Question

2. Design a square-tied reinforced concrete column to support the design loads shown below. Assume that the column must be designed with the reinforcement placed around the edges and y = 0.8. Take f.' = 4,000 psi and fy - 60,000 psi. Assume you have to use #9 bars for the longitudinal reinforcement. (35 pts.) P, - 400 kips M. -125 ft-kips

Answer 1

Given

Factored load Pu=400kips

Factore Moment Mu=125ft-kips

f'c=4000psi

fy=60000psi

$\gamma$=0.8

Bars to be used = #9

Now to design a square tied rienforced concrete column with the rienforcement placed all around the edges we need to calculate

1)Dimension

2)Longitudinal rienforcement

3)Transverse rienforcement(Lateral Ties)

4)Detailed Drawing

1)Calculation of Dimension

Assume the column will have an average compressive stress of about =0.6f'c=0.6*4000=2400psi=2.4K

Now Total Gross Area required for the column Ag=Factored Load/compressive stress

Ag=400/2.4=166.66in²

Let us try a section of 14*14(Ag=196in²)

2)calculation of Longitudinal rienforcement

We have As=$\rho$Ag

$\rho$= rienforcement ratio which can be estimated by using interation graph but to estimate $\rho$ we need to calculate first Kn and Rn which is given by as follows

Rn Pn * e f'c* Ag *h
and   
Kn= Pn f'c * Ag

Where Pn =Pu/$\phi$ ($\phi$=0.65 for tied column)

Pn=400/0.65 =615.4K

e=Mu/Pu =125*12/400

e=3.75in

$Rn=\frac{615.4*3.75}{4*196*14}$  

$Rn=0.21$

$Kn=\frac{615.4}{4*196}$

$Kn=0.8$

Now with $\gamma$ =0.8 ,f'c=4000psi,fy=60000 psi we can select a Graph of A.8 from ACI interaction Graph for rienforcement on all edges

Plot Kn and Rn value to get $\rho$ as shown below

Kn→ 100% たか25 4 102 105 oob かり ･ 0.05 010 0/5 0•20 0025 0.30 Rn>

we get $\rho$ =0.045 ($\rho$ is greater than $\rho$ _min =0.01 and less than $\rho$ _max=0.08 as per code hence OK)

Now calculate As=$\rho$Ag = 0.045*196

As=8.82in²

select 9#9 bars As=9.00in² from Table A.4(Areas of group of standard bars in in² )

3)Calculation Lateral Ties from code we have

Ties shall not be less than #3 when longitudinal bars are #10 or smaller

So assume ties of #3

Spacing shall not be more than 16 times the Diameter of the longitudinal bars 16 x9/8 =18in

or 48 times the diameter of the ties = 48 x 3/8 =18in

or least lateral dimension=14in.

use  #3 ties at 14in.

4)Detailed Drawing

Use cover of 2.5in

2.5in ulb * 25 in k x KIS.Z * ull E# 6 #b alb * asia K ull