design of the isolated or split footing for a column

The design of the isolated or split footing of the column is responsible for the fixed compression

The split footing is used to carry a pressing load from a single column. The design of the stand-alone bases mainly includes the thickness of the stone foundation, the size of the foundations, if necessary, the length of the development of the main column reinforcement, and other safety-related tests.

Problem 1: Design a single layer column of 300 mm × 450 mm, carrying a service load of 700 kN. The column reinforcement details are: 8 - 16 mm Ø with equal ties 8mmØ to 200 mm c / c. The safe carrying capacity of the soil is 150kN / m2. Use M20 integration with Fe415 grade steel.

Solution:

Step 1:Determine the size of the firm system

Let the weight of the feet be 10% of the load from the column. Therefore, the total service load on the feet should be = 1.1 × 700 = 770 kN

Required walking distance = 770 150 = 5.13 sq.m

let's choose a size like 2.2 m × 2.35 m. (Guess the base of the foot from the surface of the column on either side should be the same. This is because the distribution of soil pressure upwards evenly on the foot slide and removed with two different times in respect of both directions)

The area provided is 5.17 sq.m> required area 5.13 sq.m

Step 2: Determine the size of the foot lab

The final soil pressure applied to the foot stage is calculated as follows:

Last soil pressure = 1.5 × 770 2.2 × 2.35 = 223.40 kN / m2or 0.223 N / mm2

The size of the slab structure should be removed by three force actions:

a) a moment on the soles of the feet due to the pressure of the soil going upwards

b) shaving in one way

c) pair cutting or punching

 a) (Foot slide size based on seconds)

The temporary critical section is on the column surface. Since the appearance of the slide is similar on both sides, the minute is almost the same with both axes on the surface of the column. Therefore, the minute in the face of the column is approx

= 223.40 × 2.35 × 0.952 / 2 = 236.90 kN-m.

Construction time is Mud = 236.90 kNm.

Therefore, the thickness of the moving slide can be obtained from a known formula e.g.

Mud = 0.36 fck (xumax / d) × (1 -0.42 (xumax / d)) × bd2

236.90 × 106 = 0.36 × 20 × 0.48 × (1 -0.42 × 0.48) × 2350d2

Therefore, d = 191.14 mm

However, in terms of clause 34.1.2 of code IS 456-2000, the size of the moving slide has been 300 mm.

Case b: (The size of the base of the stone based on a single haircut)

In terms of clause 34.2.4.1, for a one-way haircut, the critical step should be at the effective level of the foot excavation from the surface of the column or foot if provided.

The shaving power of the foot shaver is = 0.223 × 2350 × (950 -d) N

Let's take 0.3% slide stabilization percentage

From table 19 of code, for mixing M20 with pt = 0.3%, the shear strength of the design is obtained as 0.384 N / mm2

Stock resistance is = 0.384 × 2350 × d = 902.4d N

Measuring the strength of both

902.4d = 0.223 × 2350 × (950 -d)

resolution d = 349. 01 mm

Therefore, from ‘case b’ the required depth is 350 mm.

Case c: (Size of stone base based on double hair action)

Critical section for performing two haircuts in terms of clause 34.2.4.1

& 31.6.1 to be part of the working size of the footing slab eg d / 2from the face of the support. In this case the thickness of the slide is obtained by measuring the shaving power and designing the shaving power by two cutting methods.

The strength of the two methods is calculated as follows:

= 0.223 × (2350 × 2200 - (300 + d) × (450 + d))

The opposing force is calculated from clause 31.6.3.1 of the code which provides ks × τc

Here ks = 0.5 + βc but not greater than 1

(Βc is nothing but the measure of the short side and the long side of the column).

τc = 0.25 × click

Therefore, column ks became 1 and shear strength design by two cutting methods

1.25 N / mm2

After that the resistance to the stock will be

= 0.25 × (2 × ((450 + d) + (300 + d)) × d)

Measuring both strengths provides

d = 316.2 mm

Therefore, the effective thickness of the moving slide is greater than the three cases i.e. 350 mm from a single shear action. Total depth = 350 + 75 = 425 mm

In terms of clause 26.4.2.2 the minimum cover must be 50 mm. Provide a functional cover as 75 mm in this case.

Step 3: Calculation of foot reinforcement

Foot stabilization is achieved by a momentary or percentage perception of a single shear reinforcement.

we know, the minute on the surface of the column is 236.90 kN m. Corresponding reinforcement can be obtained with

Mud = 0.87fyAst × (1-𝑓𝑦𝐴𝑠𝑡 𝑓𝑐𝑘𝑏𝑑) d è236.90 × 106 = 0.87 × 415 × Ast × (1-415 × 𝐴𝑠𝑡 20 × 2350 × 350) 350

Therefore, Ast = 1972.88 mm2, but from case (b) A reinforcement percentage of 0.3%. Therefore, Ast = 0.3 / 100 × 350 × 2350 = 2467.5 mm2.

Therefore, Therequiredreinforcementis2467.5mm2.

Let's choose 16mmØ bars. The bar space has been

= 2350 × Τ𝜋4 162 2467.5 = 191.48mm

Provide 20mmØ @ 180mm c / c bars on both sides.

Step 4: Check the load capacity of the column

According to clause 34.4 of the code, the load bearing must be less than 0.45fck × √ (A1 / A2) of the design state limit. Here the A1 is the support area and the A2means column area.

Carrying pressure on the column surface be = 1.5 × 700 × 1000 300 × 450 = 7.77N / mm2

√ (A1 / A2) = √ (300 × 450 300 × 450) = 1 <2 hence, 1 item

Allowable carrying capacity = 0.45 × 20 × 1 = 9 N / mm2> 7.77 N / mm2

Therefore, it is safe to carry energy. There is no need to provide foundations. However, a downhill line is preferred for economic reasons.

Step 5: Check the progress of the development

It is necessary to improve the bond between the column and the foot and thus the strength of the limb. Therefore, the maximum reinforcement of the column bars will be extended to the moving slide until the requirements for length of development are satisfied.

Column development 16mmØ bras is = 0.87 × 415 × 16 5 × 1.6 × 1.2 = 601.75 mm.

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