RC II (CEng 3111) Chapter 1 column design examples Page 1 Example 4.1. (Classification of columns as short 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.
Solution: Moments of inertia Girders: 8 4 3 10 54 12 600 300 mm x x Ig Columns: 8 4 3 10 16 12 400 300 mm x x IDE 10.71875 10 . 12 350 300 8 4 3 mm x x IEF Stiffness Coefficients:
. 10 2 . 7 7500 10 54 . 10 6 9000 10 54 : 5 8 5 8 E x x E K K E x x E K K L EI K Girders FI EH cF BE g g g Columns:
E x x E K E x x x E K L EI K EF DE c c c 5 3 8 5 3 8 10 82 . 2 10 8 . 3 10 71875 . 10 10 21 . 4 10 8 . 3 10 16The column being considered is column EF.
F E I H G E D C B A M2 = 45 KNm M1 = 30 KNm 7.5 m 9 m 3.80 m 3.80 m 400 600 600 350 300 x 400 300 x 350 525 KN F 300 300
RC II (CEng 3111) Chapter 1 column design examples Page 2 Rotational stiffnesses at joints E and F.
eff g f col eff g f col L I L I L EI L EI / / / / Joint E: 0.53 10 2 . 7 10 6 10 82 . 2 10 21 . 4 5 5 5 5 x x x x K K K K EH BE DE EF E Joint F: 0.21 10 2 . 7 10 6 10 82 . 2 5 5 5 x x x K K K FI CF EF F 0.37 2 21 . 0 53 . 0 2 E F m For a braced column (Non sway structure ) for design
0.66 0.7 8 . 0 37 . 0 4 . 0 37 . 0 8 . 0 4 . 0 m m e L L Le= (0.7) (3.8) = 2.66m = 2660mm The slenderness ratio:
10.71875 10
300 350
2660 8 x x A I L i Le e . ! 66 . 66 45 30 25 50 . 327 . 26 short is colum The ok Example 2: A column resting on an independent footing supports a flat slab. The super imposed factored load transferred from the slab is 1000 kN. Design the column assuming a gross steel ratio of (a) 0.01. Use concrete C30, steel S300 and class I works. Assume column height h = 4 m.
Solution: fcd = 13.6 MPa; fyd = 260.87 MPa Pdu = Ag [fcd (1 – ρ) + ρ fyd]
(a) For ρ = 0.01 and Pd = 1000 kN,
] ) 1 ( [ cd yd d g f f P A S2 = ) 87 . 260 ( 01 . 0 ) 01 . 0 1 ( 6 . 13 10 * 1000 3 S = 249 mm
Use 250 mm × 250 mm cross section Ast = ρ Ag = 0.01 (250)2 = 625 mm2
RC II (CEng 3111) Chapter 1 column design examples Page 3 Use 4 numbers of 16 mm dia rods; Ast provided = 804 mm2
Ties: d ≥ 6 mm (or) S ≤ 12* dia of main bar = 192 mm
≥ Dia of main bar/4 = 16/4 = 4 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
400 40 ' h d = 0.1 and d = 400-40 = 360mm ea >= 300 e L = 300 7000 * 7 . 0 = 16.33 or 20mm Therefore; ea=20mm
Check for second order effect
-
λ = A I Le = 12 400 4900 2 = 42.4-
λmax = 50-25( 2 1 M M) ; here first order moment is constant throughout the column.
Therefore; λmax= 50-25=25
As λ > λmax, 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 Page 4 For C-30 concrete; fck= 24; fcd= s ck f 85 . 0 = 5 . 1 24 * 85 . 0 = 13.6MPa
f
yd=
s yk f = 1.15 460 = 400MPa νsd = cd c sd f A N = 6 . 13 * 400 10 * 1650 2 3 = 0.76 h f A M cd c sd sd = 400 * 6 . 13 * 400 10 * 163 2 6 =0.187Using chart no- 2; for νsd = 0.76 and sd= 0.187; ω = 0.32; bal= 0.25
K2 = bal sd = 25 . 0 187 . 0 = 0.75, r 1 = K2 ( d 5 ) 10-3 = 0.75( 3 10 * 360 5 = 10.42*10-6
e2 = ( 10 2 1Le K r 1 ) here K1 = 1 for λ > 35 = (10.42*10 ) 10 ) 4900 ( 1 6 2 = 25mm e tot = ee + ea+e2 = 20 25 123.8mm 1650 10 * 130 3 Msd =Nsd* etot = 1650* 1000 8 . 123 = 204.3kNm , 6 . 13 * 400 10 * 3 . 204 3 3 = 0.236 implies ω=0.45 Recalculating k2, μbal=0.3 k2= 3 . 0 235 . 0 = 0.78 , r 1 = 0.78( )*10 3 360 5 = 10.8*10-6 e2= 26mm etot = 124.8 Msd = 1650* 1000 8 . 124 = 205.09 kNm , h f A M cd c sd sd = 400 * 6 . 13 * 400 10 * 1 . 205 2 6 = 0.236 ω = 0.45
RC II (CEng 3111) Chapter 1 column design examples Page 5 400 6 . 13 * 400 * 45 . 0 2 st A = 2448mm2
Use 8 number of 20mm diameter rods.
As provided = 2512, compare the result with minimum and maximum code requirements >0.008*4002=1280
<0.08*4002=12800 Hence ok
Lateral ties:
ø
6 or 20/4=5 Hence useø
10 bar {
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; h d ' = 400 40
= 0.1 use uniaxail chart no-2 ee 0.6eo2+0.4eo1 or 0.4eo2
eo2= *1000 1280 155 =121.1mm eo1= *1000 1280 82 =-64.1mm { ea 300 le = 300 7500 * 75 . 0 =18.75mm or 20mm; use ea =20mm
RC II (CEng 3111) Chapter 1 column design examples Page 6 Check for e2; λ= 12 400 7500 * 75 . 0 2 =48.7 ; λmax=50-25(155 82 )=63.2
λ < λmax; therefore; neglect second order eccentricity etot=eo2 +ea =121.1+20=141.1mm Msd= Nsd*etot=1280* 1000 1 . 141 =180.6kNm; fcd= 13.6 ; fyd=347.8 ν = 400 * 300 * 6 . 13 10 * 1280 3 bh f N cd sd =0.78 and μ= 2 6 2 400 * 300 * 6 . 13 10 * 61 . 180 bh f M cd sd =0.28 implis ώ=0.6 As= yd cd c f f A * * = 8 . 347 6 . 13 * 300 * 400 * 6 . 0 = 2815.4mm; use 8ø22mm bar Aspov= 8* 4 * 222 =3041mm2 < Asmax= 0.08*Ag=9600mm2 >Asmin=0.008Ag=960mm2 Lateral reinforcement
Ø 6 or 22/4 S12* 22 =264 or 300 Use 6mm Ø ties at 260mm spacing.
Example 5
Design a column to sustain a factored design axial load of 900KN and biaxial moments of Mdx=270KNM and Mdy=180KNm including all other effects. Use C-30, S-300 class I works.
Solution: fck= 24MPa ;fcd=13.6MPa; fyd=260.87MPa Assume b*h = 400*600mm and b b' = h h' =0.1, Nsd= 900kN Mh=Mdx=270kNm Mb=Mdy=180kNm ν= 600 * 400 * 6 . 13 10 * 900 3 =0.28(between0.2 and0.4) b A f M c cd b b = 2 6 400 * 600 * 6 . 13 10 * 180 =0.14 and h A f M c cd h h = 2 6 600 * 400 * 6 . 13 10 * 270 =0.14 Using biaxial chart no- 9 thus:
for =0.2; h=0.14 , b=0.14; ώ=0.4
for =0.4; h=0.14, b=0.14; ώ=0.4
RC II (CEng 3111) Chapter 1 column design examples Page 7 As= yd cd c f f A * * = 87 . 260 6 . 13 * 600 * 400 * 4 . 0 =5005mm2 < Asmax= 0.08*Ag=19200mm2
>Asmin=0.008Ag=1920mm2 use 8ø30mm bar Lateral reinforcement Ø 6 or 30/4 {