This document discusses the flexural behavior of reinforced concrete beams under service loads. It provides assumptions and equations used to analyze beams in their elastic range when both concrete and steel are within their proportional limits. The key points are: 1) Plane sections remain plane after bending. Strains in steel and concrete are equal due to bond. 2) Cracks form when tension stresses exceed concrete's tensile strength, but steel reinforcement carries load. 3) Compression stresses are limited to 0.85 times concrete's compressive strength. Strain diagrams and equations for moment capacity are derived.

1. Reinforced Concrete
Design-I
CE 01344
Flexural Analysis and
Design of Beams

2. Flexural Behavior of Beams Under Service Load
When loads are applied on the beam, stresses are produced
in concrete and steel reinforcement.
 If stress in steel bars is less than yield strength, steel is in elastic
range.
 If stress in concrete is less than 0.6fc’ concrete is assumed to be
with in elastic range.
Following are important points related to Elastic Range:
 Loads are un-factored
 Materials are in elastic range
 Allowable stress analysis and design is applicable.
Reinforced Concrete Design-I

3. Assumption for the Study of Flexural Behavior
 Plane sections of the beam remain plane after bending.
 The material of the beam is homogeneous and obeys hooks law
Stress  Strain
 Perfect bond exists between steel & concrete so whatever strain
is produced in concrete same is produced in steel.
 All the applied loads up to to failure are in equilibrium with the
internal forces developed in the material.
 At the strain of 0.003concrete is crushed.
Reinforced Concrete Design-I

4. Assumption for the Study of Flexural Behavior (contd…)
 When cracks appear on the tension face of beam the concrete
capacity to resist tension is considered zero.
 Stress and strain diagrams for steel and concrete are simplified.
Strain
Stress
Steel
Strain
0.6fc’
Stress
Concrete
Reinforced Concrete Design-I

5. Step # 4Show location of internal resultant forces.
T
Step #5 Write down the equilibrium equations or use geometry of
strain diagram to arrive at various results.
Flexural Behavior Beams
General Procedure for the Derivation of Formula
Step # 1 Draw the cross section of beam with reinforcement.
Step # 2Draw the strain diagram for the cross section.
Step # 3Draw the stress diagram.
C
la
Reinforced Concrete Design-I

6. Flexural Behavior Beams (contd…)
1. When Both Steel and Concrete are in Elastic Range
C
T
la
N.A.
Strain Diagram Stress Diagram Resultant Force
Diagram
Both steel and concrete are resisting to applied
action
fc
fs
εc
εs
Reinforced Concrete Design-I

7. Flexural Behavior Beams (contd…)
2. When Cracks are Appeared on tension Side
C
T
la
N.A.
fc
εc
fs
Strain Diagram Stress Diagram
Resultant Force
Diagram
When the tension side is cracked the concrete becomes ineffective
but the strains goes on increasing. The steel comes in to action to
take the tension.
εs
Reinforced Concrete Design-I

8. Flexural Behavior Beams (contd…)
3. When Compression Stresses
Cross Elastic Range
C
T
la
N.A.
0.85fc
εc
fs
Resultant Force
Diagram
It is clear that the stress diagram is in fact obtained by rotating the stress
strain diagram of concrete.
Strains keeps on changing linearly in all three cases.
ε
Strain Diagrasm Stress Diagram
fc’ 0.85fc’
Stress
Strain
Reinforced Concrete Design-I

9. Flexural Behavior Beams (contd…)
Final Equation for Calculating Moment Capacity
Mr = T x la = C x la
Reinforced Concrete Design-I

10. Flexural/Bending Stress Formula
f = ±My/I (Valid in Elastic Range Only)
f = ±M/(I/y)
f = ±M/S
f = Flexural Stress
S = Elastic Section Modulus
Reinforced Concrete Design-I

11. Shear Stress Formula
τ = VAy/(I b)
(Valid in Elastic Range Only)
τ = VQ/(I b)
τ = ShearStress
Q = First moment ofarea
b
d
h
First Moment of Shaded Area, Q = (b x d ) h
Reinforced Concrete Design-I

12. Notation
b As’ (Compression Face) b
As (Tension Face)
h d
bw
hf
d, Effective Depth
fc = concrete stress at any load level at any distance form the N.A
fc’= 28days cylinder strength
εc= Strain in concrete any load level
εcu = Ultimate concrete strain, 0.003
Reinforced Concrete Design-I

13. Notation (contd…)
fy = Yield strength of steel
fs = Steel stress at a particular load level
εs = strain in steel at a particular level, εs= fs/Es
εy = Yield strain in steel
Es = Modulus of elasticity of steel
Ec = Modulus of elasticity of concrete
ρ (rho) = Steel Ratio, ρ = As/Aec = As/(bd)
T = Resultant tensile force
C = Resultant compressive force
Reinforced Concrete Design-I

14. Notation (contd…)
N.A
h kd
d
C
T
la = jd
jd = Lever arm j= la /d (valid for elastic range)
kd = Depth of N.A. from compression face, k = c/d
jand k are always less than1.
b
Reinforced Concrete Design-I

15. Different Types of Cracks
1. Pure Flexural Cracks
Flexural cracks start appearing at the section of maximum
bending moment. These vertical cracks initiate from the
tension face and move towards N.A.
Reinforced Concrete Design-I

16. Different Types of Cracks (contd…)
2. Pure Shear/Web Shear Cracks
These inclined cracks appear at the N.A due to shear stress
and propagate in both direction
45o
τ
σ1 σ4
45o
σn = f
σ3 σ2
σ1and σ2are Major Principle Stresses
Reinforced Concrete Design-I

17. Pure Shear/Web Shear Cracks
Reinforced Concrete Design-I

18. Different Types of Cracks (contd…)
3. Flexural Shear Cracks
In the regions of high shear due to diagonal tension,
the inclined cracks develop as an extension of flexural
cracks and are termed as Flexural Shear Cracks.
Flexural Shear Cracks.
Reinforced Concrete Design-I

19. Tensile Strength of Concrete
There are considerable experimental difficulties in determining
the true tensile strength of concrete. In direct tension test
following are the difficulties:
1.When concrete is gripped by the machine it may be crushed
due to the large stress concentration at the grip.
2.Concrete samples of different sizes and diameters show large
variation in results.
3.If there are some voids in sample the test may show very low
strength.
4.If there is some initial misalignment in fixing the sample, the
results are not accurate.
Reinforced Concrete Design-I

20. Following are the few indirect methods through
with tensile strength of concrete is estimated.
A. Split cylinder Test
This test is performed by loading a standard 150mm x
300mmcylinder by a line load perpendicular to its
longitudinal axis with cylinder placed horizontally on the
testing machine platen.
The tensile strength can be defined as
ft = 2P / (DL)
Where
P = Total value of load registered by machine
D = Diameter of concretecylinder
L = Cylinder height
Reinforced Concrete Design-I

21. Tensile Strength of Concrete (contd…)
B. Double Punch Test
In this test a concrete cylinder is placed vertically between the loading
platens of the machine and is compressed by two steel punches placed
parallel to top and bottom end surfaces. The sample splits across many
vertical diametrical planes radiating from central axis.
Tensile strength can be defined as
ft = Q / [ (1.2bH-a2)]
Q = Crushing Load
H
Q
Q
2b
2a
Reinforced Concrete Design-I

22. Tensile Strength of Concrete (contd…)
C. Modulus of Rupture Test
For many years, tensile strength has been measured in term of the
modulus of rupture fr, the computed flexural tensile stress at which a
test beam of plain concrete fractures.
Because this nominal stress is computed on the assumption that
concrete is an elastic material, and because this bending stress is
localized at the outermost surface, it is larger than the strength of
concrete in uniform axial tension.
It is a measure of, but not identical with the real axial tensile strength.
Two point loading
Reinforced Concrete Design-I

23. There are some relationships which relate modulus of rupture, fr, with
compressive strength of concrete
fr = 0.69 √fc’
fc’ and fr are in MPa. It also varies between 10 to 15% of fc’.
ACI code gives a formula for fr for strength calculations:
fr = 0.5√ fc’
And for deflection control:
fr = 0.62√ fc’
Where  = 1 for normal weight concrete.
Mean split cylinder strength = 0.53√ fc’
Reinforced Concrete Design-I

24. Tensile strength  square root of compressive strength.
True tensile strength varies between 8to 15% of fc’.
Few empirical formulae are developed for the tensile strength of
concrete:
ft = 0.25√ fc’ to 0.42√ fc’
Modular Ratio (n): “The ratio of modulus of elasticity of steel
to modulus of elasticity of concrete is known as modular Ratio”.
n = Es / Ec
 Normally the value of n is 8 to 10
 It is a unit-less quantity
Reinforced Concrete Design-I

25. Transformed Section
Beam is a combination of concrete and steel. As a whole it is not a
homogeneous material. In transformed section the steel area is replaced
by an equivalent concrete area in order to calculate the section
properties.
d
Transformed Section
•Width of the extended area is same as diameter of steel bar and its
distance from compression face remains same.
Reinforced Concrete Design-I

26. Uncracked Transformed Section
When both steel and concrete are in elastic range and tensile stress
at the tension face of concrete is less than tensile strength of
concrete the section is un-cracked.
Within the elastic range, perfect bond (no slippage) exists between
concrete and steel, so
εs= εc
fs/Es = fc / Ec
fs = (Es/ Ec) fc
fs = n fc
Using this relationship, stress in steel can be calculated if
stress in concrete and modular ratio are known.
Reinforced Concrete Design-I

27. Consider a beam having steel area As . In order to obtain a
transformed section, the area of steel (As) is replaced by an
equivalent area of concrete so that equal force is developed in
both.
Ag
Total tensile force, T
= Ac + As
= fc Ac + fsAs
= fc (Ag – As) + n fc As
= fc (Ag – As + n As)
= fc Ag + (n – 1) As
nAs/2 nAs/2
b
h

28. The equivalent area nAs/2 is shown on either side but steel inside the
beam is removed which creates a space that is filled by area of concrete,
thus the equivalent area on either side becomes
nAs/2 –As/2
(n-1)As / 2
•Once the transformed section
has been formed the sectional
properties (A, Location of N.A., I, S
etc) are calculated in usual manner
(n-1)As/2 (n-1)As/2
Transformed Section
Total Area of transformed section = b x h + (n-1)As= Ag + (n-1)As
Reinforced Concrete Design-I

29. Example 2.1
A rectangular beam of size 250  650
mm, with effective depth equal to 590
mm, is reinforced with three No. 25 US
customary bars. C 28 concrete and
Grade 420 steel are to be used.
Determine the stresses at the top,
bottom and level of reinforcement
caused by a bending moment of 50
kN-m. The member is within its elastic
range.

30.  As = 3  510 = 1530 mm2
 Es = 200,000 MPa
 f
c
 = 28 MPa
 Ec = 24,870 MPa
 n = Es / Ec  8
 Additional steel area = (n – 1) As
= 10,710 mm2

31. b = 250
d = 590
h = 650
y
Fig. 2.7. Transformed
Area for Example 2.1.
y =
173,210
(250)(650)(325) (10,710)(590)
= 341 mm
I =
12
(250)(650)3
+ (250)(650)(16)2
+ (10,710)(249)2
= 642,700  104 mm4
(fc)top =
M y
=
I
50 106
341
642,700104
= 2.65 MPa (compressive)

32. b = 250
d = 590
h = 650
y
(fc)bot =
I
M y
=
642,700104
50 106
309
= 2.40 MPa (tensile)
fs =
I
50106
(590  341)
n
M y
= 8 
642,700104
= 15.50 MPa
fr for strength = 0.5 f
c
 = 2.65 MPa
The bottom concrete stress in tension is lesser than its modulus of
rupture indicating that the section is really uncracked. Further, the
stresses are much lesser than f
c
/ 2 and fy / 2 showing elastic behavior.
Hence, the assumption of uncracked transformed section is justified.