Methods of RCC Design & Design Philosophies:
RCC Structural Elements are designed by two methods or two Design Philosophies,
 Working Stress Method (WSM)
 Limit State Method (LSM)
Working Stress method is the old way of designing, in present era structures are generally designed using Limit State Method. In WSM concrete is considered as elastic whereas in LSM concrete is generally assumed as plastic.
Difference b/w WSM & LSM:
As discussed above there are two different methods of RCC Design which can be distinguish as follows :
Working Stress Method  Limit State Method 
















Design Example of Structural Member (Beam) by WSM:
Given Data :
 Eff. Span = 8 m
 Live Load = 12 KN/m
 Breadth of the Beam = 300 mm
 Grade of concrete, fck = M20
 fy = Fe415
Solution :
Design of Beam Depth :
For M20 & Fe415 , Sigma_cbc = 7 N/mm2
Let us assume, Overall Depth D = 550 mm
Eff. Cover = 50 mm
Effective depth, d = 550 – 50 = 500 mm
Self wt. of Beam = 0.3 * 0.55 * 25 = 4.125 KN/m
Total Load (w) = 4.125 + 12 = 16.125 KN/m
Moment (M) = w * l^2 / 8 = 16.125 *(8)^2/8 = 129 KN/m
For Balanced Section , Q = Mrb / (Sigma_cbc * b * d^2) = 0.1304
= 129*(10^6)/(7*300*d^2) = 0.1304
d = 686.35 > 750 mm ( i.e. Insufficient Depth )
So, Provide Depth = 750 mm
D = 750 + 50 =800 mm
Self Wt. = 0.3 * 0.8 * 25 = 6 KN/m
Total Load = 6+12 = 18 KN/m
Moment = w * l^2 / 8 = 18 *(8^2)/8 = 144 KN/m
144 * 10^6 / 7*300*d^2 = 0.1304
d = 725.16 < 750 mm
Reinforcement :
pt(bal) = (Ast(bal) / bd) *100 = 0.438
Ast (bal) = 985.5 mm2
Bar Dia. = 20 mm
Bar Nos. = 985.5 / (pi()/4*20^2) = 3.13 = 4 (say)
Bottom Reinforcement = 4 # 20 mm Dia. Bar
For Top Reinf.(Tie Bars) Ast = (35/100)* (pi()/4*20^2)*4 = 439.82 mm2
Bar Dia. = 12 mm
Bar No. = 439.82 / (pi()/4*12^2) = 3.13 = 4 (say)
Top Reinforcement = Provide 4#12 mm Dia. Bar
Design Example of Structural Member (Beam) by LSM:
Given Data :
 Breadth of the Beam = 300 mm
 B.M = 200 KN/m
 Factored B.M. = 1.5 * 200 = 300 KN/m
 fck = M20
 fy = Fe415
Solution :
Ultimate Moment of Resistance (Mu lim) = 0.138*fck*b*d^2
Mu lim ≥ 300 KN/m
0.138 * (20) * (300) *d^2 ≥ 300 *10^6
d ≥ 601.93 mm
Design of Beam Depth :
For M20 & Fe415 , Sigma_cbc = 7 N/mm2
Let us assume, Overall Depth D = 550 mm
Clear Cover = 25 mm
Assume , Dia. of Bar = 20 mm
Stirrups Dia. = 8 mm
D = 601.93 + 25 + 8 + 1/2 *20 = 644.93 mm
Provided , D = 650 mm
Eff. Depth of Beam(d) = 650 – 25 – 8 – 1/2 * 20 = 607 mm
Thus Section provide is 300 x 650 mm.
Reinforcement :
pt / 100 = Ast / bd
= 0.5 * fck / fy *(1 – sqrt(14.5*M/(b*d^2)/fck))
= 0.5 * 20 / 415 *( 1 sqrt ( 1 4.5 * 300 * 10^6 / 300 * 607^2*20)
= 0.93 %
Ast = 0.93 * b * d/100 = 0.93 * 300 * 607 /100 = 1696.99 mm2
Bar Dia . = 20 mm
Bar Nos. = 1696.99/(pi()/4*20^2) = 5.4 ≈ 6 (say)
Bottom Reinforcement = Provide 6#20 mm Dia. Bars
For Top Reinf.(Tie Bars) Ast = (35/100)* (pi()/4*20^2)*4 = 659.73 mm2
Bar Dia. = 12 mm
Bar No. = 659.73 / (pi()/4*16^2) = 3.13 = 4 (say)
Top Reinforcement = Provide 4#16 mm Dia. Bar