Lec 11 12 -flexural analysis and design of beams

Plain & Reinforced
Concrete-1
CE3601
Lecture # 11 &12
1st to 6th March 2012
Flexural Analysis and
Design of Beams
(Ultimate Strength Design of Beams)

Plain & Reinforced Concrete-1
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 Section
N.A. position is fixed, means “la”
remains constant. Only “T” and
“Cc” increase with the increase of
load

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
T
a/2
la

Derivation for ρ
Design Moment Capacity
abnb TM l







2
a
dfAΦMΦ ysbnb
For tension controlled section Φ = 0.9







2
a
df0.9AMΦ ysnb
And
b'0.85f
fA
a
c
ys

(1)
(2)

Put value of “a” from (1) to (2)








b'0.85f2
fA
df0.9AMΦ
c
ys
ysnb









b'0.85f2
fρbd
dfρbd0.9MΦ
c
y
ynb
For economical design
unb MMΦ 









'0.85f2
fρ
1fρbd0.9M
c
y
y
2
u







'0.85f
f
2
ρ
1f0.9ρ
bd
M
c
y
y2
u

Let
(MPa)R
bd
M
2
u
 ω
f
0.85fc'
y
And
Hence







2ω
ρ
1f0.9ρR y







2ω
ρ
1ρ
0.9f
R
y
2ω
ρ
-ρ
0.9f
R 2
y

0
0.9f
R2ω
ρ2ω-ρ
y
2



0
fc'
fc'
0.85
0.85
0.9f
R2ω
ρ2ω-ρ
y
2



0
'0.3825f
Rω
ρ2ω-ρ
c
2
2



2
0.3825fc'
ωR
44ω2ωρ
2
2 


By simplification









0.3825fc'
R
11ωρ
We have to use –ve sign for under reinforced sections. So









fc'
2.614R
11ωρ
Reason
For under reinforced section ρ <ρb
If we use positive sign ρ will
become greater than ρb, leading to
brittle failure.
y
1
y
c
b
f600
600
β
f
'f
0.85ρ


< 1.0
ω
ωρb 

Plotting of R -ρ
ω
d
ρ
ωb
ρbd
b'0.85f
fA
a
c
ys








2
a
dρbdfΦM ybu 






2d
a
1fρΦ
bd
M
yb2
u







2d
a
1fρ9.0R y
(3)
(4)
ρ
R
ρ
R

Trial Method for the determination of “As”
b'0.85f
fA
a
c
ys
 (A)







2
a
df0.9AM ysu (B)








2
a
d0.9f
M
A
y
u
s (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.

Is The Section is Under-Reinforced or NOT ?
1. Calculate ρ and if it is less than ρmax, section is
under reinforced
2. 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
2.614R
11ωρ
c

Is The Section is Under-Reinforced or NOT ?
(contd…)
1. For tension controlled section, εt = 0.005, d
8
3
βa 1
Using formula of Mn from concrete side
acbnbu CΦMΦM l







2
a
dba'0.85f9.0M cu



















2
d
8
3
0.85
dd
8
3
0.85b'0.85f0.9M cu
2
cu bd'0.205fM 
b'0.205f
M
d
c
u
min


If we keep d > dmin
the resulting section
will be under-
reinforced.
d > dmin means that
section is stronger in
compression.

Over-Reinforced Failure
Cc
T
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

Stage-III, Concrete reaches strain of 0.003 but steel not
yielding (contd…)







2
a
-dba'.85f00.65M cnb
aCcMnb l
“a” is unknown as “fs” is not known
b'0.85f
fA
a
c
ss

(i)
(ii)

Stage-III, Concrete reaches strain of 0.003 but steel not yielding
(contd…)
εcu= 0.003
Strain Diagram
εs
c
B C
A
E D
d- c
Comparing ΔABC & ΔADE
c
cd
0.003
εs 

1
1s
a
d
0.003
ε


a






 

a
aβ
0.003ε 1
s
ss εEf 





 

a
aβ
0.003200,000f 1
s





 

a
aβ
600f 1
s
(iii)
(iv)
Eq # (iv) is applicable
when εs < εy

Stage-III, Concrete reaches strain of 0.003 but steel not yielding
(contd…)
Putting value of “fs” from (iv) to (ii)
b'0.85f
a
adβ
600A
a
c
1
s 




 


(v)
Eq. # (v) is quadratic equation in term of “a”.
Flexural Capacity













2
'85.0
2
a
dbaf
a
dCM cbcbnb













2
a
dfAΦ
2
a
dTΦMΦ ssbbnb
Calculate “a” from (v)
and “fs” from (iv) to
calculate flexural
capacity from these
equations

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
≥ 0.004
Under-
Reinforced
(minimum strain
for beams)
0.65 to
0.9
≥ 0.005 Tension
Controlled
0.9
≥ 0.0075
Redistribution
is allowed
0.9










yf600
600










y
1
f600
600
β










yy
c
1
f600
600
f
'0.85f
β










yf600
600










y
1
f600
600
β 









yy
c
1
f600
600
f
'0.85f
β
7
3

7
3
β1
7
3
f
'0.85f
β
y
c
1 
8
3

8
3
β1
8
3
f
'0.85f
β
y
c
1 
7
2

7
2
β1 7
2
f
'0.85f
β
y
c
1 

Capacity Analysis of Singly Reinforced
Rectangular Beam by Strength Design method
Data
1. Dimensions, b, h, d and L (span)
2. fc’, fy, Ec, Es
3. As
Required
1. ΦbMn
2. Load Carrying Capacity

Rectangular Beam by Strength Design method (contd…)
Solution
Step # 1Calculte the depth of N.A assuming the section as
under-reinforced
ys ff  ys εε and
b'0.85f
fA
a
c
ys

1β
a
c and

Solution
Step # 2 Calculate εs and check the assumption of step# 1
c
cd
0.003εε ts

 For extreme point
If εs ≥ εy, the assumption is correct
If εs ≤ εy, the section is under-reinforced. So “a” is to be calculated
again by the formula of over reinforced section
b'0.85f
a
adβ
600A
a
c
1
s 




 



Solution
Step # 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
a
dfAM ysbnb







2
a
dba'f85.0M cbnb
For under-reinforced Section
For over-reinforced Section

Alternate Method
Step # 1 to step # 3 are for deciding whether the section is
tension 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

Selection of Steel Bars for Beams
1. 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 market
6. Small diameter (as far as possible) bars are easy to cut
and bend and place.

Selection of Steel Bars for Beams (contd…)
7. ACI Code Requirements
There 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.

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.

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.

Concrete Cover to Reinforcement (contd…)
ACI Code Minimum Clear Cover Requirements
1. Concrete permanently exposed to earth, 75 mm
2. 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

Number of Bars in a Single Layer (for beams)
4.1b02.0N wb 
Rounded to lower whole number
bw = width of web of beam
For I-Shape beam width of bottom flange should be used in
place of bw .

Example
A singly reinforced rectangular beam has a width of 228
mm and effective depth of 450 mm. fc’ = 17.25 MPa, fy =
420 MPa. Calculate flexural capacity for the following three
cases.
1. 2 # 25 bars (SI size)
2. 3 # 25 + 2 # 15 (SI)
3. Capacity for balanced steel

Lec 11 12 -flexural analysis and design of beams

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