1. SINGLY REINFORCED BEAM (R.C.C)
By limit state method of design
Prepared by
Mr. Ashok Kumar
Lect. In civil engg.

2. DIFFERENT METHODS OF DESIGN OF CONCRETE
Working Stress Method
Limit State Method
Ultimate Load Method
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
Singly reinforced beam
Doubly reinforced beam
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
TENSION
SUPPORT D d
SECTION A - A
CLEAR SPAN

8. Reinforcement in a cantilever beam
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!
0.446 fck 
0.0035  X2 X2 a  X1 X1

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

14. Moment of resistance with respect to concrete 
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 
Mlim with respect to steel
= 0.36 fck b x z
= 0.36 fck b xm (d – 0.42 xm)
= 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
b) Analysis of a section
d) Design of a section

20. The moment of resistance is calculated by following equation:
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 –
g) For balanced section, the moment of resistance is calculated by the following equation:
Mu = 0.87 fy At
( d – 0.42xm)
k) 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) effective depth (d) effective cover
Force of compression
= 250 mm
= 310 mm
= 40 mm
= 0.36 fck b x
= 0.36 X 20 X 250x
= 1800x N

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

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

25. Design of a section

26. Assume ratio of overall depth to breadth of the beam equal to 2. 
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. Grade for mild steel is Fe250
For balanced section,
Moment of resistance Mu
= 0.36 fck b xm(d xm)  fy xm
250
0.53d
415
0.48d 
Grade for mild steel is Fe250 
For Fe250 steel, xm
= 0.53d
= 0.36 fck b (0.53 d) (1 – 0.42 X 0.53) d
= 2.22bd Mu
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

28. Area of tensile steel At =
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 =
10.08 cm > 9.62 cm

30. !