# RCC & Prestressed Concrete - Design of Column & Footings Complete Study Notes

By Sidharth Jain|Updated : February 28th, 2022

Complete coverage of the APPSC AE Exam syllabus is a very important aspect for any competitive examination but before that important subjects and their concept must be covered thoroughly. In this article, we are going to discuss the Design of Column & Footings topic which is very useful for APPSC AE Exams.

## Design of Column

#### Working Stress Method

Slenderness ratio (λ)

If λ > 12 then the column is long.

Load carrying capacity for short column

where, AC = Area of concrete,

σSC Stress in compression steel

σCC Stress in concrete

Ag Total gross cross-sectional area

ASC Area of compression steel

Load carrying capacity for long column

where,

Cr = Reduction factor

where, leff = Effective length of column

B = Least lateral dimension

imin = Least radius of gyration and

where, l = Moment of inertia and A = Cross-sectional area

Effective length of column

Effective length of Compression Members

Column with helical reinforcement

Strength of the column is increased by 5%

for short column

for long column

Longitudinal reinforcement

(a) Minimum area of steel = 0.8% of the gross area of column

(b) Maximum area of steel

(i) When bars are not lapped Amax = 6% of the gross area of column

(ii) When bars are lapped Amax = 4% of the gross area of column

Minimum number of bars for reinforcement

For rectangular column  4

For circular column  6

Minimum diameter of bar = 12 mm

Maximum distance between longitudinal bar = 300 mm

Pedestal: It is a short length whose effective length is not more than 3 times of lest lateral dimension.

Transverse reinforcement (Ties)

where  dia of main logitudnal bar

φ = dia of bar for transverse reinforcement

Pitch (p)

where, φmin = minimum dia of main longitudinal bar

Helical reinforcement

(i) Diameters of helical reinforcement is selected such that

(ii) Pitch of helical reinforcement: (p)

where,

dC = Core diameter = dg – 2 × clear cover to helical reinforcement

AG = Gross area

dg = Gross diameter

Vh = Volume of helical reinforcement in unit length of column

φh = Diameter of steel bar forming the helix

dh = centre to centre dia of helix

= dg – 2 clear cover - φh

φh = diameter of the steel bar forming the helix

Some others IS recommendations

(a) Slenderness limit

(i) Unsupported length between end restrains  60 times least lateral dimension.

(ii) If in any given plane one end of column is unrestrained than its unsupported length

(b) All column should be designed for a minimum eccentricity of

### Limit state method

1. Slenderness ratio (λ)
if
λ<12 Short column
1. Eccentricity

If  then it is a short axially loaded column.
where, Pu = axial load on the column
2. Short axially loaded column with helical reinforcement
3. Some others IS code Recommendations

(a) Slenderness limit

(i) Unsupported length between end restrains  60 times least lateral dimension.

(ii) If in any given plane one end of column is unrestrained than its unsupported length

(b) All column should be designed for a minimum eccentricity of

Where e = 0, i.e., the column is truly axially loaded.

B. Footing/Foundation

INTRODUCTION

Foundation is important part of any superstructure. It transfers load from superstructure to soil.

Bearing capacity of soil

Bearing capacity of soil governs the dimensions and depth of foundation. Under no case the loading on foundation can be greater than bearing capacity of foundation

(A) Gross Bearing capacity:  Total bearing capacity at based on foundation which includes weight of foundation, super structure load, earth lying over footing.

(B) Net Bearing capacity: It can be defined as follows

Net bearing capacity= Gross Bearing capacity - W

W= weight of soil at level of footing before trench was made for footing

Depth of foundation

following formula must be use to find the depth of foundation

Types of Foundation

Based on depth it can be into two parts

(i) Shallow foundation- if total depth (D) of footing is less than width (B) of foundation then foundation is called shallow foundation.

(ii) Deep foundation- If total depth (D) is more than width (B) of foundation than foundation is called deep foundation.

Nominal cover as per IS 456:200

Minimum Nominal cover as per exposure condition

 Member Mild (mm) Moderate (mm) Severe (mm) Very severe (mm) Extreme(mm) Foundation 40 50 50 50 75

DESIGN OF FOOTING

Let’s take an example to understand important aspect of footing

Example: Design a square footing using LSM for a column load of 1000 kN. If bearing capacity, density of soil is 150 KN/m2 and 20 kN/m3. Use M30 /Fe 415.

Dimension of column= 500 ×500 mm

Sol.

Axial load P1 = 1000 kN

Weight of footing P2 = 0.20 × P1 = 200 kN

Note = P2 can be assumed to 10 to 20% of

Total load = PT = 1200 kN

(i) Area of footing required

Area = PT/q0

q0 = bearing capacity of soil

Area =  = 8 m2

Assume 4 m × 4 m square footing is provided

Area provided = 16 m2 > Area required

(ii) Net soil pressure

w=P1 /A= 62.5 kN/m2

For LSM design w40 = 1.5 × w0 = 93.75 kN/m2

1.5 → Partial factor of safety

(iii) Check for bending moment

As per our design

a = b = 500 mm

L = B = 4 m

{In case of rectangular footing a, b, L, B}

will not be same

for 1 m width → about xx

= 143.55 kNm

For 1 m width

In case Mux and Muy comes differently then take maximum value

(iv) depth of footing

(v) Check for shear

Critical section for one way shear is at Q from face of column

τc = permissible shear stress

τc = 0.29 → for M30

Vu = 93.75 × 103 × 1×[(4-0.5)/2-0.3]=

= 13594 kN

For B = 1m

τ=Vu/Bd  = 0.54

τV > τc → Not safe in shear increasing depth (G) of footing to 600 mm

Vu = 93.75 × 103 × 1×[(4-0.5)/2-0.6]=

= 107.81 kN

τV = 0.17 < τc → safe

now so new depth of footing = 600 mm

overall depth of footing = 660 mm

(vi) Check for punching shear / two way shear

Note →

Critical section for punching shear is at d/2 for five of column all-around

Pnet = Pu – Wu (a + d) (b + d)

Pnet = 1.5 × P – Wu (a + d) (b + d)

d = 0.6

ks = 0.5 + βc

= 0.5 + (b/a)

= 0.5 + 1 = 1.5

maximum possible value of ks =1 so

ks = 1

τcp = 1.36 N/mm2 > τvp → safe

so d = 600 mm

D = 660 mm

(v) Area of steel

(3) Total area of steel

= L × Ast = 4 × 560 = 2240 mm2

Total no of  bars for 16 mm ϕ

n= 2240/[(π/4)×162]=12

(4) number of central bend

∴  for width all value remains same

(5) check for bearing

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