Week 3 Challenge

Question 1:-

Effective length factor determination problem

• Height of column is 10 m and it is effectively restrained in mid height in one direction (in Z direction) but free to move in other direction (Y direction). Bottom of column is Fixed and top of column is Free.
• Calculate effective length factor of columns in following cases

Given data:-

Height of column = 10m

Column is effectively restrained in mid height in one direction (in Z direction) but free to move in other direction (Y direction).

Therefore the actual length (L) is 5m.

a. Braced frame and shear wall:

As per IS 800 : 2007, Table 11.

The effective length factor for Bottom of column is fixed and Top of column is free = 2.0L

L_e = 2.0 x 5m = 10m

b. Unbraced frame with a pinned base:

As per IS 800 : 2007, Table 11.

The effective length factor for Bottom of column is pinned and Top of column is free = 2.0L

L_e = 2.0 x 5m = 10m

c. Unbraced frame with a fixed base:

As per IS 800 : 2007, Table 11.

The effective length factor for Bottom of column is fixed and Top of column is free = 1.2L

L_e = 1.2 x 5m = 6m

Question 2:-

Design Axial strength determination of a column with its end conditions

Calculate the Axial compressive capacity of Column made up of Hot rolled ISMB 300 with flange cover plate of 200x12 mm on each flange having length of beam as 3 m. Assume the bottom of the column is Fixed and Top as Roller (Rotation Fixed, translation free). Yield strength of material fy= 250 N/mm²

Tip Steps:

• Determine new radius of gyration of compound section
• Using Buckling curve classification & slenderness ratio using table 9c in code, calculate compressive strength
• Using compressive strength, calculate design strength multiplying the cross sectional area.

Given data:-

• Section - Hot rolled ISMB 300 with flange cover plate of 200 x 12 mm on each flange.
• Unsupported Length of column is 3 m.
• Bottom of Column is fixed and Top as Roller (Rotation fixed, translation free)
• F_y is 250 N/mm²

Answer:-

Step 1:-

Properties of built-up section:

The built-up column consist of one ISMB 300 and Two cover plates of 200 x 12 mm as shown above.

From steel section tables (SP 6 - Part 1),

Cross section area of ISMB 300 = 5626 mm²

I_xx of ISMB 300 = 8603.6 x 10⁴ mm⁴

I_yy of ISMB 300 = 453.9 x 10⁴ mm⁴

Cross section area of Plate = 2400 mm²

Moment of inertia of built-up section about xx-axis

I_xx = [8603.6 x 10⁴ + 2 x 200 x 12³ x (1/12) + 2 x Ah²]

= [8603.6 x 10⁴ + 5.76 x 10⁴ + 2 x 2400 x (150+12/2)]

= [8603.6 x 10⁴ + 5.76 x 10⁴ + 74.88 x 10⁴]

I_xx = 8684.24 x 10⁴ mm⁴

Moment of inertia of built-up section about yy-axis

I_yy = [453.9 x 10⁴ + 2 x 12 x 200³ x (1/12)]

= [453.9 x 10⁴ + 1600 x 10⁴ ]

I_yy = 2053.9 x 10⁴ mm⁴

Step 2:-

Slenderness ratio:

I_yy is less than I_xx. Therefore r_yy is Minimum.

Minimum radius of gyration r_min = (I_yy / A)^1/2

= (2053.9 x 10⁴ / 10426)^1/2

r_min = 44.38 mm

Unsupported Length (L) = 3 m = 3000 mm.

Effective Length (L_e) = 1.2 x L = 3600mm

Slenderness ratio of the column = (L/r_min ) = 3600/44.38 = 81.12

Step 3:-

Safe Load:

From IS 800 : 2007

For L/r = 81.12 and the steel having yeild stress, F_y is 250 N/mm²

Allowable working stress in compression

f_cd = 138 N/mm²

Safe load carrying capacity of builtu-up section P = f_cd x A = 138 x 10426 = 1438788 N

P = 1438.788 kN

Result:-

Design Strength = 1438.788 kN

Question 3:-

Design circular tube strut member for certain load

Design 2 m long tubular strut member to carry factored axial load of 400 kN. Assume the ends of the member are simply supported and the yield strength fy=210 Mpa. Also find Equivalent Square hollow sections.

Tip Steps:

• Calculate the area required from design compressive stress and Factored load on strut
• Choose a section satisfying the area calculated.
• Check limiting width to thickness ratio
• Find buckling class & slenderness ratio
• Using Buckling curve classification & slenderness ratio using table 9c in code, calculate compressive strength
• Using compressive strength, calculate design strength multiplying the cross sectional area.

Given data:-

Length of strut member = 2 m

Factored axial load = 400 kN

F_y = 210 N/mm²

Assume the ends of member are simply supported

Hence L_e = 1.0 x L = 2 m = 2000 mm.

Answer:-

Required area of cross section = P / F_y

= 400 x 10³ / 210

=1904.76 mm²

From IS 4923 : 1997, Table 1

For a circular hollow section of 125 outer bore dia = 139.7 thickness = 4.5, Area of circular section = 1910 mm²

For a square hollow section of 100x100x6 has a Area of square section = 2163 mm²

From IS 800 : 2007, Table 2

Limiting Width to Thickness ratio of circular section should be lesser than 88 e²

e = (250/F_y)^1/2 = 1.09

D/t = 139.7 / 4.5 = 31.04

88 e² = 95.92

Width to Thickness ratio of circular section is lesser than 88 e²

Hence safe.

From IS 800 : 2007, Table 10

Square tube Sections comes under class a

For circular section Slenderness ratio λ=L/r_min = 2000/47.8 = 41.84 < 180

For square section Slenderness ratio λ= L/r_min =2000/37.9

L/r_min = 52.77 < 180

Hence safe.

From table 9a of the code for f_y = 210

For circular section f_cd = 180 N/mm²

For square section f_cd = 174.1 N/mm²

Hence, the design strength for circular section = 180 x 19.10 = 343.8kN.

Question 4:-

Design of laced column

Design a single laced column made up of two channels back to back of 10 m height to carry factored axial load of 1000kN. The lacings are bolted. Design of bolted connection not required.

Tip Steps:

• Calculate required Cross section area
• Assume a channel section and calculate the compressive capacity
• Find the appropriate spacing of channel
• Find spacing of lacings
• Calculate design forces –Transverse, compressive force in lacing bars
• Check if the assumed lacings are safe for calculated design forces.
• Check if effective slenderness ratio is within limit

Given data:-

Height = 10m

Factored axial load = 1000kN

Assume design stress of 125 N/mm² (0.5 fy)

Required area =1000x10³ / 125=8000 mm²

Answer:-

Assume the column to be built up of two ISMC 300 channels placed back-to-back

Cross section area of each channel = 4564 mm²

Cross section area of two columns (A) = 2 x 4564 = 9128 mm²

r_xx = 118.1 mm

r_yy = 26.1 mm

t_f = 13.6 mm

t_w = 7.6 mm

c_yy = 23.6 mm

I_xx for two channel = 2 x 6362.8 x 10⁴ mm⁴ = 12725.2 x 10⁴ mm⁴

I_yy for one channel = 310.8 x 10⁴ mm⁴

The distance back to back 'S' between the channels is so adjusted as to make r_xx = r_yy

L/r_xx = 10 x 10³ / 118.1 = 84.67

L/r of the built-up section should be taken as 1.05 x 84.67 = 88.90

f_cd taken from code for given r_xx = 120 N/mm²

Permissible axial load = 9128 x 120 = 1095 kN > 1000 kN.

Hence the section is safe.

Spacing of channels = 12725.2 x 10⁴ = 2 (I_yy + A (S/2 +C_yy)²)

=2 x (310.8 x 10⁴ + 4564 (S/2 +23.6)²)

solving the above equation S = 183.1mm

Design of Lacing:

Using the single lacing bar inclined at 45 degree with the column axis.

Lacing system of the built-up section using single lacing system with the inclination of lacing bar gauge length g = 50 mm for 90 mm flange

Spacing of lacing bars L_o= 2 x (183 + 50 x 2) cot 45

= 2 x 283 x 1 = 566 mm

For a single column element, slenderness ratio L_o / r_yy should be < 0.7 x L/r of whole column

566 / 26 = 21.76 < 0.7 x 88.98=62.3

Hence safe.

Transverse shear force (V) = (2.5 /100) x axial load = (2.5/100) x 1000 x 10³ = 25 kN = 25000 N

Transverse shear in each panel = V/N = 25000/2 = 12500 N

Compressive force in the lacing bar =(V/N) cosec45

= 12500 x1.414

=17675 N

Question 5:-

Design of battened column

Design a battened column made up of two channels back to back of 5 m height to carry factored axial load of 2000kN. The Battens are welded. Design of weld not required. Use Hot rolled channel sections.

Tip Steps:

• Calculate required Cross section area
• Assume a channel section and calculate the compressive capacity
• Find the appropriate spacing of channel
• Find spacing of battens & size of end/intermediate battens
• Calculate design forces –Transverse, longitudinal shear, Moment
• Check if the assumed battens are safe for calculated design forces.

Given data:-

Height = 5m

Axial load= 2000 kN

Assume design stress of 150 N/mm² (0.6 fy)

Required area =2000 x 10³ / 150 = 13333 mm²

Assume the column to be built up of two ISMC 350 channels placed back-to-back

Cross section area of each channel = 5366 mm²

Cross section area of two columns (A) = 2 x 5366 = 10732 mm²

r_xx = 136.6 mm

r_yy = 28.3 mm

t_f = 13.5 mm

t_w = 8.1 mm

c_yy = 24.4 mm

I_xx for two channel = 2 x 10008 x 10⁴ mm⁴ = 20016 x 10⁴ mm⁴

I_yy for one channel = 430.6 x 10⁴ mm⁴

The distance back to back 'S' between the channels is so adjusted as to make r_xx = r_yy

L/r_xx = 5 x 10³ / 136.6 = 36.60

f_cd taken from IS 800 : 2007 Table 9c for given r_xx = 202.4 N/mm²

Permissible axial load = 10732 x 202.4 = 2172 kN > 2000 kN.

Hence the section is safe.

Spacing of channels = 20016 x 10⁴ = 2 (I_yy + A (S/2 +C_yy)²)

=2 x (430.6 x 10⁴ + 5366 (S/2 +24.4)²)

solving the above equation S = 333.4 mm

Spacing of battens L_o/r_yy should be less than 50 or 0.7x L/r_yy

L_o/r_yy < 0.7 x 36.6

Lo < 0.7 x 36.6 x 28.3 = 725 mm

Also, Lo/ryy should be less than 50

Lo/ryy < 50

Lo < 50 x ryy = 50 x 28.3 = 1415 mm

Hence provide batten at spacing 725 mm

Size of end battens

Overall depth of batten = 334 + 2 x C_yy

= 334 + 2 x 24.4= 383 mm

Provide 60 mm overlap of batten on channel flange for welding

Length of batten = 334 x 2 x 60 = 454 mm = 460 mm

Thickness of battens = 1/50 x 460 = 9.2 mm

Provide batten size = 460 x 390 x 9.2 mm

Size of intermediate battens

Overall depth = 3/4 x 390 = 293 mm

Provide 460 x300x 9.2 mm.