yielding and φb = 0.90.

bffAA

c

yss

′

′−

85.0)(

3002085.0300)5683060(

×××−

ccd −003.0

172172525003.0 −

εs′= = = 0.00195

> εy, the compression steel is yielding.

Mn = (As − As′) fy (d − a / 2) + As′ fy (d − d′)= [(3060 − 568) × 300 × (525 − 146.6 / 2)

+ 568 × 300 × (525 − 60)] / 106

= 416.9 kN-mMu = φb Mn

= 0.9 × 416.9 = 375.2 kN-m

cdc ′−003.0

17260172003.0 −

Case (ii)

As′ = 568 mm2

As = 3060 mm2

b = 300 mm d′ = 60 mmd = 225 mmfy = 300 MPafc′ = 35 MPa Es = 200,000 MPaφb Mn = ?

ρ = = 0.0453

ρ′ = = 0.00841

For fc′ > 28 MPa, β1 = 1.05 − 0.00714 fc′≤ 0.85 = 0. 8

ρmax = singly reinforced sections

= = 0.0298

)225)(300()3060(

)225)(300()568(

y

c

ff ′

8385.0 1β

30035

838.085.0 ×××

= ρmax + ρ′assuming the compression steel to be yielding= 0.0298 + 0.00841 = 0.03821

= + ρ′

= + 0.00841

= 0.0613

maxρ

bρ

+′

yy

c

fff

60060085.0 1β

+×××

300600600

300358.085.0

= 0.0507ρ > ρmax for singly reinforced section, analyze as doubly reinforced section.ρ < , compression steel is not yielding.ρ > , extreme tensile strain is not greater than 0.005.ρ ≤ , tension steel is yielding.

ρβρ ′+

−′′

=yy

ccy fd

dff

60060085.0 1

00841.0300600

60022560

300358.085.0 +

−××××=

cyρ

maxρ

bρ

Confirmation of above results (Are not always needed)Considering both the tension and the compression steels to be yielding, calculate depth of equivalent rectangular stress block (a) and the depth of N.A. (c).

a = = = 83.8 mm

c = a / β1 = 83.8 / 0.80 = 104.8 mmεy = fy / Es = 300 / 200,000 = 0.0015

bffAA

c

yss

′

′−

85.0)(

3003585.0300)5683060(

×××−

εs = = = 0.00344

> εy, tension steel is yielding

εs′ = = = 0.00128

< εy, the compression steel is not yielding.

ccd −003.0

8.1048.104225003.0 −

cdc ′−003.0

8.104608.104003.0 −

Calculation of revised value of a:

fs′ = =

T = Cs + Cc ⇒ tension failureAs fy = As′ fs′ + 0.85 fc′ b a

3060 × 300 = 568 × As′ + 0.85 × 35 × 300 × a918,000 a = 340,800 a − 16,358,400 + 8925 a2

a2 − 64.67 a − 1832.9 = 0a = 86 mmc = a / β1 = 86 / 0.80 = 107.5 mm

ada ′− 1600

βa

a 608.0600 ×−

εs = = 0.00328

> εy = 0.0015, tension steel is still yielding

φb = 0.65 + (εt − εy) = 0.78

fs′ = = = 265.12 MPa

Mn = (As fy − As′ fs′) (d − a / 2) + As′ fs′ (d − d′)= [(3060 × 300 − 568 × 265.12) × (225 − 86 / 2)

+ 568 × 265.12 × (225 − 60)] / 106 = 164.5 kN-m

Mu = φb Mn = 0.78 × 164.5 = 128.3 kN-m

5.1075.107225003.0 −

yε−005.025.0

ada ′− 1600

β

864886600 −

Continued on # 14

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Plain & Reinforced Concrete-1 · Plain & Reinforced Concrete-1 Behavior Doubly Reinforced Beams (contd…) Case-I Both Tension & Compression steel are yielding at ultimate condition