Singly reinforced beam(Limit state method of design)

Singly reinforced beam(Limit state method of design)

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

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

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 
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 –(fy At/fck b) 
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 
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.

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)

Limiting moment of resistance values, N mm

Design of a section 
Design of  rectangular beam to resist a bending moment equal to 45 kNm using (i) M15 mix and mild steel. 
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 
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 
Adopt D = 450 mm , b = 250 mm ,d = 415mm 

  =(0.85x250x415)/250
                                       =                    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

Design a rectangular reinforced concrete beam for loads given below. The simply-supported beamhas a span ` = 18 ft and excessive deflections will cause damage. The superimposed dead load(SDL) is 1.15 kip/ft with other given quantities below.Given: f0c = 4.5 kip/in2 wL = 0.8 kip/ftγc = 150 lb/ft3fy = 60 kip/in2