School of Civil & Urban Eng., IOTec.
Hawassa University
E x ample 4.1. ( C lassi lassiff i cation cation of of columns columns as as shor shor t or long) The frame shown in figure below is composed of members with rectangular cross sections. All members are constructed of the same strength concrete (E is the same for both beams and columns). Considering bending in the plane of the frame only, classify column EF as long or short if the frame is braced. All girders are 300 x 600 mm. 525 KN
600
F
C
I
F
M1 = 30 KNm
E
M2 = 45 KNm
300
350 300
3.80 m
300 x 350
600 B
H
E 400
3.80 m 300 x 400 A
D
G
9m
7.5 m
Solution: Moments of inertia Girders: I g Columns: I EF
300 x600
I DE
8
54 x10
12
300 x 4003
300 x350
3
12
4
16 x108 mm4
3 8
12
mm
4
10.71875 x10 mm .
Stiffness Coefficients:
E 54 x108 6 x105 E . K BE K cF EI g 9000 Girders : K g L g E 54 x108 7.2 x105 E . K EH K FI 7500
Columns:
K c
EI c Lc
K K
E 16 x10
8
DE
3
4.21 x10 E 5
3.8 x10
E 10.7187510
8
EF
3
2.82 x10 E 5
3.8 x10
The column being considered is column EF. RC II (CEng 3111)
Chapter 1
column design examples
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School of Civil & Urban Eng., IOTec.
Hawassa University
Rotational stiffnesses at joints E and F. EI col / L I col / L
f EI g / Leff
E
Joint E: Joint F:
F
m
E
K EF
f I g / Leff
5
4.21 x10
K DE
K BE K EH
2
5
7.2 x10
0.53
5
K FI
F
5
6 x10
5
2.82 x10
2.82 x10
K EF K CF
5
6 x10
0.53 0.21 2
5
7.2 x10
0.21
0.37
For a braced column (Non sway structure ) for design Le L
0.4
0.8
m
m
0.37 0.4 0.37 0.8
0.66 0.7
Le= (0.7) (3.8) = 2.66m = 2660mm The slenderness ratio:
Le i
Le
2660
I A
10.71875 x10 300 x 350 8
26.327. 30 66.66 ok ! 45
50 25
The
colum is short .
resting on an independent footing supports a flat flat slab. slab. The super super Example 2: A column resting imposed factored load load transferred from from the slab is 1000 kN. Design the column assuming a gross steel ratio ratio of (a) 0.01. Use concrete C30, steel steel S300 and class I works. Assume column height h = 4 m. Solution:
fcd = 13.6 MPa;
fyd = 260.87 MPa
P du f f cd (1 – ρ) ρ) + ρ f ρ f yd ] du = A g [ cd (1 – (a) For (a) For ρ = 0.01 and Pd = 1000 kN ,
A g
P d [ f cd (1 ) f yd ] 3
2
S =
1000 *10
13.6(1 0.01) 0.01(260.87)
S = 249 mm
Use 250 mm × 250 mm cross c ross section A st = ρ A ρ A g = 0.01 (250)2 = 625 mm2 RC II (CEng 3111)
Chapter 1
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School of Civil & Urban Eng., IOTec.
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A st provided provided = 804 mm2
Use 4 numbers of 16 mm dia rods; Ties: d ≥ 6 mm mm (or) ≥ Dia of main bar/4 = 16/4 = 4 mm
S ≤ 12* dia of main bar = 192 mm ≤ Least lateral dimension = 250 mm ≤ 300 mm Therefore, use 6 mm dia rods at 190 mm center to center
Example 3
Design a slender braced (non-sway) column subjected to uniaxail bending. Given: - factored load=1650KN -factored 1st order equivalent constant Moment=130KNm -Geometric length: L=7m and Le=0.7L -Material data; C-30, S-460 class I work -Assume Column size b = 400mm; h = 400mm; Required: - quantity of reinforcement. Solution Assume cover = 20mm; ølong = 20mm and ølat. = 10mm d ' h
40
400
ea >=
= 0.1 and d = 400-40 = 360mm
Le
300
=
0.7 * 7000 300
= 16.33
or
20mm
Therefore; ea=20mm Check for second order effect
- λ=
Le I
=
4900 400
A
2
= 42.4
12
M - λ max max = 50-25(
1
M 2
) ; here first order moment is constant throughout the column.
Therefore; λ max max= 50-25=25 As λ > λ max, max, second order effect has to be considered Msd = etot*Nsd=(ee+ea) Nsd =ee* Nsd+ ea* Nsd =first order moment + moment due to ea = 130+ (1650*0.02) =163kNm
RC II (CEng 3111)
Chapter 1
column design examples
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School of Civil & Urban Eng., IOTec.
For C-30 concrete; f ck ck = 24;
Hawassa University
f cd cd=
0.85 f ck
=
0.85 * 24 1.5
s
f yk
f yd yd = νsd =
sd
1650 *10
=
Ac f cd M sd
Ac f cd h
= 400MPa
1.15
s
N sd
460
=
3
2
400 * 13.6
=
= 0.76
163 *10
6
2
400 *13.6 * 400
=0.187
Using chart no- 2; for νsd = 0.76 and K 2 =
sd
=
0.187 0.25
bal
e2 =
K 1 Le
10
2
(
1
1
= 0.75,
sd =
= K 2 (
r
5
d
0.187; ω = 0.32; ) 10-3 = 0.75(
bal =
5
*10
0.25 = 10.42*10-6
3
360
) here K 1 = 1 for λ > 35
r
1(4900) 2
=
= 13.6MPa
10
(10.42 *10 6 ) = 25mm
e tot = ee + ea+e2 =
Msd =Nsd* etot = 1650*
130 *10 1650
123.8 1000
3
20 25 123.8mm
= 204.3kNm
,
204.3 *10
3
= 0.236 implies
3
400 *13.6
ω=0.45 Recalculating k 2, μ bal=0.3 k 2=
0 .235 0.3
= 0.78 ,
1
= 0.78(
r
5
) * 10 = 10.8*10-6 3
360
e2= 26mm etot = 124.8 Msd = 1650*
124.8 1000
= 205.09 kNm ,
sd
M sd Ac f cd h
=
205.1 *10 2
6
400 *13.6 * 400
= 0.236
ω = 0.45 Interaction can be stopped.
RC II (CEng 3111)
Chapter 1
column design examples
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School of Civil & Urban Eng., IOTec.
Hawassa University
2
0.45 * 400 *13.6
A st
400
= 2448mm2
Use 8 number of 20mm diameter rods. As provided = 2512, com compare pare the result with minimum and maximum code requirements >0.008*4002=1280 <0.08*4002=12800 Lateral ties: ø
Hence ok Hence use ø 10 bar
6 or 20/4=5
12 240 ø=12∗20 = {400− 400 − 300
Use 10mm diameter bar @240mm c/c. Example 4
A uni-axial column is to be constructed from a materials C-30, S-400 class I works. If the diagram for 1st order end moment and axial force are as shown, determine the area of reinforcement assuming non-sway frame system.(use b/h =300/400 and Le=0.75L, with L=7.5m)
Soln: Assume d’= 40mm; ee
0.6eo2+0.4eo1
eo2= eo1=
d ' h
=
40 400
= 0.1
use uniaxail chart no-2
or 0.4eo2
155 * 1000 =121.1mm 1280
82
* 1000 =-64.1mm 1280
+0.4∗−64. 1 =47. 0 2: ≥ 0.6∗121.0.41∗121. ℎ =48.44 1=48.44: ea
le
300
=
0.75 * 7500 300
RC II (CEng 3111)
=18.75mm or 20mm; use ea =20mm
Chapter 1
column design examples
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School of Civil & Urban Eng., IOTec. 0.75 * 7500
Check for e2; λ=
400
2
Hawassa University
=48.7
; λ max max=50-25(
82 155
)=63.2
12
λ < λ max; max; therefore; neglect second order eccentricity etot=eo2 +ea =121.1+20=141.1mm Msd= Nsd*etot=1280* ν =
N sd f cd bh
141 .1 1000
=180.6kNm;
1280 *10
fcd= 13.6
3
13.6 * 300 * 400
=0.78 and μ=
; fyd=347.8
M sd
180.61 *10
f cd bh 2
6
13.6 * 300 * 400
2
=0.28
implis ώ=0.6 As=
*
Ac * f cd f yd
=
0.6 * 400 * 300 * 13.6 347.8
= 2815.4mm; use 8ø22mm bar
2
As pov= 8*
22 * 4
=3041mm2
< Asmax= 0.08*Ag=9600mm2 >Asmin=0.008Ag=960mm2 Lateral reinforcement Ø 6 or 22/4 S 12* 22 =264 or 300 Use 6mm Ø ties at 260mm spacing. Example 5
Design a column to sustain a factored design ax ial load of 900KN and biaxial bi axial moments of Mdx=270KNM and Mdy=180KNm including all other effects. Use C-30, S -300 class I works. Solution:
f ck ck = 24MPa ;f cd cd=13.6MPa; fyd=260.87MPa
Assume b*h = 400*600mm and
b'
b
=
h' h
=0.1,
Mh=Mdx=270kNm M b=Mdy=180kNm 900 *10 3 ν= =0.28(between0.2 and0.4) 13.6 * 400 * 600 M b 180 *10 6 b = =0.14 and h f cd Ac b 13.6 * 600 * 400 2 Using biaxial chart no- 9 thus: for =0.2; h =0.14 , b =0.14; ώ=0.4
for
=0.4; h
=0.14,
By interpolation for
RC II (CEng 3111)
b =0.14;
=0.28;
Chapter 1
Nsd= 900kN
M h f cd Ac h
=
270 *10
6
13.6 * 400 * 600
2
=0.14
ώ=0.4 ώ=0.4 ώ=0.4
column design examples
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School of Civil & Urban Eng., IOTec.
As=
* Ac
*
f cd
f yd
=
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0.4 * 400 * 600 * 13.6 260.87
=5005mm2
< Asmax= 0.08*Ag=19200mm2 >Asmin=0.008Ag=1920mm2 use 8ø30mm bar Lateral reinforcement Ø
6
or 30/4
3030 = 360 {12 ∗=400 300
Use 8mm Ø ties at 300mm spacing.
RC II (CEng 3111)
Chapter 1
column design examples
Page 7
