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1.0 INTRODUCTION Column is the structural component that transfers compression load from beams to the footings. In other words, column is a compression member. Columns are divided into two classification namely short column or long column. This classification depends on the slenderness ratio of the column. Short column will fail by crushing while long column will fail in buckling. Thus, it is important to determine the classification of the column as long column is prone to buckling. When the column buckles, it will exert additional moment to the section which will eventually leads to failure. This report only covers the design of short column using Eurocode 2 column design chart.
2.0 PROCEDURE (ONLY FOR SHORT COLUMN) Step 1. Determine whether the column is a short or long column by computing slenderness ratio (π)
π=
π0 π
=
π0
[Equation 1]
βπΌ/π΄
Where π0 is the effective height of the column. π is the radius of gyration about the axis considered. πΌ is the second moment of area of the section about the axis. π΄ is the cross-sectional area of column
- π0 for braced members (Note that only braced members is covered in the course)can be computed with the following formula: π0
π
π
1 2 = 0.5 π β(1 + 0.45+π ) (1 + 0.45+π ) 1
[Equation 2]
2
Where π is the clear height of the column between end restraint. π1 and π2 is the relative flexibilities of the rotational restraints at ends β1β and β2β of the column respectively. π
2 For column connected to footing, assume (1 + 0.45+π )=2 2
k=
ππππ’ππ π π‘ππππππ π β ππππ π π‘ππππππ π
(πΈπΌ/π)ππππ’ππ
=β
2(πΈπΌ/π)ππππ
(πΌ/π)ππππ’ππ
=β
2(πΌ/π)ππππ
1
- Then, compare π with ππππ .
ππππ = 26.2 / βππΈπ· /π΄π πππ
[Equation 3]
- If π > ππππ , then the column is a long column. - If π < ππππ , then the column is a short column. Step 2. Find the loadings on the column. - First, the load distribution of the slab must be determined. - Then using the load distributed to the beam from the slab, half of the load will be transferred to the column, while the other half of the load is transferred to the adjacent column. Half length
Half length
Figure 1 Load transfer from beam to column - Note that to have the worst case scenario, for interior column, the longer beam has to be under maximum loading (1.35Gk + 1.5Qk) while the short beam has to be under minimum loading (1.35Gk).
Max (1.35Gk + 1.5Qk)
Max (1.35Gk + 1.5Qk) Min (1.35Gk)
Corner Column
Interior Column
Figure 2 worst case scenarios for corner column and interior column 2
Step 3. Find the moment (ππ§ πππ ππ¦ ) acting on the column - Using the load on the beam, find Fixed End Moment (FEM) = wπΏ2 /12.
Max (1.35Gk + 1.5Qk) Min (1.35Gk)
FEM = wπΏ2 /12
FEM = wπΏ2 /12
- For an example, we start from the z-z axis. - Using the net moment, multiply it with the distribution factor, k and add k=
ππππ’ππ π π‘ππππππ π β ππππ π π‘ππππππ π
(πΈπΌ/π)ππππ’ππ
=β
2(πΈπΌ/π)ππππ
ππΈπ· π0 400
to get ππ§ .
(πΌ/π)ππππ’ππ
=β
2(πΌ/π)ππππ
- Repeat the same steps for the y-y axis to get ππ¦ Step 3. Check for biaxial bending - Note that for a normal frame, moment will always come from two different axis (z or y-axis) - Thus, it is important to check for biaxial bending. - First, find the eccentricities of both axis (ππ§ πππ ππ¦ )
ππ§ =
ππ§ ππΈπ·,π§
,
ππ¦ =
ππ¦
[Equation 4]
ππΈπ·,π¦
- Then, check the ratio of the corresponding eccentricities that satisfies one of the conditions. Either
ππ§ β
/
ππ¦ π
β€ 0.2
or
ππ¦ π
/
ππ§ β
β€ 0.2
- If either one of the condition is satisfied, then there is a biaxial bending.
3
Step 4. Compute π΄π¬π« Y
dβ
h Z
Z
hβ
Mz
bβ
d2
b My Y
Figure 3 Section of column with biaxial bending
- Then check for the dominating moment: If
ππ§ ββ²
β₯
ππ¦ πβ²
Then ππΈπ· = π β² π§ = ππ§ + π½
If
ππ§ ββ²
β€
ββ² πβ²
Γ ππ¦
[Equation 5]
Γ ππ§
[Equation 6]
ππ¦ πβ²
Then ππΈπ· = π β² π¦ = ππ¦ + π½
Where π½ = 1 β
πβ² ββ²
ππΈπ· πβπππ
4
Step 5. Compute π¨π,ππππππππ - Using values of
ππΈπ· πβπππ
and
ππΈπ· πβ2 π
ππ
, to find
π΄π ππ¦π πβπππ
from the column chart.
- To determine which chart to use, you have to check dβ/h ratio, each ratio has a different chart. π΄π ππ¦π
- Once obtain the value of πβπ , π΄π can be computed by substituting all the other known ππ
values. π΄π ππ¦π
- If πβπ = 0, then π΄π = minimum area of reinforcement allowed = 0.002bh. ππ - Do check for maximum area of reinforcement = 0.08bh.
Figure 4 Column Design Chart for d2/h = 0.25
5
3.0 DATA & RESULT
B
A 6.069m
C 6.069m
D 6.069m
1 3.069m
2 3.069m
3 Figure 5 Plan View
beam
300mm
v 600mm
b =300mm
Assume h =300mm
h =3m
Figure 6 Cross Section of Column and Beam
Figure 7. Section of Column
- πππ = 30π/ππ2 ππ¦π = 500π/ππ2 Diameter of main steel bar = 16mm - It can be seen that there are four columns to be designed: 1. Column A1, A3, D1, D3 2. Column B1, B3, C1, C3 3. Column A2 and D2 4. Column B2 and C2
6
3.1 Design of Column A1, A3, D1, D3 - Note that the loading on the beams are taken from progress report 2. 6.069m
3.069m
38.96kN/m
28.39kN/m
Z-Z axis
Y-Y axis Figure 8 Loading on Column A1, A3, D1, D3
Step 1. Determine whether the column is a short or long column. Although column is has a square cross section; however the beam connected to the column from Z-Z axis (6.069m) is longer than the beam from Y-Y axis (3.069m). Thus, the more critical case will be from Z-Z axis. Therefore the checking is done for Z-Z axis of the column.
π = 3.0 β 0.6 = 2.4m πΌππππ’ππ
πβ3 = 12
=
πΌππππ
300 Γ3003
πβ3 = 12 =
12 π
= πππ Γ ππ ππ
π
300 Γ6003 12
= ππππ Γ πππ πππ
(πΌ/π)ππππ’ππ
π1 = β =
2(πΌ/π)ππππ πππΓπππ /2400
2(5400Γ106 ππ4 )/6069
= 0.158
7
π0
π
π
1 2 = 0.5 π β(1 + 0.45+π ) (1 + 0.45+π ) 1
= 0.5 (2400)β(1 +
π = βπΌ/π΄
2
0.158 0.45+0.158
= β675 Γ 106 /(300 Γ 300) = 86.60mm
) (2)
= 1904.84mm Assume that (1 +
π= =
π0
π2 0.45+π2
) = 2 as
the column is connected to footing.
π 1904.84 86.60
= 22
ππΈπ·,π =
38.96 Γ 6.069
2 = 118.22kN
ππΈπ·,π =
28.39 Γ 3.069
2 = 43.56kN
ππΈπ· = ππΈπ·,π + ππΈπ·,π = 118.22 + 43.56 = 161.78kN
ππππ = 26.2 / βππΈπ· /π΄π πππ 30
= 26.2 / β161.78 Γ 103 /300 Γ 300 Γ (0.85 Γ 1.5) = 80.57 > π = 22 β΄ The column is a short column.
8
Step 2. Find the loadings on the column.
ππΈπ· = 161.78kN (found from step 1) Step 3. Find the moment (π΄π πππ π΄π ) acting on the column 6.069m
3.069m
38.96kN/m
28.39kN/m
FEM = wπΏ2 /12
FEM = wπΏ2 /12
Z-Z axis
Y-Y axis Figure 9 FEM from beam to Column A1, A3, D1, D3
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ = 675 Γ 106 /2400 = 281250
πππππ = 0.5(πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/6069 = 444883.84
Distribution Factor for the column = =
πππππ’ππ βπ
281250 281250+ 444883.84
= 0.387 πΉπΈππβπ =
=
wπΏ2 12
(38.96)(6.069)2
12 = 119.58kNm
ππ§ = πΉπΈππβπ Γ Distribution Factor + = 119.58 Γ 0.387 +
ππΈπ·,π§ π0
400 118.22(1904.84 Γ 10β3 ) 400
= 46.84kNm 9
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ = 675 Γ 106 /2400 = 281250
πππππ =0.5 (πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/3069 = 879,765.40
Distribution Factor for the column =
πππππ’ππ
=
βπ
281250 281250+ 879,765.40
= 0.242 wπΏ2
πΉπΈππβπ =
=
ππ = πΉπΈππβπ Γ Distribution Factor +
12
(28.39)(3.069)2
= 22.28 Γ 0.242 +
12
= 22.28kNm
ππΈπ·,π¦ π0
400 43.56(1904.84 Γ 10β3 ) 400
= 5.60kNm
Step 3. Check for biaxial bending
ππ§ =
=
ππ§ β
ππ§
ππ¦ =
ππΈπ·,π
46.84
=
118.22
ππ¦ ππΈπ·,π
5.60 43.56
= 0.396m
= 0.129m
= 396mm
= 129mm
/
ππ¦ π
=
396 300
/
129
ππ¦
300
π
ππ§ β
/
ππ¦ π
> 0.2 and
ππ§ β
=
129 300
/
396 300
= 0.33 >0.2
= 3.07 > 0.2 β΄ Since
/
ππ¦ π
/
ππ§ β
> 0.2, thus there is biaxial bending.
10
Step 4. Compute π΄π¬π« Y
dβ
300mm Z
Z
hβ
Mz
π2
bβ 300mm
My Y
Figure 3 Section of column with biaxial bending (re-illustrated for easy reference) dβ = π2 = Cover + β /2 = 55 + 16/2 = 63mm
ππ§
=
ββ² ππ¦ πβ²
As
=
ππ§ ββ²
46.84 Γ106 237 5.60 Γ106 237
β₯
ππ¦ πβ²
bβ = b -π2
hβ = h β dβ = 300 -63 = 237mm
= 300 β 63 = 237mm
= 197.64 Γ πππN
= 23.63 Γ πππ N
, thus:
ππΈπ· = πβ² π§ = ππ§ + π½
ββ² πβ²
Γ ππ¦
= 46.84 + 0.94 = 52.10kNm
237 237
Γ 5.60
π½=1β
ππΈπ· πβπππ
=1β
161.78Γ103 (300)(300)30
= 0.94
11
Step 5. Compute As, required. ππΈπ· πβπππ
161.78Γ103 = (300)(300)30
= 0.06 52.10Γ106 = πβ2 πππ (300)(3002 )30 ππΈπ·
= 0.06 Check: π β² /β = 63/300 = 0.21 β΄ Use πβ² /π = 0.25 chart
From Column Design Chart π΄π ππ¦π πβπππ
= 0.10
π΄π ,ππππ’ππππ =
0.10Γ 300 Γ300Γ 30
= 540ππ
500
π
π΄π ,πππ = minimum area of reinforcement allowed = 0.002(300)(300) = 180ππ2 π΄π ,πππ₯ = maximum area of reinforcement allowed = 0.08(300)(300) = 7200ππ2 Number of steel bars to be provided: For rectangular column, a minimum of four bars is required (one in each corner).Bar diameter should not be less than 12mm. β΄ As π΄π ,ππππ’ππππ is so small, 4 bars of steel bar (H16) is adequate. π΄π ,ππππ£ππππ = 4 Γ π 162 /4 = 804.25πππ Maximum shear link spacing = min {20 Γ size of compression bar; least lateral dimension of the column; 400mm} = min {20 Γ 16 = 320mm; 300mm; 400mm} = 300mm β΄ Provide Shear link (H10) with spacing 300mm c/c. 12
3.2 Design of Column B1, B3, C1, C3 - Note that the loading on the beams are taken from progress report 2.
6.069m 3.069m
6.069m 38.96kN/m 28.43kN/m
56.78kN/m
Z-Z axis
Y-Y axis
Figure 10 Loading on Column B1, B3, C1, C3 Step 1. Determine whether the column is a short or long column. Although column is has a square cross section; however the beam connected to the column from Z-Z axis (6.069m) is longer than the beam from Y-Y axis (3.069m). Thus, the more critical case will be from Z-Z axis. Therefore the checking is done for Z-Z axis of the column.
π = 3.0 β 0.6 = 2.4m πΌππππ’ππ
πβ3 = 12
=
πΌππππ
300 Γ3003
πβ3 = 12 =
12
= πππ Γ πππ πππ
300 Γ6003 12
=ππππ Γ πππ πππ
(πΌ/π)ππππ’ππ
π1 = β =
2(πΌ/π)ππππ
πππΓπππ /2400
2[
2(5400Γ106 ππ4 ) 6069
]
= 0.08 13
π0
π
π
1 2 = 0.5 π β(1 + 0.45+π ) (1 + 0.45+π ) 1
= 0.5 (2400)β(1 +
π = βπΌ/π΄
2
0.08 0.45+0.08
= β675 Γ 106 /(300 Γ 300) = 86.60mm
) (2)
= 1820.64mm Assume that (1 +
π= =
π0
π2 0.45+π2
) = 2 as
the column is connected to footing.
π 1820.64 86.60
= 21.02
ππΈπ·,π = =
π€1 πΏ
+
π€2 πΏ
2 2 38.96 Γ 6.069
2 = 204.5kN
ππΈπ·,π =
+
28.43 Γ 6.069 2
=
π€πΏ 2 56.78 Γ 3.069
2 = 87.13kN
ππΈπ· = ππΈπ·,π + ππΈπ·,π = 204.5 + 87.13 = 291.63kN
ππππ = 26.2 / βππΈπ· /π΄π πππ 30
= 26.2 / β291.63 Γ 103 /300 Γ 300 Γ (0.85 Γ 1.5) = 60 > π = 21.02 β΄ The column is a short column.
14
Step 2. Find the loadings on the column.
ππΈπ· = 291.63kN (found from step 1) Step 3. Find the moment (π΄π πππ π΄π ) acting on the column 6.069m
3.069m 6.069m 56.78kN/m
38.96kN/m 28.43kN/m
FEM = wπΏ2 /12 FEM = wπΏ2 /12
FEM = wπΏ2 /12
Z-Z axis
Y-Y axis Figure 11 FEM from beam to Column B1, B3, C1, C3
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ = 675 Γ 106 /2400 = 281250
πππππ = 0.5(πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/6069 = 444,883.84
Distribution Factor for the column =
πππππ’ππ βπ
=
281250 281250+ 2 Γ 444,883.84
= 0.24 πΉπΈππβπ = =
w1 πΏ2 w2 πΏ2 12
-
ππ§ = πΉπΈππβπ Γ Distribution Factor +
12
(38.96)(6.069)2
12 = 32.32kNm
β
(28.43)(6.069) 12
2
= 32.32 Γ 0.24 +
ππΈπ·,π§ π0
400 204.5 Γ1820.64 Γ 10β3 400
= 8.69kNm 15
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ
πππππ = (πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/3069 = 879,765.40
6
= 675 Γ 10 /2400 = 281250
Distribution Factor for the column =
πππππ’ππ βπ 281250
=
281250+ 879,765.40
= 0.24 wπΏ2
πΉπΈππβπ =
ππ = πΉπΈππβπ Γ Distribution Factor +
12
(56.78)(3.069)2
=
12
= 44.57 Γ 0.24 +
= 44.57kNm
ππΈπ·,π¦ π0
400 87.13 Γ1820.64 Γ 10β3 400
= 11.09kNm
Step 3. Check for biaxial bending
ππ§ = =
ππ§
ππ¦ =
ππΈπ·,π§ 8.69
=
204.5
= 0.04m
β
/
ππ¦ π
= ππ§
40 300
β
/
87.13
= 127mm
/
127
ππ¦
300
π
ππ¦ π
> 0.2 and
/
ππ§ β
=
127 300
/
40 300
= 3.175 > 0.2
= 0.31 > 0.2 β΄ Since
ππΈπ·,π¦ 11.09
= 0.127m
= 40mm ππ§
ππ¦
ππ¦ π
/
ππ§ β
> 0.2, thus there is biaxial bending.
16
Step 4. Compute π΄π¬π« Y
dβ
300mm Z
Z
hβ
Mz
π2
bβ 300mm
My Y
Figure 3 Section of column with biaxial bending (re-illustrated for easy reference) dβ = π2 = Cover + β /2 = 55 + 16/2 = 63mm
ππ§ ββ² ππ¦ πβ²
As
= =
ππ¦ πβ²
8.69 Γ106 237
β₯ πββ²π§
= 300 β 63 = 237mm
= 36.67Γ πππ N
11.09 Γ106 237
bβ = b -π2
hβ = h β dβ = 300 -63 = 237mm
= 46.79 Γ πππN
thus:
ππΈπ· = πβ² π¦ = ππ¦ + π½
πβ² ββ²
Γ ππ§
237
= 11.09 + 0.89 237 Γ 8.69 = 18.82kNm
π½=1β
ππΈπ· πβπππ
=1β
291.63Γ103 (300)(300)30
= 0.89
17
Step 5. Compute As, required. 291.63Γ103
ππΈπ·
= (300)(300)30
πβπππ
= 0.11 18.82Γ106
ππΈπ· πβ2 πππ
= (300)(3002 )30 = 0.02
Check: π β² /β = 63/300 = 0.21 β΄ Use πβ² /π = 0.25 chart
From Column Design Chart π΄π ππ¦π πβπππ
=0
π΄π ,ππππ’ππππ = minimum area of reinforcement allowed = 0.002(300)(300) = 180πππ π΄π ,πππ₯ = maximum area of reinforcement allowed = 0.08(300)(300) = 7200ππ2 Number of steel bars to be provided: For rectangular column, a minimum of four bars is required (one in each corner).Bar diameter should not be less than 12mm β΄ As π΄π ,ππππ’ππππ is so small, 4 bars of steel bar (H16) is adequate. π΄π ,ππππ£ππππ = 4 Γ π 162 /4 = 804.25πππ
Maximum shear link spacing = min {20 Γ size of compression bar; least lateral dimension of the column; 400mm} = min {20 Γ 16 = 320mm; 300mm; 400mm} = 300mm β΄ Provide Shear link (H10) with spacing 300mm c/c.
18
3.3 Design of Column A2 and D2 - Note that the loading on the beams are taken from progress report 2.
3.069m
6.069m
3.069m 28.39kN/m
77.91kN/m
20.72kN/m
Z-Z axis
Y-Y axis Figure 12 Loading on Column A2 and D2
Step 1. Determine whether the column is a short or long column. Although column is has a square cross section; however the beam connected to the column from Z-Z axis (6.069m) is longer than the beam from Y-Y axis (3.069m). Thus, the more critical case will be from Z-Z axis. Therefore the checking is done for Z-Z axis of the column.
π = 3.0 β 0.6 = 2.4m πΌππππ’ππ
πβ3 = 12
=
300 Γ3003 12
= πππ Γ πππ πππ
πΌππππ
πβ3 = 12 =
300 Γ6003 12
=ππππ Γ πππ πππ
(πΌ/π)ππππ’ππ
π1 = β =
2(πΌ/π)ππππ
πππΓπππ /2400 2(5400Γ106 ππ4 ) 6069
= 0.158 19
π0
π
π = βπΌ/π΄
π
1 2 = 0.5 π β(1 + 0.45+π ) (1 + 0.45+π ) 1
= 0.5 (2400)β(1 +
2
0.158 0.45+0.158
= β675 Γ 106 /(300 Γ 300) = 86.60mm
) (2)
= 1904.84mm Assume that (1 +
π= =
π0
π2 0.45+π2
) = 2 as
the column is connected to footing.
π 1904.84 86.60
= 22
ππΈπ·,π = =
π€πΏ 2 77.91 Γ 6.069
ππΈπ·,π =
2 = 236.42kN
=
π€1 πΏ
+
π€2 πΏ
2 2 28.39 Γ 3.069
2 = 75.36kN
+
20.72 Γ 3.069 2
ππΈπ· = ππΈπ·,π + ππΈπ·,π = 236.42 + 75.36 = 311.78kN
ππππ = 26.2 / βππΈπ· /π΄π πππ 30
= 26.2 / β311.78 Γ 103 /300 Γ 300 Γ (0.85 Γ 1.5) = 58.04 > π = 21.02 β΄ The column is a short column.
20
Step 2. Find the loadings on the column.
ππΈπ· = 311.78kN (found from step 1)
Step 3. Find the moment (π΄π πππ π΄π ) acting on the column
3.069m
6.069m
3.069m 28.39kN/m
77.91kN/m
20.72kN/m
FEM = wπΏ2 /12
FEM = wπΏ2 /12 FEM = wπΏ2 /12
Z-Z axis
Y-Y axis Figure 13 FEM on Column A2 and D2
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ = 675 Γ 106 /2400 = 281250
πππππ = 0.5(πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/6069 = 444,883.84
Distribution Factor for the column =
πππππ’ππ βπ 281250
=
281250+ 444,883.84
= 0.387
21
πΉπΈππβπ = =
wπΏ2
ππ§ = πΉπΈππβπ Γ Distribution Factor +
12
(77.91)(6.069)2
= 239.14 Γ 0.387 +
12 = 239.14kN
ππΈπ·,π§ π0
400 236.42 Γ1904.84 Γ 10β3 400
=93.67kNm
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ
πππππ = (πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/3069 = 879,765.40
6
= 675 Γ 10 /2400 = 281250
Distribution Factor for the column =
πππππ’ππ βπ
=
281250 281250+ 2 Γ 879,765.40
= 0.138 w πΏ2 w2 πΏ2
1 ππΈπ·,π¦ π0 πΉπΈππβπ = 12 ππ = πΉπΈππβπ Γ Distribution Factor + 12 400 (28.39)(3.069)2 (20.72)(3.069)2 75.36 Γ1904.84 Γ 10β3 = β = 6.02 Γ 0.138 + 12 12 400 = 1.19kNm = 6.02kNm
Step 3. Check for biaxial bending
ππ§ = =
ππ§
ππ¦ =
ππΈπ·,π§ 93.67
=
236.42
= 0.396m
β
/
ππ¦ π
= ππ§
396 300
β
/
75.36
= 16mm
/
16
ππ¦
300
π
= 24.75 > 0.2 β΄ Since
ππΈπ·,π¦ 1.19
= 0.016m
= 396mm ππ§
ππ¦
ππ¦ π
/
ππ§ β
=
16 300
/
396 300
= 0.04
> 0.2, thus there is biaxial bending.
22
Step 4. Compute π΄π¬π« Y
dβ
300mm Z
Z
hβ
Mz
π2
bβ 300mm
My Y
Figure 3 Section of column with biaxial bending (re-illustrated for easy reference) dβ = π2 = Cover + β /2 = 55 + 16/2 = 63mm
ππ§
=
ββ² ππ¦ πβ²
As
=
ππ§ ββ²
93.67 Γ106 237 1.19 Γ106 237
β₯
ππ¦ πβ²
bβ = b -π2
hβ = h β dβ = 300 -63 = 237mm
= 300 β 63 = 237mm
= 395.23 Γ πππ N
= 5.02
Γ πππ N
thus:
ππΈπ· = πβ² π§ = ππ§ + π½
ββ² π
Γ ππ¦
237
= 93.67 + 0.88 237 Γ 1.19 = 94.72kNm
π½=1β
ππΈπ· πβπππ
=1β
311.78Γ103 (300)(300)30
= 0.88
23
Step 5. Compute As, required. 311.78Γ103
ππΈπ·
= (300)(300)30
πβπππ
= 0.12 94.72Γ106
ππΈπ· πβ2 πππ
= (300)(3002 )30 = 0.12
Check: π β² /β = 63/300 = 0.21 β΄ Use πβ² /π = 0.20 chart for 6 steel bar
From Column Design Chart π΄π ππ¦π πβπππ
= 0.2
π΄π ,ππππ’ππππ =
0.2Γ 300 Γ300Γ 30 500
= 1080πππ
π΄π ,πππ = minimum area of reinforcement allowed = 0.002(300)(300) = 180πππ π΄π ,πππ₯ = maximum area of reinforcement allowed = 0.08(300)(300) = 7200ππ2 Number of steel bars to be provided: According to the chart, 6 steel bars are required. For our project H16 is used. π΄π ,ππππ£ππππ = 6 Γ π 162 /4 = 1206.37πππ Maximum shear link spacing = min {20 Γ size of compression bar; least lateral dimension of the column; 400mm} = min {20 Γ 16 = 320mm; 300mm; 400mm} = 300mm β΄ Provide Shear link (H10) with spacing 300mm c/c.
24
3.4 Design of Column B2 and C2 - Note that the loading on the beams are taken from progress report 2.
6.069m 6.069m 77.91kN/m 56.85kN/m
Z-Z axis 3.069m
3.069m
56.78kN/m 41.43kN/m
Y-Y axis Figure 14 Loading on Column B2 and C2
25
Step 1. Determine whether the column is a short or long column. Although column is has a square cross section; however the beam connected to the column from Z-Z axis (6.069m) is longer than the beam from Y-Y axis (3.069m). Thus, the more critical case will be from Z-Z axis. Therefore the checking is done for Z-Z axis of the column.
π = 3.0 β 0.6 = 2.4m πΌππππ’ππ
πβ3 = 12
=
πΌππππ
300 Γ3003
πβ3 = 12 =
12 π
= πππ Γ ππ ππ
π
300 Γ6003 12
=ππππ Γ πππ πππ
(πΌ/π)ππππ’ππ
π1 = β
2(πΌ/π)ππππ
πππΓπππ /2400
=
2Γ
2(5400Γ106 ππ4 ) 6069
= 0.079
π0
π
π
1 2 = 0.5 π β(1 + 0.45+π ) (1 + 0.45+π ) 1
= 0.5 (2400)β(1 +
0.079 0.45+0.079
π = βπΌ/π΄
2
) (2)
= β675 Γ 106 /(300 Γ 300) = 86.60mm
= 1819.37mm Assume that (1 +
π= =
π0
π2 0.45+π2
) = 2 as
the column is connected to footing.
π 1819.37 86.60
= 21
26
ππΈπ·,π = =
π€1 πΏ
+
π€2 πΏ
2 2 77.91 Γ 6.069
2 = 408.93kN
ππΈπ·,π = 56.85 Γ 6.069 + 2
=
π€1 πΏ
+
π€2 πΏ
2 2 56.78 Γ 3.069
2 = 150.70kN
+
41.43 Γ 3.069 2
ππΈπ· = ππΈπ·,π + ππΈπ·,π = 408.93 + 150.70 = 559.63kN
ππππ = 26.2 / βππΈπ· /π΄π πππ 30
= 26.2 / β559.63 Γ 103 /300 Γ 300 Γ (0.85 Γ 1.5) = 43.32 > π = 21 β΄ The column is a short column.
27
Step 2. Find the loadings on the column.
ππΈπ· = 559.63kN (found from step 1)
Step 3. Find the moment (π΄π πππ π΄π ) acting on the column
6.069m
6.069m
77.91kN/m 56.85kN/m
FEM = wπΏ2 /12
Z-Z axis
3.069m
3.069m
56.78kN/m 41.43kN/m
FEM = wπΏ2 /12
FEM = wπΏ2 /12
Y-Y axis Figure 15 FEM on Column B2 and C2
28
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ = 675 Γ 106 /2400 = 281250
πππππ = 0.5(πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/6069 = 444,883.84
Distribution Factor for the column =
πππππ’ππ βπ
=
281250 281250+ 2 Γ 444,883.84
= 0.24 πΉπΈππβπ = =
w1 πΏ2 w2 πΏ2 12
-
ππ§ = πΉπΈππβπ Γ Distribution Factor +
12
(77.91)(6.069)2
12 = 64.64kN
β
(56.85)(6.069) 12
2
= 64.64 Γ 0.24 +
ππΈπ·,π§ π0
400 408.93 Γ1819.37 Γ 10β3 400
=17.37kNm
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ
πππππ = (πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/3069 = 879,765.40
6
= 675 Γ 10 /2400 = 281250
Distribution Factor for the column =
πππππ’ππ βπ
=
281250 281250+ 2 Γ 879,765.40
= 0.138 w πΏ2 w2 πΏ2
1 ππΈπ·,π¦ π0 πΉπΈππβπ = 12 ππ = πΉπΈππβπ Γ Distribution Factor + 12 400 (56.78)(3.069)2 (41.43)(3.069)2 150.70 Γ1819.37 Γ 10β3 = β = 12.05 Γ 0.138 + 12 12 400 = 2.35kNm = 12.05kNm
29
Step 3. Check for biaxial bending
ππ§ = =
ππ§
ππ¦ =
ππΈπ·,π§ 17.37
=
408.93
= 0.042m
β
/
ππ¦ π
= ππ§
42 300
β
/
150.7
= 16mm
/
16
ππ¦
300
π
/
ππ¦ π
ππ§ β
=
16 300
/
42 300
= 0.38 > 0.2
= 2.625 > 0.2 β΄ Since
ππΈπ·,π¦ 2.35
= 0.016m
= 42mm ππ§
ππ¦
> 0.2 and
ππ¦ π
/
ππ§ β
> 0.2, thus there is biaxial bending.
Step 4. Compute π΄π¬π« Y
dβ
300mm Z
Z
hβ
Mz
π2
bβ 300mm
My Y
Figure 3 Section of column with biaxial bending (re-illustrated for easy reference) dβ = π2 = Cover + β /2 = 55 + 16/2 = 63mm
hβ = h β dβ = 300 -63 = 237mm
bβ = b -π2
= 300 β 63 = 237mm
30
ππ§
=
ββ² ππ¦ πβ²
As
=
ππ§ ββ²
17.37 Γ106 237 2.35 Γ106 237
β₯
ππ¦
= 73.29 Γ πππ N
= 9.92
Γ πππ N
thus:
πβ²
ππΈπ· = πβ² π§ = ππ§ + π½
ββ² π
Γ ππ¦
237
= 17.37 + 0.79 237 Γ 2.35 = 19.23kNm
π½=1β
ππΈπ· πβπππ
=1β
559.63Γ103 (300)(300)30
= 0.79 Step 5. Compute As, required. 559.63Γ103
ππΈπ·
= (300)(300)30
πβπππ
= 0.21 19.23Γ106
ππΈπ· πβ2 πππ
= (300)(3002 )30 = 0.02
Check: π β² /β = 63/300 = 0.21 β΄ Use πβ² /π = 0.25 chart for 4 steel bar
From Column Design Chart π΄π ππ¦π πβπππ
=0
π΄π ,ππππ’ππππ = minimum area of reinforcement allowed = 0.002(300)(300) = 180πππ π΄π ,πππ₯ = maximum area of reinforcement allowed = 0.08(300)(300) = 7200ππ2
31
Number of steel bars to be provided: For rectangular column, a minimum of four bars is required (one in each corner).Bar diameter should not be less than 12mm β΄ As π΄π ,ππππ’ππππ is so small, 4 bars of steel bar (H16) is adequate. π΄π ,ππππ£ππππ = 4 Γ π 162 /4 = 804.25πππ Maximum shear link spacing = min {20 Γ size of compression bar; least lateral dimension of the column; 400mm} = min {20 Γ 16 = 320mm; 300mm; 400mm} = 300mm β΄ Provide Shear link (H10) with spacing 300mm c/c.
4.0 CONCLUSION In this report, 4 columns were designed: 1. Column A1, A3, D1, D3 2. Column B1, B3, C1, C3 3. Column A2 and D2 4. Column B2 and C2 The design of these columns is done using charts from Eurocode 2.
32
2.0 PROCEDURE (ONLY FOR SHORT COLUMN) Step 1. Determine whether the column is a short or long column by computing slenderness ratio (π)
π=
π0 π
=
π0
[Equation 1]
βπΌ/π΄
Where π0 is the effective height of the column. π is the radius of gyration about the axis considered. πΌ is the second moment of area of the section about the axis. π΄ is the cross-sectional area of column
- π0 for braced members (Note that only braced members is covered in the course)can be computed with the following formula: π0
π
π
1 2 = 0.5 π β(1 + 0.45+π ) (1 + 0.45+π ) 1
[Equation 2]
2
Where π is the clear height of the column between end restraint. π1 and π2 is the relative flexibilities of the rotational restraints at ends β1β and β2β of the column respectively. π
2 For column connected to footing, assume (1 + 0.45+π )=2 2
k=
ππππ’ππ π π‘ππππππ π β ππππ π π‘ππππππ π
(πΈπΌ/π)ππππ’ππ
=β
2(πΈπΌ/π)ππππ
(πΌ/π)ππππ’ππ
=β
2(πΌ/π)ππππ
1
- Then, compare π with ππππ .
ππππ = 26.2 / βππΈπ· /π΄π πππ
[Equation 3]
- If π > ππππ , then the column is a long column. - If π < ππππ , then the column is a short column. Step 2. Find the loadings on the column. - First, the load distribution of the slab must be determined. - Then using the load distributed to the beam from the slab, half of the load will be transferred to the column, while the other half of the load is transferred to the adjacent column. Half length
Half length
Figure 1 Load transfer from beam to column - Note that to have the worst case scenario, for interior column, the longer beam has to be under maximum loading (1.35Gk + 1.5Qk) while the short beam has to be under minimum loading (1.35Gk).
Max (1.35Gk + 1.5Qk)
Max (1.35Gk + 1.5Qk) Min (1.35Gk)
Corner Column
Interior Column
Figure 2 worst case scenarios for corner column and interior column 2
Step 3. Find the moment (ππ§ πππ ππ¦ ) acting on the column - Using the load on the beam, find Fixed End Moment (FEM) = wπΏ2 /12.
Max (1.35Gk + 1.5Qk) Min (1.35Gk)
FEM = wπΏ2 /12
FEM = wπΏ2 /12
- For an example, we start from the z-z axis. - Using the net moment, multiply it with the distribution factor, k and add k=
ππππ’ππ π π‘ππππππ π β ππππ π π‘ππππππ π
(πΈπΌ/π)ππππ’ππ
=β
2(πΈπΌ/π)ππππ
ππΈπ· π0 400
to get ππ§ .
(πΌ/π)ππππ’ππ
=β
2(πΌ/π)ππππ
- Repeat the same steps for the y-y axis to get ππ¦ Step 3. Check for biaxial bending - Note that for a normal frame, moment will always come from two different axis (z or y-axis) - Thus, it is important to check for biaxial bending. - First, find the eccentricities of both axis (ππ§ πππ ππ¦ )
ππ§ =
ππ§ ππΈπ·,π§
,
ππ¦ =
ππ¦
[Equation 4]
ππΈπ·,π¦
- Then, check the ratio of the corresponding eccentricities that satisfies one of the conditions. Either
ππ§ β
/
ππ¦ π
β€ 0.2
or
ππ¦ π
/
ππ§ β
β€ 0.2
- If either one of the condition is satisfied, then there is a biaxial bending.
3
Step 4. Compute π΄π¬π« Y
dβ
h Z
Z
hβ
Mz
bβ
d2
b My Y
Figure 3 Section of column with biaxial bending
- Then check for the dominating moment: If
ππ§ ββ²
β₯
ππ¦ πβ²
Then ππΈπ· = π β² π§ = ππ§ + π½
If
ππ§ ββ²
β€
ββ² πβ²
Γ ππ¦
[Equation 5]
Γ ππ§
[Equation 6]
ππ¦ πβ²
Then ππΈπ· = π β² π¦ = ππ¦ + π½
Where π½ = 1 β
πβ² ββ²
ππΈπ· πβπππ
4
Step 5. Compute π¨π,ππππππππ - Using values of
ππΈπ· πβπππ
and
ππΈπ· πβ2 π
ππ
, to find
π΄π ππ¦π πβπππ
from the column chart.
- To determine which chart to use, you have to check dβ/h ratio, each ratio has a different chart. π΄π ππ¦π
- Once obtain the value of πβπ , π΄π can be computed by substituting all the other known ππ
values. π΄π ππ¦π
- If πβπ = 0, then π΄π = minimum area of reinforcement allowed = 0.002bh. ππ - Do check for maximum area of reinforcement = 0.08bh.
Figure 4 Column Design Chart for d2/h = 0.25
5
3.0 DATA & RESULT
B
A 6.069m
C 6.069m
D 6.069m
1 3.069m
2 3.069m
3 Figure 5 Plan View
beam
300mm
v 600mm
b =300mm
Assume h =300mm
h =3m
Figure 6 Cross Section of Column and Beam
Figure 7. Section of Column
- πππ = 30π/ππ2 ππ¦π = 500π/ππ2 Diameter of main steel bar = 16mm - It can be seen that there are four columns to be designed: 1. Column A1, A3, D1, D3 2. Column B1, B3, C1, C3 3. Column A2 and D2 4. Column B2 and C2
6
3.1 Design of Column A1, A3, D1, D3 - Note that the loading on the beams are taken from progress report 2. 6.069m
3.069m
38.96kN/m
28.39kN/m
Z-Z axis
Y-Y axis Figure 8 Loading on Column A1, A3, D1, D3
Step 1. Determine whether the column is a short or long column. Although column is has a square cross section; however the beam connected to the column from Z-Z axis (6.069m) is longer than the beam from Y-Y axis (3.069m). Thus, the more critical case will be from Z-Z axis. Therefore the checking is done for Z-Z axis of the column.
π = 3.0 β 0.6 = 2.4m πΌππππ’ππ
πβ3 = 12
=
πΌππππ
300 Γ3003
πβ3 = 12 =
12 π
= πππ Γ ππ ππ
π
300 Γ6003 12
= ππππ Γ πππ πππ
(πΌ/π)ππππ’ππ
π1 = β =
2(πΌ/π)ππππ πππΓπππ /2400
2(5400Γ106 ππ4 )/6069
= 0.158
7
π0
π
π
1 2 = 0.5 π β(1 + 0.45+π ) (1 + 0.45+π ) 1
= 0.5 (2400)β(1 +
π = βπΌ/π΄
2
0.158 0.45+0.158
= β675 Γ 106 /(300 Γ 300) = 86.60mm
) (2)
= 1904.84mm Assume that (1 +
π= =
π0
π2 0.45+π2
) = 2 as
the column is connected to footing.
π 1904.84 86.60
= 22
ππΈπ·,π =
38.96 Γ 6.069
2 = 118.22kN
ππΈπ·,π =
28.39 Γ 3.069
2 = 43.56kN
ππΈπ· = ππΈπ·,π + ππΈπ·,π = 118.22 + 43.56 = 161.78kN
ππππ = 26.2 / βππΈπ· /π΄π πππ 30
= 26.2 / β161.78 Γ 103 /300 Γ 300 Γ (0.85 Γ 1.5) = 80.57 > π = 22 β΄ The column is a short column.
8
Step 2. Find the loadings on the column.
ππΈπ· = 161.78kN (found from step 1) Step 3. Find the moment (π΄π πππ π΄π ) acting on the column 6.069m
3.069m
38.96kN/m
28.39kN/m
FEM = wπΏ2 /12
FEM = wπΏ2 /12
Z-Z axis
Y-Y axis Figure 9 FEM from beam to Column A1, A3, D1, D3
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ = 675 Γ 106 /2400 = 281250
πππππ = 0.5(πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/6069 = 444883.84
Distribution Factor for the column = =
πππππ’ππ βπ
281250 281250+ 444883.84
= 0.387 πΉπΈππβπ =
=
wπΏ2 12
(38.96)(6.069)2
12 = 119.58kNm
ππ§ = πΉπΈππβπ Γ Distribution Factor + = 119.58 Γ 0.387 +
ππΈπ·,π§ π0
400 118.22(1904.84 Γ 10β3 ) 400
= 46.84kNm 9
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ = 675 Γ 106 /2400 = 281250
πππππ =0.5 (πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/3069 = 879,765.40
Distribution Factor for the column =
πππππ’ππ
=
βπ
281250 281250+ 879,765.40
= 0.242 wπΏ2
πΉπΈππβπ =
=
ππ = πΉπΈππβπ Γ Distribution Factor +
12
(28.39)(3.069)2
= 22.28 Γ 0.242 +
12
= 22.28kNm
ππΈπ·,π¦ π0
400 43.56(1904.84 Γ 10β3 ) 400
= 5.60kNm
Step 3. Check for biaxial bending
ππ§ =
=
ππ§ β
ππ§
ππ¦ =
ππΈπ·,π
46.84
=
118.22
ππ¦ ππΈπ·,π
5.60 43.56
= 0.396m
= 0.129m
= 396mm
= 129mm
/
ππ¦ π
=
396 300
/
129
ππ¦
300
π
ππ§ β
/
ππ¦ π
> 0.2 and
ππ§ β
=
129 300
/
396 300
= 0.33 >0.2
= 3.07 > 0.2 β΄ Since
/
ππ¦ π
/
ππ§ β
> 0.2, thus there is biaxial bending.
10
Step 4. Compute π΄π¬π« Y
dβ
300mm Z
Z
hβ
Mz
π2
bβ 300mm
My Y
Figure 3 Section of column with biaxial bending (re-illustrated for easy reference) dβ = π2 = Cover + β /2 = 55 + 16/2 = 63mm
ππ§
=
ββ² ππ¦ πβ²
As
=
ππ§ ββ²
46.84 Γ106 237 5.60 Γ106 237
β₯
ππ¦ πβ²
bβ = b -π2
hβ = h β dβ = 300 -63 = 237mm
= 300 β 63 = 237mm
= 197.64 Γ πππN
= 23.63 Γ πππ N
, thus:
ππΈπ· = πβ² π§ = ππ§ + π½
ββ² πβ²
Γ ππ¦
= 46.84 + 0.94 = 52.10kNm
237 237
Γ 5.60
π½=1β
ππΈπ· πβπππ
=1β
161.78Γ103 (300)(300)30
= 0.94
11
Step 5. Compute As, required. ππΈπ· πβπππ
161.78Γ103 = (300)(300)30
= 0.06 52.10Γ106 = πβ2 πππ (300)(3002 )30 ππΈπ·
= 0.06 Check: π β² /β = 63/300 = 0.21 β΄ Use πβ² /π = 0.25 chart
From Column Design Chart π΄π ππ¦π πβπππ
= 0.10
π΄π ,ππππ’ππππ =
0.10Γ 300 Γ300Γ 30
= 540ππ
500
π
π΄π ,πππ = minimum area of reinforcement allowed = 0.002(300)(300) = 180ππ2 π΄π ,πππ₯ = maximum area of reinforcement allowed = 0.08(300)(300) = 7200ππ2 Number of steel bars to be provided: For rectangular column, a minimum of four bars is required (one in each corner).Bar diameter should not be less than 12mm. β΄ As π΄π ,ππππ’ππππ is so small, 4 bars of steel bar (H16) is adequate. π΄π ,ππππ£ππππ = 4 Γ π 162 /4 = 804.25πππ Maximum shear link spacing = min {20 Γ size of compression bar; least lateral dimension of the column; 400mm} = min {20 Γ 16 = 320mm; 300mm; 400mm} = 300mm β΄ Provide Shear link (H10) with spacing 300mm c/c. 12
3.2 Design of Column B1, B3, C1, C3 - Note that the loading on the beams are taken from progress report 2.
6.069m 3.069m
6.069m 38.96kN/m 28.43kN/m
56.78kN/m
Z-Z axis
Y-Y axis
Figure 10 Loading on Column B1, B3, C1, C3 Step 1. Determine whether the column is a short or long column. Although column is has a square cross section; however the beam connected to the column from Z-Z axis (6.069m) is longer than the beam from Y-Y axis (3.069m). Thus, the more critical case will be from Z-Z axis. Therefore the checking is done for Z-Z axis of the column.
π = 3.0 β 0.6 = 2.4m πΌππππ’ππ
πβ3 = 12
=
πΌππππ
300 Γ3003
πβ3 = 12 =
12
= πππ Γ πππ πππ
300 Γ6003 12
=ππππ Γ πππ πππ
(πΌ/π)ππππ’ππ
π1 = β =
2(πΌ/π)ππππ
πππΓπππ /2400
2[
2(5400Γ106 ππ4 ) 6069
]
= 0.08 13
π0
π
π
1 2 = 0.5 π β(1 + 0.45+π ) (1 + 0.45+π ) 1
= 0.5 (2400)β(1 +
π = βπΌ/π΄
2
0.08 0.45+0.08
= β675 Γ 106 /(300 Γ 300) = 86.60mm
) (2)
= 1820.64mm Assume that (1 +
π= =
π0
π2 0.45+π2
) = 2 as
the column is connected to footing.
π 1820.64 86.60
= 21.02
ππΈπ·,π = =
π€1 πΏ
+
π€2 πΏ
2 2 38.96 Γ 6.069
2 = 204.5kN
ππΈπ·,π =
+
28.43 Γ 6.069 2
=
π€πΏ 2 56.78 Γ 3.069
2 = 87.13kN
ππΈπ· = ππΈπ·,π + ππΈπ·,π = 204.5 + 87.13 = 291.63kN
ππππ = 26.2 / βππΈπ· /π΄π πππ 30
= 26.2 / β291.63 Γ 103 /300 Γ 300 Γ (0.85 Γ 1.5) = 60 > π = 21.02 β΄ The column is a short column.
14
Step 2. Find the loadings on the column.
ππΈπ· = 291.63kN (found from step 1) Step 3. Find the moment (π΄π πππ π΄π ) acting on the column 6.069m
3.069m 6.069m 56.78kN/m
38.96kN/m 28.43kN/m
FEM = wπΏ2 /12 FEM = wπΏ2 /12
FEM = wπΏ2 /12
Z-Z axis
Y-Y axis Figure 11 FEM from beam to Column B1, B3, C1, C3
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ = 675 Γ 106 /2400 = 281250
πππππ = 0.5(πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/6069 = 444,883.84
Distribution Factor for the column =
πππππ’ππ βπ
=
281250 281250+ 2 Γ 444,883.84
= 0.24 πΉπΈππβπ = =
w1 πΏ2 w2 πΏ2 12
-
ππ§ = πΉπΈππβπ Γ Distribution Factor +
12
(38.96)(6.069)2
12 = 32.32kNm
β
(28.43)(6.069) 12
2
= 32.32 Γ 0.24 +
ππΈπ·,π§ π0
400 204.5 Γ1820.64 Γ 10β3 400
= 8.69kNm 15
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ
πππππ = (πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/3069 = 879,765.40
6
= 675 Γ 10 /2400 = 281250
Distribution Factor for the column =
πππππ’ππ βπ 281250
=
281250+ 879,765.40
= 0.24 wπΏ2
πΉπΈππβπ =
ππ = πΉπΈππβπ Γ Distribution Factor +
12
(56.78)(3.069)2
=
12
= 44.57 Γ 0.24 +
= 44.57kNm
ππΈπ·,π¦ π0
400 87.13 Γ1820.64 Γ 10β3 400
= 11.09kNm
Step 3. Check for biaxial bending
ππ§ = =
ππ§
ππ¦ =
ππΈπ·,π§ 8.69
=
204.5
= 0.04m
β
/
ππ¦ π
= ππ§
40 300
β
/
87.13
= 127mm
/
127
ππ¦
300
π
ππ¦ π
> 0.2 and
/
ππ§ β
=
127 300
/
40 300
= 3.175 > 0.2
= 0.31 > 0.2 β΄ Since
ππΈπ·,π¦ 11.09
= 0.127m
= 40mm ππ§
ππ¦
ππ¦ π
/
ππ§ β
> 0.2, thus there is biaxial bending.
16
Step 4. Compute π΄π¬π« Y
dβ
300mm Z
Z
hβ
Mz
π2
bβ 300mm
My Y
Figure 3 Section of column with biaxial bending (re-illustrated for easy reference) dβ = π2 = Cover + β /2 = 55 + 16/2 = 63mm
ππ§ ββ² ππ¦ πβ²
As
= =
ππ¦ πβ²
8.69 Γ106 237
β₯ πββ²π§
= 300 β 63 = 237mm
= 36.67Γ πππ N
11.09 Γ106 237
bβ = b -π2
hβ = h β dβ = 300 -63 = 237mm
= 46.79 Γ πππN
thus:
ππΈπ· = πβ² π¦ = ππ¦ + π½
πβ² ββ²
Γ ππ§
237
= 11.09 + 0.89 237 Γ 8.69 = 18.82kNm
π½=1β
ππΈπ· πβπππ
=1β
291.63Γ103 (300)(300)30
= 0.89
17
Step 5. Compute As, required. 291.63Γ103
ππΈπ·
= (300)(300)30
πβπππ
= 0.11 18.82Γ106
ππΈπ· πβ2 πππ
= (300)(3002 )30 = 0.02
Check: π β² /β = 63/300 = 0.21 β΄ Use πβ² /π = 0.25 chart
From Column Design Chart π΄π ππ¦π πβπππ
=0
π΄π ,ππππ’ππππ = minimum area of reinforcement allowed = 0.002(300)(300) = 180πππ π΄π ,πππ₯ = maximum area of reinforcement allowed = 0.08(300)(300) = 7200ππ2 Number of steel bars to be provided: For rectangular column, a minimum of four bars is required (one in each corner).Bar diameter should not be less than 12mm β΄ As π΄π ,ππππ’ππππ is so small, 4 bars of steel bar (H16) is adequate. π΄π ,ππππ£ππππ = 4 Γ π 162 /4 = 804.25πππ
Maximum shear link spacing = min {20 Γ size of compression bar; least lateral dimension of the column; 400mm} = min {20 Γ 16 = 320mm; 300mm; 400mm} = 300mm β΄ Provide Shear link (H10) with spacing 300mm c/c.
18
3.3 Design of Column A2 and D2 - Note that the loading on the beams are taken from progress report 2.
3.069m
6.069m
3.069m 28.39kN/m
77.91kN/m
20.72kN/m
Z-Z axis
Y-Y axis Figure 12 Loading on Column A2 and D2
Step 1. Determine whether the column is a short or long column. Although column is has a square cross section; however the beam connected to the column from Z-Z axis (6.069m) is longer than the beam from Y-Y axis (3.069m). Thus, the more critical case will be from Z-Z axis. Therefore the checking is done for Z-Z axis of the column.
π = 3.0 β 0.6 = 2.4m πΌππππ’ππ
πβ3 = 12
=
300 Γ3003 12
= πππ Γ πππ πππ
πΌππππ
πβ3 = 12 =
300 Γ6003 12
=ππππ Γ πππ πππ
(πΌ/π)ππππ’ππ
π1 = β =
2(πΌ/π)ππππ
πππΓπππ /2400 2(5400Γ106 ππ4 ) 6069
= 0.158 19
π0
π
π = βπΌ/π΄
π
1 2 = 0.5 π β(1 + 0.45+π ) (1 + 0.45+π ) 1
= 0.5 (2400)β(1 +
2
0.158 0.45+0.158
= β675 Γ 106 /(300 Γ 300) = 86.60mm
) (2)
= 1904.84mm Assume that (1 +
π= =
π0
π2 0.45+π2
) = 2 as
the column is connected to footing.
π 1904.84 86.60
= 22
ππΈπ·,π = =
π€πΏ 2 77.91 Γ 6.069
ππΈπ·,π =
2 = 236.42kN
=
π€1 πΏ
+
π€2 πΏ
2 2 28.39 Γ 3.069
2 = 75.36kN
+
20.72 Γ 3.069 2
ππΈπ· = ππΈπ·,π + ππΈπ·,π = 236.42 + 75.36 = 311.78kN
ππππ = 26.2 / βππΈπ· /π΄π πππ 30
= 26.2 / β311.78 Γ 103 /300 Γ 300 Γ (0.85 Γ 1.5) = 58.04 > π = 21.02 β΄ The column is a short column.
20
Step 2. Find the loadings on the column.
ππΈπ· = 311.78kN (found from step 1)
Step 3. Find the moment (π΄π πππ π΄π ) acting on the column
3.069m
6.069m
3.069m 28.39kN/m
77.91kN/m
20.72kN/m
FEM = wπΏ2 /12
FEM = wπΏ2 /12 FEM = wπΏ2 /12
Z-Z axis
Y-Y axis Figure 13 FEM on Column A2 and D2
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ = 675 Γ 106 /2400 = 281250
πππππ = 0.5(πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/6069 = 444,883.84
Distribution Factor for the column =
πππππ’ππ βπ 281250
=
281250+ 444,883.84
= 0.387
21
πΉπΈππβπ = =
wπΏ2
ππ§ = πΉπΈππβπ Γ Distribution Factor +
12
(77.91)(6.069)2
= 239.14 Γ 0.387 +
12 = 239.14kN
ππΈπ·,π§ π0
400 236.42 Γ1904.84 Γ 10β3 400
=93.67kNm
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ
πππππ = (πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/3069 = 879,765.40
6
= 675 Γ 10 /2400 = 281250
Distribution Factor for the column =
πππππ’ππ βπ
=
281250 281250+ 2 Γ 879,765.40
= 0.138 w πΏ2 w2 πΏ2
1 ππΈπ·,π¦ π0 πΉπΈππβπ = 12 ππ = πΉπΈππβπ Γ Distribution Factor + 12 400 (28.39)(3.069)2 (20.72)(3.069)2 75.36 Γ1904.84 Γ 10β3 = β = 6.02 Γ 0.138 + 12 12 400 = 1.19kNm = 6.02kNm
Step 3. Check for biaxial bending
ππ§ = =
ππ§
ππ¦ =
ππΈπ·,π§ 93.67
=
236.42
= 0.396m
β
/
ππ¦ π
= ππ§
396 300
β
/
75.36
= 16mm
/
16
ππ¦
300
π
= 24.75 > 0.2 β΄ Since
ππΈπ·,π¦ 1.19
= 0.016m
= 396mm ππ§
ππ¦
ππ¦ π
/
ππ§ β
=
16 300
/
396 300
= 0.04
> 0.2, thus there is biaxial bending.
22
Step 4. Compute π΄π¬π« Y
dβ
300mm Z
Z
hβ
Mz
π2
bβ 300mm
My Y
Figure 3 Section of column with biaxial bending (re-illustrated for easy reference) dβ = π2 = Cover + β /2 = 55 + 16/2 = 63mm
ππ§
=
ββ² ππ¦ πβ²
As
=
ππ§ ββ²
93.67 Γ106 237 1.19 Γ106 237
β₯
ππ¦ πβ²
bβ = b -π2
hβ = h β dβ = 300 -63 = 237mm
= 300 β 63 = 237mm
= 395.23 Γ πππ N
= 5.02
Γ πππ N
thus:
ππΈπ· = πβ² π§ = ππ§ + π½
ββ² π
Γ ππ¦
237
= 93.67 + 0.88 237 Γ 1.19 = 94.72kNm
π½=1β
ππΈπ· πβπππ
=1β
311.78Γ103 (300)(300)30
= 0.88
23
Step 5. Compute As, required. 311.78Γ103
ππΈπ·
= (300)(300)30
πβπππ
= 0.12 94.72Γ106
ππΈπ· πβ2 πππ
= (300)(3002 )30 = 0.12
Check: π β² /β = 63/300 = 0.21 β΄ Use πβ² /π = 0.20 chart for 6 steel bar
From Column Design Chart π΄π ππ¦π πβπππ
= 0.2
π΄π ,ππππ’ππππ =
0.2Γ 300 Γ300Γ 30 500
= 1080πππ
π΄π ,πππ = minimum area of reinforcement allowed = 0.002(300)(300) = 180πππ π΄π ,πππ₯ = maximum area of reinforcement allowed = 0.08(300)(300) = 7200ππ2 Number of steel bars to be provided: According to the chart, 6 steel bars are required. For our project H16 is used. π΄π ,ππππ£ππππ = 6 Γ π 162 /4 = 1206.37πππ Maximum shear link spacing = min {20 Γ size of compression bar; least lateral dimension of the column; 400mm} = min {20 Γ 16 = 320mm; 300mm; 400mm} = 300mm β΄ Provide Shear link (H10) with spacing 300mm c/c.
24
3.4 Design of Column B2 and C2 - Note that the loading on the beams are taken from progress report 2.
6.069m 6.069m 77.91kN/m 56.85kN/m
Z-Z axis 3.069m
3.069m
56.78kN/m 41.43kN/m
Y-Y axis Figure 14 Loading on Column B2 and C2
25
Step 1. Determine whether the column is a short or long column. Although column is has a square cross section; however the beam connected to the column from Z-Z axis (6.069m) is longer than the beam from Y-Y axis (3.069m). Thus, the more critical case will be from Z-Z axis. Therefore the checking is done for Z-Z axis of the column.
π = 3.0 β 0.6 = 2.4m πΌππππ’ππ
πβ3 = 12
=
πΌππππ
300 Γ3003
πβ3 = 12 =
12 π
= πππ Γ ππ ππ
π
300 Γ6003 12
=ππππ Γ πππ πππ
(πΌ/π)ππππ’ππ
π1 = β
2(πΌ/π)ππππ
πππΓπππ /2400
=
2Γ
2(5400Γ106 ππ4 ) 6069
= 0.079
π0
π
π
1 2 = 0.5 π β(1 + 0.45+π ) (1 + 0.45+π ) 1
= 0.5 (2400)β(1 +
0.079 0.45+0.079
π = βπΌ/π΄
2
) (2)
= β675 Γ 106 /(300 Γ 300) = 86.60mm
= 1819.37mm Assume that (1 +
π= =
π0
π2 0.45+π2
) = 2 as
the column is connected to footing.
π 1819.37 86.60
= 21
26
ππΈπ·,π = =
π€1 πΏ
+
π€2 πΏ
2 2 77.91 Γ 6.069
2 = 408.93kN
ππΈπ·,π = 56.85 Γ 6.069 + 2
=
π€1 πΏ
+
π€2 πΏ
2 2 56.78 Γ 3.069
2 = 150.70kN
+
41.43 Γ 3.069 2
ππΈπ· = ππΈπ·,π + ππΈπ·,π = 408.93 + 150.70 = 559.63kN
ππππ = 26.2 / βππΈπ· /π΄π πππ 30
= 26.2 / β559.63 Γ 103 /300 Γ 300 Γ (0.85 Γ 1.5) = 43.32 > π = 21 β΄ The column is a short column.
27
Step 2. Find the loadings on the column.
ππΈπ· = 559.63kN (found from step 1)
Step 3. Find the moment (π΄π πππ π΄π ) acting on the column
6.069m
6.069m
77.91kN/m 56.85kN/m
FEM = wπΏ2 /12
Z-Z axis
3.069m
3.069m
56.78kN/m 41.43kN/m
FEM = wπΏ2 /12
FEM = wπΏ2 /12
Y-Y axis Figure 15 FEM on Column B2 and C2
28
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ = 675 Γ 106 /2400 = 281250
πππππ = 0.5(πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/6069 = 444,883.84
Distribution Factor for the column =
πππππ’ππ βπ
=
281250 281250+ 2 Γ 444,883.84
= 0.24 πΉπΈππβπ = =
w1 πΏ2 w2 πΏ2 12
-
ππ§ = πΉπΈππβπ Γ Distribution Factor +
12
(77.91)(6.069)2
12 = 64.64kN
β
(56.85)(6.069) 12
2
= 64.64 Γ 0.24 +
ππΈπ·,π§ π0
400 408.93 Γ1819.37 Γ 10β3 400
=17.37kNm
For moment π΄π Member stiffness: πππππ’ππ = (πΌ/π)ππππ’ππ
πππππ = (πΌ/π)ππππ = 0.5(5400 Γ 106 ππ4 )/3069 = 879,765.40
6
= 675 Γ 10 /2400 = 281250
Distribution Factor for the column =
πππππ’ππ βπ
=
281250 281250+ 2 Γ 879,765.40
= 0.138 w πΏ2 w2 πΏ2
1 ππΈπ·,π¦ π0 πΉπΈππβπ = 12 ππ = πΉπΈππβπ Γ Distribution Factor + 12 400 (56.78)(3.069)2 (41.43)(3.069)2 150.70 Γ1819.37 Γ 10β3 = β = 12.05 Γ 0.138 + 12 12 400 = 2.35kNm = 12.05kNm
29
Step 3. Check for biaxial bending
ππ§ = =
ππ§
ππ¦ =
ππΈπ·,π§ 17.37
=
408.93
= 0.042m
β
/
ππ¦ π
= ππ§
42 300
β
/
150.7
= 16mm
/
16
ππ¦
300
π
/
ππ¦ π
ππ§ β
=
16 300
/
42 300
= 0.38 > 0.2
= 2.625 > 0.2 β΄ Since
ππΈπ·,π¦ 2.35
= 0.016m
= 42mm ππ§
ππ¦
> 0.2 and
ππ¦ π
/
ππ§ β
> 0.2, thus there is biaxial bending.
Step 4. Compute π΄π¬π« Y
dβ
300mm Z
Z
hβ
Mz
π2
bβ 300mm
My Y
Figure 3 Section of column with biaxial bending (re-illustrated for easy reference) dβ = π2 = Cover + β /2 = 55 + 16/2 = 63mm
hβ = h β dβ = 300 -63 = 237mm
bβ = b -π2
= 300 β 63 = 237mm
30
ππ§
=
ββ² ππ¦ πβ²
As
=
ππ§ ββ²
17.37 Γ106 237 2.35 Γ106 237
β₯
ππ¦
= 73.29 Γ πππ N
= 9.92
Γ πππ N
thus:
πβ²
ππΈπ· = πβ² π§ = ππ§ + π½
ββ² π
Γ ππ¦
237
= 17.37 + 0.79 237 Γ 2.35 = 19.23kNm
π½=1β
ππΈπ· πβπππ
=1β
559.63Γ103 (300)(300)30
= 0.79 Step 5. Compute As, required. 559.63Γ103
ππΈπ·
= (300)(300)30
πβπππ
= 0.21 19.23Γ106
ππΈπ· πβ2 πππ
= (300)(3002 )30 = 0.02
Check: π β² /β = 63/300 = 0.21 β΄ Use πβ² /π = 0.25 chart for 4 steel bar
From Column Design Chart π΄π ππ¦π πβπππ
=0
π΄π ,ππππ’ππππ = minimum area of reinforcement allowed = 0.002(300)(300) = 180πππ π΄π ,πππ₯ = maximum area of reinforcement allowed = 0.08(300)(300) = 7200ππ2
31
Number of steel bars to be provided: For rectangular column, a minimum of four bars is required (one in each corner).Bar diameter should not be less than 12mm β΄ As π΄π ,ππππ’ππππ is so small, 4 bars of steel bar (H16) is adequate. π΄π ,ππππ£ππππ = 4 Γ π 162 /4 = 804.25πππ Maximum shear link spacing = min {20 Γ size of compression bar; least lateral dimension of the column; 400mm} = min {20 Γ 16 = 320mm; 300mm; 400mm} = 300mm β΄ Provide Shear link (H10) with spacing 300mm c/c.
4.0 CONCLUSION In this report, 4 columns were designed: 1. Column A1, A3, D1, D3 2. Column B1, B3, C1, C3 3. Column A2 and D2 4. Column B2 and C2 The design of these columns is done using charts from Eurocode 2.
32
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