RCC : Design of Column & footings

Design of Column

Working Stress Method

Slenderness ratio (λ)

If λ > 12 then the column is long.

Load carrying capacity for short column

where, A_C = Area of concrete, 

σ_SC Stress in compression steel

σ_CC Stress in concrete

A_g Total gross cross-sectional area

A_SC Area of compression steel

Load carrying capacity for long column

where,

C_r = Reduction factor

where, l_eff = Effective length of column

B = Least lateral dimension

i_min = 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 A_max = 6% of the gross area of column

(ii) When bars are lapped A_max = 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, φ_min = minimum dia of main longitudinal bar

Helical reinforcement

(i) Diameters of helical reinforcement is selected such that

(ii) Pitch of helical reinforcement: (p)

where,

d_C = Core diameter = d_g – 2 × clear cover to helical reinforcement

A_G = Gross area 

d_g = Gross diameter

V_h = Volume of helical reinforcement in unit length of column

φ_h = Diameter of steel bar forming the helix

d_h = centre to centre dia of helix

= d_g – 2 clear cover - φ_h

φ_h = 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, P_u = 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

Concentrically Loaded Columns

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 less 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

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 = P_T = 1200 kN

(i) Area of footing required

Area = P_T/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

w₀ =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

Moment about y-y

For 1 m width

In case M_ux and M_uy 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

τ_c = permissible shear stress

τ_c = 0.29 → for M30

V_u = 93.75 × 10³ × 1×[(4-0.5)/2-0.3]=

= 13594 kN

For B = 1m

τ_V =V_u/Bd  = 0.54

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

V_u = 93.75 × 10³ × 1×[(4-0.5)/2-0.6]=

= 107.81 kN

τ_V = 0.17 < τ_c → 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

P_net = P_u – W_u (a + d) (b + d)   

P_net = 1.5 × P – W_u (a + d) (b + d)

d = 0.6

k_s = 0.5 + β_c

= 0.5 + (b/a)

= 0.5 + 1 = 1.5  

maximum possible value of k_s =1 so

k_s = 1

τ_cp = 1.36 N/mm² > τ_vp → safe

so d = 600 mm

D = 660 mm

(v) Area of steel

(3) Total area of steel

= L × A_st = 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

You can avail of BYJU’S Exam Prep Online classroom program for all AE & JE Exams:

BYJU’S Exam Prep Online Classroom Program for AE & JE Exams (12+ Structured LIVE Courses)

You can avail of BYJU’S Exam Prep Test series specially designed for all AE & JE Exams:

BYJU’S Exam Prep Test Series AE & JE Get Unlimited Access to all (160+ Mock Tests)

Thanks

Team BYJU’S Exam Prep

Download  BYJU’S Exam Prep APP, for the best Exam Preparation, Free Mock tests, Live Classes.