# Plain & Reinforced Concrete-1

Urban and Civil

Nov 25, 2013 (4 years and 5 months ago)

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Plain & Reinforced
Concrete
-
1

CE3601

Lecture # 11 &12

1
st

to 6
th
March 2012

Flexural Analysis and

Design of Beams

(Ultimate Strength Design of Beams)

Plain & Reinforced Concrete
-
1

Under
-
Reinforced Failure

C
c

T

Internal Force Diagram

a/2

l
a

Stage
-
II, Cracked Section

When section cracks, N.A. moves
towards compression face means

l
a” increases. “T” and “Cc” also
increase.

Stage
-
I, Un
-
cracked Section

N.A. position is fixed, means “
l
a”
remains constant. Only “T” and
“Cc” increase with the increase of

Plain & Reinforced Concrete
-
1

Under
-
Reinforced Failure
(contd…)

Stage
-
III, Yielding in Steel Occur

T = A
s
f
y

remains constant and C
c

also
remains constant.

l
a” 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

C
c

T

Internal Force Diagram

a/2

l
a

Plain & Reinforced Concrete
-
1

Under
-
Reinforced Failure
(contd…)

Derivation for ρ

Design Moment Capacity

a
b
n
b
T
M
l

2
a
d
f
A
Φ
M
Φ
y
s
b
n
b
For tension controlled section Φ = 0.9

2
a
d
f
0.9A
M
Φ
y
s
n
b
And

b
'
0.85f
f
A
a
c
y
s

(1)

(2)

Plain & Reinforced Concrete
-
1

Under
-
Reinforced Failure
(contd…)

Put value of “a” from (1) to (2)

b
'
0.85f
2
f
A
d
f
0.9A
M
Φ
c
y
s
y
s
n
b

b
'
0.85f
2
f
ρbd
d
f
ρbd
0.9
M
Φ
c
y
y
n
b
For economical design

u
n
b
M
M
Φ

'
0.85f
2
f
ρ
1
f
ρbd
0.9
M
c
y
y
2
u

'
0.85f
f
2
ρ
1
f
0.9
ρ
bd
M
c
y
y
2
u
Plain & Reinforced Concrete
-
1

Under
-
Reinforced Failure
(contd…)

Let

(MPa)

R
bd
M
2
u

ω
f
0.85fc'
y

And

Hence

2
ω
ρ
1
f
0.9
ρ
R
y

2
ω
ρ
1
ρ
0.9f
R
y
2
ω
ρ
-
ρ
0.9f
R
2
y

0
0.9f
R
2
ω
ρ
2
ω
-
ρ
y
2

0
fc'
fc'
0.85
0.85
0.9f
R
2
ω
ρ
2
ω
-
ρ
y
2

0
'
0.3825f
R
ω
ρ
2
ω
-
ρ
c
2
2

2
0.3825fc'
ω
R
4
4
ω
2
ω
ρ
2
2

Plain & Reinforced Concrete
-
1

Under
-
Reinforced Failure
(contd…)

By simplification

0.3825fc'
R
1
1
ω
ρ
We have to use

ve sign for under reinforced sections. So

fc'
2.614R
1
1
ω
ρ
Reason

For under reinforced section ρ <ρ
b

If we use positive sign

ρ will
become greater than ρ
b
brittle failure.

y
1
y
c
b
f
600
600
β

f
'
f
0.85
ρ

< 1.0

ω

ω
ρ
b

Plain & Reinforced Concrete
-
1

Plotting of R
-
ρ

ω
d
ρ
ωb
ρbd
b
'
0.85f
f
A
a
c
y
s

2
a
d
ρbdf
Φ
M
y
b
u

2d
a
1
f
ρ
Φ
bd
M
y
b
2
u

2d
a
1
f
ρ
9
.
0
R
y
(3)

(4)

ρ

R

ρ

R

Plain & Reinforced Concrete
-
1

Trial Method for the determination of “A
s

b
'
0.85f
f
A
a
c
y
s

(A)

2
a
d
f
0.9A
M
y
s
u
(B)

2
a
d
0.9f
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 “A
s

Trial # 2
, Put the calculated value
of “A
s
” in
(A)

to get “a”. Put this
“a” value in
(C)

to get “A
s

Keep on doing the trials unless
“As” from a specific trial
becomes equal to the “As”
calculated from previous trial.

THIS VALUE OF A
S

WILL BE

Plain & Reinforced Concrete
-
1

Is The Section is Under
-
Reinforced or NOT ?

1.
Calculate ρ and if it is less than ρ
max
, section is
under reinforced

2.
Using “a” and “d” calculate ε
t

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
1
1
ω
ρ
c
Plain & Reinforced Concrete
-
1

Is The Section is Under
-
Reinforced or NOT ?

(contd…)

1.
For tension controlled section, ε
t

= 0.005
,

d
8
3
β
a
1

Using formula of M
n

from concrete side

a
c
b
n
b
u
C
Φ
M
Φ
M
l

2
a
d
ba
'
0.85f
9
.
0
M
c
u

2
d
8
3
0.85
d
d
8
3
0.85
b
'
0.85f
0.9
M
c
u
2
c
u
bd
'
0.205f
M

b
'
0.205f
M
d
c
u
min

If we keep d > d
min

the resulting section
will be under
-
reinforced.

d > d
min

means that
section is stronger in
compression.

Plain & Reinforced Concrete
-
1

Over
-
Reinforced Failure

C
c

T

Internal Force Diagram

a/2

l
a

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
ε
s

< ε
y
f
s
<f
y

Plain & Reinforced Concrete
-
1

Over
-
Reinforced Failure

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

2
a
-
d
ba
'
.85f
0
0.65

M
c
n
b
a
Cc

M
n
b
l

“a” is unknown as “f
s
” is not known

b
'
0.85f
f
A
a
c
s
s

(i)

(ii)

Plain & Reinforced Concrete
-
1

Over
-
Reinforced Failure

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

ε
cu
=

0.003

Strain Diagram

ε
s

c

B

C

A

E

D

d
-

c

c
c
d
0.003
ε
s

1
1
s
a
d
0.003
ε

a

a
a
β
0.003
ε
1
s
s
s
ε
E
f

a
a
β
0.003
200,000
f
1
s

a
a
β
600
f
1
s
(iii)

(iv)

Eq # (iv) is applicable
when ε
s

< ε
y

Plain & Reinforced Concrete
-
1

Over
-
Reinforced Failure

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

Putting value of “f
s
” from (iv) to (ii)

b
'
0.85f
a
a
d
β
600
A
a
c
1
s

(v)

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

Flexural Capacity

2
'
85
.
0
2
a
d
ba
f
a
d
C
M
c
b
c
b
n
b

2
a
d
f
A
Φ
2
a
d

T

Φ
M
Φ
s
s
b
b
n
b
Calculate “a” from (v)
and

“f
s
” from (iv) to
calculate flexural
capacity from these
equations

Plain & Reinforced Concrete
-
1

Extreme
Tensile
Steel Strain

ε
t

Type of
X
-
section

c/d

a/d

ρ
max

Φ

<
ε
y

Compression
Controlled

0.65

ε
y

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

y
f
600
600

y
1
f
600
600
β

y
y
c
1
f
600
600
f
'
0.85f
β

y
f
600
600

y
1
f
600
600
β

y
y
c
1
f
600
600
f
'
0.85f
β
7
3

7
3
β
1

7
3
f
'
0.85f
β
y
c
1

8
3

8
3
β
1

8
3
f
'
0.85f
β
y
c
1

7
2

7
2
β
1

7
2
f
'
0.85f
β
y
c
1

Plain & Reinforced Concrete
-
1

Capacity Analysis of Singly Reinforced
Rectangular Beam by Strength Design method

Data

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

2.
f
c
’, f
y
, E
c
, E
s

3.
As

Required

1.
Φ
b
M
n

2.

Plain & Reinforced Concrete
-
1

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

Solution

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

y
s
f
f

y
s
ε
ε

and

b
'

0.85f
f
A
a
c
y
s

1
β
a
c

and

Plain & Reinforced Concrete
-
1

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

Solution

Step # 2 Calculate ε
s

and check the assumption of step# 1

c
c
d
0.003
ε
ε
t
s

For extreme point

If ε
s

≥ ε
y
, the assumption is correct

If ε
s

≤ ε
y
, the section is under
-
reinforced. So “a” is to be calculated
again by the formula of over reinforced section

b
'
0.85f
a
a
d
β
600
A
a
c
1
s

Plain & Reinforced Concrete
-
1

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

Solution

Step # 3 Decide Φ factor

F
or ε
s

≥ 0.005, Φ = 0.9 (Tension controlled section)

F
or ε
s

≤ ε
y
, Φ = 0.65 (Compression controlled section)

For ε
y

≤ ε
s

0.005, Interpolate value of Φ
(Transition Section)

Step # 4 Calculate
Φ
b
M
n

2
a
d
f
A
M
y
s
b
n
b

2
a
d
ba
'
f
85
.
0
M
c
b
n
b
For under
-
reinforced Section

For over
-
reinforced Section

Plain & Reinforced Concrete
-
1

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

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
ρ
max

if
ρ < ρ
max

section is under
-
reinforced.

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

Plain & Reinforced Concrete
-
1

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.

Plain & Reinforced Concrete
-
1

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.

Plain & Reinforced Concrete
-
1

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.

Plain & Reinforced Concrete
-
1

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.

Plain & Reinforced Concrete
-
1

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

Plain & Reinforced Concrete
-
1

Number of Bars in a Single Layer (for beams)

4
.
1
b
02
.
0
N
w
b

Rounded to lower whole number

b
w

= width of web of beam

For I
-
Shape beam width of bottom flange should be used in
place of b
w

.

Plain & Reinforced Concrete
-
1

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

Concluded