Question

Design a reinforced concrete steel rectangular tie column with ACI standards, total LL = 14.0625 kips, total DL = 18.95625kips, cross-section 12in by 12 in, Hight of column 12ft. Show all the steps and which table used. Assume the information which is not given with justifications. Show drawing as well. If the high is not suitable you can change it.

Answer 1

Solution:- the values given in the question are as follows:

dead load(DL)=14.0625 kips

live load(LL)=18.95625 kips

cross-section of column=12 in*12 in

length of column(L)=12 ft

let we use the grade of concrete is C28 and S420

characteristic strength of concrete(f^Ic)=4000 psi

yield strength of steel(fy)=60000 psi

slenderness ratio()=k*Lu/(leat lateral dimension)

where, k=1.0

Lu=12 ft =12*12 in

least lateral dimension=12 in

slenderness ratio()=(12*12)/12=12

slenderness ratio()=12 40 ,so the column is short column.

let we design axially(with minimum eccentricity) loaded short column

load carrying capacity of column(Pu)=0.4*fc*Ac+0.75*fy*As

where, Ac=coss-section area of concrete

As=cross-section area of steel

Ac=Ag-As

gross area(Ag)=12*12=144 in^2

Ac=144-As

load carrying capacity of column(Pu)=0.4*4000*(144-As)+0.75*60000*As , [Eq-1]

total load on column(P)=DL+LL

total load on column(P)=18.95625+14.0625

total load on column(P)=33.58125 kips

total load on column(P)=33.58125*10^3 lb

we design the column for load P

Pu=P , put in above equation-(1)

33.58125*10^3=230400-1600As+45000As

33581.25-230400=43400As

As=-196818.75/43400 in^2

According to above equation, area of steel goes in negative, but area of steel in column never goes in negative.

so we will provide area of steel greater than the minimum area of steel needed for this column.

area of steel(As)=2% of gross area

area of steel(As)=(/100)*144

area of steel(As)=2.88 in^2

provide 0.875 in diameter steel bar into the column.

number of bars(n)=As/area of single bar

area of single bar=(Pi/4)*0.875^2

area of single bar=(3.14/4)*0.875^2

area of single bar=0.6013 in^2

number of bars(n)=2.88/0.6013=4.789 5

provide 0.875 in diameter of 6 bars.

let provide 0.375 in diameter tie bars

spacing of tie bars=minimum of {B or D , 16 , 11.81 in}

spacing of tie bars=minimum of {12 in , 16*0.875=14 , 11.81 in}

spacing of tie bars=11.81 in

provide spacing of tie bars is 11 in center to center.

let provide 1 in constant clear cover, shown in figure.

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