Sunday, October 22, 2006

Design of R.C.C members

 

 Design of R.C.C members                                    

      R.C.C members can be design by two approaches

F  Working stress method

F  Limit state method

 

Working stress method

Working stress method is based on classical elastic theory and developed for purely elastic material. The method ensures satisfactory behavior of the structure under working load but doesn't give actual margin of safety at collapse. Therefore design is generally done by limit state method.

 

Limit state of collapse in flexure

Assumption:

ü      A normal section, which is plane before bending, remains plane after bending.

ü      This implies that strain at any point on the cross-section is directly proportional to it

ü      The tensile strength of concrete is ignored.

ü      The maximum strain at the outermost compression fiber is taken as 0.0035 in bending.

ü      The maximum strain in the tension reinforcement in the section at failure shall not be less than    (f y /1.15Es)+0.002

Where,

   fy = characteristic strength of steel.

   Es = modulus of elasticity of steel. 

 

Design Of Beam

 

Design of rectangular section

 

F  Singly Reinforced Beam:  Design of singly and doubly reinforced rectangular section shall be made as below the compressive stress block for concrete is represented by the design stress strain curve. The triangular distribution of strain and non-linear stress diagram in compression concrete are as shown in the design aids.

 

Properties of the singly reinforced section:

 

i. Compressive force and the centroid of compressive force: It is seen from this stress block that the centroid of the compressive force in a rectangular section lies at the distance of 0.416*Xu from the extreme compression fiber and the total force of compression is 0.36*fck*b*Xu.

 

ii.Lever arm : The lever arm, the distance between the centroid of compressive force and centroid of tensile force.

                         Zu =  (d-0.416*Xu)

 

iii. Depth of the neutral axis (Xu): Depth of neutral axis below the highly compressed edge is obtained by equating compression in concrete with tension in steel to satisfy the condition of equilibrium.

                                   C = T

                Hence,

                           Xu = (fs*Ast)/(0.36*fck*b)

                                 

                                = Pt*b*d*(0.87fy)/100*0.36*fck*b

 

                  or, Xu/d =  (Pt/10)*(0.87fy/0.36fck)

                     

                          Where, fs = 0.87f

 

iv. Ultimate moment of resistance: The ultimate moment of resistance of a singly reinforced rectangular section is obtained by taking moment of total compression (C) or about CG of steel or moment of total tension (T) about CG of compression.

 

            Moment of resistance in terms of concrete strength is given by

 

Mu = C * Zu

      

        = 0.36* fck* b* Xu *(d-0.416Xu)

 

The ultimate moment of resistance, Mu in terms of steel and concrete strength can be found by

 

Mu  = T * Zu

       

         = fs *Ast *( d- 0.416Xu)

      

         = 0.87fy(Pt*b*d/ 100)* (1-0.416Xu)

 

          =0.87fy*(Pt/100)*(1-1.005(fy/fck)(Pt/100)*b*d2

 

Substituting for Xu/d, we get

 

Mu = 0.87fy*( Pt /100) * (1-1.005(fy/fck)( Pt /100)*b*d2

 

v. The area of steel is obtain as:

 

Ast  = Mu/(0.87*fy(d-0.416*Xu))

 

Design constants for balance or critical section: A balance or critical section is that section in which strain in concrete and steel reach their ultimate values simultaneously. As the design of a beam should be under – reinforced, the value Xu for the critical section is the maximum allowable value Xul and can be obtained from strain distribution diagram as,

 

             Xul=0.0035/(0.005+0.87fy/Es)*d

 

Where, Es = 2*105N/mm2

 

Once the value of Xul is known the other design constants can be obtained from

 

Mul = 0.87fy* Ast *(d-0.416 Xul) and, Mul  = 0.36fck/(.87fy)*b* Xul

 

FDoubly reinforced beam: Doubly reinforced sections are generally adopted when the dimensions of the beam predetermined from other considerations and the design moment exceeds the moment of the resistance of a singly reinforced section. The additional moment of resistance needed is obtained by providing compression reinforcement and additional tensile reinforcement.

 

Properties of doubly reinforced section:

 

i. Depth of neutral axis (Xu): The depth of neutral axis is obtained by equating       total compression in concrete and compression steel with total tension in tension steel.

                                                

                                                  C   =   T

                                  

                                 Or,  Cs  + Cc   =    T

 

Or,  0.36 * fck* b* Xu + (fsc-fcc)*Asc = 0.87fy*Ast

 

Where,  fsc = stress in compression steel (Asc) corresponding to strain

Where,  d  = effective cover to compression steel

             fcc= stress in concrete in compression at the level of compression steel

 

ii. Moment of resistance: The ultimate moment of resistance is obtained by taking moment of compressive force in concrete and compression steel about the centroid of tension steel as:

 

             M=   Mu1 + Mu2

Where, Mu1 = Mu(lim) = K*fck*b*d2

                Mu2  = the additional moment in compression failure

            Mu  = 0.36*fck*b*Xu*(d - 0.416Xu) + (fsc-fcc)*Asc*(d-d')

 

iii.Area of steel: In the design of doubly reinforced section, the section is kept balanced to make full utilization of the resistance of concrete. The area of tension steel and compression steel required for resisting Mu1 and Mu2 can be obtained as follows:

 

1. Tension steel: Area of tension steel (Ast1) corresponding to Mul is obtained as

                

               Ast1 = Mu1/(0.87fy*(d-0.416Xul))

 

And the area of tension steel (Ast2) is obtained by equating Mu2 with the moment about compression steel.

                 

                 Ast2=Mu2/(0.87fy(d-d'))

    

Thus total area tension steel = Ast1+Ast2

 

2. Compression steel:  The area of compression steel (Asc) is obtained by equating (Mu2) with the moment of compression C2 = (fsc-fcc)*Asc about tension steel.

         Mu2 = Asc*(fsc-fcc)(d-d')

Or,    Asc= Mu2/(fsc-fcc)*(d-d')

 

Design of Flanged section: When the beam and slab cast together in one operation and the main reinforcement of the slab extend into the beam and the integral connection between the slab and beam is formed due to this part of slab lying on the compression zone of slab act along with the beam in compression zone. Then the resulting section is the flanged section.

 

Singly reinforced flanged section:

 

Properties of flanged section

 

 i. Effective width of flange:

ü      For T-Beam: b= (Lo/6 + 6*Df)

ü      For L-Beam: b= (Lo/12 + bw + 3*Df)

ü      For isolated beams, effective width of beam is given by

¯     bf = Lo/(Lo/b + 4) + bw  or actual width for T-beam

¯     bf = 0.5 Lo(Lo/b + 4) + bw  or actual width for L-beam

 

ii.Neutral axis: The neutral axis lie either on the flange or in web, which may be ascertained by comparing the value Xu and Df. Where Xu may be determined by assuming that the neutral axis lies either in the flange or web.

 

Case-1: Neutral axis lies in the flange             

               C     =    T

 

             0.36fck*bf*Xu =.87fy*Ast

 

               Xu = (.87fy*Ast)/(.36fck*bf)

 

  If Xu ≤ Df, then the neutral axis is within the flange. Otherwise, neutral axis lies in the web and section is analyzed as flanged section then the position of neutral axis may be determined from equilibrium of compressive and tensile forces as,

   

         Case 2:Neutal axis below the flange and Df/d<0.2

                        C = T

 

                  C = C1 + C2

                   

       Or,      C = 0.36fck*bw*X + 0.446fck* (bf - bw)* Df

 

      Or,     0.87fy*Ast = 0.36*fck*bw*X + 0.446*fck*(bf - bw)*Df

 

     Case 3: Neutral axis below the flange and Df/d>0.2

 

In this case, the Df in the above formula is replaced by the term Yf .

       

    where,  Yf  = (0.15 * X + 0.65 * Df) or Df  whichever is smaller.

 

iii. Moment of resistance and area of tension steel

 

For case1: Moment of resistance is similar to that for singly reinforced rectangular beam.

 

              Mu = 0.36fck*bf (d - 0.416X)

 

Area of steel can be calculated by comparing force in compression to the force in tension. i.e,    

 

         0.36fck*bf*Xu =.87fy*Ast

 

For case 2: Moment of resistance is given by ,

         

                Mu = Mu1 + Mu2

               

                Mu = 0.36fck*bw*X*(d-0.416*X)+0.446fck*(bf-bw)*Df(d-Df/2)

 

Area of steel can be calculated by comparing force in compression to the force in tension. i.e,           

    

            0.87fy*Ast = 0.36fck* bw *X + 0.446fck*(bf - bw)*Df

 

Limit State Of Shear And Torsion

 

1. Limit state of collapse in shear    

 

i. Nominal Shear stress:  The nominal shear stress in the uniform depth of beam is given by the equation,

                                  tv = Vu/bd

Where,       Vu = Shear force due to design loads

                   B = Breadth of the member

                   D = Effective depth

 

i.Shear Reinforcement: The design  shear strength  of vertical stirrups is given by the following equation,

                       Vus =  0.87*fy*Asv*d/Sv

 

Where,  Asv = the total cross-section area of the vertical legs of stirrups

              Sv = the spacing (pitch) of the stirrups.

 

2. Design for torsion

 

i.Equivalent Bending  Moment (Me): The torsion moment acting on the member is converted into equivalent shear , which is added to the B.M and shear acting at the section . The member is designed for resulting B.M and shear.

 

Me1 = Mu (bending) ± Mt (Equivalent torsion)

 

Where , Mt =Tu(1+D/b)/1.7

              Tu    =  torsion moment at the section

              M=  bending moment at the section

When, Mt> Mu, steel will be provided on compression face to resist bending moment    

                   Me2  = Mt - Mu

 

ii.Equivelent Shear,

   

Ve=Vu+1.6*Tu/b

 

iii.Equivalent Shear stress

  

  tve= Ve/(b*d)

 

iv.Maximum Shear Stress

 

tcmax is from table 20 of IS:456:2000

tve should be less than tcmax

 

v.Design of Shear Strength of concrete: tc is from table 19 of Design Aids IS: 456:2000, for an assumed value of Pt . If the equivalent nominal shear stress, tve does not exceed  tgiven in table 19 of the code , minimum shear reinforcement shall be provided as per clause 26.5.1.6 of IS: 456 : 2000.

                          

If  tve exceeds tc given in table 19, both longitudinal and transverse shall be provided in accordance with clause 41.4.

 

vi.Longitudinal reinforcement: Longitudinal  reinforcement shall be provided to resist equivalent moment.

 

vii.Transverse reinforcement:The transverse reinforcement shall consist of vertical stirrups enclosing the longitudinal reinforcement.Its design requires determination of spacing for its chosen diameter is as according to 41.43 of I.S code.

 

Design of Column:

A column is a compression member subjected to compressive force in a direction parallel to its longitudinal axis .The column are further classified into various type in with the different criteria such as

Based on shape of cross-section

ü      Rectangular

ü      Circular

Based on type of footing

ü      Axially load

Based on slenderness ratio

ü      Short

ü      Slender

 

Design criteria

 

  i.  Minimum eccentricity: According to code, Emin=L/500+D/30,subjected to minimum of 20mm.

Where, L= unsupported length of column

            D= lateral dimension of column under consideration

 

ii.Clear cover to reinforcement: When the longitudinal reinforcement bars are not exceeding to code IS 456-1978 species that d' should not less than 40mm.

 

iii.Requirement Required:

 

a. Longitudinal reinforcement :The percentage of reinforcement shouldn't be less than 0.8% and maximum should  not exceed 4% up to 6% of the gross area of the section is allowed. The minimum number of bars should not be less than 4 in rectangular and 6 in circular column.

 

b.Transverse reinforcement: If the longitudinal bars are not spaced more than 75mm on either side the transverse reinforcement  needs only to go exceeding 48 times the diameter of the purpose of providing effective lateral support.

If the longitudinal bars spaced at a distance of not exceeding 48 times the diameter of tie are effectively tied in two directions by open ties.

        Where the longitudinal reinforcement bars in compression members are placed in more one row, effective lateral support to longitudinal bars in the row may be assumed to have been provided. If transverse reinforcement is provided for the outermost row and no of bar of inner row is closer to the nearest compression face three times the diameter of largest longitudinal bar in inner row.

 Where the longitudinal bars in the compression member are grouped and each adequately tied with transverse reinforcement for compression members as a whole may be provided on the assumption that each group is a single longitudinal for the purpose of determining the pitch and the diameter of the transverse reinforcement. The diameter of such traverse reinforcement need not however exceed 20mm.

 

FAxially loaded column: According to IS 456-1978 code following formula is used.

                

                    Pu= 0.4fck*Ac+0.6*fy*Asc

              

                Where,  Pu = Axial load

                             Ac =Area of concrete

                             Asc = Area of reinforcement

 

Column with unaxial bending : The formula used in this case is same as that of axially loaded column.

 

Column with biaxial bending: The theoretical limit strength of a section under bi-axial bending and compression is a function of three variable Pu, Mx,My which may be expressed in terms of axial load placing at eccentricity,

                           ex = Mx/Pu,                           ey =My/Pu

Moment Mx and My are acting along the x-axis and y-axis respectively.

The code IS 456-1978 utilize the load contour method and the general known dimensional method.

Equation for contour at constant P may be expressed as

(M x /M ux) æ+(Mu/Muy)æ≤1

where, Mx=Pu*ex

           My=Pu*ey

           Mux=maximum moment capacity for bending along x-axis at axial load Pa

           Muy=maximum moment capacity for bending along y-axis load Pa

æ= an exponent that depends upon the dimension of cross section, amount of reinforcement, concrete strength and yield stress of steel.

 

æ=.445+.155Pu/Puz

 

where,  Pu=.45fck.Ac+.155Pu/Puz

            Puz=.45fck.Ac+.75fy.Asc

 

DETAILING: Ductility is capacity to take additional displacement (without collapse) even after yield.

 

Ductility increases in:

ü      as concrete grade increases

ü      as steel grade increases

Structural ductility can be enhanced by

ü        Regular configuration

ü      More redundancy

ü      Avoiding yielding of column

ü      Avoiding failure of foundation

     So, ductility design and detailing is done in this project work which is based on IS:13920-1993.

 

i. Detailing of flexural member:

Requirements;

ü      The factor axial stress on the member under earthquake loading shall not exceed 0.1fck

ü       Members shall have a width to depth ratio of more than 3

ü      The width of member shall  not less than 200mm.

ü      The overall depth of member not more than ¼ of clear spans

 

Anchorage of beam bar: In external joint anchorage length beyond inner face of the column equal to development length in tension plus ten times bar diameters minus allowance for 90 degree bend.

 

Lap splice: Lap length should be greater than development length in tension

 

Lap splice shall not be provided

(a)    within a joints

(b)    within 2d from joint face

(c)    within ¼ of member lengthwhere flexure yielding may occur during earthquake.

 

Shear reinforcement

Hook length >10Ф or 75mm whichever is more

 Minimum diameter = 6mm

Minimum diameter = 8mm for clear span greater than 5m

Spacing as per ductility: At distance of 2d from face of column spacing should be more than 1/4d .at mid span spacing should not be more than 1/2d.

 

ii.Detailing of compression member:

-         minimum dimension should be greater than 300mm

-         B/D >0.4

-         Lap splice shall be provided only in the central half of the member length.

-         Spacing of hoop within lap splice<150m

-          

Requirement of longitudinal reinforcement

ü      The longitudinal reinforcement shall not be less than 0.8% or more than 4%of the gross cross-sectional area of column.

ü      The minimum number of longitudinal bars provided in a column shall be four in rectangular columns and six in a circular.

ü      Spacing of longitudinal bars measured along periphery of the column shall not exceed 300mm.

ü       

 Requirement ofTransverse reinforcement as per ductility design

ü      Up to distance as per ductility design 450mm, spacing should not be more than 100mm.

ü      At middle spacing should be less than ½ width of column

ü      Lap should be provided at distance more than ¼ clear height

 

Minimum and Maximum Sizes of Reinforcement: The minimum size of bars for column not less than 12mm in diameter.The diameter of the polygonal links  or ties shall  not be less than ¼ of the diameter of the largest longitudinal bar, and in no case less than 5mm.When a column terminates into a raft special confining reinforcement shall extend at least 300mm into raft.

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