# Brief explanation of wall footing design

Wall footing design is performed on the basis of some theories which must be taken into consideration at the time of designing. The first theory is to examine the soil pressure that is circulated linearly all through design process. This can be obtained, if resultant of soil pressure conforms with the resultant of soil force. Due to this, rotation of footing can be prevented. Usually, the following strength design method is applied while designing the footings.

**WALL FOOTING DESIGN EXAMPLE STATEMENT**

A 10” thick wall bears a service dead load of 8k/ft and service live load of 9k/ft. At the base of footing the permissible soil pressure is 5000psf and base of footing is 5’ underneath the present ground surface. Now your responsibility is to design the wall footing for;

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Concrete compressive strength= f’c = 3ksi

Yield strength of steel = fy = 60ksi.

Soil density = 120lb/ft3.

**SOLUTION:**

Note: Typically low strength of concrete in footing is applied in columns. It necessitates the use of dowel to adapt this modification in strength.

**STEP 1: ESTIMATE THE SIZE OF FOOTING AND FACTORED NET PRESSURE.**

At the time of designing wall footing, generally one feet strip of the wall and footing is undertaken to simplify the calculation process. The permissible soil pressure is 5ksf. The permissible net soil pressure is 5ksf. As, the thickness of footing is unknown at this stage a acceptable thickness of footing equivalent to (1-1.5)x The wall thickness can be guessed.

1.5 x 9’’ = 13.5’’ take 13’’

**STEP 2: FIND THE PERMISSIBLE SOIL PRESSURE**

qn=5−[(1.083times0.150)+(5−1.083)times0.120]

[q_{n}=4.367 ksf]

[Area-of-footing=frac{8+9}{4.3676}]

Try 47’’ (3.91’) Wide footing

Factored Net Pressure = qnet = 6.138ksf

**STEP 3: CHECK THE TOLERABILITY OF FOOTING DEPTH AGAINST SHEAR**

Only one way shear or beam shear is substantial in wall footings. The crucial section for this type of shear is at distance ‘d’ from the face of wall. Where ‘d’ denotes the actual depth of footing.

Supposing that one will use #4 bar.

d = 13 – 3 – ¼ = 9.75″

The Tributary area for one way shear

Tributary area = Shaded area for one way shear

D = H – CLEAR COVER – 1/2 BAR DIAMETER

d = 13 – 3 -1/4

d = 9.75″

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