cllqfkd=rkabo=^uf^i=il^a=^ka=jljbkq=ER. KlingnerF=A reinfoced concrete footing needs to bedesigned to support column A3 in theindicated drawing. According to thepreliminary structural analysis that wasconducted, the column is subjected to bothaxial loads as well as moments. Forreasons pertaining to ductwork, plumbingand other utilites that run under thebuilding, the length (L) of the footingneeds to be restricted to the dimensionspecified. It would be fair to assume afooting load approximately 10% of thefactored axial load transfered from thecolumn.

L 10.75ft≡γconc 150

lbf

ft3:= ASP 5

kip

ft2:=

b 30in:=

f'c 4.ksi:= fy 60ksi:= h 30in:= cover 3in:=

PDini 300kip:= PD PDini 1.2⋅:= PD 360 kip=

PLini 80kip:= PL PLini 1.6⋅:= PL 128 kip=

Pu PD PL+:= Pu 488 kip=

MD1ini 0k':= MD1 MD1ini 1.2⋅:= MD1 0 k'=

MD2ini 25k':= MD2 MD2ini 1.2⋅:= MD2 30 k'=

ML1ini 600k':= ML1 ML1ini 1.6⋅:= ML1 960 k'=

ML2ini 100k':= ML2 ML2ini 1.6⋅:= ML2 160 k'=

Mu1 MD1 ML1+:= Mu1 960 k'=

Mu2 MD2 ML2+:= Mu2 190 k'=

1 Estimate size of footing (ASP usually governs):

qservPDini PLini+( )

Agqft_ini+

MD1ini ML1ini+( ) C1⋅ I1

+MD2ini ML2ini+( ) C2⋅

I2+=

Assuming a value for B: Here we can try a series of values until the ASP condition is met..Try:

B 14ft:=

I1L B3

⋅

12:= I1 2458.17 ft4= I2

B L3⋅

12:= I2 1449.35 ft4= Ag L B⋅:= Ag 150.5 ft2=

C1B2

:= C1 7 ft= C2L2

:= C2 5.38 ft=

Given this initial stress applied we candeduce that this 10% stress caused by thefooting would correspond to a depth(including cover) of about..

qft_iniPDini PLini+( ) 0.1( )⋅

Ag:= qft_ini 0.2525

kip

ft2=

hini 20.2 in=

qservPDini PLini+( )

Agqft_ini+

MD1ini ML1ini+( ) C1⋅ I1

+MD2ini ML2ini+( ) C2⋅

I2+

:=

qserv 4.9496kip

ft2=ASPcondition "OK"=

Verify that soil is in compression throughout the area of the footing:

qminPDini PLini+( )

Agqft_ini+

MD1ini ML1ini+( ) C1⋅ I1

MD2ini ML2ini+( ) C2⋅ I2

+

−:=

qmin_condition "OK"= qmin 0.6052kip

ft2=

Note: Punching shear depends on axial loads only2 Assume that thickness is controlled by punching shear:

We will solve by trial and error:a) Assume a depth db) Calculate β0c) Verify that ΦVc>Vn required Note: Vc>2Vn does not apply to slabs and footings (11.5.6.1(a))d) revise d and re-iterateIt is generally suggested that a good starting point for the depth of a footing to use a rule of thumbmeasurement of 1.5*(Column width or diameter). However, with a relatively low ASP, we maywant to minimize the load we shall apply to the soil. We can initially address the condition ofVu<ΦVc and iterate the process with a series of values for d. Another process that we can followis the already given condition above where we established an initial q value for the footing. Thatcan yield a maximum acceptable value. See formula below. After iteration we assume a minimumacceptable value for knowing that we can not surpass the value given as d_max.

hmaxASP qserv qft_ini−( )−

γconc:= hmax 24.23 in= dmax hmax cover−:= dmax 21.23 in=

Try: d 14in:=qft γconc d⋅:= qft 0.18

kip

ft2=

qnetPuAg

:= qnet 3.24kip

ft2=

β0 2 2d h+ b+( ):= β0 14.67 ft=

Aext_col b d+( )2:= Aext_col 13.44 ft2=

Vu2way qnet Ag Aext_col−( )⋅:= Vu2way 444.41 kip=

Given the following: Φ .75:= βcbh

:= βc 1= αs 40:=

ΦVc1 Φ 24βc

+

⋅ f'c⋅ β0⋅ d⋅= ΦVc1 701.27 kip=

ΦVc2 Φαs d⋅

β02+

⋅ f'c⋅ β0 d⋅( )⋅= ΦVc2 605.64 kip=

ΦVc3 Φ 4( )⋅ f'c⋅ β0⋅ d⋅= ΦVc3 467.51 kip=

ΦVc2way 467.51 kip=

Punch_Shearcondition "Approved"=

3 Verify thickness chosen is good for 1-way shear:

We check for the moments applied on each direction separately:a) First on the major axis,b) Then on the secondary axis.

Vu1wayq' qmax+( )

2

L⋅B h−( )

2d−

⋅=We start with the main axis:

q'PD PL+( )

B L⋅

Mu1 dh2

+

⋅

I1+:= q' 4.19

kip

ft2= qmax

PD PL+( )B L⋅

Mu1 C1⋅

I1+:= qmax 5.98

kip

ft2=

Vu1wayq' qmax+( )

2

L⋅B h−( )

2d−

⋅:= Vu1way 250.36 kip=

If higher than Vu, it is approved.ΦVc1way Φ 2⋅ f'c L⋅ d⋅= ΦVc1way 171.332 kip=

OneWayShearcondition "Not Approved"=

Assuming a new value for "d".......Try: d 19in:=

q'PD PL+( )

B L⋅

Mu1 dh2

+

⋅

I1+:= q' 4.35

kip

ft2= qmax

PD PL+( )B L⋅

Mu1 C1⋅

I1+:= qmax 5.98

kip

ft2=

Vu1wayq' qmax+( )

2

L⋅B h−( )

2d−

⋅:= Vu1way 231.244 kip=

ΦVc1way Φ 2⋅ f'c L⋅ d⋅= If higher than Vu, it is approved.

ΦVc1way 232.522 kip=

OneWayShearcondition "Approved"=

Now we check the behavior about the secondary axis:

q'PD PL+( )

B L⋅

Mu2 dh2

+

⋅

I2+:= q' 3.61

kip

ft2= qmax

PD PL+( )B L⋅

Mu2 C2⋅

I2+:= qmax 3.95

kip

ft2=

Vu1wayq' qmax+( )

2

B⋅L h−( )

2d−

⋅:= Vu1way 134.525 kip=

If higher than Vu, it is approved.ΦVc1way Φ 2⋅

f'cpsi

1ksi( )⋅ B⋅d

1000⋅:= ΦVc1way 302.82 kip=

OneWayShearcondition if Vu1way ΦVc1way< "Approved", "Not Approved", ( ):=

OneWayShearcondition "Approved"=

4 Check Flexure - in each direction:

Critical section occurs at column face. In each direction, only axial load and moment in thatdirection will contribute.

Starting with the main axis: Φ .9:=

q''PD PL+( )

B L⋅

Mu1h2

⋅

I1+:= q'' 3.73

kip

ft2= qmax

PD PL+( )B L⋅

Mu1 C1⋅

I1+:= qmax 5.98

kip

ft2=

distance alongfootingmomentarm

Mu L q''

B h−

2

2

2⋅

qmax q''−( )2

B h−

2

3⋅

B h−

2

2⋅+

⋅:= Mu 929.027 k'=

As_ini15Mu12fy d

:= As_ini 12.22 in2=

To be more precise and follow the ACI code we can apply the following formula for slabs as thefooting is essentially a slab. Please note that the value "b" refers to the geometric base of theelement we are analyzing, thus "b" is the length L in this case:

Φ Mn⋅ Φ ω⋅ b⋅ d2⋅ f'c⋅ 1 0.59 ω⋅−( )⋅=

Given that ΦMn Mu≥ ΦMnMu0.9

≥MuΦ

1032.25 k'=

We can assume the minimum value for Mn

OR MnMuΦ

:= Mn 1032.25 k'= Mn 12387.03 k''=

We end up with a quadratic equation If we solve for ω.The value "b" is changed to "L" from here on:

ω 0.59ω2−

Mn

L d2⋅ f'c⋅

=Mn

L d2⋅ f'c⋅

0.0665=

ω 0.59ω2− 0.0665= solve ω,

1.6255788036424443171

0.069336450594843818446

→ OR

ω5

59 L f'c⋅⋅

10 L⋅ f'c⋅ d⋅ 2 25 L2⋅ f'c

2⋅ d2

⋅ 59 L⋅ f'c⋅ Mn⋅−⋅−

d⋅:= ω 0.069=

Continuing with the process we apply the following formula to determine the area of steelneeded. Please note that the term "b" in the formula, again should replaced by our geometricbase of "L". not the "b" of the column side.

As ωf'cfy

⋅ b⋅ d⋅= Replacing the "b" to "L": As ω

f'cfy

⋅ L⋅ d⋅:= As 11.329 in2

=

ACI 10.5.3 & 7.12.2.2 determine maximum spacing. 18" is the max we can use. Ten #9 bars willnot quite do it. We can use ten #10 bars at 12" O.C.

Now let's determine the total depth "t" of the footing cover 3in:= db 1.128in:=

t rounddin

db2in

+cover

in+

in:= t 23 in=

Thus we can precisely determine the actual "d" values for the long and the short directions

dlong t cover−db2

−:= dlong 19.44 in= dshort t cover− 1.5 db⋅−:= dshort 18.31 in=

Now we can solve for the secondary axis:

q'PD PL+( )

B L⋅

Mu2h2

⋅

I2+:= q' 3.41

kip

ft2= qmax

PD PL+( )B L⋅

Mu2 C2⋅

I2+:= qmax 3.95

kip

ft2=

distance alongfootingmomentarm

Mu B q'

L h−

2

2

2⋅

qmax q'−( )2

L h−

2

3⋅

L h−

2

2⋅+

⋅:= Mu 448.673 k'=

MnMuΦ

:= Mn 498.53 k'= OR Mn 5982.31 k''=

We end up with a quadratic equation If we solve for ω.The value "b" is changed to "B" from here on:

ω 0.59ω2−

Mn

B dshort2

⋅ f'c⋅=

Mn

B dshort2

⋅ f'c⋅0.026559=

ω5

59 B f'c⋅⋅

10 B⋅ f'c⋅ dshort⋅ 2 25 B2⋅ f'c

2⋅ dshort

2⋅ 59 B⋅ f'c⋅ Mn⋅−⋅−

d⋅:=

ω 0.02601=

As ωf'cfy

⋅ B⋅ dshort⋅:= As 5.333 in2

=

This value however is very low. According to ACI 10.5.3, As_min=0.0018*b*t. Therefore:

As 0.0018 B⋅ t⋅:= As 6.955 in2=

For minor axis of a rectangular beam we have to establish a band within which most steel shouldbe placed. According to ACI 15.4.4.2. a band width equal to the span of the short directionreceives a certain minimum percentage of the reinforcement. The formula is the following

Reinforcement_in_Band_Width 2Total_Reinforcement_in_short_direction( )

β 1+⋅=

that is considering that "L" is the length of the long direction and "B" the span of theshort, but in the case the terms are reversed we can set a default formula with the terms"min" and "max" applied so that the process is automated to give a value beta equal orhigher than 1.

βLB

=

βmax L B, ( )min B L, ( )

:= β 1.302=

Therefore, within a band as wide as the short side of the footing the area of steel should be asfollows: As_central_band

2 As⋅

β 1+:= As_central_band 6.042 in2

=

As_rest As As_central_band−:= As_rest 0.913 in2=

5 Verify soil pressure maximum is not exceeded:

w t L⋅ B⋅ γconc⋅:= w 43.269 kip=

qsoilPDini PLini+ w+( )

Ag

MD1ini ML1ini+( ) C1⋅ I1

+MD2ini ML2ini+( ) C2⋅

I2+:= qsoil 4.985

kip

ft2=

ASPcondition if qsoil ASP< "OK", "Redesign", ( ):= ASPcondition "OK"=