Framed str beam

1. By limit state method of design

2. Different methods of design of
concrete
1. Working Stress Method
2. Limit State Method
3. Ultimate Load Method
4. Probabilistic Method of Design

3. Limit state method of design
The object of the design based on the limit state concept
is to achieve an acceptable probability, that a structure
will not become unsuitable in it’s lifetime for the use for
which it is intended,
i.e. It will not reach a limit state
A structure with appropriate degree of reliability should be
able to withstand safely.
All loads, that are reliable to act on it throughout it’s life
and it should also satisfy the subs ability requirements,
such as limitations on deflection and cracking.

4. It should also be able to maintain the required
structural integrity, during and after accident,
such as fires, explosion & local failure.
i.e. limit sate must be consider in design to
ensure an adequate degree of safety and
serviceability
The most important of these limit states, which
must be examine in design are as follows
Limit state of collapse
- Flexure
- Compression
- Shear
- Torsion
This state corresponds to the maximum load
carrying capacity.

5. Types of reinforced concrete
beams
a) Singly reinforced beam
b) Doubly reinforced beam
c) Singly or Doubly reinforced flanged beams

6. Singly reinforced beam
 In singly reinforced simply supported beams or
slabs reinforcing steel bars are placed near the
bottom of the beam or slabs where they are most
effective in resisting the tensile stresses.

7.  Reinforcement in simply supported beam
COMPRESSION b
STEEL REINFORCEMENT D d
TENSION
SUPPORT SECTION A - A
CLEAR SPAN

8.  Reinforcement in a cantilever beam
A
TENSION
D
d
COMPRESSION SECTION A - A
A
CLEAR COVER

9.  STRESS – STRAIN CURVE FOR CONCRETE
fck
STRESS
.20 % .35%
STRAIN

10.  STRESS ― STARIN CURVE FOR STEEL

11. STRESS BLOCK PARAMETERS
!
 0.0035 0.446 fck
 X2 X2 a

 X1 X1

12.  x = Depth of Neutral axis
 b = breadth of section
 d = effective depth of section
 The depth of neutral axis can be obtained by
considering the equilibrium of the normal forces
, that is,
 Resultant force of compression = average stress X area
= 0.36 fck bx
 Resultant force of tension = 0.87 fy At
 Force of compression should be equal to force of tension,
0.36 fck bx = 0.87 fy At
x =
Where At = area of tension steel
0.87 fy At
0.36 fck b

13. The distance between the lines of action of two
forces C & T is called the lever arm and is
denoted by z.
Lever arm z = d – 0.42 x
 z = d – 0.42
 z = d –
0.87 fy At
0.36 fck b
fy At
fck b

14.  Moment of resistance with respect to concrete = compressive force x lever arm
= 0.36 fck b x z
 Moment of resistance with respect to steel = tensile force x lever arm
= 0.87 fy At z

15. Maximum depth of neutral axis
 A compression failure is brittle failure.
 The maximum depth of neutral axis is limited to ensure that tensile steel will
reach its yield stress before concrete fails in compression, thus a brittle
failure is avoided.
 The limiting values of the depth of neutral axis xm for different grades of steel
from strain diagram.

16. Maximum depth of neutral axis
fy N/mm2 xm
250 0.53 d
415 0.48 d
500 0.46 d

17. Limiting value of tension steel
and moment of resistance
Since the maximum depth of neutral axis is limited,
the maximum value of moment of resistance is also
limited.
Mlim with respect to concrete = 0.36 fck b x z
 = 0.36 fck b xm (d – 0.42 xm)
Mlim with respect to steel = 0.87 fck At (d – 0.42 xm)

18. Limiting moment of resistance
values, N mm
Grade of
concrete
Grade of steel
Fe 250 steel Fe 450 steel Fe 500 steel
General 0.148 fck bd 0.138 fck bd 0.133 fck bd
M20 2.96 bd 2.76 bd 2.66 bd
M25 3.70 bd 3.45 bd 3.33 bd
M30 4.44 bd 4.14 bd 3.99 bd

19. Types of problem
a) Analysis of a section
b) Design of a section

20. a) For under reinforced section, the value of x/d is less than xm/d value.
The moment of resistance is calculated by following equation:
Mu = 0.87 fy At d –
a) For balanced section, the moment of resistance is calculated by the
following equation:
Mu = 0.87 fy At ( d – 0.42xm)
a) For over reinforced section, the value of x/D is limited to xm/d and the
moment of resistance is computed based on concrete:
Mu = 0.36 fck b xm ( d – 0.42 xm )

21.  Analysis of section

22.  Determine the moment of resistance for the section shown in figure.
(i) fck = 20 N/mm , fy = 415 N/mm
Solution:
(i) fck = 20 N/mm , fy = 415 N/mm
breadth (b) = 250 mm
effective depth (d) = 310 mm
effective cover = 40 mm
Force of compression = 0.36 fck b x
= 0.36 X 20 X 250x
= 1800x N

23. Area of tension steel At = 3 X 113 mm
Force of Tension = 0.87 fy At
= 0.87 X 415 X 3 X 113
= 122400 N
 Force of Tension = Force of compression
 122400 = 1800x
 x = 68 mm
 xm = 0.48d
 = 0.48 X 310
 = 148.8 mm
 148.8 mm > 68 mm
 Therefore,
 Depth of neutral axis = 68 mm
fy xm
415 0.48d
500 0.46d

24.  Lever arm z = d – 0.42x
= 310 – 0.42 X 68
= 281 mm
As x < xm ( It is under reinforced )
o
o Since this is an under reinforced section, moment of resistance is
governed by steel.
o Moment of resistance w.r.t steel = tensile force X z
o Mu = 0.87fy At z
o = 0.87 X 415 X 3 X 113 X 281
o Mu = 34.40kNm

25. Design of a section

26.  Question : Design a rectangular beam to resist a bending moment equal to 45 kNm
using (i) M15 mix and mild steel
 Solution :
 The beam will be designed so that under the applied moment both materials
reach their maximum stresses.
 Assume ratio of overall depth to breadth of the beam equal to 2.
Breadth of the beam = b
Overall depth of beam = D
therefore , D/b = 2
For a balanced design,
Factored BM = moment of resistance with respect to concrete
= moment of resistance with respect to steel
= load factor X B.M
= 1.5 X 45
= 67.5 kNm

27.  For balanced section,
 Moment of resistance Mu = 0.36 fck b xm(d - 0.42 xm)
 Grade for mild steel is Fe250
 For Fe250 steel,
xm = 0.53d
Mu = 0.36 fck b (0.53 d) (1 – 0.42 X 0.53) d
= 2.22bd
Since D/b =2 or, d/b = 2 or, b=d/2
Mu = 1.11 d
Mu = 67.5 X 10 Nmm
d=394 mm and b= 200mm
fy xm
250 0.53d
415 0.48d

28.  Adopt D = 450 mm
 b = 250 mm
 d = 415mm
 Area of tensile steel At =
 =
 = 962 mm
 = 9.62 cm
 Minimum area of steel Ao= 0.85

29. =
= 353 mm
353 mm < 962 mm
In beams the diameter of main reinforced bars is usually selected between
12 mm and 25 mm.
Provide 2-20mm and 1-22mm bars giving total area
= 6.28 + 3.80
= 10.08 cm > 9.62 cm

30.  !