V. IS456 2000-Circular Column

Verify the designed results produced by the program with hand calculated design results for an axially loaded circular column section as per IS456:2000.

References

Bureau of Indian Standards. SP 16 - Design aids for reinforced concrete to IS: 456-1978. 1980. New Dehli: Indian Standards Institution. p. 141.

Details

The fixed-base, 2.5 m cantilever column is subjected to axial force of 800 kN and a moment at the top of 162.5 kN·m. The section is 500 mm in diameter. The concrete strength is 20 N/mm² and the yield strength of steel is 415 N/mm². Use a clear cover of 50 mm. The main reinforcement is 20 mm and the lateral ties are 10 mm.

Validation

Slenderness Ratio

From Table. 28 Clause E-3 of IS 456 : 2000, the effective length of the compression member:

(L_eff ) = 2 × L = 2 × 2,500 = 5,000 mm

Least lateral dimension of column = (5,000/500) = 10 < 12

Large lateral dimension of column = (5,000/500) = 10 < 12

So, we can consider and design as a short column.

Check for Minimum Eccentricity of Column

As per Clause 25.4 of IS 456 : 2000, calculate the minimum eccentricity:

For Width e_min = (2,500/500) + (500/30) = 21.7 mm > 20 mm. (OK)

Hence Minimum Eccentricity is ok.

Check for Maximum Eccentricity of Column

As per Clause 39.3 of IS 456 : 2000 calculate the maximum eccentricity:

For Width e_max = 0.05 × 500 = 25 mm. Hence, e_min < e_max(OK)

Hence Maximum Eccentricity is ok.

Design for Flexure

Determine the non-dimenstional parameters to use with interaction chart 56 of SP 16:

P_{u} / f_{ck} D^{2} = 800 (10^{3}) / (20 \times 500^{2}) = 0.16

M_{u} / f_{ck} D^{3} = 162.5 (10^{6}) / (20 \times 500^{3}) = 0.065

From the chart, $P / f_{ck} = 0.06$ , and thus $P = 20 \times 0.06 = 1.2$

The area of longitudinal reinforcement required then is given by:

A_{sc} = \frac{P \times π \times D^{2}}{400} = \frac{1.2 \times π \times 500^{2}}{400} = 2,355 {mm}^{2}

Let assume 20 mm dia bar. So, A_st = (3.14/4) × 25² = 314 mm² .

Number of bars = (2,355/314) = 7.5 ≈ 8 bars

So, the actual reinforcement will be = 8 × 314 mm² = 2,513 mm²

As per Clause 26.5.3.2 (c) of IS 456 : 2000 Bar Diameter of Lateral Ties:

¼ × Main Bar dia = ¼ × 20 = 5 mm

or, 6 mm

So, Provide 10 mm dia. bar for lateral ties.

As per Clause 26.5.3.2 (c) of IS 456 : 2000 Pitch of Lateral Ties:

Least lateral dimension = 500 mm
16 × 20 = 320 mm
300 mm

Provide 10 mm dia. bar @ 300 mm for lateral ties.

Results

Result Type	Reference	STAAD.Pro	Difference	Comments
Reqd. area of steel (mm²)	2,355	2,367.71	negligible
Spacing of rectangular ties reinforcement (mm)	300	300	none

STAAD.Pro Input

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\10 Concrete Design\India\IS456 2000-Circular Column.STD is typically installed with the program.

STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 09-Dec-20
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 2.5 0;
MEMBER INCIDENCES
1 1 2;
DEFINE MATERIAL START
ISOTROPIC CONCRETE
E 2.17185e+07
POISSON 0.17
DENSITY 23.5616
ALPHA 1e-05
DAMP 0.05
TYPE CONCRETE
STRENGTH FCU 27579
END DEFINE MATERIAL
MEMBER PROPERTY
1 PRIS YD 0.5
CONSTANTS
MATERIAL CONCRETE ALL
SUPPORTS
1 FIXED
LOAD 1 LOADTYPE Dead  TITLE LOAD CASE 1
JOINT LOAD
2 FY -800
LOAD 2 LOADTYPE None  TITLE LOAD CASE 2
JOINT LOAD
2 MZ 162.5
LOAD COMB 3 COMBINATION LOAD CASE 3
1 1.0 2 1.0 
PERFORM ANALYSIS PRINT ALL
LOAD LIST 3
START CONCRETE DESIGN
CODE INDIAN
FC 20000 ALL
FYMAIN 415000 ALL
FYSEC 415000 ALL
CLEAR 0.050 ALL
MAXMAIN 20 ALL
MAXSEC 10 ALL
MINMAIN 20 ALL
MINSEC 10 ALL
TRACK 2 ALL
DESIGN COLUMN 1
END CONCRETE DESIGN
FINISH

STAAD.Pro Output

 COLUMN 
            =========================================================
            IS 456 - 2000   C O L U M N   D E S I G N   R E S U L T S
            =========================================================
   ============================================================================
                IS-456    L I M I T    S T A T E    D E S I G N
            C O L U M N   N O.      1  D E S I G N   R E S U L T S
         M20                    Fe415 (Main)               Fe415 (Sec.)
   LENGTH:  2500.0 mm   CROSS SECTION:  500.0 mm dia.  COVER: 50.0 mm
   ** GUIDING LOAD CASE:    3 END JOINT:     1  SHORT COLUMN
   DESIGN FORCES (KNS-MET)
   -----------------------
   DESIGN AXIAL FORCE (Pu)             :     800.00
                                            About Z            About Y
   INITIAL MOMENTS                     :     162.50               0.00
   MOMENTS DUE TO MINIMUM ECC.         :      17.33              17.33
   SLENDERNESS RATIOS                  :     -                  -     
   MOMENTS DUE TO SLENDERNESS EFFECT   :     -                  -     
   MOMENT REDUCTION FACTORS            :     -                  -     
   ADDITION MOMENTS (Maz and May)      :     -                  -     
   TOTAL DESIGN MOMENTS                :     162.50              17.33
   REQD. STEEL AREA   :     2367.71 Sq.mm.
   REQD. CONCRETE AREA:   193981.83 Sq.mm.
   MAIN REINFORCEMENT : Provide   8 - 20 dia. (1.28%,   2513.27 Sq.mm.)
                        (Equally distributed)                   
   TIE REINFORCEMENT  : Provide 10 mm dia. circular ties @ 300 mm c/c
   SECTION CAPACITY BASED ON REINFORCEMENT REQUIRED (KNS-MET)
   ----------------------------------------------------------
   Puz :   2482.79   Muz1 :    173.76   Muy1 :    173.76
   INTERACTION RATIO: 0.98 (as per Cl. 39.6, IS456:2000)
      STAAD SPACE                                              -- PAGE NO.    7
   SECTION CAPACITY BASED ON REINFORCEMENT PROVIDED (KNS-MET)
   ----------------------------------------------------------
   WORST LOAD CASE:     3
   END JOINT:     2 Puz :   2526.78   Muz :    176.84   Muy :    176.84   IR: 0.97
   ============================================================================
   ********************END OF COLUMN DESIGN RESULTS********************