RCC Design Philosophies | Comparison of Working Stress Method and Limit State Method | Example for WSM & LSM

Methods of RCC Design & Design Philosophies:

RCC Structural Elements are designed by two methods or two Design Philosophies,

  1. Working Stress Method (WSM)
  2. 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.

Reinforced Cement Columns: Working Stress

Reinforced Cement Columns: Working Stress

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

  • Stress- Strain behavior is linear.
  • Stress-Strain behavior is non-linear.
  • WSM is also known as plastic design.
  • LSM is also known as Elastic design.
  • WSM is Stress based method of RCC design.
  • LSM is Strain based method of RCC design.
  • In WSM , the material follows Hooke’s law as stress is not allowed to cross the yield limit.
  • In LSM , stress is allowed to cross the yield limit.
  • In WSM , Safety Factor is considered.
  • In LSM , Partial Safety Factor is considered.
  • In RCC Design by WSM Tensile Stresses considered.
  • In RCC Design by LSM Tensile Stresses are not considered.
  • A section which is plane before bending remains plane after bending.
  • A section which is plane  normal to the axis of structural component remains plane after bending.
  • Modular Ratio (m = 280/(3*Sigma_cbc) )  is generally taken into consideration to calculate allowable stresses.
  • The Ultimate Stresses are considered as allowable permissible stresses.
Waffle Slab Construction

Waffle Slab Construction

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(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 (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

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