## INTRODUCTION:-

Grillage foundation is a type of isolated footing that is used when heavy loads are transferred from the column to the foundation. This foundation consists of one, two or more layers of I-beams that are placed at right angles on top of each other and they are all encased in concrete. The column sits on a base plate which transfers the load to the layer of I-beams below it. The top layer of I-beams spread the load to the bottom layer of I-beams which in turn transfer the load safely to the soil. This foundation is characterized by being light weight and economical. The I-beam layers are often called Tier.

**Advantages of Grillage Foundation:**

1-This foundation is suitable for low bearing capacity soils since it allows for better distribution of point loads due to the arrangement of the I-beam layers.

2-When heavy columns are placed on the foundation it becomes very difficult to design the foundation with ordinary reinforcement bars.

** Grillage foundation **

**Design example:**

A concrete encased steel column carries 10000KN, the allowable bearing capacity of the soil is 2000Kpa. The base plate is 60cm x 60cm, Find the steel section for the beams if it is assumed that the foundation consists two layers of I-beams, 3 beams at the top and 5 beams at the bottom (the section is show above in Fig 1).

Solution:

1-Find the size of the footing.

Area=P/Qall=10000/2000=5m

the length of the beams will be 5m in both layers and the cover used is 100mm. Hence the dimension of the footing is 5+2*0.1=5.2m

2-The loads acting on the top beam must be found.

The total load coming from the column is 10000KN. since there are three beams, the load on each column is 10000/3=3333.33KN

The length of the base plate is 60cm, therefore the distributed load on each top beam is

3333.33/0.6=5555.55KN/m

Since the load is transferred from the top layers to the bottom layers. The bottom layers generate upward reaction against the top beams.

The reaction generated by the bottom beams=3333.33/5=666.66KN

Top beam

Now the maximum bending moment must be calculated in order to find the required steel section. The maximum bending moment will be located in the center of the beam.

M=666.66*2.5+666.66*1.25-5555.55*0.3*0.15

M=2249.97 KN.m

0.3m is half of the width of the base plate

Now a section must be selected that can resist the internal bending moment. This can be done by calculating the required section modulus and the choosing a section that has an equal or greater section modulus.

M=maximum bending moment.

Z=section modulus.

σ=allowable stress in the steel.

Z=M/σ

σ=250000 Kpa

Zrequired=2249.97/250000=0.00899988^{}m^{3}=8999.88 cm^{3}

The the section chosen has a Z=9962 cm^{3}

Since it's greater than the required section modulus then we can use that section for our top beams design which can be seen below.

Top beams section

2-The loads acting on the bottom beams must be found.

Let's assume that the spacing between the top beams is 60cm.

The load that is transferred from each top beam to bottom beam is 666.66KN

The total load one a single bottom beam=666.66*3=2000KN

The reaction generated by the soil below=2000/5=400KN/m

Similarly the maximum bending moment is in the middle of the beam, which mean that we will have to take a section.

M=400*2.5*1.25-666.66*0.6=850KN.m

Zrequired=850/250000=0.0034m^{3}=3400 cm^{3}

The the section chosen has a Z=3538 cm^{3}

Since it's greater than the required section modulus then we can use that section for our bottom beams design which can be seen below.

**Bottom beams section**

## REFERENCE:-

Ruwan Rajapakse.Geotechnical Engineering Calculations and Rules of Thumb.

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### Writer Profile

Mohammad Barzan

BSc- Geotechnical Engineering- Koya University.