# Designing a Strap Foundation to Eurocode

The strap foundation belongs to the family of spread foundation. It is very useful where the base for an exterior column must not project beyond the property line of a building. This often requires the exterior base to be placed eccentrically under its column so that it doesn’t project beyond the property line. This kind of arrangement (figure 1) will lead to a strong uneven pressure distribution which could lead to the tilting of the base to one side. To counteract this effect a strap beam is used to connect the exterior footing to a nearby interior footing. Thus the duty of the strap beam is to restrain the exterior footing from overturning due to the high eccentricity induced.

In designing a strap foundation, both footings are proportioned in such a way that the bearing pressure under the footings is equal. To achieve this, it is necessary that the centroid of the combined area coincides with the resultant of the column loads as illustrated in figure 2.

### Procedure for Design

The process of designing a strap foundation is often an interactive process. Several trial designs may be necessary to achieve economical sizes for the footing.

#### Proportioning the Footings

1. Choose a trial B for the rectangular exterior footing estimate the weight of the footings W1 & W2 for the footing Ws for the strap beam

2. Determine the areas of the footing required to spread the load without exceeding the permissible bearing pressure using expression 1

{ A }_{ T }=\frac { { N }_{ 1 }+{ N }_{ 2 }+{ W }_{ 1 }+{ W }_{ 2 }+{ W }_{ s } }{ bearing\quad resistance } ——-(1)
where\quad { A }_{ T }=BD+{ S }^{ 2 }———–(2)

3. Determine the value of x by taking a moment of the loads about the centerline of the interior column

x=\frac { { N }_{ 1 }L+{ W }_{ 1 }(L-0.5D)+{ 0.5W }_{ s }L }{ { N }_{ 1 }+{ N }_{ 2 }+{ W }_{ 1 }+{ W }_{ 2 }+{ W }_{ s } } —–(3)

4. The centre of gravity must coincide with the resultant of the applied loads. Therefore take moments of the footing areas about the centreline of the interior column.

x=\frac { BD\left( L+f-0.5D \right) }{ BL+{ S }^{ 2 } } ———-(4)

5. Equate expression 3 & 4, solve the ensuing quadratic equation to determine D and calculate S from equation 2.

#### Designing the Strap Beam & Bases

5. Using the ultimate limit state, determine the bearing pressure using expression 5.

{ p }_{ u }=\frac { { N }_{ Ed,1 }+{ N }_{ Ed.2 } }{ BD+{ S }^{ 2 } }———–(5)

6. Design the interior footing as a square pad base with bending in both directions.

7. Design the exterior footing as a pad base with bending in one direction supported by the strap beam.

8. The strap beam itself is a rectangular beam and normal procedures observed for beam design applies. The bending moment occurs at the point of zero shears as shown in figure 3 and the shear on the beam is virtually constant. Tension steel is provided at the top of the beam while compression steel is provided based on the minimum area of steel at the bottom. The stirrups should extend into the footings over the support in order to ensure a monolithic construction.

x=\frac { { N }_{ Ed,1 } }{ { p }_{ u }B }
{ M }_{ Ed }={ N }_{ Ed,1 }\left( x-f \right) +\frac { { p }_{ u }B{ x }^{ 2 } }{ 2 }
{ V }_{ Ed }={ N }_{ Ed }-{ p }_{ u }BD

### Worked Example

A strap foundation is required for a proposed office building whose external columns must not project beyond the property line of the structure. The columns are spaced at 6.0m centres. The interior column is 350mm squares and carries a characteristic permanent and variable actions of 900kN & 600kN respectively. The exterior column carries 500kN & 350kN respectively. Design the strap footing completely assuming a presumed bearing resistance of 200kN/m2 using C25/30 concrete with 460mpa steel bars.

##### Proportioning the Footings
{ N }_{ 1 }=1.0{ G }_{ k }+1.0{ Q }_{ k }=1.0(500)+1.0(350)=850kN
{ N }_{ 2 }=1.0{ G }_{ k }+1.0{ Q }_{ k }=1.0(900)+1.0(600)=1500kN

Assume the self-weight of the footing as 10% of the characteristic axial loads from the columns.

{ W }_{ 1 }=0.1\times 850=85kN\quad ;\quad { W }_{ 2 }=0.1\times 1500=150kN

Assuming a strap beam with dimensions of 350x750mm

{ W }_{ 3 }=0.35\times 0.75\times 6.0\times 25=39.375kN
{ A }_{ T }=\frac { 850+1500+85+150+39.375 }{ 200 } =13.12{ m }^{ 2 }

from equation 3 we have:

x=\frac { 850(6)+85(6-0.5D)+0.5(39.375\times 6) }{ 2624.375 }

Assuming B=3.0m and substituting into equation 4 we have :

x=\frac { 3.0D(6+0.15-0.5D) }{ 13.12 } ———(b)

Equating expression (a) and (b) will result in the quadratic equation (c)

1.5{ D }^{ 2 }-18.65{ D }+28.60=0———(c)

Substitute the values of B & D into equation 2 and determine the size of the interior footing as

S=\sqrt { 13.12-BD } =\sqrt { 13.12-3.0(1.8) } =2.78m

Now that we have successfully proportion the base for even pressure distribution, we are going to proceed to the design at ultimate limit state.

#### Ultimate Limit State

{ N }_{ Ed,1 }={ 1.35G }_{ K }+{ 1.5Q }_{ K }=1.35(500)+1.5(350)=1200kN
{ N }_{ Ed,2 }=1.35{ G }_{ K }+1.5{ Q }_{ K }=1.35(900)+(1.5(600)=2115kN
p_{ u }=\frac { { N }_{ Ed,1 }+{ N }_{ Ed,2 } }{ BD+{ S }^{ 2 } } =\frac { 1200+2115 }{ (3.0\times 1.8)+{ 2.8 }^{ 2 } } =250.4kN/{ m }^{ 2 }

#### Exterior Base Design

##### Flexural Design

The base is designed one way and the critical section in bending is taken at the column face.

{ M }_{ Ed }=250.4{ \left( \frac { 3.0-0.3 }{ 2 } \right) }^{ 2 }=456.35kN.m/m
d=600-(50+20/2)=540mm
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 456.34\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times { 540 }^{ 2 }\times 25 } =0.063
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
z=0.94d=0.94\times 540=507.6mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 456.34\times { 10 }^{ 6 } }{ 0.87\times 460\times507.6 } =2246.4{ mm }^{ 2 }/m
##### Shear Verification

The critical section for shear is taken at d from the face of the column in the longer direction

{ V }_{ Ed }=250.4\left( \frac { 3.0-0.3 }{ 2 } -0.54 \right) =202.82kN/m
{ V }_{ Rd,c }=0.12k\left( 100\rho { f }_{ ck } \right) ^{ 1/3 }bd\ge { V }_{ Rd,min }
k=1+\sqrt { \frac { 200 }{ d } } 1+\sqrt { \frac { 200 }{ 540 } } =1.61<2
\rho =\frac { { A }_{ s } }{ bd } =\frac { 2093 }{ { 10 }^{ 3 }\times 540 } =0.00388
{ V }_{ Rd,c }=0.12\times 1.61(100\times 0.00388\times 25)^{ 1/3 }\cdot { 10 }^{ 3 }\times 540=225.4kN
{ V }_{ Rd,min }=0.035\times { 1.61 }^{ 3/2 }\sqrt { 25 } \cdot { 10 }^{ 3 }\times 540=193.75kN
225.4kN>({ V }_{ Ed }=202.82kN)

Therefore since the concrete shear resistance is greater than the applied shear force, no shear reinforcement is required. Punching shear is also not critical due to the presence of the beam.

#### Interior Footing

##### Flexural Design

The base is designed two-ways and similarly the critical section in bending is taken at the column face.

{ M }_{ Ed }=250.4{ \left( \frac { 2.8-0.35 }{ 2 } \right) }^{ 2 }=375.76kN.m/m
d=600-(50+20/2)=540mm
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 375.76\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times { 540 }^{ 2 }\times 25 } =0.051
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
z=0.95d=0.95\times 540=513mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 375.76\times { 10 }^{ 6 } }{ 0.87\times 460\times 513 } =1830.3{ mm }^{ 2 }/m
##### Shear Verification

The critical section in shear is taken at d from the column face in any direction.

{ V }_{ Ed }=250.4\left( \frac { 2.8-0.35 }{ 2 } -0.54 \right) =175.52kN/m
{ V }_{ Rd,c }=0.12k\left( 100\rho { f }_{ ck } \right) ^{ 1/3 }bd\ge { V }_{ Rd,min }
k=1+\sqrt { \frac { 200 }{ d } } 1+\sqrt { \frac { 200 }{ 540 } } =1.61<2
\rho =\frac { { A }_{ s } }{ bd } =\frac { 2093 }{ { 10 }^{ 3 }\times 540 } =0.0039
{ V }_{ Rd,c }=0.12\times 1.61(100\times 0.00339\times 25)^{ 1/3 }\cdot { 10 }^{ 3 }\times 540=222.87kN
{ V }_{ Rd,min }=0.035\times { 1.61 }^{ 3/2 }\sqrt { 25 } \cdot { 10 }^{ 3 }\times 540=193.05kN
222.87kN>({ V }_{ Ed }=175.52kN)

similarly the concrete shear resistance is greater than the applied shear force, no shear reinforcement is required. Punching shear is also not critical due to the presence of the beam.

#### Strap Beam

The critical section in bending and shear for the strap beam is shown in figure 3.

##### Flexural Design

The critical section in bending occurs at the point of zero shears as illustrated in figure 3. The point of zero shear x is given as:

x=\frac { { N }_{ Ed,1 } }{ { p }_{ u }B } =\frac { 1200 }{ 250.4\times 3.0 } =1.60m
{ M }_{ Ed }={ N }_{ Ed,1 }(x-f)-\frac { { P }_{ u }B{ x }^{ 2 } }{ 2 }
=1200(1.6-0.15)-\frac { 250.4\times 3\times { 1.6}^{ 2 } }{ 2 } =778.5kN.m

As the moment is very high, we can increase the beam depth to 900mm Assume two layers of 25mm bars with 10mm links

d=900-(50+25+25/2+10)=802.5mm
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 778.5\times { 10 }^{ 6 } }{ 350\times { 802.5 }^{ 2 }\times 25 } =0.138<0.168
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
z=0.86d=0.86\times 802.5=690.15mm
{ A }_{ S }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac {778.5\times { 10 }^{ 6 } }{ 0.87\times 460\times 690.15 } =2818.6{ mm }^{ 2 }
##### Shear Design
{ V }_{ Ed }={ p }_{ u }BD-{ N }_{ Ed }
=(250.4\times 3.0\times 1.8)-1200=152.16kN=152.16kN

By observation, the shear force on the beam is reasonably low and not critical. Therefore provide minimum area of shear reinforcement

\frac { { A }_{ s,min } }{ { s }_{ v } } =\frac { 0.08\sqrt { { f }_{ ck } } { b }_{ w } }{ { f }_{ yk } } =\frac { 0.08\times \sqrt { 25 } \times 400 }{ 460 } =0.35
{ s }_{ v }=0.75d=0.75\times 802.5=601.875mm

#### Minimum Areas

To conclude, check the minimum areas of reinforcement required in the footings as well as the strap beam.

##### Minimum Areas in Footings
{ A }_{ s,min }=0.26\frac { { f }_{ ctm } }{ { f }_{ yk } } bd\ge 0.0013bd
{ f }_{ ctm }=0.3{ { { { f }_{ ck } }^{ 1/3 } } }=0.3\left( { 25 }^{ 1/3 } \right) =2.56mpa
=\frac { 0.26\times 2.56 }{ 460 } \cdot { 10 }^{ 3 }\times 540\ge 0.0013\cdot { 10 }^{ 3 }\times 540
{ A }_{ s,min }=718.85{ mm }^{ 2 }
##### Minimum Areas in Strap Beam
{ A }_{ s,min }=0.26\frac { { f }_{ ctm } }{ { f }_{ yk } } bd\ge 0.0013bd
{ f }_{ ctm }=0.3{ { { { f }_{ ck } }^{ 1/3 } } }=0.3\left( { 25 }^{ 1/3 } \right) =2.56mpa
=\frac { 0.26\times 2.56 }{ 460 } \times 300\times 802.5\ge 0.0013\times 300\times 802.5
{ A }_{ s,min }=348.35{ mm }^{ 2 }