Reinforcement Calculator for Beam, Slab or footing by limit State method

 Beam Footing or Slab design by Limit State Method

Note:- Above design applicable to M35 grade only , for grades above M35, various coefficients changes as shown in following Design Procedure.

For Shear Design Flow Diagram as follows-

Procedure for DESIGN CHECKS

For a given cross-section, the ultimate axial load carrying capacity can be checked for uniform strain of e_c2 if the applied force is pure compressive  and for 0.9 εs  if the applied axial force is pure tensile.  The ultimate  axial force  carrying  capacity  of  the  section  for  given bending moment shall always be between these two limits. The position of neutral axis, x for which the ultimate force carrying capacity of the section matches with·the factored axial force acting on the section is obtained by solving the following equilibrium condition.

f(x) = Pus (x) + Puc (x) +  N = 0                                                                            

;Where N is the factored axial force applied and Pus (x) and Puc (x) are the contributions from steel and concrete respectively in the ultimate resistance of the section and can be express as

Pus (x) = Σσ_s(x) A_s

Puc= ʃσ_d A_c

The Direct  Solution  of Rectangular  Sections                                                                          

In case of direct solution, magnitude of compressive  force (Cu)  and its location is obtained from concrete stress block in compression. The ultimate failure of concrete is caused by crushing of concrete which requires the limiting strain (Ecu) on extreme compressed fiber of concrete. Following two cases are discussed here:

1)             Neutral axis within the section

2)             Neutral axis outside the section

1)                              Neutral axis within the section

a)             Parabolic - Rectangular stress block

When neutral axis is within the section, strain in extreme compressed fiber is limited to £uc 2 and corresponding stress fcct· For rectangular section of width b and neutral axis depth x the resultant force Cu is expressed by

Cu=β₁f_cdbx

And its position, measured from extreme compressed edge is defined by βx2  .  The expressions for β₁  and β₂,   as function of strain ε_c, are:

Table C 8.1 Values of β₁ and β₂

fck (N /mm)2	Upto 60	70	75	90	100	115
β1	0.8095	0.74194	0.69496	0.63719	0.59936	0.58333
Β2	0.416	0.39191	0.37723	0.36201	0.35294	0.35294

b)            Rectangular stress block

Instead of parabolic - rectangular stress block, more simplified rectangular stress block can be used for evaluating compressive force and its line of action. The expressions for β₁ and β₂ are simplified to,

                                    β_{1 = λη}             β_{2 = λ/2}

Table C8.2 Values of for Rectangular Stress Block

fck (N /mm) 2	Upto 60	70	75	90	100	115
β1	0.8	0.76781	0.73625	0.675	0.61625	0.56
Β2	0.4	0.39375	0.3875	0.375	0.3625	0.35

 

2)                            Neutral axis outside the section

a)             Parabolic - Rectangular stress block

2

The adoption of the assumptions in Clause 8.2.1 of the Code leads to the range of possible strain diagrams at ultimate limit states subjected to different forces. Numerous conditions of neutral axis outside the section arise between two cases of strain distribution, one with uniform εc2 ,  over section for uniformneutral axis. For this condition strain diagram is defined by assuming that compressive strain εcu2  for extreme compressed edge and 0 at neutral axis.

b)            Rectangular stress blocK

Table C8.4 Values of _λ, η and k

fck (N /mm) 2	λ	η	k
< 60	0.8	1	0.75
70	0.7875	0.975	0.73016
75	0.775	0.95	0.70968
90	0.75	0.9	0.66667
100	0.725	0.85	0.62069
115	0.7	0.8	0.57143

 Table C 8.3. Values  of b3 and b4

Parabolic Rectangle Constitutive Law
x/h	fck = 60 N/mm2		fck = 70 N/mm2		fck = 75 N/mm2		fck = 90 N/mm2		fck = 100 N/mm2		fck = 115 N/mm2
	b3	b4	b3	b4	b3	b4	b3	b4	b3	b4	b3	b4
1	0.80952	0.41597	0.74194	0.39191	0.69496	0.37723	0.63719	0.36201	0.59936	0.35482	0.58333	0.35294
1.2	0.89549	0.45832	0.83288	0.43765	0.78714	0.42436	0.72968	0.41022	0.69249	0.40355	0.6772	0.40186
1.4	0.93409	0.4748	0.88197	0.45841	0.84129	0.44724	0.78831	0.43492	0.75381	0.42907	0.73986	0.42761
1.6	0.95468	0.48304	0.91168	0.4699	0.87615	0.46046	0.82826	0.44975	0.79679	0.44461	0.78422	0.44335
1.8	0.96693	0.48779	0.93113	0.47702	0.90007	0.46895	0.85695	0.45954	0.82834	0.45499	0.81702	0.45389
2	0.97481	0.49077	0.9446	0.48178	0.9173	0.47478	0.87838	0.46644	0.85234	0.46237	0.84211	0.4614
2.5	0.9855	0.49475	0.96464	0.48861	0.9442	0.48347	0.91348	0.47705	0.89255	0.47385	0.88448	0.47311
5	0.99702	0.49893	0.9906	0.49705	0.98285	0.49512	0.96937	0.49234	0.95972	0.49089	0.95622	0.49057

Table C 8.5 Values of b3 and b4 for Rectangular Stress Block

Rectangle Constitutive Law
x/h	fck = 60 N/mm2		fck = 70 N/mm2		fck = 75 N/mm2		fck = 90N/mm2		fck = 100N/mm2		fck = 115N/mm2
	b3	b4	b3	b4	b3	b4	b3	b4	b3	b4	b3	b4
1	0.8	0.4	0.76781	0.39375	0.73625	0.3875	0.675	0.375	0.61625	0.3625	0.56	0.35
1.2	0.88889	0.4444	0.85601	0.43898	0.82344	0.43339	0.75938	0.42188	0.69695	0.40997	0.63636	0.39773
1.4	0.92308	0.4615	0.89154	0.4572	0.86011	0.45269	0.79773	0.44318	0.73623	0.43308	0.67586	0.42241
1.6	0.94118	0.47059	0.91073	0.46704	0.8803	0.4633	0.81964	0.45536	0.75946	0.44674	0.7	0.4375
1.8	0.9524	0.47619	0.9227	0.4732	0.89308	0.47004	0.8338	0.4632	0.7748	0.45577	0.71628	0.44767
2	0.96	0.48	0.931	0.4774	0.9019	0.4747	0.8438	0.4688	0.78572	0.46219	0.728	0.455
2.5	0.97143	0.4857	0.9434	0.4838	0.91534	0.48176	0.85909	0.4773	0.80282	0.47255	0.74667	0.46667
5	0.9882	0.4941	0.9619	0.4933	0.9355	0.4924	0.8827	0.4904	0.8298	0.48809	0.74677	0.48548

Above Tables are Taken from IRC.

Estimate the area of steel = 

Example : Ultimate moment = 150 kN/m, cover 50mm concrete M45 grade steel Fe depth provided is 250mm. Determine the steel required assuming 20mm bars. 

Step 1:

    Depth required =    

    D = 142.2 + 10 + 50 = 203mm < 250m. Hence depth provided is adequate.

    deff =250 – 50 – 10 = 190mm.

 

Step 2:    

    = 0.290 < 0.617

 Hence steel yields

 

It can be noted that an increase of 47mm in depth, the x/d > ratio comes down from 0.617 to 0.290.

 

Step 3: 

Suppose anybody assumes the balanced section lever arm itself and estimates the reinforcement then

 18 % Variation in steel. The Neutral axis depth changes from 117.23m to 55.1m and the ratio x/d from 0.617 to 0.290.

Hence it is better to estimate the neutral axis and arrive at the steel, in case of depth provided is more than the requirement.