Question

A short axially loaded RC column of size 425 mm x 425 mm is reinforced with 4 bars of 25mm diameter. Consider M 20 grade concrete, Fe 415
grade steel. The design axial load carrying capacity of said column is​

Answer 1

Answer:

The design axial load carrying capacity of said column is​ 1974.973 KN.

Explanation:

Given that:

Size of rectangular column = 425mm×425 mm

Grade of concrete = M20

$f_{ck}$ = characteristics of comprehensive strength of concrete = 20 $\frac{N}{mm^{2} }$

Grade of steel = Fe415

$f_{y}$= yield strength or ultimate tensile strength of Steel = 415 $\frac{N}{mm^{2} }$

Number of steel bar = 4

Diameter of Steel bar = 25 mm

axial load carrying capacity of column formula

                $P_{u}$ = 0.4$f_{ck} A_{c}$ + 0.67$f_{y} A_{SC}$

where, $P_{u}$ = ultimate axial load carrying capacity of column

            $f_{ck}$= characteristics of comprehensive strength of concrete                  

             $A_{c}$ = area of concrete in column which will be calculate

           $A_{SC}$ = area of Steel in column which will be calculated

Solve:

1) First we have to calculate gross cross sectional area of column:

$A_{g}$= gross cross sectional area of column

Size of column = 425×425 mm

Ag = 425x425 $mm^{2}$

Ag = 180625 $mm^{2}$

2) Second we have to calculate area of Steel in column:

$A_{SC}$= area of Steel in column

No. of steel bar = 4

Diameter of Steel bar D = 25 mm

Area of bar = 4×$\frac{\pi }{4}$× $D^{2}$

Where π = 3.14

$A_{SC}$ = 4× (3.14/4) 25x25 $mm^{2}$

$A_{SC}$ = 1962.5 $mm^{2}$

3) Now we calculate area of concrete in column:

$A_{C}$ = area of concrete in column

We know that gross cross sectional area of column is equal to area of concrete in column and area of Steel in column

$A_{g} = A_{c} + A_{SC}$

$A_{c} = A_{g} - A_{SC}$

Putting the value of gross area of column and steel area subtracting both we get concrete in column

$A_{c}$ =( 180625 - 1962.5) $mm^{2}$

$A_{c}$ = 178662.5 $mm^{2}$

Putting all the value in formula for ultimate axial load carrying capacity of column

             $P_{u}$ = 0.4$f_{ck} A_{c}$ + 0.67$f_{y} A_{SC}$

Where , $f_{ck}$= 20$\frac{N}{mm^{2} }$

              $A_{c}$ = 178662.5 $mm^{2}$

               $f_{y}$ = 415 $\frac{N}{mm^{2} }$

           $A_{SC}$ = 1962.5 $mm^{2}$

$P_{u}$ = (0.4× 20×178662.5) N +(0.67×415×1962.5) N

 $P_{u}$ = 1429300 N + 545673.125 N

$P_{u}$= 1974973.125 N

Converting into kilo Newton we have to divide by 1000

$P_{u}$ = 1974973.125  N/1000 = 1974.973 KN

Hence, about 1974.973 KN ultimate axial load carrying capacity of column for above this calculation.