How to design Residential Building- Civil Engineering/ Construction (2)

     

Beams Design

   Beam is the structural member that carrys  loads come from ribs and other loads come from stone at the faces of the building and transfers it to columns.This chapter

includes analysis and design for flexure, including  dimensioning of the concrete cross section , the selection and placement of reinforcing steel ,in addetion to shear desigen including diameter and spasing of sterups.

       Beams, are  usually found in slabs and its main function is to carry loads comes from ribs and other loads come from stone faces of the building. Beams are bigger than ribs and have a larger inertia, so that ribs are carried by beams, and these beams transfer its loads to columns, which are supported by.

There are basics types of beams can be used in the projects:

ü Drop beam:  carries, in addition to its own weight, slab  dead and live loads,

           And h is greater than 310mm .

ü hidden beam: carries, in addition to its own weight, slab dead and live loads with restricted  h=310mm

                                                       

ü T-beam: carries, in addition to its own weight, slab  dead and live loads,

           it consists of  two parts : web which is the drop part , and    flang  which is the  hidden part. T-section  has aditional advantage that is its risestance to shear is  high .

ü Secondary beam: an edge beam, which is carries stone load in addition to its own weight.

Analysis of beams:

 

    Slab loads, dead and live ; transfer to beam as shear force, where the summation of left and right shear forms reaction.

Analysis of beams is similar to that for ribs, the only deferent is the estimation of beam own weight since the beam width (b) is not known. This width is one of the required unknown, which must found to support the loads carried by beam.

Before the analyses started the loads on the beams must be determined, this loads can be calculated by knowing the strips carried by this particular beam, since the beam is the support of these strips, the load on it will be the reaction of the strips on this support, the reaction can be found from the analyses of the strips. These loads are per rib not per meter length of beam, so the beam load is then calculated as:

¯      

                                         

(Where: unit area of rib = 0.52 m)

     

 The analysis is done by the same software “Prokon Structural Analysis”, and the analyses results with the final beams section are shown next.

Design of Beams:

 

·       Drawing shear and moment diagram for each beam by loaded it by dead and live loads which are the reaction of ribs (from shear diagram of ribs) in addition to its own weight and in some beams the weight of stones.

·       Get maximum M_u from the moment diagram, take ρ = 0.5ρ_max, then compute ω.

·       Use Mu/Φfc`bd²,to compute the width of beam (b) ,(d=310 mm for all beams).

·       Then for each value of Mu use Mu/Φfc`bd², to compute by using table of moment strength of rectangular section with tension reinforcement only (PCA).

·            calculate ,then A_s=ρ*b*d

œ  Flexural Design of Beams:

·               Firstly  , beam width (b) must be assumed (500 mm) as a initial value.

                                                                          

Determining the loads acting on the beam .

·     the maximum moment must be determined for the assumed section, which will be used to find b.

·     Assume .

·     Find the reinforcement index ()from equation :

·      find the value of the term .

·     From this value and by rearrangement of the last term find b as:                  .

·     Compare this value of b with the assumed one:

Ä If b assumed b calculated      b is ok.

Ä If b assumed > b calculated b is large, try again using smaller b.

Ä If b assumed < b calculated b is small, try again using bigger b.

·               Find steel area by using the equation:

             

·     Choose an appropriate steel diameter to satisfy the steel area required.

·     For negative moment at support, ACI-99 gives the following value for negative moment where the support is a beam:

                                                                                                                

   œ  Shear Design Procedure:

ü Calculate the concrete shear capacity of the beam:

                                                  

                                                                

ü Determine the maximum shear force (Vu) acting there on the beam at distance=(d/2) from the face of support .if   Vu <ØVc     is no need for stirrups but we should provide minimum reinforcement as follow:

   

             S                                      

ü Assume required stirrup area for the section. (for one stirrup, legs number = 2).

ü  calculate the spacing between the stirrups as follow:

          Vs=Av* Fyv *d/S

                     

                 

    ¯   the limits of the spacing (s) is given as the smallest value of the following:

               

(use the smallest value)

Hidden beams analysis:

 

                                

   

Beam  ( 1 )

ρ _balance  = 0.85 * β (f_c / f_y) [ 600 / 600 + f_y]

             = 0.85 * 0.85 (25 / 414) * [600 / 600 +414]

             = 0.025816

 ρ _max  = 0.75 ρ _balance

           = 0.75 * 0.025816

           = 0.019362

        ρ =0.5*ρ _max  

        ρ=0.009681

Shear & Moment Diagram :

¯   Asume b= 500 mm ..........

Maximum moment = 177KN.m

w = ρ *  f_y  /  f_y

    = 0.009681 * 414 / 25

    = 0.160317

w( 1 -0.59 * w) = 0.2168 ( 1 – 0.59 *0.16031)

                          = 0.14515

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

177* 10⁶ / [ 0.9 * 25 * b * 285² ] = 0.14515

b = 682.303 mm  ………    (use   b = 700 mm)

Mu = Ф * f_c  * b * d^{2  *}w ( 1 – 0.59 w )

      = 0.9*25*700*285²*0.14515

      = 185.689 >> 177 KN.m       ok

Span (1) :

@ M +ve= 93.7 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

 93.7 * 10⁶ / [0 .9 * 25 * 700 * 285² ] =  0.07324= w ( 1 – 0.59 w )           

                                                                                 w = 0.07671588

ρ =  w (f_c / f_y) = 0.07671588 ( 25/ 414) = 0.004632

    

As = ρ * b * d   = 0.004632* 700 * 285  = 924 mm²                    

                                                          

( use 6 Ф 14 = 924  mm²  )

Span (2) :

@ M +ve  = 72.5 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

72.5 * 10⁶ / [ 0.9 * 25 * 700 * 285² ] =  0.05667189 = w ( 1 – 0.59 w ) 

                                                                               w = 0.058705

ρ =  w (f_c / f_y) = 0.058705 ( 25 / 414) = 0.003545

   

As = ρ * b * d   = 0.003545 * 700 * 285 = 707.227 mm²                    

( use 5 Ф 14 = 770  mm²  )

Span (3) :

@ M +ve  = 53.7 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

53.7 * 10⁶ / [ 0.9 * 25 *700 * 285² ] =  0.041976 = w ( 1 – 0.59 w ) 

                                                                                 w = 0.0430707

ρ =  w (f_c / f_y) = 0.0430707 ( 25 / 414) = 0.0026008

   

As = ρ * b * d   = 0.003381* 700 * 285 = 674.63 mm²   

                

( use 5 Ф 14 = 770 mm²  )

Span (4) :

@ M +ve  = 166 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

166 * 10⁶ / [ 0.9 * 25 * 700 * 285² ] =  0.12975 = w ( 1 – 0.59 w ) 

                                                                                 w = 0.14158

ρ =  w (f_c / f_y) = 0.14158 ( 25 / 414) = 0.00855

   

As = ρ * b * d   = 0.00855 * 700 * 285 = 1705.7 mm²                    

( use 7 Ф 18 = 1781  mm²  )

Ø  Negative moment:

Support (2) :

@ M -ve= 122 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

 122 * 10⁶ / [0 .9 * 25 * 700* 285² ] =  0.095365 = w ( 1 – 0.59 w )           

                                                                               w = 0.101435

ρ =  w (f_c / f_y) = 0.101435 ( 25 / 414) = 0.0061253

    

As = ρ * b * d   = 0.0061253 * 700 * 285  = 1222mm²                    

( use 8 Ф 14 = 1232  mm²  )

Support (3) :

@ M -ve= 67.2 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

 67.2 * 10⁶ / [0 .9 * 25 * 700 * 285² ] =  0.052528 = w ( 1 – 0.59 w )           

                                   w = 0.054266

ρ =  w (f_c / f_y) = 0.054266 ( 25/ 414) = 0.003276

    

As = ρ * b * d   = 0.003381 * 700 * 285  = 674.6 mm²                    

( use 5Ф 14 = 770  mm²  )

Support (4) :

@ M -ve= 181 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

 181 * 10⁶ / [0 .9 * 25 * 700* 285² ] =  0.141484 = w ( 1 – 0.59 w )           

                                                                                w = 0.155807

ρ =  w (f_c / f_y) = 0.155807 ( 25 / 414) = 0.0094086

    

As = ρ * b * d   = 0.0094086 * 700* 285  = 1877.02 mm²                    

( use 6 Ф 20 = 1885  mm²  )

 

Span (1) :

                        

                       =   0.85 * (25)^0.5 * 700 * 285 / 6                                                                   

                       =  141.31 KN

Vu = 173.1 KN

           Ü     Vu > fVc  

Vu > fVc

 V_s = (V_u -ØV_c)/Ø= ( 173.1 – 141.31 KN) /0.85

     = 37.4 KN

Assume 4-leg Ø 10 mm (A_v = 314 mm²)

Min reinforcement must be used (Smax)

S= 3 * A_v*f_yv / b 

  =3 *314 *280 / 700 = 376.8 mm

or

S=A_v*f_yv*d/V_s =314 *280*285/37.4  = 670 mm

S =  0.75h =0.75*285= 213.75  mm 

            Ü          Use  2f 10 / 200 mm

Span (2) :

                        

                       =   0.85 * (25)^0.5 * 700 * 285 / 6                                                                   

                       =  141.31 KN

Vu = 159.1 KN

           Ü     Vu > fVc  

 Vu > fVc

 V_s = (V_u -ØV_c)/Ø= ( 159.1 – 141.31 KN) /0.85

     = 20.93  KN

Assume 4-leg Ø 10 mm (A_v = 314 mm²)

Min reinforcement must be used (Smax)

S= 3 * A_v*f_yv / b 

  =3 *314 *280 / 700 = 376.8 mm

or

S=A_v*f_yv*d/V_s =314 *280*285/20.93  = 1197.2 mm

S =  0.75h =0.75*285= 213.75  mm 

            Ü          Use  2f 10 / 200 mm

Span (3) :

                       

                       =   0.85 * (25)^0.5 * 700 * 285 / 6                                                                   

                       =  141.31 KN

Vu = 149.1 KN

           Ü     Vu > fVc  

Vu > fVc

 V_s = (V_u -ØV_c)/Ø= ( 149.1 – 141.31 KN) /0.85

     = 9.45  KN

Assume 4-leg Ø 10 mm (A_v = 314 mm²)

Min reinforcement must be used (Smax)

S= 3 * A_v*f_yv / b 

  =3 *314 *280 / 700 = 376.8 mm

or

S=A_v*f_yv*d/V_s =314 *280*285/9.45  =265.16 mm

S =  0.75h =0.75*285= 213.75  mm 

            Ü          Use  2f 10 / 200 mm

Span (4) :

                        

                       =   0.85 * (25)^0.5 * 700 * 285 / 6                                                                   

                       =  141.31 KN

Vu = 163.3 KN

           Ü     Vu > fVc  

 Vu > fVc

 V_s = (V_u -ØV_c)/Ø= ( 163.3 – 141.31 KN) /0.85

     = 25.84 KN

Assume 4-leg Ø 10 mm (A_v = 314 mm²)

Min reinforcement must be used (Smax)

S= 3 * A_v*f_yv / b 

  =3 *314 *280 / 700 = 376.8 mm

or

S=A_v*f_yv*d/V_s =314 *280*285/25.84  = 283.6 mm

S =  0.75h =0.75*285= 213.75  mm 

            Ü          Use  2f 10 / 200 mm

Beam  ( 2 )

ρ _balance  = 0.85 * β (f_c / f_y) [ 600 / 600 + f_y]

             = 0.85 * 0.85 (25 / 414) * [600 / 600 +414]

             = 0.025816

 ρ _max  = 0.75 ρ _balance

           = 0.75 * 0.025816

           = 0.0193620

        ρ =0.5*ρ _max  

        ρ=0.00968104

¯ Asume b= 500 mm ........

Maximum moment = 147 KN.m

w = ρ *  f_y  /  f_y

    = 0.00968104 * 414 / 25

    = 0.160318

w( 1 -0.59 * w) = 0.160318 ( 1 – 0.59 *0.160318)

                          = 0.1451538

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

147* 10⁶ / [ 0.9 * 25 * b * 285² ] = 0.1451538

b = 554.136 mm                 use   b =600 mm

Mu = Ф * f_c  * b * d^{2  *}w ( 1 – 0.59 w )

      = 0.9*25*600*285²*0.14515

      = 159.16 >> 147 KN.m       ok

Span (1) :

@ M +ve= 33.1 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

 33.1 * 10⁶ / [0 .9 * 25 * 600 * 285² ] =  0.030185= w ( 1 – 0.59 w )           

                                                                                  w = 0.030743

ρ =  w (f_c / f_y) = 0.030743 ( 25 / 414) = 0.001856

ρ < ρ _min          so that we use minimum steel

    

As = ρ * b * d   = 0.003381* 600* 285  = 578.26 mm²                    

( use 4 Ф 14 = 616 mm²  )

Span (2) :

@ M +ve  = 61.3 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

61.3* 10⁶ / [ 0.9 * 25 * 600 * 285² ] =  0.05590 = w ( 1 – 0.59 w ) 

                                                                                w = 0.05787

ρ =  w (f_c / f_y) = 0.05787 ( 25 / 414) = 0.003495

   

As = ρ * b * d   = 0.003495 * 600 * 285 = 597.67 mm²                    

( use 7 Ф 12 =792  mm²  )

Span (3) :

@ M +ve= 36.2 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

36.2 * 10⁶ / [0 .9 * 25 * 600 * 285² ] =  0.057635 = w ( 1 – 0.59 w )           

                                            w = 0.059741

ρ =  w (f_c / f_y) = 0.059741 ( 25 / 414) = 0.0036075

ρ < ρ _min          so that we use minimum steel

    

As = ρ * b * d   = 0.0036075 * 600 * 285  = 616.8 mm²                    

( use 7 Ф 12 = 792 mm²  )

Span (4) :

         

@ M +ve= 138 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

 138 * 10⁶ / [0 .9 * 25* 600 * 285² ] =  0.12585 = w ( 1 – 0.59 w )           

                                                                                  w = 0.136909

ρ =  w (f_c / f_y) = 0.136909 ( 25 / 414) = 0.0082675

    

As = ρ * b * d   = 0.0082675* 600 * 285  = 1413.7 mm²                    

( use 8 Ф 16 = 1608  mm²  )

Ø  Negative moment:

Support (2) :

@ M -ve  = 73.2 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

73.2 * 10⁶ / [ 0.9 * 25 * 600 * 285² ] =  0. 066755 = w ( 1 – 0.59 w ) 

                                                                                  w = 0.069614

ρ =  w (f_c / f_y) = 0.069614 ( 25 / 414) = 0.004203

   

As = ρ * b * d   = 0.004203* 600 * 285 = 718.8 mm²                    

( use 7 Ф 12 = 792  mm²  )

Support (3) :

@ M -ve  = 61.7 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

61.7 * 10⁶ / [ 0.9 * 25* 600 * 285² ] = 0.056268= w ( 1 – 0.59 w ) 

                                         w =0.058271

ρ =  w (f_c / f_y) = 0.058271 ( 25 / 414) = 0.0035188

   

As = ρ * b * d   = 0.0035188 * 600 * 285 = 601.7 mm²                    

( use 6 Ф 12 = 679  mm²  )

Support (4) :

@ M -ve  = 149 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

149 * 10⁶ / [ 0.9 * 25 * 600 * 285² ] =  0.1358822 = w ( 1 – 0.59 w ) 

                                                                                  w = 0.148976

ρ =  w (f_c / f_y) = 0.148976 ( 25 / 414) = 0.0089961

   

As = ρ * b * d   = 0.0089961 * 600* 285 = 1538.3 mm²                    

( use 8 Ф 16 = 1608  mm²  )

Span (1) :

                        

                       =   0.85 * (25)^0.5 * 600 * 285 / 6                                                                   

                       =  121.125 KN

Vu = 262.1 KN

           Ü      Vu > fVc

 V_s = (V_u -ØV_c)/Ø= ( 262.1 – 121.125 KN) /0.85

     = 39.1 KN

Assume 4-leg Ø 10 mm (A_v = 314 mm²)

Min reinforcement must be used (Smax)

S= 3 * A_v*f_yv / b 

  =3 *314 *280 / 1000 = 263.76 mm

or

S=A_v*f_yv*d/V_s =314 *280*285/39.1  = 640.85 mm

S =  0.75h =0.75*285= 213.75 mm 

Ü       Use  2f 10 / 200 mm

Span (2) :

                        

                       =   0.85 * (30)^0.5 * 600 * 285 / 6                                                                   

                       =  121.125 KN

Vu = 338 KN

             Ü    Vu > fVc

 V_s = (V_u -ØV_c)/Ø= (338 – 121.125 KN) /0.85

     = 129.4 KN

Assume 4-leg Ø 10 mm (A_v = 314 mm²)

Min reinforcement must be used (Smax)

S= 3 * A_v*f_yv / b 

  =3 *314 *280 / 1000 = 263.76 mm

or

S=A_v*f_yv*d/V_s =314 *280*285/129.4  = 200.4 mm

S =  0.75h =0.75*285= 213.75 mm 

Ü       Use  2f 10 / 200 mm

Span (3) :

                        

                       =   0.85 * (25)^0.5 * 600 * 285 / 6                                                                   

                       =  121.125 KN

Vu = 162.1 KN

           Ü      Vu > fVc

 V_s = (V_u -ØV_c)/Ø= ( 162.1 – 121.125 KN) /0.85

     = 48.2 KN

Assume 4-leg Ø 10 mm (A_v = 314 mm²)

Min reinforcement must be used (Smax)

S= 3 * A_v*f_yv / b 

  =3 *314 *280 / 1000 = 263.76 mm

or

S=A_v*f_yv*d/V_s =314 *280*285/48.2  = 519.8 mm

S =  0.75h =0.75*285= 213.75  mm 

            Ü          Use  2f 10 / 200 mm

Span (4) :

                        

                       =   0.85 * (25)^0.5 * 600 * 285 / 6                                                                   

                       =  121.125 KN

Vu = 190.6   KN

           Ü      Vu > fVc

 V_s = (V_u -ØV_c)/Ø= ( 190.6 – 121.125 KN) /0.85

     = 81.74 KN

Assume 4-leg Ø 10 mm (A_v = 314 mm²)

Min reinforcement must be used (Smax)

S= 3 * A_v*f_yv / b 

  =3 *314 *280 / 1000 = 263.76 mm

or

S=A_v*f_yv*d/V_s =314 *280*285/81.74  = 306.5 mm

S =  0.75h =0.75*285= 213.75 mm 

        Ü              Use  2f 10 / 200 mm

Beam  ( 3 )

ρ _balance  = 0.85 * β (f_c / f_y) [ 600 / 600 + f_y]

             = 0.85 * 0.85 (25 / 414) * [600 / 600 +414]

             = 0.025816

 ρ _max  = 0.75 ρ _balance

           = 0.75 * 0.025816

           = 0.019362083

        ρ =0.5*ρ _max  

        ρ=0.009681041

¯  Asume b= 250 mm .........

Maximum moment = 50.8 KN.m

w = ρ *  f_y  /  f_y

    = 0.00968104 * 414 / 25

    = 0.160318

w( 1 -0.59 * w) = 0.160318 ( 1 – 0.59 *0.160318)

                          = 0.1451538

Mu / Ф * f_c  * b * d²   =  w ( 1 – 0.59 w )

     50.8*10⁶/ [0.9*25*b* 285² ]  =   0.1451538

 b= 191.49 mm                      (  Use    b= 300 mm  )               

Mu = Ф * f_c  * b * d^{2  *}w ( 1 – 0.59 w )

      = 0.9*25*300*285²*0.14515

      = 79.58 >> 50.8 KN.m  …….. ok

Span (1) :

@ M +ve= 36.9  KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

 36.9 * 10⁶ / [0 .9 * 25 * 300 * 285² ] =  0.067302= w ( 1 – 0.59 w )           

                                                                                  w = 0.070211

ρ =  w (f_c / f_y) = 0.070211 ( 25 / 414) = 0.0042398

    

As = ρ * b * d   = 0.0042398* 300* 285  = 362.5 mm²                    

( use 3 Ф 14 = 462  mm²  )

  

Span (2) :

@ M +ve  = 31.5 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

31.5* 10⁶ / [ 0.9 * 25 * 300 * 285² ] =  0.057453 = w ( 1 – 0.59 w ) 

                                                                                w = 0.0595455

ρ =  w (f_c / f_y) = 0.0595455 ( 25 / 414) = 0.0035957

 

As = ρ * b * d   = 0.0035957 * 300 * 285 = 307.4 mm²                    

( use  3 Ф 14 = 462 mm²  )

Ø  Negative moment:

Support (2) :

@ M -ve  = 51.1 KN . m

Mu  / Ф * f_c  * b * d²  = w ( 1 – 0.59 w )

51.1 * 10⁶ / [ 0.9 * 25 * 300 * 285² ] =  0. 093202 = w ( 1 – 0.59 w ) 

                                                                                  w = 0.098983

ρ =  w (f_c / f_y) = 0.098983 ( 25 / 414) = 0.005977

   

As = ρ * b * d   = 0.005977* 300 * 285 = 511.05 mm²                    

( use 4 Ф 14 = 616  mm²  )

Span (1) :

                        

                       =   0.85 * (25)^0.5 * 300 * 285 / 6                                                                   

                       =  60.56 KN

Vu = 52.5 KN

      Ü                Vu < fVc

So that we use minimum reinforcement

Assume 2-leg Ø 10 mm (A_v =157 mm²)

Min reinforcement must be used (Smax)

S= 3 * A_v*f_yv / b 

  =3 *157 *280 / 200 = 659.4 mm

S=0.75d=0.75*300 =225 mm

S=500mm

Ü       Use 1 f 10 / 200

Span (2) :

                        

                       =   0.85 * (25)^0.5 * 300 * 285 / 6                                                                   

                       =  60.56 KN

Vu = 52.8 KN

            Ü          Vu < fVc

So that we use minimum reinforcement

Assume 2-leg Ø 10 mm (A_v =157 mm²)

Min reinforcement must be used (Smax)

S= 3 * A_v*f_yv / b 

  =3 *157 *280 / 300 = 659.4 mm

S=0.75d=0.75*285 = 213.75 mm

S=500mm

           Ü                Use 1 f 10 / 200

Columns Design

Columns Design

      Introduction: 

          Columns are defined as members that carry loads chiefly in compression.   Generally, columns are referred to as compression members, because the compression forces dominate their behavior.

     According to (ACI code 10.3.5) the useful design strength of an axially loaded column is to be found based on the equation { Pn=0.85fćAc + fyAs} with the introduction of certain strength reduction factor.

ÄColumns may bedevided  into two broad  catagories :

     ¯short columns :  for which the strength is governed by the the strength of the materials and the geometry of the cross section.

    ¯slender columns: for which the strength may be significantly reduced by lateral deflections.

   To know whether the column is short or slender we should no if the column braced or un   braced.

Structural frames whose joints are restrained against lateral displacement by attachment to rigid elements or by bracing are called braced or non sway frames.

ØThere are three basic types of reinforced concrete columns:

1.Tied columns,  are reinforced with longitudinal bar which are enclosed by horizontal, or lateral, ties placed at specified spacing.

2. Spiral columns, are reinforced with longitudinal bars enclosed by a continuous, rather closely spaced, steel spiral. The spiral is made up of either wire or bar and is formed in the shaped of a helix.

3. composite columns, This type encompasses compression members reinforce longitudinally with structural steel shapes, pipes or tubes with or without longitudinal bars.

  O Our discussion will be limited to the first type only – tied columns. Tied columns are generally square to rectangular while spiral columns are normally circular.

However, this is not a hard-and-fast rule, since square spirally reinforced column and circular tied columns do exist as well as do other shapes, such as octagonal and L-shape columns.

Design procedure:

 

       To Design the Columns  we find load from the “Prokon Structural Analysis”, Programme by determining the reactions of the beams on the columns.

The ultimate column load can be calculated according ACI-99 code, which gives the axial load strength for nonprestresses members using tie reinforcement as follow steps:

Æ Calculate gravity load axial forces.

Æ  Calculate gravity load bending moments.

Æ  Calculate lateral load axial forces and bending moments.

Æ Proportion section and select rebar for strength using load factors, strength reduction factors and load equations.

Ä  

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 where: A_g is the gross column cross section area.

For safe design:

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                      Pu = f b [ 0.85 * f_c * Ac + As * f_y ]

 Where:

     f =  0.80    for tied columns

      =  0.85    for spiral columns

  b =  0.70      for tied columns

       =   0.75     for spiral columns

   Ac: area of concrete which equal to (Ag – Ast)

where:  P_u is the max factor load subjected to column.

   ¯ When the steel area found, the steel ratio then calculated     by:

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Ø This steel ratio should be in the range:       

And The spacing between bars Is the smallest value of:

1.          The least dimension of the column.

2.          16  d_b, where d_b is the diameter of the major bars.

3.          48  d_s, where d_s is the diameter of the stirrup.

   

 

   O    Because there is a large number of  columns in the building, it is empirical to divide the columns to categories according to their loads, the following table In calculation of columns shows the columns categories and Their designed load.

Calculation of columns:

 

     Design of columns depends on the load come from beams, so that divide the column into three groups based on the load that will carry.

For design of columns use  fc = 25 Mpa &   f_y = 414 Mpa

Group	Section	Pu (KN)	Number of Column
1	200*300	250<Pu<500	Two columns
2	200*400	500<Pu<900	Four columns
3	300*400	900<Pu<1300	Three columns
4	300*500	1300<Pu<1800	Three columns

First Column (1) :

250 < P < 500 KN

P_max. =300.08 KN

Assume r = 0.01

r = Ast / Ag   use   r = 0.01         Ü      Ast = 0.01 Ag

Pu = ( P_max  + own weight of column ) for all floors

     = 300.08 + 0.3 * 0.6 * 3 * 4 * 24 * 1.4

     =  372.66 KN

Pu = f b [ 0.85 * f_c * Ac + As * f_y ]

Where:

   f =  0.80 ,  b =  0.70       (for tied columns)

      

Ø  Ag = Pu /  f b [ 0.85 * f_c  + r ( f_y  - 0.85 * f_c ) ]

          Ag = 372.66* 10³ / 0.8 * 0.7 [ 0.85 * 25+ 0.01 ( 414 – 0.85 * 25)]

                           =  372.66 * 10³ / 14.1

                           = 26429.79  mm²

Suppose  Gross Area ( Ag ) = 200 * 300= 60000 mm²

                    Ø  Ag  suppose = 60000 mm²  > Ag = 26429.79 mm²       

      Ø Ast = 0.01 * Ag = 0.01 * 60000 = 600 mm²       

(Use 4 F 14 = 616 mm²)

Design Tie of  (C1)

Use minimum spacing from following:

S = minimum dimension  = 200 mm

S = 16 d_b  = 16 * 14  = 244 mm

S = 48 ds  = 48 * 8    = 384 mm

                   

Ü          (Use  1 F 8 / 200 mm)

No need for other ties because the spacing between main steel bars is less than  300 mm .

Second  Column (2) :

500 < P < 900 KN

P_max. = 896.96 KN

Assume r = 0.01

r = Ast / Ag   use   r = 0.01        Ü           Ast = 0.01 Ag

Pu = ( P_max  + own weight of column ) for all floors

     = 896.96  + 0.3 * 0.6 * 3 * 4 * 24 * 1.4

     = 969.54 KN

Pu = f b [ 0.85 * f_c * Ac + As * f_y ]

Where:

   f =  0.80 ,  b =  0.70       (for tied columns)

Ü      Ag = Pu /  f b [0.85 * f_c  + r (fy  - 0.85 * f_c ) ]

        Ag = 969.54 * 10³ / 0.8 * 0.7 [0.85 * 25 + 0.01 (414 – 0.85 * 25)]

                           = 969.54 * 10³ / 14.1

                           = 68761.42  mm²

Suppose Gross Area ( Ag ) = 200 * 400 = 80000 mm²

 Ø     Ag   suppose = 80000 mm²  > Ag = 73190.35 mm²       

Ü                Ast = 0.01 * Ag = 0.01 * 80000 = 800 mm²       

(Use 4 F 16 = 804 mm²)

Design Tie of  (C2)

Use minimum spacing from following:

S = minimum dimension  = 200 mm

S = 16 d_b  = 16 * 16  = 169 mm

S = 48 ds  = 48 * 8    = 384 mm

                   

Ü           (Use 1 F 8 / 200 mm)

No need for other ties because the spacing between main steel bars is less than  300 mm .

Third  Column (3) :

900 < P < 1300 KN

P_max. = 1114.24 KN

Assume r = 0.01

r = Ast / Ag   use   r = 0.01       Ü      Ast = 0.01 Ag

Pu = ( P_max  + own weight of column ) for all floors

     = 1114.24   + 0.3 * 0.6 * 3 * 4 * 24 * 1.4

     = 1186.82 KN

Pu = f b [ 0.85 * f_c * Ac + As * f_y ]

Where:

   f =  0.80 ,  b =  0.70       (for tied columns)

Ø   Ag = Pu /  f b [0.85 * f_c  + r (fy  - 0.85 * f_c ) ]

 Ag = 1186.82 *10³ / 0.8 * 0.7 [0.85 * 25 + 0.01 (414 – 0.85 * 25)]

                           = 1186.82 * 10³ / 14.1

                           = 84171.63  mm²

Suppose Gross Area ( Ag ) = 300* 400 = 120000 mm²

   Ø  Ag   suppose = 120000 mm²  > Ag = 84171.63 mm²       

         Ü    Ast = 0.01 * Ag = 0.01 * 120000 = 1200 mm²       

(Use 6 F 16 = 1206 mm²)

Design Tie of  (C3)

Use minimum spacing from following:

S = minimum dimension  = 300 mm

S = 16 d_b  = 16 * 18  = 288 mm

S = 48 ds  = 48 * 8    = 384 mm

                   

Ü          (Use 2 F 8 / 200 mm)

There is a need for other ties because the spacing between main steel bars is more than  300 mm . 

Fourth Column (4) :

1300< P < 1800 KN

P_max. = 1783 KN

Assume r = 0.01

r = Ast / Ag   use   r = 0.01   Ü          Ast = 0.01 Ag

Pu = ( P_max  + own weight of column ) for all floors

     = 1783   + 0.3 * 0.6 * 3 * 4 * 24 * 1.4

     = 1855.58 KN

Pu = f b [ 0.85 * f_c * Ac + As * f_y ]

Where:

   f =  0.80 ,  b =  0.70       (for tied columns)

Ø   Ag = Pu /  f b [0.85 * f_c  + r (fy  - 0.85 * f_c ) ]

    Ag = 1855.58 *10³ / 0.8 * 0.7 [0.85 * 25 + 0.01 (414 –        0.85 * 25)]

                           = 1855.58 * 10³ / 14.1

                           = 131601.135  mm²

Suppose Gross Area ( Ag ) = 300* 500 = 150000 mm²

 ØAg   suppose = 150000 mm²  > Ag = 131601.135 mm²       

Ü  Ast = 0.01 * Ag = 0.01 * 150000 = 1500 mm²   

   

(Use 6 F 18 = 1527 mm²)

Design Tie of  (C4)

Use minimum spacing from following:

S = minimum dimension  = 300 mm

S = 16 d_b  = 16 * 18  = 288 mm

S = 48 ds  = 48 * 8    = 384 mm

                   

Ü         (Use  2 F 8 / 200 mm)

There is a  need for other ties because the spacing between main steel bars is more than  300 mm .