example 4.1 [uni-axial column design] - · pdf fileexample 4.2 [biaxial column design]...

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 𝑑′

𝐻= 0.1 and section 270 𝑚𝑚 ∗ 450 𝑚𝑚

Solution

Step 1- Material

𝑓𝑐𝑑 =0.85 ∗ 25

1.5= 14.16667 𝑚𝑝𝑎

𝑓𝑦𝑑 =460

1.15= 400 𝑚𝑝𝑎

Step 2-Determine the normalized axial and bending moment value

𝑣𝑠𝑑 =𝑁𝑠𝑑

𝑓𝑐𝑑𝑏ℎ=

1100 ∗ 103

14.1667 ∗ 270 ∗ 450= 0.639

𝜇𝑠𝑑 =𝑚𝑠𝑑

𝑓𝑐𝑑𝑏𝑑2=

160 ∗ 106

14.1667 ∗ 270 ∗ 4502= 0.2065

Step 3-Using 𝑑′

𝐻= 0.1 read the mechanical steel ratio from uniaxial interaction chart for

𝑉𝑠𝑑 = 0.639 𝜇𝑠𝑑 = 0.2065

From interaction chart 𝜔 = 0.3

𝐴𝑠,𝑡𝑜𝑡 =𝜔𝑓𝑐𝑑𝑏𝑑

𝑓𝑦𝑑=

0.3 ∗ 14.166 ∗ 270 ∗ 450

400= 1290.937 𝑚𝑚2

𝐴 =𝐴𝑠,𝑡𝑜𝑡

2=

1290.937

2= 645.468 𝑚𝑚2

Check with maximum and minimum reinforcement limit

𝐴𝑠,𝑚𝑖𝑛 = 𝑚𝑎𝑥 {

0.1 𝑁𝐸𝐷

𝑓𝑦𝑑

0.002𝐴𝑐

= 275 𝑚𝑚2 𝑂𝐾!

𝐴𝑠,𝑚𝑎𝑥 = 0.08 𝐴𝑐 = 0.08 ∗ 270 ∗ 450 = 𝑂𝐾!

Step 4- Detailing

𝑈𝑠𝑒 4∅16 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑓𝑎𝑐𝑒

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

𝑃 = 1360 𝐾𝑁 𝑀𝑠𝑑,ℎ = 200 𝐾𝑁𝑚 𝑀𝑠𝑑,𝑏 = 100 𝐾𝑁𝑚

Use C25/30 and S460 class 1 work take ℎ′

ℎ=

𝑏′

𝑏= 0.1

Solution

Step 1- Material

𝑓𝑐𝑑 =0.85 ∗ 25

1.5= 14.16667 𝑚𝑝𝑎

𝑓𝑦𝑑 =460

1.15= 400 𝑚𝑝𝑎

Step 2- Determine the normalized axial and bending moment value

𝑣𝑠𝑑 =𝑝

𝑓𝑐𝑑𝑏ℎ=

1360∗103

14.1667∗400∗400= 0.6

𝜇𝑠𝑑,ℎ =𝑚𝑠𝑑

𝑓𝑐𝑑𝑏𝑑2=

200 ∗ 106

14.1667 ∗ 400 ∗ 4002= 0.2206

𝜇𝑠𝑑,𝑏 =𝑚𝑠𝑑

𝑓𝑐𝑑𝑑𝑏2=

100 ∗ 106

14.1667 ∗ 400 ∗ 4002= 0.1103

Step 3- Find 𝜔 using 𝑑′

𝑑=

𝑏′

𝐵= 0.1 , 𝑉𝑠𝑑 = 0.6 ,

𝜇𝑠𝑑,ℎ = 0.2206 ,

𝜇𝑠𝑑,𝑏 = 0.1103

From biaxial chart 𝜔 = 0.6

𝐴𝑠,𝑡𝑜𝑡 =𝜔𝑓𝑐𝑑𝑏𝑑

𝑓𝑦𝑑=

0.6 ∗ 14.166 ∗ 400 ∗ 400

400= 3400 𝑚𝑚2

Check with maximum and minimum reinforcement limit

𝐴𝑠,𝑚𝑖𝑛 = 𝑚𝑎𝑥 {

0.1 𝑁𝐸𝐷

𝑓𝑦𝑑

0.002𝐴𝑐

= 340 𝑚𝑚2 𝑶𝑲!

𝐴𝑠,𝑚𝑎𝑥 = 0.08 𝐴𝑐 = 0.08 ∗ 400 ∗ 400 = 12800 𝑶𝑲!

Step 4- Detailing

𝑢𝑠𝑖𝑛𝑔 ∅24 𝑛 =3400

452.16= 7.52 𝑠𝑜 𝑈𝑠𝑒 8∅24

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 𝑓𝑐𝑑 = 14.1667 𝑚𝑝𝑎

Solution

Step 1- Slenderness limit

ƛ𝑙𝑖𝑚 = 20𝐴𝐵𝐶√𝑛

⁄ 𝑡𝑎𝑘𝑒 𝐴 = 0.7 𝐵 = 1.1 𝐶 = 1.7 − 𝑟𝑚

𝑤ℎ𝑒𝑟𝑒 𝑟𝑚 =𝑚01

𝑚02=

−34.4

68.8= −0.5 𝐶 = 1.7 − (−0.5) = 2.2

𝑛 =𝑁𝑒𝑑

𝐴𝑐𝑓𝑐𝑑=

516 ∗ 103

14.1667 ∗ 400 ∗ 400= 0.227647

ƛ𝑙𝑖𝑚 =20 ∗ 0.7 ∗ 1.1 ∗ 2.2

√0.227647= 71.0088

Step 2-Find the slenderness of the column

2.1 Effective length

For Braced member

𝑙0 = 0.5𝑙√[1 +𝐾1

0.5 + 𝐾1] [1 +

𝐾2

0.5 + 𝐾2]

𝐾 =𝐶𝑜𝑙𝑢𝑚𝑛 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠

∑ 𝑏𝑒𝑎𝑚 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠

𝐾𝑖 =(𝐸𝐼

𝑙⁄ )𝑐𝑜𝑙𝑢𝑚𝑛

2(2 𝐸𝐼𝑙⁄ )𝑏𝑒𝑎𝑚

𝐼𝑐𝑜𝑙𝑢𝑚𝑛 =400 ∗ 4003

12= 2133333333 𝑚𝑚4

𝐼𝑏𝑒𝑎𝑚 =250 ∗ 5003

12= 2604166667 𝑚𝑚4

𝐾1 =

2133333333𝐸6

2(2 ∗ 2604166667𝐸

6)

= 0.2048

For fixed restraint K=0 but in reality we cannot provide a fully fixed support so use 𝐾2 = 0.1

𝑙0 = 0.5 ∗ 6000√[1 +0.2048

0.5 + 0.2048] [1 +

0.1

0.5 + 0.1] = 3736.7187 𝑚𝑚

ƛ =𝑙𝑜

𝑖 𝑖 = √

𝐼

𝐴= √

2133333333

160000= 115.470 𝑚𝑚

ƛ =3736.7187

115.470= 32.361 𝑚𝑚

ƛ < ƛ𝑙𝑖𝑚 𝑆ℎ𝑜𝑟𝑡 𝑐𝑜𝑙𝑢𝑚𝑛

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

𝑓𝑐𝑑 =0.85 ∗ 25

1.5= 14.16667 𝑚𝑝𝑎

𝑓𝑦𝑑 =460

1.15= 400 𝑚𝑝𝑎

Step 2- Check slenderness limit

ƛ𝑙𝑖𝑚 = 20𝐴𝐵𝐶√𝑛

⁄ 𝑡𝑎𝑘𝑒 𝐴 = 0.7 𝐵 = 1.1 𝐶 = 1.7 − 𝑟𝑚

𝑤ℎ𝑒𝑟𝑒 𝑟𝑚 =𝑚01

𝑚02=

−95

115= −0.826 𝐶 = 1.7 − (−0.826) = 2.526

𝑛 =𝑁𝑒𝑑

𝐴𝑐𝑓𝑐𝑑=

950 ∗ 103

14.1667 ∗ 300 ∗ 300= 0.745

ƛ𝑙𝑖𝑚 =20 ∗ 0.7 ∗ 1.1 ∗ 2.526

√0.745= 45.066

Step 3-Slenderess

ƛ =𝑙𝑜

𝑖 𝑖 = √

𝐼

𝐴= √

675000000

90000= 86.6025 𝑚𝑚

ƛ =0.66 ∗ 5500

86.6025= 41.915 𝑚𝑚

ƛ < ƛ𝑙𝑖𝑚 𝑺𝒉𝒐𝒓𝒕 𝒄𝒐𝒍𝒖𝒎𝒏 𝒏𝒆𝒈𝒆𝒍𝒆𝒄𝒕 𝒔𝒆𝒄𝒐𝒏𝒅 𝒐𝒓𝒅𝒆𝒓 𝒆𝒇𝒇𝒆𝒄𝒕

Step 4- accidental eccentricity

𝑒𝑎 =𝑙𝑜

400=

3630

400= 9.075 𝑚𝑚

Step 5- Equivalent first order eccentricity

𝑒𝑒 = 𝑚𝑎𝑥 {0.6𝑒02 + 0.4 𝑒01

0.4𝑒02

𝑒02 =𝑀02

𝑁𝑠𝑑=

115 ∗ 106

950 ∗ 103= 121.052 𝑚𝑚

𝑒01 =𝑀01

𝑁𝑠𝑑=

95 ∗ 106

950 ∗ 103= 100 𝑚𝑚

𝑒𝑒 = 𝑚𝑎𝑥 {0.6𝑒02 + 0.4 𝑒01

0.4𝑒02 = 48.4208 mm

𝑒𝑡𝑜𝑡 = 𝑒𝑜 + 𝑒𝑒 + 𝑒02 = 9.075 + 48.4208 + 0 = 57.4958 𝑚𝑚

𝑐ℎ𝑒𝑐𝑘 𝑤𝑖𝑡ℎ 𝑒 = 𝑒02 + 𝑒𝑎 = 121.052 + 9.075 = 130.127 𝑚𝑚

𝑆𝑜 𝑡𝑎𝑘𝑒 𝑒𝑡𝑜𝑡 = 130.127 𝑚𝑚

Step 6- Design

𝑁𝑠𝑑 = 950 𝐾𝑁 𝑀𝑠𝑑 = 𝑁𝑠𝑑 ∗ 𝑒𝑡𝑜𝑡 = 123.62 𝐾𝑁𝑚

𝑣𝑠𝑑 =𝑝

𝑓𝑐𝑑𝑏ℎ=

950 ∗ 103

14.1667 ∗ 300 ∗ 300= 0.745

𝜇𝑠𝑑 =𝑚𝑠𝑑

𝑓𝑐𝑑𝑏𝑑2=

123.62 ∗ 106

14.1667 ∗ 300 ∗ 3002= 0.3232

Using 𝑑′

𝑑= 0.15 read the mechanical steel ratio from uniaxial interaction chart for

𝑉𝑠𝑑 = 0.745

𝜇𝑠𝑑 = 0.3232

𝜔 = 0.79

𝐴𝑠,𝑡𝑜𝑡 =𝜔𝑓𝑐𝑑𝑏𝑑

𝑓𝑦𝑑=

0.79 ∗ 14.166 ∗ 300 ∗ 300

400= 2518.125 𝑚𝑚2

𝐴 =𝐴𝑠,𝑡𝑜𝑡

2=

2518.125

2= 1259.0625 𝑚𝑚2

Check with maximum and minimum reinforcement limit

𝐴𝑠,𝑚𝑖𝑛 = 𝑚𝑎𝑥 {

0.1 𝑁𝐸𝐷

𝑓𝑦𝑑

0.002𝐴𝑐

= 237.5 𝑚𝑚2 𝑂𝐾!

𝐴𝑠,𝑚𝑎𝑥 = 0.08 𝐴𝑐 = 0.08 ∗ 300 ∗ 300 = 7200 𝑚𝑚2 𝑶𝑲!

Using ∅20 𝑝𝑟𝑜𝑣𝑖𝑑𝑒 4∅20 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑓𝑎𝑐𝑒

Step 7- Detailing

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 𝑑′

𝑑= 0.1 𝑢𝑠𝑒 ∅(∞, 𝑡𝑜) = 2

Solution

Step 1. Material data

𝑓𝑐𝑑 =0.85 ∗ 25

1.5= 14.16667 𝑚𝑝𝑎

𝑓𝑦𝑑 =460

1.15= 400 𝑚𝑝𝑎

𝐸𝑐𝑚 = 30 𝐺𝑝𝑎

𝐸𝑠 = 200 𝐺𝑝𝑎

Step 2- Check slenderness limit

ƛ𝑙𝑖𝑚 = 20𝐴𝐵𝐶√𝑛

⁄ 𝑡𝑎𝑘𝑒 𝐴 = 0.7 𝐵 = 1.1 𝐶 = 1.7 − 𝑟𝑚

𝑤ℎ𝑒𝑟𝑒 𝑟𝑚 =𝑚01

𝑚02=

140

140= 1 𝐶 = 1.7 − (1) = 0.7

𝑛 =𝑁𝑒𝑑

𝐴𝑐𝑓𝑐𝑑=

1650 ∗ 103

14.1667 ∗ 400 ∗ 400= 0.72794

ƛ𝑙𝑖𝑚 =20 ∗ 0.7 ∗ 1.1 ∗ 0.7

√0.72794= 12.63487

Step 3-slenderess

ƛ =𝑙𝑜

𝑖 𝑖 = √

𝐼

𝐴= √

2133333333

160000= 115.470 𝑚𝑚

ƛ =0.7 ∗ 6000

115.470= 36.373 𝑚𝑚

ƛ > ƛ𝑙𝑖𝑚 𝑺𝒍𝒆𝒏𝒅𝒆𝒓 𝒄𝒐𝒍𝒖𝒎𝒏 𝒄𝒐𝒏𝒔𝒊𝒅𝒆𝒓 𝒔𝒆𝒄𝒐𝒏𝒅 𝒐𝒓𝒅𝒆𝒓 𝒆𝒇𝒇𝒆𝒄𝒕

Step 4- accidental eccentricity

𝑒𝑎 =𝑙𝑜

400=

4200

400= 10.5 𝑚𝑚

Step 5- Equivalent first order eccentricity

𝑒𝑒 = 𝑚𝑎𝑥 {0.6𝑒02 + 0.4 𝑒01

0.4𝑒02

𝑒02 =𝑀02

𝑁𝑠𝑑=

140 ∗ 106

1650 ∗ 103= 84.848 𝑚𝑚

𝑒01 =𝑀01

𝑁𝑠𝑑=

140 ∗ 106

1650 ∗ 103= 84.848 𝑚𝑚

𝑒𝑒 = 𝑚𝑎𝑥 {0.6𝑒02 + 0.4 𝑒01

0.4𝑒02 = 84.848 mm

Step 6- Calculate the second order moment

a) Using Nominal curvature method

𝑒2 =1

𝑟𝑙𝑜

2

𝐶⁄ 𝑪 = 𝟏𝟎 𝑭𝒐𝒓 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒄𝒓𝒐𝒔𝒔 − 𝒔𝒆𝒄𝒕𝒊𝒐𝒏

1

𝑟= 𝐾𝑟𝐾∅

1

𝑟𝑜 𝐾∅ = 1 + 𝛽∅𝑒𝑓𝑓

∅𝑒𝑓𝑓 = 0 𝑖𝑓 {

∅(∞, 𝑡𝑜) ≤ 2 𝑶𝑲 ƛ ≤ 75 𝑶𝑲 𝑀𝑜𝑒𝑑

𝑁𝑒𝑑≥ ℎ 𝑵𝒐𝒕 𝑶𝑲

∅𝑒𝑓𝑓 = ∅(∞, 𝑡𝑜)𝑀𝑜𝑒𝑞𝑝

𝑀𝑜𝑒𝑑

𝑤ℎ𝑒𝑟𝑒 𝑀𝑜𝑒𝑞𝑝 = 𝑓𝑖𝑟𝑠𝑡 𝑜𝑟𝑑𝑒𝑟 𝑚𝑜𝑚𝑒𝑛𝑡 𝑖 𝑛𝑞𝑢𝑎𝑠𝑖 − 𝑝𝑒𝑟𝑚𝑎𝑛𝑒𝑛𝑡 𝑙𝑜𝑎𝑑(𝑆𝐿𝑆)

𝑀𝑜𝑒𝑑 = 𝐹𝑖𝑟𝑠𝑡 𝑜𝑟𝑑𝑒𝑟 𝑚𝑜𝑚𝑒𝑛𝑡 𝑖𝑛 𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 (𝑈𝐿𝑆)

∅𝑒𝑓𝑓 = ∅(∞, 𝑡𝑜)𝑀𝑜𝑒𝑞𝑝

𝑀𝑜𝑒𝑑= 2 ∗

70

140= 1

𝛽 = 0.35 +𝑓𝑐𝑘

200⁄ − ƛ

150⁄ = 0.35 + 25200⁄ − 36.375

150⁄ = 0.2325

𝐾∅ = 1 + 𝛽∅𝑒𝑓𝑓 = 1 + 0.2325 ∗ 1 = 1.2325

1

𝑟𝑜=

𝜀𝑦𝑑

0.45𝑑=

2 ∗ 10−3

360= 123456 ∗ 10−5 𝑑 = 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ

𝐾𝑟 =(𝑛𝑢 − 𝑛)

(𝑛𝑢 − 𝑛𝑏𝑎𝑙) 𝑛𝑢 = 1 + 𝜔 𝑛 =

𝑁𝑒𝑑

𝐴𝑐𝑓𝑐𝑑=

1650 ∗ 103

14.166 ∗ 4002

= 0.72974 𝑛𝑏𝑎𝑙 = 0.4

For first iteration take 𝒆𝟐 = 𝟎

𝑒𝑡𝑜𝑡 = 𝑒𝑜 + 𝑒𝑒 + 𝑒02 = 10.5 + 84.848 + 0 = 95.348 𝑚𝑚

𝑁𝑠𝑑 = 1650 𝐾𝑁 𝑀𝑠𝑑 = 𝑁𝑠𝑑 ∗ 𝑒𝑡𝑜𝑡 = 157.3242 𝐾𝑁𝑚

𝑣𝑠𝑑 =𝑁𝑠𝑑

𝑓𝑐𝑑𝑏ℎ=

1650 ∗ 103

14.1667 ∗ 400 ∗ 400= 0.72794

𝜇𝑠𝑑 =𝑚𝑠𝑑

𝑓𝑐𝑑𝑏𝑑2=

157.3242 ∗ 106

14.1667 ∗ 400 ∗ 4002= 0.17352

𝑈𝑠𝑖𝑛𝑔 𝑣𝑠𝑑 = 0.72794 𝜇𝑠𝑑 = 0.1752 𝑑′

𝑑= 0.1 𝜔 = 0.25

So

𝑛𝑢 = 1 + 𝜔 = 1.25 𝐾𝑟 =(𝑛𝑢 − 𝑛)

(𝑛𝑢 − 𝑛𝑏𝑎𝑙)=

(1.25 − 0.72794)

(1.25 − 0.4)= 0.614

1

𝑟= 0.614 ∗ 1.2325 ∗ 1.23456 ∗ 10−5 = 9.3456 ∗ 10−6

𝑒2 =1

𝑟42002

10⁄ = 16.485 𝑚𝑚

For Second iteration take 𝒆𝟐 = 𝟏𝟔. 𝟒𝟖𝟓 𝒎𝒎

𝑒𝑡𝑜𝑡 = 𝑒𝑜 + 𝑒𝑒 + 𝑒02 = 10.5 + 84.848 + 16.485 = 111.833 𝑚𝑚

𝑁𝑠𝑑 = 1650 𝐾𝑁 𝑀𝑠𝑑 = 𝑁𝑠𝑑 ∗ 𝑒𝑡𝑜𝑡 = 184.525 𝐾𝑁𝑚

𝑈𝑠𝑖𝑛𝑔 𝑣𝑠𝑑 = 0.72794 𝑎𝑛𝑑 𝜇𝑠𝑑 = 0.2035 𝜔 = 0.35

𝑛𝑢 = 1 + 𝜔 = 1.35 𝐾𝑟 =(𝑛𝑢 − 𝑛)

(𝑛𝑢 − 𝑛𝑏𝑎𝑙)=

(1.35 − 0.72794)

(1.35 − 0.4)= 0.6548

1

𝑟= 0.6548 ∗ 1.2325 ∗ 1.23456 ∗ 10−5 = 9.9634 ∗ 10−6

𝑒2 =1

𝑟42002

10⁄ = 17.575 𝑚𝑚

For their iteration take 𝒆𝟐 = 𝟏𝟕. 𝟓𝟕𝟓 𝒎𝒎

𝑒𝑡𝑜𝑡 = 𝑒𝑜 + 𝑒𝑒 + 𝑒02 = 10.5 + 84.848 + 17.575 = 112.923 𝑚𝑚

𝑁𝑠𝑑 = 1650 𝐾𝑁 𝑀𝑠𝑑 = 𝑁𝑠𝑑 ∗ 𝑒𝑡𝑜𝑡 = 186.323 𝐾𝑁𝑚

𝑈𝑠𝑖𝑛𝑔 𝑣𝑠𝑑 = 0.72794 𝑎𝑛𝑑 𝜇𝑠𝑑 = 0.2055 𝜔 = 0.35

The iteration converges with similar mechanical steel ratio 𝝎 = 𝟎. 𝟑𝟓 .

𝐴𝑠,𝑡𝑜𝑡 =𝜔𝑓𝑐𝑑𝑏𝑑

𝑓𝑦𝑑=

0.35 ∗ 14.166 ∗ 400 ∗ 400

400= 1983.33 𝑚𝑚2

𝐴 =𝐴𝑠,𝑡𝑜𝑡

2=

1983.33

2= 991.666 𝑚𝑚2

Check with maximum and minimum reinforcement limit

𝐴𝑠,𝑚𝑖𝑛 = 𝑚𝑎𝑥 {

0.1 𝑁𝐸𝐷

𝑓𝑦𝑑

0.002𝐴𝑐

= 412.5 𝑚𝑚2 𝑶𝑲!

𝐴𝑠,𝑚𝑎𝑥 = 0.08 𝐴𝑐 = 0.08 ∗ 400 ∗ 400 = 12800 𝑚𝑚2 𝑶𝑲!

Using ∅𝟏𝟔 𝒑𝒓𝒐𝒗𝒊𝒅𝒆 𝟓∅𝟏𝟔 𝒐𝒏 𝒆𝒂𝒄𝒉 𝒇𝒂𝒄𝒆

.

b) Using Nominal stiffness method

𝐸𝐼 = 𝐾𝑐𝐸𝑐𝑑𝐼𝑐 + 𝐾𝑠𝐸𝑠𝐼𝑠

𝐸𝑐𝑑 =𝐸𝑐𝑚

𝛾𝑐𝑒=

30

1.2= 25 𝐺𝑝𝑎

Initially let us assume 𝝆 ≥ 𝟎. 𝟎𝟏 𝒕𝒐 𝒖𝒔𝒆 𝒕𝒉𝒆 𝐬𝐢𝐦𝐩𝐥𝐢𝐟𝐢𝐞𝐝 𝒎𝒆𝒕𝒉𝒐𝒅

𝐾𝑠 = 0 𝐾𝑐 = 0.31 + 0.5∅𝑒𝑓𝑓

⁄

∅𝑒𝑓𝑓 = 1 𝐾𝑐 = 0.2

𝐼𝑐 =𝑏ℎ3

12= 2133333333 𝑚𝑚4

𝐸𝐼 = 𝐾𝑐𝐸𝑐𝑑𝐼𝑐 + 𝐾𝑠𝐸𝑠𝐼𝑠 = 0.2 ∗ 25 ∗ 103 ∗ 2133333333 = 1.06667 ∗ 1013

𝑁𝑏 =П2𝐸𝐼

𝑙𝑜2 =

П2 ∗ 1.06667 ∗ 1013

42002= 5968.01475 𝐾𝑁

Nb= Buckling load based on nominal stiffness

Design moment 𝑀𝑒𝑑 = 𝑀𝑜𝑒𝑑 [1 +𝛽

(𝑁𝑏

𝑁𝑒𝑑⁄ )−1

]

𝛽 = П2𝑐𝑜

⁄

𝑐𝑜 = 8

𝑀𝑜𝑒𝑑 = 140 𝐾𝑁𝑚

𝑀𝑒𝑑 = 𝑀𝑜𝑒𝑑 [1 +𝛽

(𝑁𝑏

𝑁𝑒𝑑⁄ ) − 1

] = 140 ∗ [1 +1.2337

(5968.014751650⁄ ) − 1

]

= 205.999 𝐾𝑁𝑚

For first iteration neglecting accidental eccentricity

𝑣𝑠𝑑 =𝑁𝑠𝑑

𝑓𝑐𝑑𝑏ℎ=

1650 ∗ 103

14.1667 ∗ 400 ∗ 400= 0.72794

𝜇𝑠𝑑 =𝑚𝑠𝑑

𝑓𝑐𝑑𝑏𝑑2=

205.999 ∗ 106

14.1667 ∗ 400 ∗ 4002= 0.2272

𝑈𝑠𝑖𝑛𝑔 𝑣𝑠𝑑 = 0.72794 𝜇𝑠𝑑 = 0.2272 𝑑′

𝑑= 0.1 𝜔 = 0.4

𝐴𝑠,𝑡𝑜𝑡 =𝜔𝑓𝑐𝑑𝑏𝑑

𝑓𝑦𝑑=

0.4 ∗ 14.166 ∗ 400 ∗ 400

400= 2266.666 𝑚𝑚2

𝐴 =𝐴𝑠,𝑡𝑜𝑡

2=

2266.666

2= 1133.333 𝑚𝑚2

𝜌 =𝐴𝑠

𝑏𝑑=

1133.333

400 ∗ 360= 7.870 ∗ 10−3 𝜌 < 0.01

We cannot use the simplified method

𝑓𝑜𝑟 𝜌 > 0.002 𝐾𝑠 = 1

𝐾𝑐 =𝐾1𝐾2

(1 + ∅𝑒𝑓𝑓)⁄

𝐾1 = √𝑓𝑐𝑘20

⁄ = 1.11803

𝐾2 = 𝑛ƛ

170≤ 0.2

𝑛 =𝑁𝑒𝑑

𝐴𝑐𝑓𝑐𝑑= 0.72794

𝐾2 = 0.2 𝑆𝑜 𝐾𝑐 = 0.111803

𝐼𝑐 =𝑏ℎ3

12= 2133333333 𝑚𝑚4

𝐼𝑠 = 2 ∗ [1133.333(360 − 200)2] = 58026666.67 𝑚𝑚2

𝐸𝐼 = 𝐾𝑐𝐸𝑐𝑑𝐼𝑐 + 𝐾𝑠𝐸𝑠𝐼𝑠

= 0.111803 ∗ 25 ∗ 103 ∗ 2133333333 + 1 ∗ 200 ∗ 103

∗ 58026666.6

𝐸𝐼 = 1.756816 ∗ 1013

𝑁𝑏 =П2𝐸𝐼

𝑙𝑜2 =

П2 ∗ 1.756816 ∗ 1013

42002= 9829.40982 𝐾𝑁

𝑀𝑒𝑑 = 𝑀𝑜𝑒𝑑 [1 +𝛽

(𝑁𝑏

𝑁𝑒𝑑⁄ ) − 1

] = 140 ∗ [1 +1.2337

(9829.409821650⁄ ) − 1

]

= 174.842 𝐾𝑁𝑚

Design moment including accidental eccentricity is given by

𝑀 = 174.842 + 1650 ∗ (10.5

1000) = 192.167 𝐾𝑁𝑀

𝑣𝑠𝑑 =𝑁𝑠𝑑

𝑓𝑐𝑑𝑏ℎ=

1650 ∗ 103

14.1667 ∗ 400 ∗ 400= 0.72794

𝜇𝑠𝑑 =𝑚𝑠𝑑

𝑓𝑐𝑑𝑏𝑑2=

192.167 ∗ 106

14.1667 ∗ 400 ∗ 4002= 0.212

𝑈𝑠𝑖𝑛𝑔 𝑣𝑠𝑑 = 0.72794 𝜇𝑠𝑑 = 0.212 𝑑′

𝑑= 0.1 𝜔 = 0.35

𝐴𝑠,𝑡𝑜𝑡 =𝜔𝑓𝑐𝑑𝑏𝑑

𝑓𝑦𝑑=

0.35 ∗ 14.166 ∗ 400 ∗ 400

400= 1983.333 𝑚𝑚2

𝐴 =𝐴𝑠,𝑡𝑜𝑡

2=

1983.33

2= 991.666 𝑚𝑚2

𝜌 =𝐴𝑠

𝑏𝑑=

991.666

400 ∗ 360= 6.886 ∗ 10−3 𝜌 > 0.002

Second Iteration

𝐼𝑐 =𝑏ℎ3

12= 2133333333 𝑚𝑚4

𝐼𝑠 = 2 ∗ [991.666(360 − 200)2] = 50773333.33 𝑚𝑚2

𝐸𝐼 = 𝐾𝑐𝐸𝑐𝑑𝐼𝑐 + 𝐾𝑠𝐸𝑠𝐼𝑠

= 0.111803 ∗ 25 ∗ 103 ∗ 2133333333 + 1 ∗ 200 ∗ 103

∗ 50773333.3

𝐸𝐼 = 1.611749 ∗ 1013

𝑁𝑏 =П2𝐸𝐼

𝑙𝑜2 =

П2 ∗ 1.611749 ∗ 1013

42002= 9017.7598 𝐾𝑁

𝑀𝑒𝑑 = 𝑀𝑜𝑒𝑑 [1 +𝛽

(𝑁𝑏

𝑁𝑒𝑑⁄ ) − 1

] = 140 ∗ [1 +1.2337

(9017.75981650⁄ ) − 1

]

= 𝟏𝟕𝟖. 𝟔𝟕𝟗 𝑲𝑵𝒎

Design moment including accidental eccentricity is given by

𝑀 = 178.679 + 1650 ∗ (10.5

1000) = 196.004 𝐾𝑁𝑀

𝑣𝑠𝑑 =𝑁𝑠𝑑

𝑓𝑐𝑑𝑏ℎ=

1650 ∗ 103

14.1667 ∗ 400 ∗ 400= 0.72794

𝜇𝑠𝑑 =𝑚𝑠𝑑

𝑓𝑐𝑑𝑏𝑑2=

196.004 ∗ 106

14.1667 ∗ 400 ∗ 4002= 0.216

𝑈𝑠𝑖𝑛𝑔 𝑣𝑠𝑑 = 0.72794 𝜇𝑠𝑑 = 0.216 𝑑′

𝑑= 0.1 𝜔 = 0.35

The iteration converges with similar mechanical steel ratio 𝜔 = 0.35 .

𝐴𝑠,𝑡𝑜𝑡 =𝜔𝑓𝑐𝑑𝑏𝑑

𝑓𝑦𝑑=

0.35 ∗ 14.166 ∗ 400 ∗ 400

400= 1983.33 𝑚𝑚2

𝐴 =𝐴𝑠,𝑡𝑜𝑡

2=

1983.33

2= 991.666 𝑚𝑚2

Check with maximum and minimum reinforcement limit

𝐴𝑠,𝑚𝑖𝑛 = 𝑚𝑎𝑥 {

0.1 𝑁𝐸𝐷

𝑓𝑦𝑑

0.002𝐴𝑐

= 412.5 𝑚𝑚2 𝑶𝑲!

𝐴𝑠,𝑚𝑎𝑥 = 0.08 𝐴𝑐 = 0.08 ∗ 400 ∗ 400 = 12800 𝑚𝑚2 𝑶𝑲!

Using ∅𝟏𝟔 𝒑𝒓𝒐𝒗𝒊𝒅𝒆 𝟓∅𝟏𝟔 𝒐𝒏 𝒆𝒂𝒄𝒉 𝒇𝒂𝒄𝒆

Step 7- Detailing