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Combined footings

Contents:

Introduction

Combined footings and strap footings are normal used when one of columns is subjected to large eccentric loadings.  When two columns are reasonably close, a combined footing is designed for both columns as shown in Figure 3.1.  When two columns are far apart, a strap is designed to transfer eccentric moment between two columns as shown in Figure 3.1.  The goal is to have uniform bearing pressure and to minimize differential settlement between columns.

 

Figure 3.1 Combined footing and strap footing

Design procedure:

Service load design:

  1. Determine the size of combined footing. 

Structural analysis:

  1. Perform structural analysis to determine moment and shear in various section of  the footing. 

Reinforced concrete design:

  1. Check punching shear & direct shear
  2. Design longitudinal reinforcements. 
  3. Design transverse reinforcements. 
  4. Design column dowels.

Service load design: determine size of footing

The size of the footing shall be determined to have uniform bearing pressure under the footing so that differential settlement is minimized. The resultant of bearing pressures needs to coincide with the resultant of column loads.  The procedures are as follows:

  1. Determine the location of the resultant of column loads.
  2. Calculated the required length of footing. The length of the footing is twice the distance from the edge footing of the exterior column to the resultant of column loads. 
  3. Calculate the width of footing.  The required area of footing is the total column load divided by allowable net soil bearing pressure. The width of footing is the required footing area divided by the length of footing.

Example 3.1: Determine size of a combined footing

Given:

  • Column information:

  • Column A: Live load = 40 kips, Dead load = 50 kips

  • Column B: Live load = 80 kips, Dead load = 100 kips.

  • Distance between two columns: 15 ft.

  • Footing information:

  • Allowable soil bearing capacity; 3000 psf

  • Distance from column A to edge of footing: 1 ft.

  • Allowable soil bearing capacity = 3000 psf

  • Weight of soil above footing = 120 psf

  • Depth of footing= 24”

  • Depth of soil above footing = 12”

Requirements: Determine the size of a combined footing.

Solution:

Total column load of A = 40+50=90 kips

Total column load of B = 80+100 = 180 kips

Take moment about A,

Location of resultant from A= 180*15/(90+180) = 10 ft.

The length of footing = 2*(10+1) = 22 ft                                   Use 22 ft

Net soil bearing capacity = 3000-2*150-120=2580 psf

Required footing area = (90+180)/2.58=104.7 ft2.

Required width of footing = 104.7/22=4.8’                               Use 5 ft

 

Structural analysis:

Structural analysis of a combined footing is the same as analyzing an invert simply support beam supported by two columns with factored soil pressure as loading.  The procedures are as follows:

  1. Calculate factored footing pressure.
  2. Calculate maximum shear at  an effective depth from the face of column
  3. Calculate maximum positive and negative moment in the footing.  Maximum positive moment occurs at face of column.  Maximum negative moment occurs between two columns at zero-shear.

It is worth to mention that because of load factors, the centroid of factored column loads does not necessary located at the center of the footing.  It means that the factored footing pressure is no longer uniform.  The correct way to solve the problem to analyze the footing with trapezoid shape of factored footing pressure.

Example 3.2. Determine maximum shear and moment of a combined footing

Given:

  • A combined footing as shown in Example 3.1

  • Column size: 1 ft by 1ft for both A & B

Design code: ACI 318-05

Requirements: Determine maximum shear and moment in longitudinal direction.

Solution:

Factored column loads:

Column A: Pua = 1.2*50+1.6*40=124 kips

Column B: Pub = 1.2*100+1.6*80=248 kips

Location of resultant from column A= 248*15/(124+248)=10 ft.

Since the location of resultant is at center of footing, factored footing pressure is uniform.

Factored footing pressure per linear foot of footing, Qu = (124+248)/22=16.9 k/ft

Shear diagram:

At point 1: Vu = 16.9*1.5-124= -98.7 kips

At point 3: Vu = 16.9*(1.5+14)-124=138 kips

At point 4: Vu = 16.9*(1.5+14+1)-124-248= -93.2 kips

Moment diagram:

At point 1: Mu = 16.9*1.52/2-124*0.5= -43 ft-kips

At point 2:

Location of point 2:  from triangular relation between point 1 and point 3 in shear diagram

X = 14*98.7/(98.7+122.9)=6.24’                from inside face of column A

Mu = 16.9*(1.5+6.24)2/2-124*(0.5+6.24)=-329.5 ft-kips

At point 3: Mu = 16.9*(1.5+14)2/2-124*(0.5+14)=232.1 ft-kips

At point 4: Mu = 16.9*5.52/2=255.6 ft-kips

 

Reinforced concrete design:

Design procedure:

  1. Check both punching shear and direct shear.  The critical section of punching shear is at ½ effective depth from face of column.  The critical section of direct shear is at one effective depth of column.  For column at the edge of footing the critical section of punching shear only has three sides along the column.  Critical sections of punching shear and direct shear are shown below.

  1. Design longitudinal reinforcements.  Longitudinal reinforcements are design based on the maximum moments from structural analysis.  Reinforcement for negative moment should be placed near top face of the footing.  Positive reinforcement should be placed near bottom face of the footing.
  2. Design transverse reinforcements.  Transverse reinforcements are designed based on moment in the transverse direction at face of column.  They should be placed near bottom face of the footing.
  3. Design column dowels.

Example 3.3: Reinforced concrete design of a combined footing

Given:

  • A combined footing with loading, shear, and moment as shown in example 3.1 & 3.2

  • Compressive strength of concrete for footing at 28 days: 4000 psi

  • Yield strength of rebar: 60 ksi

Design code; ACI 318-05

Requirement: Check shear stresses and design flexural reinforcements.

Solution:

a. Check punching shear for column A

Assume the reinforcements are #6 bars, the effective depth

d = 24" - 3" (cover) - 0.75" (one bar size) = 20.3 " = 1.7'

Factored footing pressure = (124+248)/(22*5)=3.38 kips/ft2.

The perimeter of punching shear is

P = 2*(6”+12”+20.3”/2)+(12”+20.3”)=88.6”

The punch shear stress can be calculated as

vu = [124-(3.38)(1+1.7)(0.5+1+1.7/2)](1000)/(20.3*88.6) =57 psi

The shear strength of concrete is

f vc = 0.75 x 4 x 4000 = 189.7 psi                                                                              O.K.

Check punching shear for column B

The perimeter of punching shear is

P = 4*(12+20.3)=129.2”

The punch shear stress can be calculated as

vu = [248-(3.38)(1+1.7)2](1000)/(20.3*129.2) = 85.2 psi     < 189.7 psi    O.K.

b. Check direct shear:

The critical section of direct shear is at one effective depth from the face of column.  From Example 3.2, the maximum direction shear is 138 kips at inside face of column B. The distance from zero shear (point 2) to the maximum direct shear (point 3) is 14-6.24= 7.76’. From triangular relationship, the direct shear at critical section is

Vu = 138*(7.76-1.2)/7.76= 116.7 kips

The shear strength of concrete for footing section,

f Vc = f vc*b*d = (0.75 x 2 x 4000)*60*20.3/1000=115.6 kips »  116.7 kips     (less than 1% difference) O.K.

c. Determine Maximum positive reinforcement in longitudinal direction

The maximum positive moment at the face of the column B is

Mu = 255.6 k-ft.          for 5’ width of footing

Use trail method for reinforcement design

Assume depth of stress block, a = 0.9". 

T = Mu/[f(d-a/2)] = [(255.6)(12)]/[(0.9)(20.4-0.9/2)] = 170.8 kips

Calculate new a,

a = T/[(0.85)(fc')(b)] = 170.8/[(0.85)(4)(60)] = 0.84    » 0.9”

As = T/fy = 170.8 / 60 = 2.84 in2        .

The reinforcement ratio is

r = As/bd = 2.84/(60)(20.4) = 0.0023

Minimum reinforcement ratio,

r min= 0.0033  > rmin =( (4/3)*0.0023=0.0031

Use rmin = 0.0031,

As = (0.0031 )(60)(20.4) = 3.8 in2.

Use 9-#6 bars in both directions, As = 3.96 in2. 

d. Maximum negative reinforcement in longitudinal direction

The maximum negative moment between column is

Mu = 329.5 k-ft.          for 5’ width of footing

Use trail method for reinforcement design

Assume a = 1.2". 

T = (329.5)(12)/[(0.9)(20.4-1.2/2)] = 221.8 kip

Calculate new a,

a = 221.8/[(0.85)(4)(60)] = 1.1"    » 1.2”

As = 221.8 /60 = 3.7 in2          .

The reinforcement ratio is

r = As/bd = 3.7/(60)(20.4) = 0.0031

Minimum reinforcement ratio,

r min= 0.0033  < rmin =( (4/3)*0.0031=0.0041

Use r = 0.0033,

As = (0.0033 )(60)(20.4) = 4.0 in2.

Use 10-#6 bars in both directions, As = 4.4 in2. 

e. Determine reinforcement in transverse direction

The distance from face of column to the edge of the footing is

l = (5– 1)/2 =2'

The factored moment at the face of the column is

Mu = (3.38)(2)2/2 = 6.76 k-ft. per foot width of footing

Use trail method for reinforcement design

Assume a = 0.1". 

T = (6.76)(12)/[(0.9)(20.4-0.1/2)] = 4.4 kip

Calculate new a,

a = 4.4/[(0.85)(4)(12)] = 0.11 » 0.1” assumed

As = 4.4/60 = 0.073  for one foot section.

The reinforcement ratio is

r = As/bd = 0.073/(12)(20.4) = 0.0003

Minimum reinforcement ratio,

r min= 0.0033  < rmin =( (4/3)*0.0003=0.0004

Use rmin =0.0004

As = (0.0004 )(22)(12)(20.4) = 2.2 in2.

Use 16 #4 bars, As = 0.2*15= 3 in2.

Maximum spacing = (22*12-3-3)/15= 17.2”                <Maximum spacing, 18” O.K.

e. Designing column dowels. 

The bearing capacity of concrete at column base is

Pc = (0.7)(0.85)(4)(12)(12) = 342.7 kips

Which is greater than factored column loads of both A and B.

The minimum dowel area is

As,min = (0.0005)(12)(12) = 0.72 in2

Use 4 - #4 dowels As = 0.8 in2

The footing is shown in below