Flexural design of Beam...PRC-I

Plain & Reinforced Concrete-1

By Engr. Rafia Firdous

Flexural Analysis and Design of Beams

(Ultimate Strength Design of Beams)

1


Ultimate Strength Design of Beams(Strength Design of Beams)

Strength design method is based on the philosophy of dividing F.O.S. in such a way that Bigger part is applied on loads and smaller part is applied on material strength.

fc’

0.85fc’

Stress

Strain

Crushing Strength

0.003

favg

favg = Area under curve/0.003

If fc’ ≤ 30 MPa

favg = 0.72 fc’

β1 = Average Strength/Crushing Strength

β1 = 0.72fc’ / 0.85 fc’ = 0.85 2


Ultimate Strength Design of Beams (contd…)

Cc

T = Asfs

la = d – a/2

N.A.

εcu=

0.003

Strain Diagram

Actual Stress

Diagram

Internal Force Diagram

In ultimate strength design method the section is always taken as cracked.

c = Depth of N.A from the extreme compression face at ultimate stage

a = Depth of equivalent rectangular stress diagram.

εs

hc

d

b 0.85fc

fs

0.85fca

Equivalent Stress

Diagram/ Whitney’s

Stress Diagram

a/2

fs

3



Actual Stress

Diagram

0.85fc 0.85fc

Equivalent Stress Diagram/

Whitney’s Stress Diagram

cCc Cca

• The resultant of concrete compressive force Cc, acts at the centriod of parabolic stress diagram.

• Equivalent stress diagram is made in such a way that it has the same area as that of actual stress diagram. Thus the Cc, will remain unchanged.

a/2

ab'0.85fcbf cav

a'0.85fc'0.72f cc

c'0.85f

'0.72fa

c

c

cβa 1 4



Factor β1

β1 = 0.85 for fc’ ≤ 28 MPa

Value of β1 decreases by 0.05 for every 7 MPa increase in strength with a minimum of 0.65

0.65 '0.00714f1.064β c1 85.0

5


Determination of N.A. Location at Ultimate ConditionCASE-I: Tension Steel is Yielding at Ultimate Condition

ys εε orys ff

CASE-II: Tension Steel is Not Yielding at Ultimate Condition

yf

yεsε

ysε orys ff

0.0015200,000

300

E

fε y

y 0.0021200,000

420

E

fε y

y

For 300 grade steel

For 420 grade steel

6


CASE-I: Tension Steel is Yielding at Ultimate Condition

ysss fAfAT ab'0.85fC cc

2

ada l

abffA cys '85.0

For longitudinal Equilibrium

T = Cc

bf

fAa

c

ys

'85.0

1β

ac and

Cc

T = Asfs

Internal Force Diagram

a/2

la

7


CASE-I: Tension Steel is Yielding at Ultimate Condition (contd…)

Nominal Moment Capacity, Mn depending on steel = T x la

2Mn

adfA ys

Design Moment Capacity

2M bnb

adfA ys

Nominal Moment Capacity, Mn depending on Concrete = Cc x la

2

adab0.85fc'Mn

2

adab0.85fc'M bnb

8


Minimum Depth for Deflection Control

I

1 α Δ

3(Depth)

1 α Δ

For UDL4ωL α Δ

3LωL α ΔDeflection Depends upon Span, end conditions, Loads and fy of steel. For high strength steel deflection is more and more depth is required. 9


Minimum Depth for Deflection Control (Contd…)

ACI 318, Table 9.5(a)

Steel Grade

Simply Supported

One End Continuous

Both End Continuous

Cantilever

300 L/20 L/23 L/26 L/10

420 L/16 L/18.5 L/21 L/8

520 L/14 L/16 L/18 L/7

10


Balanced Steel Ratio, ρb It is corresponding to that amount of steel which will cause yielding of steel at the same time when concrete crushes.At ultimate stage:

ys εε ys ff and

0.003εcu From the internal Force diagram

bc ab'0.85fC c

ab = depth of equivalent rectangular stress block when balanced steel ratio is used.

εcu=

0.003

Strain Diagram

εy

cb

d- cb

11


Balanced Steel Ratio, ρb (contd…)

ybys fd)b(ρfAT

For the longitudinal equilibrium

cCT

bcyb ab'0.85ffd)b(ρ

d

a

f

'f0.85ρ b

y

cb (1)

12


Balanced Steel Ratio, ρb (contd…) From the strain diagram εcu=

0.003

Strain Diagram

εy

cb

d- cb

B C

A

E D

ADE & ABC Δs

b

y

b cd

ε

c

0.003

yb ε0.003

0.003dc

s

s

s

yb E

Ed

Ef

0.003

0.003c

13


Balanced Steel Ratio, ρb (contd…)

df600

600c

yb

df0.003E

0.003Ec

ys

sb

b1b Cβa As we know

df600

600βa

y1b

(2

)Put (2) in (1)

y1

y

cb f600

600β

f

'f0.85ρ

14


Types of Cross Sections w.r.t. Flexure at Ultimate Load Level1. Tension Controlled SectionA section in which the net tensile strain in the extreme tension steel is greater than or equal to 0.005 when the corresponding concrete strain at the compression face is 0.003.

εcu=

0.003

Strain Diagram

εs=>0.005

c

d- c

cd

0.005

c

0.003

d0.008

0.003c

d8

3c d

8

3βa 1and

15


Types of Cross Sections w.r.t. Flexure at Ultimate Load Level (contd…)

2. Transition Section The section in which net tensile strain in the extreme tension steel is greater than εy but less than 0.005 when

corresponding concrete strain is 0.003. εcu=

0.003

Strain Diagram

εy<εs<0.005

c

d- cd8

3βa 1 baa

16



2. Transition Section (contd…)

εcu=

0.003

Strain Diagram

0.004

c

d- c

To ensure under-reinforced behavior, ACI code establishes a minimum net tensile strain of 0.004 at the ultimate stage.

cd

0.004

c

0.003

d7

3c d

7

3βa 1

17



Both the “Tension Controlled Section” and “Transition Section” are “Under-Reinforced Section”

In Under-Reinforced Sections steel starts yielding before the crushing of concrete and:

ρ < ρ b

It is always desirable that the section is under-reinforced otherwise the failure will initiate by the crushing of concrete. As concrete is a brittle material so this type of failure will be sudden which is NOT DESIREABLE.

18



3. Compression Controlled Section (over-reinforced section)

The section in which net steel strain in the extreme tension steel is lesser than εy when corresponding concrete strain is 0.003.

Capacity of steel remain unutilized. It gives brittle failure without warning.

baa bCC

19


Strength Reduction Factor (Resistance Factor), Φ It is reciprocal of minor part of overall factor of safety that is applied on strength of member to obtain its design strength.

Tension Controlled Section, Φ = 0.9 Compression Controlled Section

Member with lateral ties, Φ = 0.65

Members with spiral reinforcement, Φ = 0.75

Transition Section

For transition section Φ is permitted to be linearly interpolated between 0.65 or 0.75 to 0.9. 20


Strength Reduction Factor (Resistance Factor), Φ Transition Section (contd…)

yty

εεε0.005

0.250.65Φ

For members with ties

For members with Spirals

yty

εεε0.005

0.150.75Φ

21


Maximum Steel Ratio, ρmax

For T = C

ab'0.85ffA cys

ab'0.85ffbdρ cy

d

a

f

'f0.85ρ

y

c

For tension controlled section d8

3βa 0.005 ε 1s

So

1

y

cmax β

8

3

f

'f0.85ρ

22


Maximum Steel Ratio, ρmax (contd…)

For transition section

d7

3βa 0.004 ε 1s

1

y

cmax β

7

3

f

'f0.85ρ

23


Minimum Reinforcement of Flexural Members(ACI – 10.5.1)

The minimum steel is always provided in structural members because when concrete is cracked then all load comes on steel, so there should be a minimum amount of steel to resist that load to avoid sudden failure.

For slabs this formula gives a margin of 1.1 to 1.5. This formula is not used for slabs.

yy

cmin f

1.4

4f

'fρ

24


Under-Reinforced Failure

Cc

T Internal Force Diagram

a/2

la

Stage-II, Cracked Section

When section cracks, N.A. moves towards compression face means “la” increases. “T” and “Cc” also increase.

Stage-I, Un-cracked SectionN.A. position is fixed, means “la” remains constant. Only “T” and “Cc” increase with the increase of load

25


Under-Reinforced Failure (contd…)

Stage-III, Yielding in Steel Occur

T = Asfy remains constant and Cc also remains constant. “la” increases as the N.A. moves towards compression face because cracking continues.

Failure initiates by the yielding of steel but final failure is still by crushing of concrete

Cc


a/2

la

26



Derivation for ρ Design Moment Capacity

abnb TM l

2

adfAΦMΦ ysbnb

For tension controlled section Φ = 0.9

2

adf0.9AMΦ ysnb

And

b'0.85f

fAa

c

ys

(1)

(2)

27



Put value of “a” from (1) to (2)

b'0.85f2

fAdf0.9AMΦ

c

ysysnb

b'0.85f2

fρbddfρbd0.9MΦ

c

yynb

For economical design unb MMΦ

'0.85f2

fρ1fρbd0.9M

c

yy

2u

'0.85f

f

2

ρ1f0.9ρ

bd

M

c

yy2

u

28



Let

(MPa) Rbd

M2u ω

f

0.85fc'

y

And

Hence

2ω

ρ1f0.9ρR y

2ω

ρ1ρ

0.9f

R

y

2ω

ρ-ρ

0.9f

R 2

y

00.9f

R2ωρ2ω-ρ

y

2

0fc'

fc'

0.85

0.85

0.9f

R2ωρ2ω-ρ

y

2

0'0.3825f

Rωρ2ω-ρ

c

22

20.3825fc'

ωR44ω2ωρ

22

29



By simplification

0.3825fc'

R11ωρ

We have to use –ve sign for under reinforced sections. So

fc'

2.614R11ωρ

Reason

For under reinforced section ρ <ρb

If we use positive sign ρ will become greater than ρb, leading to brittle failure.

y1

y

cb f600

600β

f

'f0.85ρ

< 1.0

ω

ωρb 30


Trial Method for the determination of “As”

b'0.85f

fAa

c

ys (A)

2

adf0.9AM ysu (B)

2

ad0.9f

MA

y

us (C)

Trial # 1, Assume some value of “a” e.g. d/3 or d/4 or any other reasonable value, and put in (C) to get “As”

Trial # 2, Put the calculated value of “As” in (A) to get “a”. Put this “a” value in (C) to get “As”Keep on doing the trials unless “As” from a specific trial becomes equal to the “As” calculated from previous trial.

THIS VALUE OF AS WILL BE THE FINAL ANSWER.

31


Is The Section is Under-Reinforced or NOT ?1. Calculate ρ and if it is less than ρmax, section is

under reinforced2. Using “a” and “d” calculate εt if it is ≥ 0.005,

section is under-reinforced (tension controlled)

3. If section is over-reinforced than in the following equation –ve term will appear in the under-root.

'f

2.614R11ωρ

c

32


Is The Section is Under-Reinforced or NOT ?

(contd…)

1. For tension controlled section, εt = 0.005, d

8

3βa 1

Using formula of Mn from concrete side

acbnbu CΦMΦM l

2

adba'0.85f9.0M cu

2

d83

0.85dd

8

30.85b'0.85f0.9M cu

2cu bd'0.205fM

b'0.205f

Md

c

umin

If we keep d > dmin the resulting section will be under-reinforced.

d > dmin means that section is stronger in compression. 33


Over-Reinforced Failure

Cc


a/2

la

Stage-II, Cracked Section

These two stages are same as in under-reinforced section.

Stage-I, Un-cracked Section

Stage-III, Concrete reaches strain of 0.003 but steel not yielding

We never prefer to design a beam as over-reinforced (compression controlled) as it will show sudden failure.

Φ = 0.65 εs < εy fs<fy

34



Stage-III, Concrete reaches strain of 0.003 but steel not yielding (contd…)

2

a-dba'.85f00.65 M cnb

aCc Mnb l

“a” is unknown as “fs” is not known

b'0.85f

fAa

c

ss

(i)

(ii)

35




εcu=

0.003

Strain Diagram

εs

cB C

A

E D

d- c

Comparing ΔABC & ΔADE

c

cd

0.003

εs

1

1s

a

d

0.003

ε

a

a

aβ0.003ε 1

s

ss εEf

a

aβ0.003200,000f 1

s

a

aβ600f 1

s

(iii) (iv)Eq # (iv) is applicable when εs < εy

36




Putting value of “fs” from (iv) to (ii)

b'0.85fa

adβ600A

ac

1s

(v)

Eq. # (v) is quadratic equation in term of “a”.

Flexural Capacity

2'85.0

2

adbaf

adCM cbcbnb

2

adfAΦ

2

ad T ΦMΦ ssbbnb

Calculate “a” from (v) and “fs” from (iv) to calculate flexural capacity from these equations

37


Extreme Tensile Steel Strain

εt

Type of X-

section c/d a/d ρmax Φ

< εy

Compression Controlled

0.65

≥ εy

Transition Section

(Under-Reinforced)

0.65 to 0.9

≥ 0.004Under-

Reinforced(minimum strain for beams)

0.65 to 0.9

≥ 0.005 Tension Controlle

d

0.9

≥ 0.0075Redistribution is allowed

0.9

yf600

600

y1 f600

600β

yy

c1 f600

600

f

'0.85fβ

yf600

600

y1 f600

600β

yy

c1 f600

600

f

'0.85fβ

7

3

7

3β1

7

3

f

'0.85fβ

y

c1

8

3

8

3β1

8

3

f

'0.85fβ

y

c1

7

2

7

2β1

7

2

f

'0.85fβ

y

c1 38


Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design methodData

1. Dimensions, b, h, d and L (span)

2. fc’, fy, Ec, Es

3. As

Required1. ΦbMn

2. Load Carrying Capacity39


Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…)

SolutionStep # 1Calculte the depth of N.A assuming the section as under-reinforced

ys ff ys εε and

b' 0.85f

fAa

c

ys1β

ac and

40



SolutionStep # 2 Calculate εs and check the assumption of step# 1

c

cd0.003εε ts

For extreme

pointIf εs ≥ εy, the assumption is correct

If εs ≤ εy, the section is not under-reinforced. So “a” is to be calculated again by the formula of over reinforced section

b'0.85fa

adβ600A

ac

1s

41



SolutionStep # 3 Decide Φ factor

For εs ≥ 0.005, Φ = 0.9 (Tension controlled section)

For εs ≤ εy, Φ = 0.65 (Compression controlled section) For εy ≤ εs ≤ 0.005, Interpolate value of Φ (Transition Section)

Step # 4 Calculate ΦbMn

2

adfAM ysbnb

2

adba'f85.0M cbnb

For under-reinforced Section

For over-reinforced Section

42



Alternate MethodStep # 1 to step # 3 are for deciding whether the section is over reinforced or under-reinforced. Alternatively it can be done in the following manner.

1. Calculate ρ and ρmax if ρ < ρmax section is under-reinforced.

2. Calculate dmin, if d ≥ dmin, section is tension controlled

43


Selection of Steel Bars for Beams1. When different diameters are selected the

maximum difference can be a gap of one size.2. Minimum number of bars must be at least two,

one in each corner.3. Always Place the steel symmetrically.4. Preferably steel may be placed in a single layer

but it is allowed to use 2 to 3 layers.5. Selected sizes should be easily available in

market6. Small diameter (as far as possible) bars are

easy to cut and bend and place.44


Selection of Steel Bars for Beams (contd…)

7. ACI Code RequirementsThere must be a minimum clearance between bars (only

exception is bundled bars). Concrete must be able to flow through the reinforcement. Bond strength between concrete and steel must be fully developed.

Minimum spacing must be lesser of the following

Nominal diameter of bars 25mm in beams & 40mm in columns 1.33 times the maximum size of aggregate used.

We can also give an additional margin of 5 mm. 45


Selection of Steel Bars for Beams (contd…)

8. A minimum clear gap of 25 mm is to be provided between different layers of steel

9. The spacing between bars must not exceed a maximum value for crack control, usually applicable for slabs

What is Detailing? Deciding diameter of bars Deciding no. of bars Deciding location of bent-up and curtailment of bars making sketches of reinforcements.

25mm

Not O.K.

46


Concrete Cover to Reinforcement Measured as clear thickness outside the outer most steel bar.

Purpose

To prevent corrosion of steel To improve the bond strength To improve the fire rating of a building It reduces the wear of steel and attack of

chemicals specially in factories.

47


Concrete Cover to Reinforcement (contd…)

ACI Code Minimum Clear Cover Requirements

1. Concrete permanently exposed to earth, 75 mm2. Concrete occasionally exposed to earth,

# 19 to # 57 bars 50 mm # 16 and smaller bars 40 mm

3. Sheltered Concrete Slabs and Walls 20 mm Beams and Columns 40 mm

48


Load Carried by the BeamBeam Supporting One-way Slab

lx

lx

Exterior BeamInterior Beam

Width of slab supported by interior beam = lx

Width of slab supported by exterior beam = lx/2 + Cantilever width

ly

(ly/lx > 2)


Load Carrie by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2)

lx

lx

ly ly

Exterior Long Beam

Interior Long Beam

Exterior Short Beam

Interior Short Beam

45o


Load Carrie by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2) contd…

45olx/2

lx/2

o45cos

2/xl

2

o45cos2/x

lArea of

Square

Shorter Beams

For simplification this triangular load on both the sides is to be replaced by equivalent UDL, which gives same Mmax as for the actual triangular load.

2

2

xl


Load Carried by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2) contd…

45o45o

Equivalent Rectangular Area

22

x3

2

2

x

3

4l

l

Factor of 4/3 convert this VDL into UDL.

Equivalent width supported by interior short beam

lx

x

x32

l

l 2

x3

2l

Equivalent width supported by exterior short beam

3

xlCantilever

x3

2l


Load Carried by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2)contd…

lx

lx

ly

Exterior Long Beam

2

2

xxy

2

x

l

lll

Supported Area

lx/2 lx/2ly - lx

4

x

2

x

2

yx 22 llll

4

x

2

yx 2lll

2

xyx

2

1 2lll



Exterior Long Beam

2R1

3R1

F

2

y

xR

ll

where

Factor F converts trapezoidal load into equivalent UDL for maximum B.M. at center of simply supported beam. For Square panel

R = 1 and F = 4/3



Exterior Long Beam

Equivalent width

lx/2 lx/2ly - lx

Equivalent width

Length...Span

F)Supported..Area(

lyy

1

2R1

3R1

2

xyx

2

1 22

ll

ll

2R1

3R1

2

x1

2

x 2lyll

3R1

2

x 2l+ Cantilever (if present)


Load Carried by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2) contd…

Interior Long Beam

Equivalent width

lx/2 lx/2ly - lx

Equivalent width

ly

3R1x 2l


Wall Load (if present) on Beam

tw (mm)

H (m)1000

81.91930

1000

twHm1

UDL on beam

Htw019.0 (kN/m)

Htw019.0


Wall Load on the Lintel

Equivalent UDL on lintel if height of slab above lintel is greater than 0.866L

0.866L

60o 60o

LLtw11.0UDL kN/m

tw = wall thickness in “mm”

L = Opening size in “m”If the height of slab above lintel is less than 0.866L

Total Wall Load + Load from slab in case of load bearing wall

UDL = (Equivalent width of slab supported) x (Slab load per unit area)

= m x kN/m2 = kN/m


Slab Load per Unit AreaTop Roof

Slab Thickness = 125 mmEarth Filling = 100 mmBrick Tiles = 38 mm

Dead Load 2m/kg3002400

1000

125Self wt. of R.C. slab

Earth Filling 2m/kg1801800

1000

100

Brick Tiles 2m/kg741930

1000

38

554 kg/m2Total Dead Load, Wd =


Slab Load per Unit Area (contd…)

Top Roof

Live Load WL = 200 kg/m2

Ldu W6.1W2.1W

Total Factored Load, Wu

1000

81.92006.15542.1Wu

2u m/kN66.9W



Intermediate Floor Slab Thickness = 150 mmScreed (brick ballast + 25% sand) = 75 mmP.C.C. = 40 mmTerrazzo Floor = 20 mm

Dead Load2m/kg3602400

1000

150Self wt. of R.C. slab

Screed 2m/kg13518001000

75

Terrazzo + P.C.C 2m/kg13823001000

)4020(

633 kg/m2Total Dead Load, Wd =



Intermediate Floor Live Load

Occupancy Live Load = 250 kg/m2

Moveable Partition Load = 150 kg/m2

WL = 250 + 150 = 400 kg/m2

Ldu W6.1W2.1W

Total Factored Load, Wu

1000

81.94006.16332.1Wu

2u m/kN73.13W



Self Weight of Beam

Service Self Wight of Beam = b x h x 1m x 2400

2L11.112400m118

L

12

L Kg/m

Factored Self Wight of Beam

22 L131.01000

81.92.1L11.11 kN/m

Self weight of beam is required to be calculated in at the stage of analysis, when the beam sizes are not yet decided, so approximate self weight is computed using above formula.


Design of Singly Reinforced Beam by Strength Method (for flexure only)

Data: Load, Span (SFD, BMD) fc’, fy, Es

Architectural depth, if any

Required: Dimensions, b & h Area of steel Detailing (bar bending schedule)

Plain & Reinforced Concrete-1Design of Singly Reinforced Beam by Strength Method

(contd…)

Procedure:

1. Select reasonable steel ratio between ρmin and ρmax. Then find b, h and As.

2. Select reasonable values of b, h and then calculate ρ and As.


(contd…)

2. Using Trial DimensionsI. Calculate loads acting on the beam.II. Calculate total factored loads and plot SFD

and BMD. Determine Vumax and Mumax.

III. Select suitable value of beam width ‘b’. Usually between L/20 to L/15. preferably a multiple of 75mm or 114 mm.

IV. Calculate dmin.b'f205.0

Md

c

umin

hmin = dmin + 60 mm for single layer of steel

hmin = dmin + 75mm for double layer of steelRound to upper 75 mm


(contd…)

V. Decide the final depth.

minhh For strength

minhh For deflection

ahh Architectural depth

12hh

Preferably “h” should be multiple of 75mm.

Recalculate “d” for the new value of “h”


(contd…)

VI. Calculate “ρ” and “As”.

fc'

2.614R11wρ

Four methods

y

c

f

'f0.85w 2

u

bd

MR

Design Table

Design curves

Using trial Method

a)

b)

c)

d)


(contd…)

VII. Check As ≥ As min.

As min = ρmin bd (ρmin = 1.4/fy to fc’ ≤ 30 MPa)

VIII. Carry out detailing

IX. Prepare detailed sketches/drawings.

X. Prepare bar bending schedule.


Design of Singly Reinforced Beam by Strength Method (contd…)

1. Using Steel RatioI. Step I and II are same as in previous method. III. Calculate ρmax and ρmin & select some suitable

“ρ”.IV. Calculate bd2 from the formula of moment

V. Select such values of “b” and “d” that “bd2” value is satisfied.

VI. Calculate As.VII. Remaining steps are same as of previous

method.

'1.7f

ρf1fρbd0.9MΦM

c

yy

2nbu

Flexural design of Beam...PRC-I

Design

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