Step by step on how to design a distillation column with all equations and formulas. Provides in depth analysis to the procedure, application and method involved.
Mome Moment nt in the the dir direc ecti tion on of Larg Larger er dime dimens nsio ion n (To (Top) p)
= 1760
x 1.0 = 1760
kN.M.
Moment in the direction of Larger dimension (Bott)
= 1760
x 1.0 = 1760
kN.M.
Mome Moment nt in the the dire direct ctio ion n of of Shor Shorte terr dim dimen ensi sion on (Top (Top))
= 200
x 1.0 = 200.0
kN.M.
Momen Momentt in in the the direc directio tion n of of Sho Short rter er dime dimens nsio ion n (Bo (Bott) tt)
= 200
x 1.0 = 200.0
kN.M.
Axial Load on column
Effective Le Length parallel to to Larger Dimensions
l ex
= 6200
mm
Effective Length Parallel to Shorter Dimensions
l ey
= 6200
mm
= 6200
mm
Unsupported Length
b
Concrete Grade
f ck
= 35
N/mm2
Steel Grade
f y
= 415
N/mm2
Dia of bars used
= 20
mm
Cover to reinforcement
= 40
mm
= 0.8
%
p
Assumed reinforcement percentage
Check for slenderness
lex / D = ley / b =
D
Mx
My
Equally on all the four sides.
4.8
< 12 Sh Short Column
7.3
< 12
Short Column
Therefore the Column is Short Column Additional Moment due to minimum eccentricity to be considered: t r o h S $ # ' s l x . 2 0 7 4 5 5 9 3 / 4 1 6 2 h c t a r c s / p m t / n o i s r e v n o c / d b i r c s / t p o / /
The above actual moments should be compared with those calculated from minimum eccentricity consideration ( see 25.4 of the IS code 456:2000) and greater value is to be taken as the initial moment for adding the additional moments. Min. eccentricity
ex ey
= l /500 /500 + D/30 or 20mm = = l /500 /500 + b/30 or 20mm =
55.73
mm
least of the two
40.73
mm
least of the two
Moments due to e x or ey or Both are Greater than 20mm. Mux
=
1312.6
x 0.0557
=
73.2
kN.m
Muy
=
1312.6
x 0.0407
=
53.5
kN.m
1760
kN.m
Total moments for which the column is to be designed are : Mux = Max. of (1760.07, 1760.07, = 39554702.xls / Short
abc
Project
Delhi Metro MC1B Calculations for
Design of RCC Column
Discipline
: Civil
File No.
59120 / X /
Calc by
: RK
Date
29-Aug-02
Date
29-Aug-02
Checked by :
: e l i f '
Muy
=
Max. of (200, 200, 53.4682
=
200
kN.m
Section design for axial force & biaxial bending.
Pu / f ck b D
=
0.034
p / f ck
=
0.023
Puz
=
0.45 f ck . Ac + 0.75 f y . Asc
=
[ 0.45*35*(1-0.0008) + 0.75*415*0.008 ] * 11050 N
=
20016 kN Pu/Puz
For x-x axis
Pu / Puz
=
0.07
α n
=
1.00
d'/D ratio
For Y-Y axis
=
IF(M59<0.201,1,IF(M59>0.799,2,AG61))
0.038
Mu / f ck bD2
=
Mux1
=
2262
Mux / Mux1
=
0.78
d'/D ratio
=
Mu / f ck bD2
=
0.045
Muy1
=
1479
Muy / Muy1
=
0.14
Referring chart for d'/D =0.05
0.045
kN.m
0.059
Referring chart for d'/D =0.10
kN.m
Now, with References to IS 456:2000, Page 71, Clause 39.6 and Clause 39.7. (Mux / Mux1)
n
α
+ (Muy / Muy1)
n
α
=
0.78^1.00 + 0.14^1.00
=
0.913
Therefore Area of Steel A s
=
0.8 % of (850 x 1300)
Provided reinf.
=
39554702.xls / Short
20 mm Dia Bar
Column design is OK
29 Nos.
= 8840
mm2
= 9111
mm2
Sheet
Project
Delhi Metro MC1B Calculations for
Design of RCC Column
abc Discipline
: Civil
File No.
59120 / X /
Calc by
: RK
Date
29-Aug-02
Date
29-Aug-02
Checked by :
39554702.xls / Short
Sheet
abc
Project
Delhi Metro MC1B Calculations for
Design of RCC Column
Discipline
: Civil
File No.
59120 / X /
Calc by
: RK
Date
02-Aug-02
Date
02-Aug-02
Checked by :
Design data
b x D = 300 mm
Size of Column
P u
x
600 mm
x 1.0 = 1000
kN
Moment in the direction of Larger dimension (Top) M1x = 150 Moment in the direction of Larger dimension (Bott)M2x = 85
x 1.0 = 150.0
kN.M.
x 1.0 = 85.0
kN.M.
Moment in the direction of Shorter dimension (Top M1Y = 10 Moment in the direction of Shorter dimension (BottM2Y = 10
x 1.0 = 10.0
kN.M.
x 1.0 = 10.0
kN.M.
Axial Load on column
= 1000
Effective Length parallel to Larger Dimensions
l ex
= 6840
mm
Effective Length Parallel to Shorter Dimensions
l ey
= 6840
mm
= 6840
mm
Unsupported Length
b
Concrete Grade
f ck
= 35
N/mm2
Steel Grade
f y
= 415
N/mm2
Dia of bars used
= 20
mm
Cover to reinforcement
= 40
mm
= 1.3
%
p
Assumed reinforcement percentage Member is bent in SINGLE / DOUBLE curvature
= Double
Design shear force--Factored
= 85
Check for slenderness
lex / D = ley / b =
D
Mx
My
Equally on all the four sides.
kN
11.4
< 12 Short Column
22.8
> 12
Slender Column
Column is slender about Minor Axis Additional Moment due to slendernesss: g n o L $ # ' s l x . 2 0 7 4 5 5 9 3 / 4 1 6 2 h c t a r c s / p m t / n o i s r e v n o c / d b i r c s / t p o / / /
Max
2 = (Pu*D / 2000) ( L ex / D )
= 0.0
kN.m
May
2 = (Pu*b / 2000) ( L ey / b )
= 78.0
kN.m
The above moments is reduced by a multiplying factor (K), as calculated belowK = (Puz - Pu) / (Puz - P bx) IS-456:2000, 38.7.1.1 where
Puz
=
0.45 f ck . Ac + 0.75 f y . Asc
=
[ 0.45*35*(1-0.0013) + 0.75*415*0.013 ] * 18000 N
=
###
kN
d' / D (about xx-axis)
=
0.083
Use Chart for d' / D
0.1
d' / b (about yy-axis)
=
0.17
Use Chart for d' / D
0.2
Calculation of P b :
From Table 60, Design Aid of IS:456-1978 Page - 171
39554702.xls / Long
Sheet
abc
Project
Delhi Metro MC1B Calculations for
Design of RCC Column
Discipline
: Civil
File No.
59120 / X /
Calc by
: RK
Date
02-Aug-02
Date
02-Aug-02
Checked by :
: e l i f '
P b (about xx-axis)
P b (about yy-axis)
P bx
=
(k1 + k2 * p / fck) * fck * b * D
k 1
=
0.21
=
1381
k 2
=
0.33
=
(k1 + k2 * p / fck) * fck * b * D
k 1 =
0.18
=
1166
k 2 =
0.03
k x
=
(Puz - Pu) / (Puz - P bx)
=
1.18
k y
=
(Puz - Pu) / (Puz - P by)
=
1.07
P by
kN
kN
Sheet
The Additional slender moments calculated earlier will now be multiplied by the above value of K ( K<=1) Max = 0.0 1.00 = 0.0 kN.m x May
=
78.0
x
1.00
=
78.0
kN.m
The Additional moments due to slenderness effects should be added to the initial moments after modifying the initial moments as follows (see section 39.7.1 of the code IS 456:2000) Mux = 0.6 x 150.0 - 0.4 x 85.0 = 56.0 kN.m but not less than
0.4* Max (M1x & M2x) =
=
Muy but not less than
=
60.0
Max of the above two
=
60.0
kN.m
0.6 x 10.0 - 0.4 x 10.0
=
2.0
kN.m
0.4* Max (M1Y & M2Y) =
=
kN.m
Max of the above two
4.0 =
kN.m
4.0
kN.m
Additional Moment due to minimum eccentricity to be considered:
The above actual moments should be compared with those calculated from minimum eccentricity consideration ( see 25.4 of the IS code 456:2000) and greater value is to be taken as the initial moment for adding the additional moments. Min. eccentricity
ex ey
= l /500 + D/30 or 20mm = = l /500 + b/30 or 20mm =
33.68
mm
least of the two
23.68
mm
least of the two
Moments due to e x OR ey OR Both are Greater than 20mm. Mux
=
1000.0
x 0.0337
=
33.7
kN.m
Muy
=
1000.0
x 0.0237
=
23.7
kN.m
Total moments for which the column is to be designed are : Mux = 150.0 + 0.0
=
150.0
kN.m
39554702.xls / Long
abc
Project
Delhi Metro MC1B Calculations for
Design of RCC Column
Discipline
: Civil
File No.
59120 / X /
Calc by
: RK
Date
02-Aug-02
Date
02-Aug-02
Checked by :
Muy
101.7 kN.m = 23.7 = + 78.0 But this moment cannot be less than the initinal end moments M 1x & M2x / M1Y & M2Y Hence, Design moments
Mux
=
150.0
kN.m
Muy
=
101.7
kN.m
Section design for axial force & biaxial bending.
For x-x axis
Pu / f ck b D
=
0.159
p / f ck
=
0.037
Pu / Puz
=
0.28
α n
=
1.14
d'/D ratio
For Y-Y axis
=
0.100
Mu / f ck bD2
=
Mux1
=
227
Mux / Mux1
=
0.66
d'/D ratio
=
Mu / f ck bD2
=
0.060
Muy1
=
113
Muy / Muy1
=
0.90
(Design Aid page No 130 )
0.060
kN.m
0.200
(Design Aid page No 131 )
kN.m
Now, with References to IS 456:2000, Page 71, Clause 39.6 and Clause 39.7. (Mux / Mux1)
n
α
+ (Muy / Muy1)
n
α
=
0.66^1.14 + 0.90^1.14
=
1.507
Therefore Area of Steel A s
=
1.3 % of (300 x 600)
Provided reinf.
=
20 mm Dia Bar - 8
REDESIGN Column
Nos.
= 2340
mm2
= 2513
mm2
1.40 %
Shear Design
Shear force
V Provided tension reinf percentage pt Nominal shear stress
τ
v
=
85
kN
=
1.40
%
=
5.05
N/mm2
Design shear strength of concrete τc Vus Shear to be taken by Stirrup
