How is rcc column design?

Answer 1

DESIGN OF AXIALLY LOADED COLUMN (SQUARE)

15.1 Data :

Column No. : ………. Load on column = …………..KN.

Design constant a) Grade of concrete = ……….. b) Grade of steel = ………..

L = Unsupported length of the column in mm = …………mm

Design load Pu = 1.5 x …………. = ………….. KN.

Assume percentage of comp. steel between 0.8 to 6% of gross c/s area

Assume Asc = … …%Ag = , Ac =………%Ag

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION ♦ 31

Design of structures (RCC) Exercise No. 1

15.2 Find the section of the column (Ag):

Pu = 0.4 fck Ac +0.67 fy Asc

=………………………..

Ag = ………………… mm2.

Adopt square column. Size of column =……….x…………. mm Say ……… X ………mm.

15.3 Find the No. of bars (not less than 4)

Asc = ……. % Ag

Calculate no. of bars

For ∅ = …….. No of Bar = N1 = = ------- =

For ∅ = …….. No of Bar = N2 = = …………… =

Use diameter of bars not less than 12 mm

15.4 Lateral ties.

a) Select diameter of lateral ties least of

5 mm or 1/4 diameter of the largest longitudinal bar not less than16 mm diameter.

Diameter of the lateral ties / link = ………. mm

b) Spacing of lateral ties / link Least of

i) Least lateral dimension of column =…………..mm

ii) Sixteen times the smallest dia. of bar = 16 x ……=………..mm and

iii) 300 mm

Provide … mm dia. @ ………. C/c

15.5 Check for minimum eccentricity

e min >/

0.05 D =……………..= ………….mm > e min

e min = 20 mm

L = Unsupported length of the column in mm

D = Lateral dimension of column in the direction under consideration in mm

32 ♦ MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION

Exercise No. 1 Design of structures (RCC)

15.6 Summary of design :

1. Column size ……………x…………. mm

2. Longitudinal Steel ………………………………………….

3. Lateral Steel ……………………………………………

Note : Draw the diagram showing the reinforcement details for the above column on

drawing sheet with the guidance of teacher.

16.0 DESIGN OF PLINTH BEAM

16.1 Data :

Sr. No. (from load calculation sheet) = beam mark =

Design constant a) Grade of concrete = b) Grade of steel =

span = -------m.

Singly Reinforced Rectangular Beam

Maximum load = ----------kN

Factored load = -----------kN

Given Fact B.M. = ..................................kN. m.

Fact S.F. = ...................................KN.

16.2 Find dimension of rectangular beam

Mu lim = Mu

Mu = 0.148 fck bd 2 for Fe 250 steel

= 0.138 fck bd2 for Fe 415 steel

= 0.133 fck bd2 for Fe 500 steel

Assume b = ……………. (Generally equal to width of the wall or 230 mm)

Mu = ....................................

=

d2 =

d = ...............................mm Say ------- mm

(Assume diameter of bar)

Over all depth (D) = d+ + cover ( Assume 25 mm cover)

D = ..................mm

Round up the value of D say ………..mm

d = D - - cover

d = .........................mm

:. b = ..........................mm

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION ♦ 33

Design of structures (RCC) Exercise No. 1

16.3 Find Longitudinal steel.

= ………………………………………

=................................m m2

Calculate no. of bars

For ∅ = …….. No of Bar = N1 = = ------- =

For ∅ = …….. No of Bar = N2 = = …………… =

Provide --- # -- mm ∅ bars

16.4 Design of shear reinforcement.

1. Find nominal shear stress τv = = …………….

= .......................N/mm2

2. Find shear strength of concrete (τc)

With referring IS 456 - 2000 clause

Pt =

Pt = ...................... %

find τc =

τc = .................. N/mm2

3. Compare

a) if τv < τc Provide nominal shear reinforcement.

b) if τv > τc Design shear reinforcement.

a) When τv < τc Provide nominal shear reinforcement as fallows :

Assuming -- mm Ø two legged M.s. stirrups.

Asv =2 x π/4 x d2 = ……….=………..

Spacing of stirrups Sv ≤

= ……………….. = ………….. mm

34 ♦ MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION

Exercise No. 1 Design of structures (RCC)

Spacing should be less than …….

1. as calculated = ………..mm

2. 0.75 d = 0.75 x ……. = …….mm

3. 450 mm.

Provide -- mm ∅ two legged M.S. stirrups @ ----mm c/c.

b) when τv > τc Design shear reinforcement as fallows :

Find shear to be resisted Vus

Vus = Vu - τc bd

= ............ - ...........

Vus = ......................N.

Find shear resisted by bent up bars

Vub = 0.87 fy Asv. Sin 450

= 0.87 x ............x................sin 450

Vub =....................... N

Contribution of bent up bars should not be greater Vus / 2

Shear to be resisted by stirrups Vus

V'us = Vus - (Vub or Vus / 2 whichever is lesser)

= ……… - ………..

=…………

Assuming -- mm Ø two or …. legged M.s. stirrups.

Asv =2 x ð/4 x d2 = ……….=……….. mm2

Spacing of stirrups Sv ≤

= ……………….. = ………….. mm

Spacing should be less than …….

1. as calculated = ………..mm

2. 0.75 d = 0.75 x ……. = …….mm

3. 450 mm.

Provide -- mm ∅ two or …. legged M.S. stirrups @ ----mm c/c.