﻿ Straped footing design CE-REF.COM

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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 assumptions

1. Strap does not provide bearing.
2. Strap is ridge enough to transfer moment from one footing to the other.

## Design procedure

1. Determine the length of exterior footing and its eccentricity, e.
2. Determine eccentric moment, M = Pa*e.
3. Determine shear force, V = M / L
4. Determine footing reaction, Ra = Pa+V, and Rb = Pb-V
5. Determine footing sizes for both A & B.

Structural analysis

1. Calculate factored column loads, Pua & Pub
2. Calculate factored eccentric moment Mu = Pua*e
3. Calculate factored shear, Vu = Mu / L
4. Determine factored reactions, Rua & Rub.
5. Perform structural analysis, determine factored shear and moment on footings and strap.

Reinforced concrete design

1. Design exterior footing.  Check shear stresses and design flexural reinforcement.
3. Design footing strap as a reinforced concrete beam.

### Design procedure:

1. Determine the length of exterior footing and its eccentricity, e.
2. Determine eccentric moment, M = Pa*e.
3. Determine shear force, V = M / L
4. Determine footing reaction, Ra = Pa+V, and Rb = Pb-V
5. Determine footing sizes for both A & B.

### Example 3.4. Determine sizes of strap footing

Given:

• Column information:

• Distance between two columns: 22 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 footing A & B.

Solution: Assume a footing width of 6 ft, the eccentricity of footing A is e = 6/2-1=2’.

The distance between footing reaction, L = 22-2=20’

The eccentric moment is M = (40+50)*2=180 ft-kips

The shear produced by M is, V = 180/20=9 kips

Reaction at footing A = 40+50+9 =99 kips

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

Required footing area of A = 99/2.58=38.4 ft2.

Use 6’ by 6.5’ footing, area = 39 ft2.

Reaction at footing B = 100+80-9=171 kips

Required footing area = 171/2.58=66.3 ft2.

Use 8’ by 8.5’ footing, A = 68 ft2. ## Structural analysis of strap footing

### Procedures

1. Calculate factored column loads, Pua & Pub
2. Calculate factored eccentric moment Mu = Pua*e
3. Calculate factored shear, Vu = Mu / L
4. Determine factored reactions, Rua & Rub.
5. Determine factored footing reactions.
6. Perform structural analysis; determine factored shear and moment on footings and strap.

### Example 3.5:Determine moment and shear in a strap footing

Given: The strap footing in example 3.4

Design code: ACI 318-05

Requirement: Determine maximum factored shears and moment in the footings and strap.

Solution:

Factored column load of A = 1.2*50+1.6*40=124 kips

Factored column load of B = 1.2*100+1.6*80=248 kips

Factored eccentric moment, Mu = 124*2=248 ft-kips

Factored shear, Vu = 248/20=12.4 kips

Factored footing reaction at A = 124+12.4= 136.4 kips

Factored footing pressure per linear foot of A = 136.4/6=22.7 k/ft

Factored footing reaction at B = 248-12.4= 235.6 kips

Factored footing pressure per linear foot at B = 235.6/8=29.5 k/ft.

Shear diagram:

At point 1: Vu = 22.7*1.5-124= -90 kips

At point 2: Vu = 22.7*6-124= 12.2 kips

At point 3: Vu = 22.7*6-124=12.2 kips

At point 4: Vu = 12.4+29.5*3.5= 115.7 kips

At point 5: Vu = 29.5*-3.5=-103.5 kips

Moment diagram:

At point 1: Mu = 22.7*1.52/2-124*0.5= -36.5 ft-kips

At point 2: Mu = 22.7*62/2-124*5= -211.4 ft-kips

At point 3: Mu = 22.7*6*(6/2+13)-124*(5+13)=-52.8 ft-kips

At point 4: Mu = 22.7*6*(6/2+13+3.5)-124*(5+13+3.5)+29.5*3.52/2=183 ft-kips

At point 5: Mu = 29.5*3.52/2= 180.7 ft-kips ## Reinforced concrete design of strap footing

Design procedure:

1. Design exterior footing.  Check shear stresses and design flexural reinforcement.
3. Design footing strap as a reinforced concrete beam.

### Example 3.6 Reinforced concrete design of a strap footing

Given:

• A strap footing with loading, shear, and moment as shown in example 3.4 & 3.5

• Compressive strength of concrete for footing at 28 days: 3000 psi

• Yield strength of rebars: 60 ksi

Design code: ACI 318-05

Requirement: design footing depth and flexural reinforcements.

Solution:

1. Design footing strap

Assume a 2’-6” by 2’ footing strap and the reinforcement is #8 bars, with 2” top cover the effective depth, d = 24-2-1=21”

a. Check direct shear

From Example 3.5, the factored shear force on footing strap, Vu = 12.4 kips

Factored shear strength of concrete,

f Vc = f vc*b*d = (0.75 x 2 x 3000)*30*21/1000=51.7 kips

Minimum shear strength of concrete without shear reinforcement.

1/2f Vc =0.5*51.7=25.8 kips  > 12.4 kips no shear reinforcement is required

b. Design flexural reinforcement

From Example 3.5, the maximum factored moment at point 2, Mu =211.4 ft-kips

Use trail method for reinforcement design

Assume a = 1.8 ".

T = Mu/f(d-a/2) = (211.4)(12)/[(0.9)(21-1.8/2)]=140.2 kip

Calculate new a,

a = T/0.85fc'b =  140.2/[(0.85)(3)(30)] = 1.83 in  » 1.8” assumed

As = T/fy = 140.2/60 = 2.33 in2.

The reinforcement ratio is

r = As/bd = 2.33/[(21)(30)] = 0.0037

Minimum reinforcement ratio,

rmin = 0.0033

Use 4#7 bars, As = 0.6*4= 2.4 in2.

#### 2. Design footing for column A

a. Check punching shear

Assume a 16” depth of footing and #6 bars, the effective depth

d = 16" - 3" (bottom cover) – 0.75 (one bar size) = 12.3 " = 1.02’

Factored footing pressure = 22.7/6.5=3.49 kips/ft2.

The perimeter of punching shear is at one half effective depth from face of column.

Since the distance from exterior face of column to exterior face of footing, 6", is less than effective depth, d,

the perimeter of punching shear is only 3 sides.  At the interior face of column the length is column width, 12" plus d.

At the other two faces is column width + d + 6".

P = 2*(6”+12”+12.3”/2)+(12”+12.3”)=72.6”

The punch shear stress can be calculated as

vu = [124-(3.49)(1+1.02)(0.5+1+1.02/2](1000)/[(12.3)(72.6)]= 123 psi

The shear strength of concrete is

f vc = 0.75 x 4 x Ö3000 = 164.3 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 3.4, the maximum direction shear is –90 kips at inside face of column.

The factored shear at one effective depth, d, from the face of the column is

Vu = 90*(4.5-1.02)/4.5=69.6 kips

Factored shear strength of concrete,

f Vc = f vc*b*d = (0.75 x 2 x Ö3000)*6.5*12*12.3/1000 = 78.8 kips > 69.6 kips O.K.

c. Determine maximum negative reinforcement in longitudinal direction

The location of zero shear is at

X = 4.5*90/(90+12.2) = 3.96’ from inside face of the column

The maximum negative moment is at zero shear, at 3.96’ from inside face of column, or 5.46’ from exterior end of footing.

Mu = 22.7*5.462/2-124*(0.5+3.96)=-214.7 ft-kips

Use trail method for reinforcement design

Assume a = 1.3".

T = Mu/f(d-a/2) = (214.7)(12)/[(0.9)(12.3-1.3/2)]= 245.7 kip

Calculate new a,

a = T/0.85fc'b =  245.7/[(0.85)(3)(6.5*12)] = 1.24 in  » 1.3” assumed

As = T/fy = 245.7/60 = 4.1 in2.

The reinforcement ratio is

r = As/bd = 4.1/[(78)(12.3)] = 0.0043

Minimum reinforcement ratio,

rmin = 0.0033

Use 8-#7 bars, 4#7 extended from footing strap, 2 #7 in each side of footing,

As = 0.6*8=4.8 in2.  Place reinforcement at top face of footing.

d. Determine reinforcement in transverse direction

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

l = (6.5– 1)/2 =2.75'

The factored moment at the face of the column is

Mu = (3.49)(2.75)2/2 = 13.2 k-ft.        per foot width of footing

Use trail method for reinforcement design

Assume a = 0.5".

T = Mu/f(d-a/2) = (14.7)(12)/[(0.9)(12.3-0.5/2)]= 16.3 kip

Calculate new a,

a = T/0.85fc'b =  16.3/[(0.85)(3)(12)] = 0.53 in  » 0.5” assumed

As = T/fy = 16.3/60 = 0.27 in2.  per one foot section.

The reinforcement ratio is

r = As/bd = 0.27/[(12)(12.3)] = 0.0018

Minimum reinforcement ratio,

rmin = 0.0033   > rmin =(4/3)*0.0018=0.0024

Use rmin =0.0024

As = (0.0024 )(6)(12)(12.3) = 2.1 in2.

Use 5 #6 bars, As = 0.44*5= 2.2 in2.

Place reinforcement at bottom face of footing.

#### 2. Design footing for column B

Assume a footing depth of 20” and #8 bars, the effective depth =20-3-1=16”

The factored footing pressure = 29.5/8.5= 3.47 ksf

a. Check punching shear

The perimeter of punching shear is at d/2 beyond faces of column

P = 4*(12+16)=112”

The punch shear stress can be calculated as

vu = [248-(3.47)(1+1.33)2](1000)/(16)(112) = 127.9 psi    <186 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 3.4, the maximum direct shear is 115.7 kips at inside face of column.

The factored shear at one effective depth from the face of the column is

Vu = 12.2+(115.7-12.2)*(3.5-1.33)/3.5= 76.3 kips

Factored shear strength of concrete,

f Vc = f vc*b*d = (0.75 x 2 x Ö3000)*8.5*12*16/1000= 134.1 kips > 85 kips O.K.

c. Determine maximum positive reinforcement in longitudinal direction

Mu = 180.7 ft-kips

Use trail method for reinforcement design

Assume a = 0.6".

T = Mu/f(d-a/2) = (180.7)(12)/[(0.9)(16-0.6/2)]= 153.4 kip

Calculate new a,

a = T/0.85fc'b =  153.4/[(0.85)(3)(8.5*12)] = 0.59 in  » 0.6” assumed

As = T/fy = 153.4/60 = 2.6 in2.

The reinforcement ratio is

r = As/bd = 2.6/[(8.5*12)(16)] = 0.0016

Minimum reinforcement ratio,

rmin = 0.0033  >rmin =(4/3)*0.0016=0.0021

Use rmin =0.0021

As = (0.0021 )(8.5)(12)(16) = 3.5 in2.

Use 6-#7 bars, As = 0.6*6= 3.6 in2.

d. Determine reinforcement in transverse direction

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

l = (8.5– 1)/2 =3.75'

The factored moment at the face of the column is

Mu = (3.47)(3.75)2/2 = 24.4 k-ft.        per foot width of footing

Use trail method for reinforcement design

Assume a = 0.7".

T = Mu/f(d-a/2) = (24.4)(12)/[(0.9)(16-0.7/2)]= 20.8 kip

Calculate new a,

a = T/0.85fc'b =  20.8/[(0.85)(3)(12)] = 0.68 in  » 0.7” assumed

As = T/fy = 20.8/60 = 0.35 in2 per one foot section.

The reinforcement ratio is

r = As/bd = 0.35/[(12)(16)] = 0.0018

Minimum reinforcement ratio,

rmin = 0.0033 > rmin =(4/3)*0.0018=0.0024

Use rmin =0.0024

As = (0.0024)(8)(12)(16) = 3.7 in2.

Use 7 #7 bars, As = 0.6*7= 4.2 in2.

4. 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 