Page 1
Example 4.1 [Uni-axial Column Design]
1. Design the braced short column to sustain a design load of 1100 KN and a design moment of
160KNm which include all other effects .Use C25/30 and S460 class 1 works
Take πβ²
π»= 0.1 and section 270 ππ β 450 ππ
Solution
Step 1- Material
πππ =0.85 β 25
1.5= 14.16667 πππ
ππ¦π =460
1.15= 400 πππ
Step 2-Determine the normalized axial and bending moment value
π£π π =ππ π
ππππβ=
1100 β 103
14.1667 β 270 β 450= 0.639
ππ π =ππ π
πππππ2=
160 β 106
14.1667 β 270 β 4502= 0.2065
Step 3-Using πβ²
π»= 0.1 read the mechanical steel ratio from uniaxial interaction chart for
ππ π = 0.639 ππ π = 0.2065
From interaction chart π = 0.3
Page 2
π΄π ,π‘ππ‘ =ππππππ
ππ¦π=
0.3 β 14.166 β 270 β 450
400= 1290.937 ππ2
π΄ =π΄π ,π‘ππ‘
2=
1290.937
2= 645.468 ππ2
Check with maximum and minimum reinforcement limit
π΄π ,πππ = πππ₯ {
0.1 ππΈπ·
ππ¦π
0.002π΄π
= 275 ππ2 ππΎ!
π΄π ,πππ₯ = 0.08 π΄π = 0.08 β 270 β 450 = ππΎ!
Step 4- Detailing
ππ π 4β
16 ππ πππβ ππππ
Page 3
Example 4.2 [Biaxial column design]
Determine the longitudinal reinforcement for corner column of size 400*400 mm and the design
factored moment and axial force of
π = 1360 πΎπ ππ π,β = 200 πΎππ ππ π,π = 100 πΎππ
Use C25/30 and S460 class 1 work take ββ²
β=
πβ²
π= 0.1
Solution
Step 1- Material
πππ =0.85 β 25
1.5= 14.16667 πππ
ππ¦π =460
1.15= 400 πππ
Step 2- Determine the normalized axial and bending moment value
π£π π =π
ππππβ=
1360β103
14.1667β400β400= 0.6
ππ π,β =ππ π
πππππ2=
200 β 106
14.1667 β 400 β 4002= 0.2206
ππ π,π =ππ π
πππππ2=
100 β 106
14.1667 β 400 β 4002= 0.1103
Step 3- Find π using πβ²
π=
πβ²
π΅= 0.1 , ππ π = 0.6 ,
ππ π,β = 0.2206 ,
ππ π,π = 0.1103
From biaxial chart π = 0.6
π΄π ,π‘ππ‘ =ππππππ
ππ¦π=
0.6 β 14.166 β 400 β 400
400= 3400 ππ2
Check with maximum and minimum reinforcement limit
π΄π ,πππ = πππ₯ {
0.1 ππΈπ·
ππ¦π
0.002π΄π
= 340 ππ2 πΆπ²!
π΄π ,πππ₯ = 0.08 π΄π = 0.08 β 400 β 400 = 12800 πΆπ²!
Page 4
Step 4- Detailing
π’π πππ β
24 π =3400
452.16= 7.52 π π ππ π 8β
24
Page 5
Example 4.3 [Column]
Determine whether the column CD is slender or not, if it is subjected t loads shown below.
Consider the frame to be non-sway
Use C25/30 πππ = 14.1667 πππ
Solution
Step 1- Slenderness limit
Ζπππ = 20π΄π΅πΆβπ
β π‘πππ π΄ = 0.7 π΅ = 1.1 πΆ = 1.7 β ππ
Page 6
π€βπππ ππ =π01
π02=
β34.4
68.8= β0.5 πΆ = 1.7 β (β0.5) = 2.2
π =πππ
π΄ππππ=
516 β 103
14.1667 β 400 β 400= 0.227647
Ζπππ =20 β 0.7 β 1.1 β 2.2
β0.227647= 71.0088
Step 2-Find the slenderness of the column
2.1 Effective length
For Braced member
π0 = 0.5πβ[1 +πΎ1
0.5 + πΎ1] [1 +
πΎ2
0.5 + πΎ2]
πΎ =πΆπππ’ππ π π‘ππππππ π
β ππππ π π‘ππππππ π
πΎπ =(πΈπΌ
πβ )ππππ’ππ
2(2 πΈπΌπβ )ππππ
πΌππππ’ππ =400 β 4003
12= 2133333333 ππ4
πΌππππ =250 β 5003
12= 2604166667 ππ4
πΎ1 =
2133333333πΈ6
2(2 β 2604166667πΈ
6)
= 0.2048
For fixed restraint K=0 but in reality we cannot provide a fully fixed support so use πΎ2 = 0.1
π0 = 0.5 β 6000β[1 +0.2048
0.5 + 0.2048] [1 +
0.1
0.5 + 0.1] = 3736.7187 ππ
Ζ =ππ
π π = β
πΌ
π΄= β
2133333333
160000= 115.470 ππ
Ζ =3736.7187
115.470= 32.361 ππ
Ζ < Ζπππ πβπππ‘ ππππ’ππ
Page 7
Example 4.4 [Column Design]
Design the braced column to resist an axial load of 950 KN and a moment of Msd =115 KNM at
the top and Msd =-95 KNM at the bottom as shown below .length of the column is 5.5 m and
cross-section of 300*300 mm use C25/30 and S460 take Le=0.66L
.
Solution
Step 1- Material
πππ =0.85 β 25
1.5= 14.16667 πππ
ππ¦π =460
1.15= 400 πππ
Step 2- Check slenderness limit
Ζπππ = 20π΄π΅πΆβπ
β π‘πππ π΄ = 0.7 π΅ = 1.1 πΆ = 1.7 β ππ
π€βπππ ππ =π01
π02=
β95
115= β0.826 πΆ = 1.7 β (β0.826) = 2.526
π =πππ
π΄ππππ=
950 β 103
14.1667 β 300 β 300= 0.745
Ζπππ =20 β 0.7 β 1.1 β 2.526
β0.745= 45.066
Step 3-Slenderess
Ζ =ππ
π π = β
πΌ
π΄= β
675000000
90000= 86.6025 ππ
Page 8
Ζ =0.66 β 5500
86.6025= 41.915 ππ
Ζ < Ζπππ πΊππππ ππππππ ππππππππ ππππππ
πππ
ππ ππππππ
Step 4- accidental eccentricity
ππ =ππ
400=
3630
400= 9.075 ππ
Step 5- Equivalent first order eccentricity
ππ = πππ₯ {0.6π02 + 0.4 π01
0.4π02
π02 =π02
ππ π=
115 β 106
950 β 103= 121.052 ππ
π01 =π01
ππ π=
95 β 106
950 β 103= 100 ππ
ππ = πππ₯ {0.6π02 + 0.4 π01
0.4π02 = 48.4208 mm
ππ‘ππ‘ = ππ + ππ + π02 = 9.075 + 48.4208 + 0 = 57.4958 ππ
πβπππ π€ππ‘β π = π02 + ππ = 121.052 + 9.075 = 130.127 ππ
ππ π‘πππ ππ‘ππ‘ = 130.127 ππ
Step 6- Design
ππ π = 950 πΎπ ππ π = ππ π β ππ‘ππ‘ = 123.62 πΎππ
π£π π =π
ππππβ=
950 β 103
14.1667 β 300 β 300= 0.745
ππ π =ππ π
πππππ2=
123.62 β 106
14.1667 β 300 β 3002= 0.3232
Using πβ²
π= 0.15 read the mechanical steel ratio from uniaxial interaction chart for
ππ π = 0.745
ππ π = 0.3232
π = 0.79
π΄π ,π‘ππ‘ =ππππππ
ππ¦π=
0.79 β 14.166 β 300 β 300
400= 2518.125 ππ2
π΄ =π΄π ,π‘ππ‘
2=
2518.125
2= 1259.0625 ππ2
Page 9
Check with maximum and minimum reinforcement limit
π΄π ,πππ = πππ₯ {
0.1 ππΈπ·
ππ¦π
0.002π΄π
= 237.5 ππ2 ππΎ!
π΄π ,πππ₯ = 0.08 π΄π = 0.08 β 300 β 300 = 7200 ππ2 πΆπ²!
Using β
20 ππππ£πππ 4β
20 ππ πππβ ππππ
Step 7- Detailing
Page 10
Example 4.5 [Column]
Design the braced column if it is subjected to the following loading .the column has total length
of 6 m. Le=0.7L
Use C25/30 and S460
If the column is slender .compute the total design moment using
a) Nominal curvature
b) Nominal stiffness
Use 400*400 mm section πβ²
π= 0.1 π’π π β
(β, π‘π) = 2
Solution
Step 1. Material data
πππ =0.85 β 25
1.5= 14.16667 πππ
ππ¦π =460
1.15= 400 πππ
πΈππ = 30 πΊππ
πΈπ = 200 πΊππ
Step 2- Check slenderness limit
Ζπππ = 20π΄π΅πΆβπ
β π‘πππ π΄ = 0.7 π΅ = 1.1 πΆ = 1.7 β ππ
Page 11
π€βπππ ππ =π01
π02=
140
140= 1 πΆ = 1.7 β (1) = 0.7
π =πππ
π΄ππππ=
1650 β 103
14.1667 β 400 β 400= 0.72794
Ζπππ =20 β 0.7 β 1.1 β 0.7
β0.72794= 12.63487
Step 3-slenderess
Ζ =ππ
π π = β
πΌ
π΄= β
2133333333
160000= 115.470 ππ
Ζ =0.7 β 6000
115.470= 36.373 ππ
Ζ > Ζπππ πΊππππ
ππ ππππππ ππππππ
ππ ππππππ
πππ
ππ ππππππ
Step 4- accidental eccentricity
ππ =ππ
400=
4200
400= 10.5 ππ
Step 5- Equivalent first order eccentricity
ππ = πππ₯ {0.6π02 + 0.4 π01
0.4π02
π02 =π02
ππ π=
140 β 106
1650 β 103= 84.848 ππ
π01 =π01
ππ π=
140 β 106
1650 β 103= 84.848 ππ
ππ = πππ₯ {0.6π02 + 0.4 π01
0.4π02 = 84.848 mm
Step 6- Calculate the second order moment
a) Using Nominal curvature method
π2 =1
πππ
2
πΆβ πͺ = ππ πππ ππππππππ πππππ β πππππππ
1
π= πΎππΎβ
1
ππ πΎβ
= 1 + π½β
πππ
β
πππ = 0 ππ {
β
(β, π‘π) β€ 2 πΆπ² Ζ β€ 75 πΆπ² ππππ
πππβ₯ β π΅ππ πΆπ²
Page 12
β
πππ = β
(β, π‘π)πππππ
ππππ
π€βπππ πππππ = ππππ π‘ πππππ ππππππ‘ π πππ’ππ π β πππππππππ‘ ππππ(ππΏπ)
ππππ = πΉπππ π‘ πππππ ππππππ‘ ππ πππ πππ ππππ ππππππππ‘πππ (ππΏπ)
β
πππ = β
(β, π‘π)πππππ
ππππ= 2 β
70
140= 1
π½ = 0.35 +πππ
200β β Ζ
150β = 0.35 + 25200β β 36.375
150β = 0.2325
πΎβ
= 1 + π½β
πππ = 1 + 0.2325 β 1 = 1.2325
1
ππ=
ππ¦π
0.45π=
2 β 10β3
360= 123456 β 10β5 π = ππππππ‘ππ£π ππππ‘β
πΎπ =(ππ’ β π)
(ππ’ β ππππ) ππ’ = 1 + π π =
πππ
π΄ππππ=
1650 β 103
14.166 β 4002
= 0.72974 ππππ = 0.4
For first iteration take ππ = π
ππ‘ππ‘ = ππ + ππ + π02 = 10.5 + 84.848 + 0 = 95.348 ππ
ππ π = 1650 πΎπ ππ π = ππ π β ππ‘ππ‘ = 157.3242 πΎππ
π£π π =ππ π
ππππβ=
1650 β 103
14.1667 β 400 β 400= 0.72794
ππ π =ππ π
πππππ2=
157.3242 β 106
14.1667 β 400 β 4002= 0.17352
ππ πππ π£π π = 0.72794 ππ π = 0.1752 πβ²
π= 0.1 π = 0.25
So
ππ’ = 1 + π = 1.25 πΎπ =(ππ’ β π)
(ππ’ β ππππ)=
(1.25 β 0.72794)
(1.25 β 0.4)= 0.614
1
π= 0.614 β 1.2325 β 1.23456 β 10β5 = 9.3456 β 10β6
π2 =1
π42002
10β = 16.485 ππ
For Second iteration take ππ = ππ. πππ ππ
ππ‘ππ‘ = ππ + ππ + π02 = 10.5 + 84.848 + 16.485 = 111.833 ππ
ππ π = 1650 πΎπ ππ π = ππ π β ππ‘ππ‘ = 184.525 πΎππ
ππ πππ π£π π = 0.72794 πππ ππ π = 0.2035 π = 0.35
ππ’ = 1 + π = 1.35 πΎπ =(ππ’ β π)
(ππ’ β ππππ)=
(1.35 β 0.72794)
(1.35 β 0.4)= 0.6548
1
π= 0.6548 β 1.2325 β 1.23456 β 10β5 = 9.9634 β 10β6
Page 13
π2 =1
π42002
10β = 17.575 ππ
For their iteration take ππ = ππ. πππ ππ
ππ‘ππ‘ = ππ + ππ + π02 = 10.5 + 84.848 + 17.575 = 112.923 ππ
ππ π = 1650 πΎπ ππ π = ππ π β ππ‘ππ‘ = 186.323 πΎππ
ππ πππ π£π π = 0.72794 πππ ππ π = 0.2055 π = 0.35
The iteration converges with similar mechanical steel ratio π = π. ππ .
π΄π ,π‘ππ‘ =ππππππ
ππ¦π=
0.35 β 14.166 β 400 β 400
400= 1983.33 ππ2
π΄ =π΄π ,π‘ππ‘
2=
1983.33
2= 991.666 ππ2
Check with maximum and minimum reinforcement limit
π΄π ,πππ = πππ₯ {
0.1 ππΈπ·
ππ¦π
0.002π΄π
= 412.5 ππ2 πΆπ²!
π΄π ,πππ₯ = 0.08 π΄π = 0.08 β 400 β 400 = 12800 ππ2 πΆπ²!
Using β
ππ ππππππ
π πβ
ππ ππ ππππ ππππ
.
b) Using Nominal stiffness method
πΈπΌ = πΎππΈπππΌπ + πΎπ πΈπ πΌπ
πΈππ =πΈππ
πΎππ=
30
1.2= 25 πΊππ
Initially let us assume π β₯ π. ππ ππ πππ πππ π¬π’π¦π©π₯π’ππ’ππ ππππππ
πΎπ = 0 πΎπ = 0.31 + 0.5β
πππ
β
Page 14
β
πππ = 1 πΎπ = 0.2
πΌπ =πβ3
12= 2133333333 ππ4
πΈπΌ = πΎππΈπππΌπ + πΎπ πΈπ πΌπ = 0.2 β 25 β 103 β 2133333333 = 1.06667 β 1013
ππ =Π2πΈπΌ
ππ2 =
Π2 β 1.06667 β 1013
42002= 5968.01475 πΎπ
Nb= Buckling load based on nominal stiffness
Design moment πππ = ππππ [1 +π½
(ππ
πππβ )β1
]
π½ = Π2ππ
β
ππ = 8
ππππ = 140 πΎππ
πππ = ππππ [1 +π½
(ππ
πππβ ) β 1
] = 140 β [1 +1.2337
(5968.014751650β ) β 1
]
= 205.999 πΎππ
For first iteration neglecting accidental eccentricity
π£π π =ππ π
ππππβ=
1650 β 103
14.1667 β 400 β 400= 0.72794
ππ π =ππ π
πππππ2=
205.999 β 106
14.1667 β 400 β 4002= 0.2272
ππ πππ π£π π = 0.72794 ππ π = 0.2272 πβ²
π= 0.1 π = 0.4
π΄π ,π‘ππ‘ =ππππππ
ππ¦π=
0.4 β 14.166 β 400 β 400
400= 2266.666 ππ2
π΄ =π΄π ,π‘ππ‘
2=
2266.666
2= 1133.333 ππ2
π =π΄π
ππ=
1133.333
400 β 360= 7.870 β 10β3 π < 0.01
We cannot use the simplified method
πππ π > 0.002 πΎπ = 1
πΎπ =πΎ1πΎ2
(1 + β
πππ)β
πΎ1 = βπππ20
β = 1.11803
Page 15
πΎ2 = πΖ
170β€ 0.2
π =πππ
π΄ππππ= 0.72794
πΎ2 = 0.2 ππ πΎπ = 0.111803
πΌπ =πβ3
12= 2133333333 ππ4
πΌπ = 2 β [1133.333(360 β 200)2] = 58026666.67 ππ2
πΈπΌ = πΎππΈπππΌπ + πΎπ πΈπ πΌπ
= 0.111803 β 25 β 103 β 2133333333 + 1 β 200 β 103
β 58026666.6
πΈπΌ = 1.756816 β 1013
ππ =Π2πΈπΌ
ππ2 =
Π2 β 1.756816 β 1013
42002= 9829.40982 πΎπ
πππ = ππππ [1 +π½
(ππ
πππβ ) β 1
] = 140 β [1 +1.2337
(9829.409821650β ) β 1
]
= 174.842 πΎππ
Design moment including accidental eccentricity is given by
π = 174.842 + 1650 β (10.5
1000) = 192.167 πΎππ
π£π π =ππ π
ππππβ=
1650 β 103
14.1667 β 400 β 400= 0.72794
ππ π =ππ π
πππππ2=
192.167 β 106
14.1667 β 400 β 4002= 0.212
ππ πππ π£π π = 0.72794 ππ π = 0.212 πβ²
π= 0.1 π = 0.35
π΄π ,π‘ππ‘ =ππππππ
ππ¦π=
0.35 β 14.166 β 400 β 400
400= 1983.333 ππ2
π΄ =π΄π ,π‘ππ‘
2=
1983.33
2= 991.666 ππ2
π =π΄π
ππ=
991.666
400 β 360= 6.886 β 10β3 π > 0.002
Second Iteration
Page 16
πΌπ =πβ3
12= 2133333333 ππ4
πΌπ = 2 β [991.666(360 β 200)2] = 50773333.33 ππ2
πΈπΌ = πΎππΈπππΌπ + πΎπ πΈπ πΌπ
= 0.111803 β 25 β 103 β 2133333333 + 1 β 200 β 103
β 50773333.3
πΈπΌ = 1.611749 β 1013
ππ =Π2πΈπΌ
ππ2 =
Π2 β 1.611749 β 1013
42002= 9017.7598 πΎπ
πππ = ππππ [1 +π½
(ππ
πππβ ) β 1
] = 140 β [1 +1.2337
(9017.75981650β ) β 1
]
= πππ. πππ π²π΅π
Design moment including accidental eccentricity is given by
π = 178.679 + 1650 β (10.5
1000) = 196.004 πΎππ
π£π π =ππ π
ππππβ=
1650 β 103
14.1667 β 400 β 400= 0.72794
ππ π =ππ π
πππππ2=
196.004 β 106
14.1667 β 400 β 4002= 0.216
ππ πππ π£π π = 0.72794 ππ π = 0.216 πβ²
π= 0.1 π = 0.35
The iteration converges with similar mechanical steel ratio π = 0.35 .
π΄π ,π‘ππ‘ =ππππππ
ππ¦π=
0.35 β 14.166 β 400 β 400
400= 1983.33 ππ2
π΄ =π΄π ,π‘ππ‘
2=
1983.33
2= 991.666 ππ2
Check with maximum and minimum reinforcement limit
π΄π ,πππ = πππ₯ {
0.1 ππΈπ·
ππ¦π
0.002π΄π
= 412.5 ππ2 πΆπ²!
π΄π ,πππ₯ = 0.08 π΄π = 0.08 β 400 β 400 = 12800 ππ2 πΆπ²!
Using β
ππ ππππππ
π πβ
ππ ππ ππππ ππππ
Step 7- Detailing
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