 # Comparison Of Plastic Design: Methods Of RCC Design

Satisfactory Essays
Page Title: Methods of RCC Design | Comparison of Working Stress Method and Limit State Method | Design Example for WSM & LSM
Keywords: RCC Design , Methods of RCC designing , WSM , LSM , Diff. b/w WSM & LSM , Working Stress Method , Limit State Method , Example of WSM , Example of LSM , Elastic Design , Plastic Design
Methods of RCC Design:
RCC Structural Elements are designed by two methods,
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
= 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(1-4.5*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