Analysis and Design of Reinforced Concrete Beams

RC Structures
Behavior
Session 3
Flexure on Beam

 Plane sections remain plane (not true for
deep beams h > 4b)
 The strain in the reinforcement is equal to
the strain in the concrete at the same level,
i.e. es = ec at same level.
 Stress in concrete & reinforcement may be
calculated from the strains using s-e curves
for concrete & steel.
Basic Assumptions in Flexure Theory

 Tensile strength of concrete is neglected for
calculation of flexural strength.
 Concrete is assumed to fail in compression,
when ec (concrete strain) = ecu (limit state) =
0.003
 Compressive s-e relationship for concrete may
be assumed to be any shape that results in an
acceptable prediction of strength.
Additional Assumptions for Simplification
in Design

Flexural Stress
The concrete may exceed the ec at the outside edge of
the compressive zone.

The compressive force is modeled as Cc = k1k3f’c b*c
at the location x = k2*c
Flexural Stress (cont’d)

The compressive coefficients
of the stress block at given
for the following shapes.
k3 is ratio of maximum stress
at fc in the compressive zone
of a beam to the cylinder
strength, fc’ (0.85 is a
typical value for common
concrete) (ACI [10.2.7])

The compressive zone is modeled with a equivalent
stress block.

The equivalent rectangular concrete stress
distribution has what is known as a b1 coefficient is
proportion of average stress distribution covers. (ACI
[10.2.7.3])
65
.
0
1000
4000
*
05
.
0
85
.
0
psi
4000
for
85
.
0
c
1
c
1








 -
-



f
f
b
b

Requirements for analysis of reinforced concrete
beams:
1. Stress-Strain Compatibility – Stress at a point in member
must correspond to strain at a point.
2. Equilibrium – Internal forces balances with external
forces.

The beam is a structural
member used to support
the internal moments and
shears. It would be called
a beam-column if a
compressive force existed.
C = T
M = C*(jd)
= T*(jd)

The stress in the block
is defined as:
s  (M*y) / I
Sxx = I / (ymax)
The equation for Sxx
modulus for calculating
maximum compressive
stress.

There are 5 stages the concrete through which the beam
goes.
Stage 1:
No external loads self weight.

Stage 2:
The external load P
cause the bottom fibers
to equal to modulus of
rupture of the concrete.
Entire concrete section
was effective, steel bar
at tension side has
same strain as
surrounding concrete.

Stage 3:
The tensile strength of
the concrete exceeds
the rupture fr and
cracks develop. The
neutral axis shifts
upward and cracks
extend to neutral axis.
Concrete loses tensile
strength and steel starts
working effectively and
resists the entire tensile
load.

Stage 4:
The reinforcement
yields.
Stage 5:
Failure of the
beam.

Stage 1:
No external loads
acting on the beam.
Stage 2:
Service loading on the
beam.
Stage 3:
Beam failure.
The three stages of the beam.

The moment-curvature
diagram show the five
stages of the beam. The
plot is of the curvature
angle, f , verse the
moment.
f = (e / y) = [ s / E ] / y
= [(My / I) / E] / y
f = M / ( E I )

The beam fails first in shear and the second beam fails
in bending moment.

There are three types of flexural failure of a structural
member:
1. Steel may reach its yield strength before the
concrete reaches its maximum. (Under-
reinforced section).
2. Steel reaches yield at same time as concrete
reaches ultimate strength. (Balanced section).
3. Concrete may fail before the the yield of steel due
to the presence of a high percentage of steel in the
section. (Over-reinforced section).
Flexural Failure

Steel may reach its yield strength before the concrete
reaches its maximum. (Under-reinforced section).
Flexural Failure (cont’d)

Steel reaches yield at same time as concrete reaches
ultimate strength. (Balanced section).

Concrete may fail before the the yield of steel due to
the presence of a high percentage of steel in the
section. (Over-reinforced section).

The flexural strain and stress distribution of beam
from a test beam.
Flexural Strain and Stress

Strain measured in test of eccentrically loaded
columns for a tied and spiral columns.
Flexural Strain and Stress (cont’d)

Flexural Stress - Example
Consider a simple rectangular beam (b x h)
reinforced with steel reinforcement of As.
1. Determine the centroid ( neutral axis, NA ) and
moment of inertia Izz of the beam for an ideal
beam (no cracks).
2. Determine the NA and moment of inertia, Izz, of
beam if the beam is cracked and tensile forces
are in the steel only.

Example - Definitions
Ec – Modulus of Elasticity - concrete
Es – Modulus of Elasticity - steel
As – Area of steel
d – distance to steel
b – width
h – height s
c
E
E
n 

Example - Mechanics of Materials
( 
i i i
i i
2
i i i i i
y n A
n A
n y n A
y
I I y

  -


 
Centroid (NA)
Moment of Inertia

Example - (uncracked)
Area yi yiA I yi - y (yi -y)2
A
Concrete bh h/2 bh2
/2 bh3
/12 (h/2-y) (h/2-y)2
bh
Steel (n-1)As d d(n-1)As --- (d-y) (d-y)2
(n-1)As
( 
( 
(  (  ( 
2
s
i i
i s
2
3
2 2
s
1
y A 2
A 1
1
12 2
i i i
bh
n A d
y
bh n A
bh h
I I y y A y bh d y n A
 -
 
 -
 
  -   -  - -
 
 


 

Example - (cracked)
For a cracked section the
concrete is in compression
and steel is in tension.
The strain in the beam is
linear.
c
s s
1
2
C yb f
T A f



Using Equilibrium
s s c
s c
s
1
2
2
T C
A f yb f
yb
f f
A


 
  
 
Example - (cracked) (cont’d)

Using Hooke’s law
s s c c
s
2
yb
E E
A
e e
 
  
 

f Ee

However, this is an indeterminate problem to
find . We will need to use a compatibility
condition.
y
c s s s
s c
2 2
E A nA
E yb yb
e
e
  
 
  
 
 

Using a compatibility condition.
s c c
s
d d
y
y y y
e e e
e
  
- -
Substitute into the first equation.

s
2nA
y
d y yb

-
Substitute in for the strain relationship.
2 s s
2 2
0
nA nA
y y d
b b
   
 - 
   
   
Rearrange the equation into a quadratic equation.

s
A
bd
 
Use a ratio of areas of concrete and steel.
Modify the equation to create a non-dimensional ratio.
2
2 2 0
y y
n n
d d
 
   
 - 
   
   
2 2
2 2 0
y n d y n d
 
 - 

Use the quadratic formula
Solve for the centroid by multiplying the result by d.
( 
( 
2
2
2 2 8
2
2
n n n
y
d
y
n n n
d
  
  
-  
 

 
 
 
  -
 
 

The moment of inertia using the parallel axis
( 
( 
( 
2
2
3
2
s
3
2
s
12 2
3
i i i
I I y y A
by y
by d y nA
by
d y nA
  -
 
   -
 
 
  -
 

For the following example find centroid and moment
of inertia for an uncracked and cracked section and
compare the results.
Es = 29000 ksi
Ec = 3625 ksi
d = 15.5 in b = 12 in. h = 18 in.
Use 4 # 7 bars for the steel.
Example

A #7 bar has an As = 0.6 in2
s
c
E 29000 ksi
8
E 3625 ksi
n   
( 
2 2
s 4 0.6 in 2.4 in
A  
Example (cont’d)

The uncracked centroid is
( 
( 
( ( 
( ( ( 
( (  ( ( 
2
s
s
2
2
2
3
2
1
2
1
12 in 18 in
8 1 2.4 in 15.5 in
2
12 in 18 in 8 1 2.4 in
2204.4 in
9.47 in
232.8 in
bh
n A d
y
bh n A
 -

 -
 -

 -
 
Example (cont’d)

The uncracked moment of inertia
(  ( 
( ( 
( ( 
(  ( ( 
2
3
2
s
3 2
2 2
4
1
12 2
12 in 18 in 18 in
9.47 in 12 in 18 in
12 2
15.5 in 9.47 in 8 1 2.4 in
6491 in
bh h
I y bh d y n A
 
  -  - -
 
 
 
  -
 
 
 - -

Example (cont’d)

The cracked centroid is defined by:
( ( 
2
s 2.4 in
0.0129
12 in 15.5 in
A
bd
   
( 
( ( 
(  ( (  ( ( 
( 
2
2
2
8 0.0129 2 8 0.0129 8 0.0129
0.3627
0.3627 15.5 in 5.62 in
y
n n n
d
y
  
 
  -
 
 
 - -

 
Example (cont’d)

The cracked moment of inertia is
( 
( ( 
(  ( ( 
3
2
s
3
2 2
4
3
1
12 in 5.62 in
3
15.5 in 5.62 in 8 2.4 in
2584.2 in
by
I d y nA
  -

 -

Example (cont’d)

Notice that the centroid changes from 9.47 in. to
5.62 in. and the moment of inertia decreases from
6491 in4 to 2584 in4. The cracked section loses
more than half of its’ strength.
Example (cont’d)

Analysis and Design of Reinforced Concrete Beams

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